Why I hate physics

Curses! Foiled again.

2012-02-29T13:37:00.002-08:00

It's the damned calculus that gets me every time. I actually ran the integration on my antenna cross-section calculation, and I got the same answer I got the first time by just guessing. Which means I'm still out of whack with the rest of the world by a factor of 2. Hard to understand.

One the one hand, you don't get that close to the right answer unless your physics is pretty sound to begin with. There's little doubt that the discrepancy is just a matter of a math error on my part. The problem is: do you go back and obsess over it until you find the error, or do you accept that you're basically on the right track and move on to other things? You'd theoretically make more progress that way, by pressing ahead; but you never know when you might miss something important. It's a tough one.

The calculus was in fact rather picturesque, but it's not really the kind of thing I that lends itself to a blog treatment. However, I suppose I ought to sketch it out, just in case anyone is wondering. So here goes:

If you recall the superposition of the two wave systems, one circular and one planar, the question becomes: how fast doe the circular waves go out of phase with the plane waves, as you move away from the axis of symmetry? It shouldn't be that hard to convince yourself that the phase is quadratic in x, where x is the radial distance. (That's because a circle is basically the same as a parabola, if you take it over a shallow enough section of the curve. Which we can always do by backing off sufficiently far from the antenna.) The assumption is going to be that we're far enough away from the receiving antenna that the circular field is much smaller than the planar field, so the binomial approximation for power density will apply: (1-x)^2 = 1-2x, where x is the in-phase component of the circular wave.

Here it gets a little tricky: we still don't know what the phase difference of the two wave patterns is, taken along the axis of symmetry. I've drawn it as though the phase difference is zero and grows as you move outwards; but in fact, it might be anything. It's not clear exactly how it has to start off in order to maximize the power absorption. Fortunately, we can cover all possibilities by two special cases: sin and cosine. Either you start off exactly in phase, or you start of 90 degrees out of phase...or it's an intermediate case which you can make up by putting together those two special cases.

Then we just have to integrate the field over the cross-section. Don't forget the factor of 2*pi*xdx which you get for the circular rings when you set up the integrals: we then have to evaluate:

and we also have to check the quadrature case,

where k is a scaling factor to be determined by the physics of the situation. I actually ran these integrals in Excel and had two surprises: first, it was the sin integral that gave me the positive result. I always thought it was the cosine integral, but that one goes to zero. It means that along the axis of symmetry, far away from the antenna, the two fields are 90 degrees out of phase. For maximum power absorption, of course. The other weird thing was that doing it numerically, I clearly came up with a simple multiple of pi. It's funny in physics how the crazy integrals that come up happen to be analytically solvable as often as they are. So I looked again, and there is an obvious substitution of u=x^2. I guess anyone would have seen that. The integral solves pretty easily after all. It's the scaling factor that takes a bit of work.

But if you really think about it, you shouldn't be that quick to accept the formal mathematical solution. Just look at the function. It grows and grows, and oscillates faster and faster. If we graph it, it's pretty nasty looking:

Doesn't that grow like crazy? Yes and no. Physically, what's happening is that as you get farther from the axis of symmetry, the two wave patterns are going in and out of phase so rapidly that there is no net effect. It's like shining to flashlight beams across each other. In theory, you should calculate the power flow by adding the wave vectors everywhere and calculating the Poynting vector, which will be going crazy. But in practise, the effect is for all the micro-fluctuations to average out to zero. All the real physics happens in the first few lumps.

But how do we handle that mathematically? I actually did a problem like this once before, when I was adding up one of those crazy Ramanujan series. It came up in my calculation of the Casimir effect, and it was something like 1+2-3+4.... which adds up to 0.25. How? You cover it over with a very gentle Gaussian that preserves the low end and gradually suppresses the high-end fluctuation. I did exactly the same thing when I did this sin integral in Excel. You just keep adjusting the width of the Gaussian until the value stabilises, and then you know you're done. Actually, the fluctuations in this integral are almost like that other series, except instead of alternating integers, it's pretty much the square roots of the integers that alternate. (Because of the x-squared inside the sine.)

Like I said, the integral turned out to be not the hard part. It was getting the right scaling factor to line it up to the physics. I did my best; I'm not going to drag you through the details, but when it was all over, I was still out by a factor of 2. It's just one of those things.

Effective Cross-Section of a Dipole Antenna

2012-02-28T14:04:00.003-08:00

Over the last couple of weeks I’vebeen doing some pretty rough calculations to figure out the radiation from anoscillating charge. I’ve been in the ballpark, but today I think I’m like to getit right on the money. This is a calculation I figured out just over twentyyears ago, and I’m pretty sure you won’t see it anywhere. I sketched it outlast year in this article on the Crystal Radio, but I never did the numbers.That will be today’s project.

So far we’ve worked on theLarmor Formula, which gives you the radiation of an accelerating charge, and we’vealso tackled the radiation resistance of a short dipole. A charge in sinusoidalmotion is just a special case of an oscillating charge, and I worked out ageometric argument which shows how to convert this case to an equivalent shortdipole antenna. So these two formulas are really different versions of the samething. Today I’m going to do a third version, which is also equivalent: theeffective cross-sectional area of a small receiving antenna. If we know thepower carried in the wave, this cross-section gives us the same information, interms of absorbed power, as the radiation resistance. So if we know one, weknow the other…in fact, we know all three.

The key to the calculation isthis picture (see below), showing the interference of two wave patterns. Theplane waves, moving from left to right, represent the incoming power from adistant transmitter. The circular waves, moving outwards, represent theresponse of the receiving antenna. What you have to understand is an antennabecomes a receiver by being a transmitter. The key is the shadow zone I’veshown, where the two wave systems are close enough to be in phase with eachother.

Everyone knows that a parabolais defined by the equation y = xˆ2, but not everyone knows this alternativedefinition: the locus of all points such that the difference of the distancesto a given point (the focus) and a line not on that point (the directrix) is aconstant. It’s called the “focus-directrix definition", and there are similardefinitions for all the conic sections: circles, ellipses, and hyperbolas. Forthe antenna system shown in the diagram, the rough outline of the shadow zoneis defined by the set of points where the phase difference of the two wavesystems becomes 180 degrees, and according to the focus-directrix definition,that outline is a perfect parabola, Actually, it’s a paraboloid of rotation, ifwe look at it in three dimensions.

The calculation I’m about to dodoesn’t rely on any details of electromagnetic theory, except two: I need to assumethat the power in a wave is proportional to the square of its amplitude, and Iwill also assume that the radiated power depends only on that component ofmotion perpendicular to the angle of viewing. Putting these two assumptionstogether, for example, we would conclude that a short dipole puts out 50% of it’speak output when viewed from a 45 degree angle. It’s a physically reasonableresult, and it happens to be true.

Here’s how it works: The wayyou receive power is that you adust the current in your receiving antenna tomaximize the depth of the shadow zone. How do you do this? There are only twothings you can adjust: the phase and amplitude of your antenna current. Youadjust the phase by tuning your antenna with a coil, and you adjust theamplitude by matching your load resistance. Tuning and matching.

We’re not going to worry aboutthe details of tuning and matching: we’re going to focus on the resultinginterference pattern. There’s a nice, quick-and-dirty way of ballparking theanswer, and that is just to assume that inside the shadow zone, we have perfectinterference; and outside, we don’t. It’s the cross-sectional area of theshadow zone that’s a bit dodgy: from the picture, you can see that it dependshow far you are back from the receiving antenna, and there’s a square root relationship.In fact, if you are, say, seven wavelengthsback of the antenna, the the cross-section of the paraboloid will be on theorder of seven square wavelengths.And that will be enough to tell us the effective absorption cross-section ofthe antenna.

Here is how it goes. We backoff reasonably far from the antenna, e.g. 20 wavelengths. The cross-section ofthe shadow is therefore on the order of 400 square wavelengths. Now we ask: howmuch energy is being scattered by the receiving antenna? Suppose the incidentfield has unit amplitude, and let the power in the wave be one watt per squarewavelength. What is the amplitude of the scattered field? Well, it can bealmost whatever we want, because we control it by how we tune and match theantenna. Let’s say, for example, that we tune the antenna so the amplitude ofthe scattered field at a distance of 20 wavelengths is 1% of the incident field.That means the power density is1/10,000 watts/(squ. wavelength), since power goes as the square of amplitude. Multiplying this over thespherical surface at a radius of 20 wavelengths, we get a scattered power of… 4*pi*400/10000,or close to 0.51 watts.

Actually, the scattered poweris a little less than that, because a short dipole does not radiate uniformlyover a sphere. There’s something called the antenna gain, which we can actuallycalculate, but for now I’m going to assume you can look it up. The bottom lineis you only radiate 2/3 as much as I calculated, so it’s actually 0.34 watts.But the real question is: where did that power come from? Answer: it wasabsorbed from the wave, and we should therefore see a weakening of the wave inthe shadow zone.

And of course we do…it’sweakened by exactly 1%, because of the superposition of the two field patterns.Now, how much power do we take out of a wave at unit strength if we diminishits amplitude by 1%? Well, power goes as amplitude squared, so we actuallyreduce it by 2%; and this prevails over the cross-sectional area of the shadow,which we took to be 20 square wavelengths. You should be able to see that wehave taken 0.4 watts out of the incoming wave.

What happened to the missing 60milliwatts, I hear you ask? That’s the magic of an antenna. You have 60milliwatts available to drive a load with, perhaps a small speaker.

Well,that’s convenient. Why don’t we gather more power, then? Instead of a paltry1%, why not go for 2%? All you have to do is let twice as much current flow inyour receiving antenna. If it’s just a question of changing resistors andtweaking a coil…

A nastything happens if you try to up your radiated amplitude to 2%. Radiated powergoes as the square of the amplitude, so instead of pumping out 0.34 Watts, you’renow wasting 1.36 Watts in all directions. But the absorbed power, which happensin the shadow zone, is different. Instead of reducing the amplitude of theincident field in the shadow zone by one percent, giving you 98% power density,you now reduce it by two percent, giving you…96% power density. Your powerabsorption is only linear with the antenna current. You only absorb 0.8 watts,which isn’t enough to cover the wasted, re-radiated power, never mind driving aspeaker.

Wheneveryou have two quantities, and one is linear and the other is quadratic, thelinear quantity dominates first, and then the quadratic takes over. Thereceiving antenna works that way. At first, the more current you allow to flow,the more useful power you absorb. But then the wasted power starts to growfaster, and it dominates. There is a maximum useful power you can draw. I havedrawn a graph of it here (I worked out the numbers so it agrees with the calcuations I've done so far:

You cansee that the maximum useful power you can capture with this antenna is 0.22Watts. SInce we are working with unit power density (1 watt/square wavelength) that also happens to be the amount of power flowing through 0.22 square wavelengths, and sowe call that the absorption cross-section. Or we can also talk about thescattering cross-section, which is basically when you short-circuit the antennaand maximize the current. That’s of course four times as much power. (You can see why from basic circuit theory and load matching!) Either way, onceyou’ve got the cross-section, you’ve implicitly determined the radiationresistance. You just have to convert wave power to volts-per-meter, mutliply bythe length of the antenna, and push your numbers around until everything agreeswith everything else.

Actually,looking it up on Wikipedia I see they give an actual cross-section (they callit “effective aperture) of 0.11 square wavelengths, which is just half of whatI got. It’s not surprising I’m out, since I used the coarsest imaginableapproximation for the effective cross-section of the shadow zone. There’s an integration that gives me an exact value, and oddly enough, in thiscase I’m actually capable of doing the calculus. It’s just that the math tendsto get in the way of the physics, and it’s so much easier to describe what you’redoing when you keep it simple.

Still,I promised I was going to try and get this one exact for a change, so maybe we’llcome back tomorrow and see if we can’t work in that missing factor of 2.

Deriving the Larmor Formula

2012-02-26T11:57:00.000-08:00

I promised to derive the Larmor Formula and yesterday Idrew up an outline of how the calculation is supposed to go. It’s based on theequilibrium between mechanical and electromagnetic energy in a cavity. Reviewingthe outline, we can see that these are the key steps:

1. Set up a rigid rotor inside the cavity with arbitraryparameters of mass, length, and charge. Let it come to thermal equilibrium insidethe cavity with an arbitrary amount of energy. By the Equipartition Theorem,that will also be the energy per modeof the electromagnetic field.

2. Pick an arbitrary size for the cavity and use theRayleigh-Jeans formula to count up how many cavity modes there are in thevicinity of the rotor frequency, as already defined by its energy anddimensions. Pick an arbitrary bandwidth and only count the number of modeswithin that bandwidth.

3. From the mode energy of the field and the size of thecavity, calculate the electric field amplitude of the mode. Add up all themodes within our arbitrary bandwidth as though they were a Fourier Series, withthe amplitudes in phase at an arbitrary point. The result will be a pulse trainof frequency bursts with a characteristic amplitude, burst length, andrepetition rate.

4. Imagine a harmonic oscillator with the same mass as ourrigid rotor, and the same characteristic frequency. Apply a single frequencyburst, as calculated in Step 3, to this oscillator at rest, and calculate theresulting amplitude of oscillation.

5. From the amplitude determined in (4), calculate theequivalent energy. By analogy with the Drunkard’s Walk, this is the averageamount of energy gained from each pulse burst. Therefore, the rate of powerabsorption is just this amount of energy multiplied by the repetition rate(from Step 3).

6. The system is in balance when the rate of absorption isequal to the rate of emission. Assume that the emission rate is proportional tothe square of (acceleration)*(charge). From all the parameters determined sofar, calculate the constant of proportionality. Compare the result to theLarmor Formula.

Shall we begin? I already did the first couple of steps theother day. Let’s pick up where we left off.

1. The rigid rotor is a single mass on the end of a swivelarm attached to a post, so it is constrained to spin about one axis only. We gaveit these arbitrary parameters:

Mass =10ˆ(-27) kg

Charge= 10ˆ(-20) Coulomb

Radius of gyration= 25 nanometers

Frequency at Equilibrium = 100 GigaHertz

From these parameters, we can also derive the energy:

Energy at Equilibrium = 1.25 x 10ˆ(-19) joules (or 125 nanopicojoules).

2. We put our rigid rotorinside a cubical cavity 30 cm on a side. From our frequency, we can calculate awavelength of 3 millimeters, so the cavity accommodates exactly 100 wavelengthseach way. Now we use the Rayleigh-Jeans formula to count up how many modesthere are between 99 GHz and 101 GHz. This is actually a calculation we didlast year, and we counted up approximately 250,000 modes. But this year we’regoing to notice that only some of those modes are effective in driving therotor. Considering three orthogonal axes, it is clear that two out of threeorientations will have no driving force on the rotor. So we will divide bythree and work with only 80,000 modes.

3. From the size of the cavityand the energy of the standing wave, we can calculate the peak electric fieldamplitude of a typical mode. Correct me if I’m wrong, but I make it 1.0 millivolt/meter.That’s not exactly an accident: I picked the rotor energy to make it come outthat way. Now comes the Fourier magic. We know that we have to add up 80,000sine waves, centered on a frequency of 100 GHz. Spreading these modes out overour chosen bandwidth (2 GHz), we find that the average spacing between modes is25,000 Hz.

The calculation is actually easier than it looks. We start off justby focussing on the middle frequency and it’s two closest sidebands:

I’ve cut the sidebands in halfso this picture represents the very familiar AM radio calculation for amodulated sine wave. We know what the result looks like:

It’s a modulated sine wave. Forpurposes of description, I’m going to pretend it’s an “equivalent”square wave,so I can use the language of duty cycle and repetition rate. The nice thingabout the calculation is that the basic repetition rate isn’t going to changeas we add more sidebands. The additional sidebands only shape the pulse withinthat framework. It’s not hard to see that the peak amplitude grows by twomillivolts each time you add another pair of sidebands; so the height isproportional to the number of sidebands. What about the burst duration? Itturns out to be just inverselyproportional to the number of sidebands. You can see that it has to be thatway, because that’s the only way the total energy of the burst will beproportional to the number of sidebands. (I explained how to do these Fourier series last year.) The result is a train of pulse bursts with the following parameters:

Amplitude: 80 volts/meter

Duration: 500 picoseconds

Repetition Rate: 25,000 persecond

4. The next thing we have to dois apply one of these pulse bursts to a harmonic oscillator initially at rest. I’m going to assume that in this situation theforce is always applied with maximum efficiency, that is: it is always in synchwith the velocity.With this assumption it is fairly easy to ballpark how muchmomentum it pumps into the oscillator: it is just ½ *(force)* (time). (Thefactor of ½ is due to the pulsating character of the driving force. The forceis of course given by the (electric field)*(charge); put the numbers in, and Iget a momentum of of 2*10ˆ(-28) Newton-seconds.

5. How much energy did thatpulse burst impart to the oscillator? Using the formula pˆ2/2m, this convertsto an equivalent energy of... 2*10ˆ(-29) joules. Since there are 25,000 pulsebursts per second, it means the oscillator will be absorbing power at a rate of of 0.5 * 10ˆ(-24) joules per second, orone-half pico-pico-watt.

Remember that funny property ofthe random walk: the amplitude-squaredgrows at a constant rate. Similarly with the driven oscillator: the random phase of the incoming wave burstcorresponds to the random directionof the drunkard’s footstep, so in this case it is the energy of the oscillator that grows at a constant rate…a rate of0.5 ppWatts.

6. Here’s where I mix metaphorsjust a little. I’ve been building up the oscillations as though I have aharmonic oscillator…a mass-on-a-spring. But now that I’m there, I’d rather applyit to the case of the rigid rotor. At the point of equilibrium, it’s the same frequencyand the same energy in either case. The way it works is that you assume theradiated power is proportional to (square of the charge) x (square of the acceleration).This is a totally straightfoward assumption that is justified by all kinds ofphysical reasoning. All we need to do is figure out the constant ofproportionality. The charge we know: 10ˆ(-20) coulombs; the acceleration iseasy to get from omega-squared-r. Put them together and square them and I getvery nearly 10ˆ(-8) in SI units. Recalling that we are putting out a radiativepower of one-half pico-pico-watts, this means the constant of proportionalityis…5 x 10ˆ(-17).

How does this compare with theLarmor Formula? It’s pretty close. Recalling the formula with the permittivityof free space and all that, we see that all the constants multiplied togethercome to…2.2 x 10ˆ(-16). I seem to be within a factor of 4 of the exact answer.

I have to say that’s not a badoutcome. With all the approximations I made and the short-cuts I took, thereare so many places I could haveslipped in an extra factor of 2…which incidentally would have given me an errorof 4x on the power. So all things considered, I’m really lucky I came out asclose as I did. I could have easily gone back and patched it up afterthe fact, to make it look like I got it perfect, but I think it’s better toleave it as it is.

Finally, we shouldn’t forgetthat having basically derived the Larmor Formula, we’ve also effectivelyderived the radiation resistance of a short dipole…since I worked out thephysical correspondence between thosetwo formulas last week. I know I did a derivation for the antenna resistancelast year by letting sheets of current flow on a sphere, but that was actually a prettysketchy calculation. Now we’ve done it right...in theory, at least. Regardless of calculation errors, the method obviously works in principle.

It turns out that I have onemore method for calculating the radiation resistance of an antenna; it’s verydifferent, and as I don’t seem to get tired of saying these days, it’s also a very cool calculation. I’m goingto post it one of these days.

Every good calculation should have a road map.

2012-02-24T16:00:00.000-08:00

This is such a cool calculation. There is so much in here, it would be a real shame if the physics gets lost in the details. So before we continue, let's take a step back and recall what we're doing.

First, we're setting up a thermal equilibrium between a mechanical oscillator and the radiation field. You might think you need to have a statistical quantity of oscillators, but you don't. It's the same equilibrium if you have just one single charged oscillator. So I've created an artificial tether-ball in the middle of an empty box. Actually, it wouldn't have hurt to have filled the box with uncharged balls the same size as the tether-ball, and let them bounce about freely. We could have given them a temperature, and they would have imparted random motion to the constrained, charged ball. Since the tether-ball has only one degree of freedom, it would take on the same average energy as all the other balls. Sometimes it would have less energy, and in those circumstances the subsequent collision would by more likely to add energy to the tetherball. Sometimes it would have more than average energy, and...well, you get the idea. When it happens to have the average energy, it is no more likely to gain than to lose with the next collision...it is in mechanical equilibrium. The real point to understand is that at that moment, it is also in equilibrium with the radiation field. You should be able to see why this has to be so. The bottom line is you only need one charged oscillator in the box to establish the equilibrium between matter and radiation.

Now we invoke the equipartition theorem, which says that the energy per mode of the radiation field has to equal the energy per mode of the mechanical system. This is the same principle that supposedly leads to the ultraviolet catastrophe, but we really don't need to worry about that here. What people forget about this principle is that it is strictly true only in the vicinity of a specific frequency. There might be different average mode energies at 200 Megahertz versus 200 Gigahertz, but at each frequency the electrical mode energy will equal the mechanical mode energy. The idea that a mechanical oscillator at 200 Gigahertz must have the same energy as one at 200 Megahertz is a misapplication of the equipartition principle.

Having said all that, for the purposes of the present calculation it is rather a moot point. We can choose our numbers so that the frequencies are all within the Rayleigh-Jeans regime where equipartition prevails in the broader sense...that is, all frequencies have the same mode energy regardless. It ought to be noted, however, that the validity of this calculation will ultimately not depend on any such assumption.

The equilibrium will be independent of any specific properties of the oscillator. That means we can give it any arbitrary properties we like. In our case, we have chosen to configure it as what we call a rigid rotor, with a certain arbitrary mass, charge, and radius of rotation. One degree of freedom, which defines its frequency and energy at equilibrium. The radiation field must then be at equilibrium with the rigid rotor, or the tether-ball as we call it.

But how can our charged tether-ball be at equilibrium if it is spinning about, radiating energy like crazy? There is only one way out: it must be also absorbing energy at the same time, and it must be absorbing energy at the same average rate as it is radiating energy. If we can possibly figure out how much energy the tether-ball is absorbing, we will automatically know how much it is radiating. And wouldn't that be something. You tie a charged ball to a string, and whirl it about your head, and now you can calculate how much power you're radiating. Without looking up any formulas. You can actually figure it out.

I'm not saying we won't use any formulas. We'll use mostly basic formulas that everyone knows, like the force on a charged body and the energy of an electric field. We'll do some stuff with Fourier Series that you can ultimately verify with high-school level trig identities. And we'll use one formula that you could theoretically figure out by yourself, but each time I try to do it I end up off by a factor of three, so I caved in and copied it from the internet. That's the famous Rayleigh-Jeans formula for counting the modes of the electromagnetic field in a rectangular box. The point is that none of these formulas tell you how much radiation you get from an accelerating charge. That's the result we're going to come up with, and ultimately we'll get it by doing nothing more than enforcing the condition of equilibrium between the electromagnetic field and the mechanical oscillator.

It's a very funny trick, and to be honest I'm still not really sure why it works. Part of the secret is that there is a fundamental asymetry between absorption and emission of radiation. Emission of radiation is a huge conceptual mystery: the charge is giving off energy as it oscillates, which means you are doing work to shake the charge: but what is the force which is reacting against the shaking charge, the force which you are working against? I'm not sure anyone really knows.

The absorption of energy is different. An oscillating electric field comes along and pushes on the charge, so the charge starts to oscillate in response to the field. It's pretty straightforward. Actually, even here there is a hidden mystery: we agree that the charge moves, gaining energy, but how does this process remove energy removed from the electromagnetic field? That's a story for another day. The point is that unlike the case of radiation, for absorption we have a pretty reasonable calculation.

It's a reasonable calculation, but it's far from a walk in the park. There are two pretty huge problems facing us. First, the electromagnetic field. What is it? Why, it's just equal to the average energy of the rigid rotor, which is an arbitrary number we said we can just pull out of our ass. No, that's not quite right: we said that energy would be the average energy per mode of the electromagnetic field. We still have to add up all the modes to get the total field.

Add up all the modes? But there are billions of modes, and they are all all kinds of crazy frequencies. Why not just take the ones that are at the same frequency as the tether-ball? Because there is not one mode out of all those billions that has the exact frequency we want. There are all kinds of modes very close, but how close is close enough? It's a horrible mess.

I was stuck on this point until I made an amazing discovery that changed everything. It's that business with Fourier Series that I mentioned earlier. You add up a bunch of modes, starting with the ones closest frequency to your target, and moving outwards as you go. And then you just stop. You truncate your series. Where do you truncate it? Anywhere you want. It turns out you're going to get the same final answer for the energy of the tetherball regardless of where you truncate your series. It's hard to believe, but I show why it works here.

The Fourier calculation converts your random, distributed field into a uniform chain of pulse trains of a specific size, frequency, and duty cycle. Depending on where you cut off your frequency band, you get different parameters for those three values of the pulse train. That's a bad sign: it means you're not calculating anything that is physically real in a measurable sense. But here's the catch: when you apply any one of those pulse trains to your tether-ball, you get the exact same final result. So who cares if they're real or not? How can you doubt that the result you get is consistent with the true physical result? (Or at least a very good approximation to it.)

But let's not get ahead of ourselves. The result of the Fourier calculation is a train of pulses - frequency bursts, actually. We apply that pulse train to the tether-ball, and it twirls about its post. How fast does it go? It turns out we can figure that out using the theory of the drunkard's walk. Everybody knows that the expactation of the distance traveled by the drunkard from the lamppost goes as the square root of the number of steps. But for an amplitude of oscillation, which is what we are dealing with here, the energy of oscillation is proportional to the square of the amplitude. Since it turns out that the amplitude of oscillation corresponds to the distance of the drunkard from the lamppost, the conclusion is that the energy of the oscillator grows linearly with time. Each pulse train adds an equal amount of energy to the oscillator!

What oscillator am I talking about? I thought we were going to drive a tether ball. That's not a harmonic oscillator, it's a rigid rotor. Well, I have to admit I'm cheating a bit here. The first time I did this calculation I applied it to a harmonic oscillator. When I came back to it the other day, I decided the rigid rotor would be a better case to work out, for a couple of reasons. I have an easier time restricting the degrees of freedom. I have a very convenient formula (the Larmor Formula) to check my work against, whereas with the harmonic oscillator I was having trouble getting it to fit into the antenna formula for the short dipole. And not least, I like the image of the tether-ball. I think it's evocative.

The problem with the tether ball is it's hard to get it started. To get it up to speed, you have to drive it through the whole spectrum of frequencies. That's a problem. The nice thing about the harmonic oscillator is that you're working with the same frequency through the whole range of amplitudes. So I end up returning to the old reliable mass-on-a-spring after all. It's kind of cheating, but you can ultimately justify it. After all, you could imagine some kind of variable-tether mechanism whereby the ball is let out gradually as it revs up, maintaining the same frequency the whole time. But more importantly, once you actually get up to speed, the equilibrium conditiions are exactly the same for the tether-ball and the mass-on-a-spring. So with apologies, I'm going to start the system off as a harmonic oscillator. It's a bit of a mixed metaphor, but it works.

Here's how it works. You have this oscillating charged mass, and you hit it with one of these standard pulse trains that you've worked out. This gives a certain impulse to speed up the oscillation...or does it? The problem is you don't know the relative phase. Isn't it just as likely that your pulse train arrives 180 degrees out of phase, so it slows down the oscillator instead of speeding it up? Or any other phase in between? In fact, you have no idea what the outcome of the interaction is going to be.

That's why it's just like the drunkards walk. He takes a step in a random direction, and you have absolutely no way of knowing whether he's getting closer or farther from the lamppost. It turns out to be slightly more probable that he will end up farther, but it's totally random. Except, that is, in one particular circumstance: the very first step. You know that if he starts out right under the lamppost, that after the first step he will be...exactly one step away from the lamppost. And the harmonic oscillator works exactly the same way!

We have to apply one of our standard frequency bursts to our mass-on-a-spring when it is perfectly at rest. In that case, and in that case only, can we calculate the outcome of the interaction. We can calculate the resulting amplitude of oscillation, and also the energy of the oscillation. With subsequent frequency bursts, the amplitude will grow more and more slowly, like the drunkard's walk. But the energy, since it is the square of the amplitude, will grow...on average...linearly with each frequency burst. Each frequency burst adds, on average, the same energy to the oscillator. And we know how big the bursts are, and how far apart they are.

Does the oscillator grow without limit? No it doesn't. Because as it oscillates, it begins to radiate. The problem is, we don't yet know the laws of radiation for a harmonic oscillator, or a rigid rotor, or any kind of accellerating charge. Well, that's not quite right. We ought to know on general grounds that the radiation is quadratic with the amplitude of oscillation; and as goes the amplitude, so go the velocity and accelertation. We will guess that the formula for radiation should be in some way proportional to the square of the acceleration.

So what are we missing? We need to calculate how much energy our frequency burst puts into a stationary oscillator. That's going to be pretty much an F=ma calculation; it shouldn't be an obstacle. Since we know how often the pulse trains are coming, we know the rate at which energy is being absorbed: in fact, it's being absorbed at a constant rate. How about the rate at which energy is being radiated? From what we just said, it's clearly proportional to the total energy of the oscillator, which is growing at a linear rate. In other words, at some point, the radiation must overtake the absorption. That is the point of equilibrium, and we already know where that point is!

At the point of equilibrium, we now know the amplitude of oscillation, and we also know how fast it is radiating. The formula which relates these two quantities is nothing other than the Larmor Formula, and our data must agree with it. All we need is to divide quantity A by quantity B to derive the constant of propotionality which links them. That will be our job for tomorrow.

Equipartition and the Larmor Formula

2012-02-24T08:47:00.001-08:00

Today I’m going to do something that’s probably never beendone before. I’m going to derive the Larmor Formula from without using Maxwell’sEquations. I’m not going to use anything except the Equipartition Theorem.Well, maybe I’m going to cheat a little. I’m going to use the Rayleigh-Jeans formulafor the number of modes in a cavity. I’ve used this formula once before. Itlooks like this:

Here is how the derivation is going to proceed. I’m going tostart with a box 30cm x 30cm x 30cm. Inside the box I’m going to construct atiny, miniature tether ball. You know what a tether ball is. It’s a ball thatyou tie to the end of a pole, and then you whack the ball so it flies aroundthe pole. Except I don’t think I want the string to wrap around the pole, so I’mgoing to have a rotating sleeve where the string is attached. So the ball cancircle indefinitely without getting wrapped up.

I need to specify the exact parameters of the tether ball.It will weigh 10ˆ-27 kilograms, and the length of the string will be 25nanometers. Oh yes, and the ball will be charged. I will give it a charge ofexactly 10ˆ(-20) coulombs. I think we’re all set.

Now we go up to the tether ball and give it a good hardwhack, so it starts spinning around the pole. Since it’s charged, it startsspraying electromagnetic radiation into the box. The radiation goes in alldirections, bouncing off the walls; at the same time, the tether ball slowsdown, because it’s losing energy. Does it finally come to rest? No…because atsome point, there is so much radiation in the box that sometimes instead oflosing energy, the tether ball actually absorbsenergy from the radiation. At some point, on average, the ball is absorbingjust as much radiation as it is emitting. That’s called thermal equilibrium. Andthat’s what I’m going to calculate, right?

Not exactly. I’m going to do something a little differenttoday. We know that the system has to come to equilibrium: what I’m going to dois make use of a peculiar fact about that equilibrium to calculate theradiative power of the tether ball, otherwise known as the Larmor Formula. It’sa very backwards way of doing it, but it ought to work.

What is the singular property of the system in equilibriumwhich will give us the key to the calculation? Simply this: that for a given frequency, the energy per mode of the electromagnetic field is equal to theenergy per mode of the mechanicalsystem.

That’s it. It’s easy to state, and not so hard to interpret.Can it really be enough derive the Larmor Formula, which tells us the amount ofradiation from an accelerating charge? Let’s give it a try.

I know what some of you are thinking: "It's wrong! The equipartition formula fails! It leads to the ultraviolet catastrophe!" Calm down. I never said that every mode of the e-m field gets the same amount of energy. I said that the equivalence between electrical and mechanical mode energies holds true at any give frequency. That's true classically, and I'm going to show that it remains true in quantum mechanics.

I’m going to make one more adjustment to the system beforewe get started. Instead of using a piece of string to attach my tether ball tothe pole, I think I’ll use a rigid massless rod, with a sleeve bearing on theend of it. So the tether ball is constrained to rotate around one axis only, ata fixed distance from the pole. I think that might simplify the calculation abit.

So here goes: we whack the ball, it spins like crazy, itfills the box with radiation, and eventually we come to some equilibrium wherethe ball stops slowing down on average. Let’s suppose that when we get to thispoint, the ball has an average velocity of 16,000 meters/sec. It could beanything, after all; it depends on how hard we whacked the ball to get itstarted. Let’s say it’s 16,000 m/sec.

It’s not too hard to convert this to equivalent energy: wealready know the mass of the ball, and if you put the numbers in it comes to 1.25x 10^(-19) joules. It’s 125 nanomicromicrojoules. Or 125 nanopicojoules,if you prefer.

The Equipartition Theorem tells us that energy distributesitself equally among all modes. The spinning of the tether ball is exactly onemode: each standing wave in an electromagnetic cavity is also a mode. So eachelectromagnetic standing wave mode should have, on average, 125 npjoules ofenergy.

How do you calculate the energy of an electric field? You take the squareof the field, divide by the impedance of free space, multiply by the volume ofthe box and divide by the speed of light. Or something like that..I think you also have to dived by two forsome reason vaguely related to RMS values. I find if I start with 1.0millivolts/meter, it comes out just right…a nice round number, isn’t it? Ofcourse, that’s no accident. I cooked up all my parameters so it would end upthis way.

Now let’s see what else I’ve cooked up. I gave the tetherball an average velocity of 16,000 m/sec, on a radius of 25 nanometers. If I’mright, this gives me an average frequency of exactly 100 GigaHertz. This givesme a radiated wavelength of exactly 3 millimeters, which fits quite nicely intomy 30-cm box.

This is starting to look like a calculation I did once before, exactly fifty-one weeks ago. I had a different angle on it back then: Iwas going to show that by calculating the equilibrium between the field and anatomic oscillator, you could prove the Equipartition Theorem. I did a lot ofpretty loose approximations and came up within a factor of ten of the exactanswer. Which was pretty good at the time, all things considered. But since Ifigured out the other day that I had been using the wrong approximation for theantenna-to-atomic-oscillator correspondence, I now had the tantalizingpossibility of recovering a missing factor of five. Hence today’s calculation.But along the way I decided to get fancy and do it backwards: instead ofdemonstrating the equipartition theorem, I’m trying to use the equipartitiontheorem to derive the radiation of an atomic oscillator.

What it all means is I’m finding a lot of corrections andadjustments in my original calculation. These things tend to equalizethemselves in the long run when you’re just doing ballpark work, which is why Iwas OK with being within a factor of ten last year. But now I’m trying to doaccurate work, and it’s stressing me out just a bit. Never mind. We mustpersevere.

We’re getting close to the real meat of the calculationright now. This is based on a brilliantinsight which I figured out last year. The problem is to calculate the effectivefield at the oscillator. But haven’t we already done that? I cooked up thenumbers so that my electric field would come to 1 millivolt per meter...whatmore do I need?

We’re not even closeyet. That 1mV/meter is the field amplitude permode. There are millions of modes, all randomly adding up with differentphases and frequencies. Yes, it would be simple if we just had to calculate themotion of a tuned oscillator driven by a 100-mHz field of known amplitude. Thatis so not what we have to do.

The brilliant thing I figured out last year was that I couldapply the paradigm of the drunkard’s walk to the driven oscillator. I canhardly begin to describe what a brilliant idea this is. (Yes, that’s the thirdtime I’ve used the word brilliant.Does that get me points on the crackpot index?)

The crux of the idea is this: I can truncate my fielddistribution anywhere I want, and I’llstill get the same result. I have a formula for the number of modes in acavity. I can take all the modes between 99 GHz and 101 GHz, or I can just takeonly the modes between 99.9 GHz and 100.1 GHz. It doesn’t matter: I’m going toget the same result when I add them up and apply them to the harmonicoscillator. I demonstrated how that works in this article from last year.

Let’s pick up this calculation again when I come backtomorrow.

The equivalent antenna of an atom

2012-02-23T20:02:00.000-08:00

Here’s the problem we were working on when I left off yesterday.I wanted to use the classical antenna equations to calculate the amount ofradiation from a hydrogen atom. Specifically, I was looking at the s-ptransition, at the point where the atom was just 1% excited. I already showedthat the classical calculation gives you the same radiated power as youcalculate Copenhagen-style with the Einstein A coefficient and Fermi’s GoldenRule. I got it to come out exactly right using the Larmor Equation for theradiation from an accelerating charge. I was just having some trouble gettingthe same result from the antenna equations.

It might seem like a small point, but it turns out to havehuge consequences for me, which I’ll get to at the end of this post. But let’sstart off by remembering where we were yesterday. The Schroedinger equation of.995*s + 0.1*p (that’s an atom excited to the extent of 1%) gives me anoscillating dipole moment of 8 picometer-electrons. I did some rough geometryto equate this to a dipole antenna 32 picometers in length. I wasn’t veryexplicit about how I got that, but it’s a fairly straightforward geometricalargument using moments. Here is a picture of how it looks:

The first picture shows the Larmor Atom, which is pretty close to the actual atom except it's oscillating in two dimensions instead of one. You can see that I've drawn an equivalent antenna with the same dipole moment; I hope it's clear why the first one has a radius of 8 picometers, and the second one has a length of 32 picometers. They both have the same oscillating dipole moment.

So if that's my equivalent antenna, it should be pretty easy to use the antenna equations. I showed yesterday that for the frequency in question (2.5*10^15 Hz) you get a radiation resistance of about 12 micro-ohms. So all you need is to put the current in and calculate i-squared-r. But that's easy. The current is just the charge times the frequency, right?

Wrong! That's what was screwing me up so bad. It's actually really tricky to figure out the equivalent current. I finally worked it out like this:

This graph represents a sinusoidal current with unit amplitude. It delivers a total charge of 2 coulombs in pi seconds. The frequency is 1/2*pi and the charge is 2. If I said that the current was just the charge times the frequency (which is what I said it would be yesterday!) I would calculate a current of 1/pi. But that means I'd be out by a factor of pi!

It's worse than that because I also forgot that the antenna formulas are written in terms of RMS currents...the electrical engineers's standard. Which throws me out by another factor of sqrt(2). It turns out that the two errors are in opposite directions, so the compound error works out to 2.2. Since current is squared in the power formula, this gives me a factor of 5. Which is exactly the amount I was out by.

There's no reason to torture you by walking you through the numbers: you can do it yourself if you really care. The point is it came out perfectly. The antenna calculation agreed with the Larmor formula, which agreed with the Fermi Golden Rule calculation. So everything was fine after all.

But then it hit me: exactly one year ago, I was trying to do a similar calculation where I equated the mode energy of the electric field in a cavity to the mode energy of a mechanical oscillator. It was a very cool calculation but no matter what I did, my numbers were still out by a factor of ten! Could it be the same error, and if so, was it in the right direction? I'd hate to go and fix it up only to find that now I was out by a factor of fifty or a hundred.

When we return, I'm going to check this out.

The antenna calculation

2012-02-22T16:05:00.000-08:00

Yesterday I said I couldn’t get the radiation emission fromhydrogen to come out correctly, using classical antenna theory. Does this meanthat atoms do not behave like classical antennas? Possibly. The otherpossibility is that I was messing up the calculation. Which could it be?

Eventually, in desperation, looked up the formula for theradiation from a charged body moving in a uniform circular orbit. It’s calledthe Larmor Formula, and it’s a little more general than that: it claims to givethe radiation for any acclerating motion. I adapted the atomic equations to thecase of uniform circular motion, and I got exactly the right amount ofradiation. The semi-classical picture agrees with the Copenhagen picture.

So why wouldn’t the antenna equations work? That’s a toughone, but I think I have the answer. The proper thing for me to do now would beto show you how its done correctly, but I can’t bring myself to do that. Forsome reason, I feel I have to first show you how I did it wrong. A though you were to somehow be in danger of making the samemistake as I did. So here goes:

Everybody knows that the radiation resistance of a half-wavedipole antenna is 73 ohms. Right? I mean, we have to start somewhere. Iactually did a calculation a couple of months ago to justify those 73 ohms. Itwasn’t really a very good calculation, but it kind of at least gets you in theballpark. I actually have a much better way of doing it, where you can actually see the physical basis for what's happening, but again it's just a ballpark calculation. The three-dimensional calculus is just too hairy, so I have to take some shortcuts. Someday I'll write it up and post it.

But for today, I'm actually going to just use a formula we can find on the internet. It's not the formula for the half-wave dipole, it's the more general formula for the short dipole, where by "short" we mean much shorter than the half-wave. Since atoms are on the order of anstroms, and they emit light in the optical or (at worst) ultraviolet spectrum, this qualifes them as "short" dipoles. And as such, we can use this formula for their radiation resistance:

You can check this formula for the case of lambda = 1/2: you see it gives you a value of about 50 ohms, which is a little on the low side. That's because the half-wave dipole is a little more focussed than the short dipole. It makes sense.

Now we can plug our numbers in and see what to expect for the power output. First, compare the radiated wavelength to the atomic dimension. We already noted that the wavelength is 120 nanometers, and the dipole moment is 8 picometers. Actually, that's not quite the right figure to use for the ratio: that's the one-sided moment, so we should double it up for full distance the charge travels. And that's not all either. What is the dipole moment of a short electrical antenna? Is it equal to its full length? Not really. In a small dipole, the current distribution is such that you get a uniform distribution of charge, equal and opposite, on the two legs of the antenna. The center-of-mass of an equal distribution is half-way along the length: so the dipole antenna which has the same moment as the atom should actually be twice as long. All together, I get a length of 32 picometers. I think this is right.

From the ratio of these two lengths, using the formula you should calculate a radiation resistance of

about 12 micro-ohms. Now we just have to figure out the current and plug everything into the formula "power equals i-squared-r". Are you ready?

The Semi-Classical Calculation

2012-02-22T06:38:00.000-08:00

Yesterday, I set up a sample of hydrogen atoms in an excited state and calculated the radiation output according to the standard Copenhagen picture. Or, at least, I walked you through Richard Fitzpatrick's lecture on the University of Texas website. I took the sample to be 1,000,000 atoms, with 1% of them in the 2p state. (I ignored the possibility that some of the excited atoms should have been in the 2s state, but I don't think that matters. Except it would have screwed up my temperature calculation.) Using these numbers, I calculated a power output of 10 microwatts.

Then I said I was going to redo the same calculation, this time treating the atoms as tiny classical antennas. Since everybody knows that Maxwell's Equations don't apply to atomic systems, we should get nonsense. Let's see what actually happens.

I'm going to admit that I sweated bullets doing this calculation. I probably did it at least twenty times and got a different answer each time. The antenna calculation just wouldn't come out right. The best I could do was still out by a factor of ten. In desperation, I looked up the formula for the radiation from an accelerating charge. It's called the Larmor Formula, and I found it on the Wolfram Alpha website:

This gives us a formula to work with: but what numbers do we put in the formula? Well, there's not too much to worry about; we've got a couple of constants, which are provided in the table; we've got the charge on the electron, which everyone knows is 1.6x10^-19. The only other thing we need is the acceleration.

That's the hard part. What is the acceleration of the electron, according to the Schroedinger picture? Fortunately, Professor Fitzpatrick has already done all the hard work for us. I said there were two alternative pictures of the physics. In the Copenhagen picture, one percent of the hydrogen atoms are in the excited state. In the Schroedinger picture, each of the atoms is in a mixed state to the extent of one percent excited. To calculate the dipole moment of such a mixed state, you do the bra-ket thing with x in the middle. Actually, the order of operations doesn't really matter in this calculation (as it sometimes does in quantum mechanics). You can just think of this as the square of the amplitude (that's the product of the bra and ket states) integrated against x to get the dipole moment. The calculation looks like this:

What exactly are we seeing here? I said that the atoms were one percent excited, but here I'm showing the P state with a ten percent amplitude. Is that a mistake? No, because don't forget you square the amplitude to get the density. And why don't I have any s^2 or p^2 terms in my result? Because there is no dipole moment for a pure state: because the s^2 and p^2 states are symmetric (even) about the origin, the integral against x (an odd function) gives you zero. No, the calcuation reduces to the dipole coupling of s versus p.

But this is the calculation which the Professor has already done for us. Recall from yesterdeay:

He's actually given us the dipole moment for all three p states in terms of the Bohr Radius of 5.3*10^-11 m. (It's the third value which pertains to our case, the one with the extra factor of root 2.) Working out all those powers, it comes to about 75%, or 0.4 Angstroms. We also mustn't forget the factor of 20% which came out of the 1% excitation: so the dipole moment in our case actually comes to 8 x 10^-12 meters. (Multiplied by the charge of the electron, of course.) It's funny how much dipole moment you actually get by mixing in just 1% of the p state, but there it is.

Most importantly, let's notice that this dipole moment is an oscillating dipole moment. Because the s and p states precess through time at different exponential frequencies. So based on the difference in those two frequencies, what is initially a positive dipole moment becomes, after one-half cycle, a negative moment. It oscillates.

We've now got almost all the numbers we need to do the Larmor calculation. We still need the acceleration of the charge. We've got the oscillation amplitude - that's the 8 picometers - and we've got the frequency from yesterday - that was 2.5 x 10^15 Hz. Now I'm going to do something pretty slick. Instead of working out the acceleration due to sinusoidal motion, I'm going to pretend I'm working with uniform cirular motion instead. That's not what the atom is doing: in fact, it will throw me off by a factor of two, because the energy output of the circular motion is just the superposition of two orthogonal harmonic oscillators. No problem...I'll just divide by two at the end. The nice thing is I get to use the omega-squared-r formula for circular motion that I still remember from high school physics. Remembering the factor of 2-pi to change from hertz to radian frequency, I get an acceleration of 2*10^21 m/sec^2.

Putting all the numbers together, I get a power output of...20 picowatts. (You can check this if you like by plugging the numbers into the applet on the Wolfram site.) Divide by two to convert circular motion into simple harmonic motion, multiply by one million for the number of atoms in the sample, and I get, incredibly, exactly the same power output that I calculated yesterday. The Copenhagen picture with its quantum leaps gives exactly the same result as the Schroedinger picture with its tiny oscillating dipoles.

Not convinced? There's one more calculation that really needs to be done to bring this argument full circle. I said originally that I was going to use antenna theory to do the calculation, and when I tried, the numbers wouldn't come out. I kept getting two microwatts instead of ten. How do you get a factor of five for an error? It didn't make sense, and it was driving me crazy. But now I've worked it out so the antenna calculation comes out right. That's a story for my next blogpost.

The Copehagen Calculation

2012-02-21T07:49:00.000-08:00

I set myself the challenge yesterday of showing that the semi-classical antenna picture of the hydrogen atom gave us the same physics as the Standard Model, the Copenhagen picture of quantum mechanics. Actually, it was more of a reverse-challenge: the idea was to find a single example whereby the antenna picture fails to give the correct physics. If classical electromagnetics is the miserable failure at the atomic level that everyone seems to say it is, it should be easy to find examples.

The spontaneous transition of hydrogen from the excited p-state to the ground s-state is as good a place as any to start. We can readily compare the predictions of the Standard Model with those of Maxwell's Equations. I haven't done the calculation yet, but we'll see how it goes. Let's open the show with the Copenhagen Version.

In Copenhagen, the hydrogen atom starts off in an excited state, and there is a certain probability that it will instantly jump to the ground state, emitting a photon in the process. Of course, this is wildly out of whack with what the semi-classical picture says; but that doesn't matter. What matters is the measurable outcome of an experiment; and in practise, the simplest way to measure this is to prepare a sample of hydrogen gas with a known energy and measure the radiation which it gives off at that transition frequency.

It has to be a pretty low energy, because we want to be dealing with essentially a two-state system. If we give it too much energy, then other higher-energy states start to become "populated", and the system gets more complicated. Let's say we have one million hydrogen atoms, and we give the system enough energy so that one percent of them are in the excited state. Then we can calculate how much they should radiate.

It's a pretty sophisticated calculation, but fortunately we have some help from Richard Fitzpatrick's excellent farside website at the University of Texas. He starts off by computing what they call the dipole coupling between the ground state and the three p-states, taking the Bohr radius a0 as the reference:

The coupling looks a little different for the first two states, but that's just because of the way physicists define the states according to z-axis spin, as opposed to the so-called "vector states" of the chemist. A chemist's Px and Py state would be a sums and differences of the first two "physicists states", and with the normalization you get back the factor of root two that appears in the third expression.

It's encouraging to note that this business of calculating the dipole moment is very similar to what we're going to do later in the semi-classical analysis. Of course, we use different words to describe what we're doing. The Copenhagen people talk about something called "Fermi's Golden Rule" for calculating these transition rates. It's hard for me to understand how people are so quick to accept something because they call it a Golden Rule, or any kind or rule for that matter. What's going to be different about the semi-classical calculation is that I'm not going to tell you that such-and-such is the "rule". We're going to be able to relate to real physics at every step of the way. But I digress...

We note that the frequency of this transition is close to 2.5 x 10^15 Hz, giving a wavelength of 120 nanometers. Furthermore, the energy of the excited state is 3/4 Rydberg, 10.2 electron volts. Fitzpatrick then uses an equation derived earlier for the transition rate:

The "lifetime" of the p-state is evidently a nanosecond or two. What does this mean in terms of the physics we can measure? Let's recall that we had a sample of 1,000,000 hydrogen atoms, with one percent of them in the excited state. I'm putting the numbers together and I get a power output of very close to 10 microwatts. That seems awfully high for such a tiny sample, but remember the excited state has a very short lifetime indeed. And also, it's worth noting that the numbers we picked correspond to a sample at quite a high temperature. The Boltzmann factor of kT is only 0.026 eV at room temperature, and to get 1% of the atoms in the excited state (energy = 10.2 eV) we need the ratio E/kT to be the natural log of 100. This pushes the temperature up to around 5000 degrees if I'm doing it right. (EDIT: Now that I try again, it comes out to 24 000 degrees K. I don't know where I got 5000. In any case it's really hot.) So 10 microwatts it is, I suppose.

Tomorrow we'll see how the calculation looks if we treat the atoms as tiny classical antennas. Does anyone want to bet we get the same answer?

Who says energy comes in lumps?

2012-02-20T08:44:00.000-08:00

Everyone says energy comes in lumps. That's the basic starting point of our whole paradigm. You'd have to be crazy to argue with it.

Call me crazy, but where are the lumps? The lumps are supposed to be necessary to derive the black-body spectrum, but just where do they come from? When we left off the other day, I was contrasting the Copenhagen interpretation with the Schroedinger interpretation. In Copenhagen, the atoms in a hot gas are each in some particular state of definite energy. According to certain probabilities, they may randomly absorb or emit a quantum of energy, thereby jumping from one state to another. Energy always appears in discrete lumps, or quanta.

In the Schroedinger picture, atom is surrounded by a charge of cloud which can have any shape at all. From our human perspective it is convenient for us to analyze those shapes as the sum of what are called eigenstates: those correspond to the definite energy states of Copenhagen. The equations show that any combination of eigenstates will generate an oscillating charge distribution, which will tend to radiate or absorb energy, depending on the state of the ambient electromagnetic field. The atoms are doing this all the time except in the exceptional circumstance when they happen to find themselves in a pure eigenstate. Only then does its charge distribution become stationary, and accordingly it stops exchanging energy with the surrounding field, if only for a moment.

How can these two radically different pictures give the same physics? The obvious answer is that they don't. Everyone knows that energy is absorbed or given off by atoms only in discrete quanta. The so-called Schroedinger picture, as I've described it, violates this fundamental rule. Therefore it must be wrong.

But why is it wrong? If it's so wrong, it should be easy to show where it falls down. Let's take a typical calculation. But what exactly should we calculate? It's easier said than done.

In situations like this, the thing to do is to back right down to the simplest possible case. In the Copenhagen Theory, you have something called the Einstein A and B coefficients. They give you the probability of an atom jumping from one state to another state. The Einstein A coefficitent has a particularly simple interpretation: it represesents an atom in an excited state which emits a photon and jumps to the ground state. For the s-p transition of hydrogen, we can actually calculate it. (Or we can just look it up.)

What does the Schroedinger picture tell us about this transition? Well, on the face of it, things don't look so good for Schroedinger. An atom in the pure p-state has a stationary electron cloud, so it doesn't radiate at all. It stays that way forever. This clearly contradicts what we know to be true from Copenhagen.

Or does it? Just how do we put an atom into a pure p-state, and how long do we have to watch it before it decays into the s-state? It's not obvious how we would do an experiment to verify the Copenhagen results. And yet we know it to be true. How do we know it? Because we take bulk measurements of the radiation intensity of a gas containing atoms in both states. Knowing the total energy of the gas, we know what percentage are in the excited state. Copenhagen predicts how often they decay, and knowing the amount of energy in each decay, we can calculate the average radiation. The experiment verifies that Copenhagen is correct.

But notice that this experiment doesn't tell us anything about individual atoms suddenly jumping from one state to another. That is a human spin that Copenhagen attaches to the physics. To compare the Schroedinger result, we ought to compare only the bottom-line physics. That means we have a gas with a certain amount of energy distributed between the ground state and the excited state, and there is a certain amount of electromagnetic energy associated with that gas. That's what we can measure, and that's what we have to calculate using the Schroedinger picture.

Let's say that 10% of the atoms are in the excited state. Of course, that's Copenhagen language: Schroedinger would have us say that each of the atoms is in a partially excited state, to the extent of 10%. Because of the mix between the s and p states, each atom has an oscillating charge distribution. This means it is a tiny antenna. The laws for calculating radiation from antennas are well known. We just need to figure out how much the atoms are oscillating, and apply the antenna equations.

Do you think we will get the same result as the Copenhagen calculation?

What's so special about pure states?

2012-02-19T06:28:00.000-08:00

When I left off yesterday, I was talking about pure states and mixed states. There is a widespread point of view that in quantum mechanics, you can only observe a system in a pure state, never in a mixed state. People go a little farther than this: they say that in a gas of hydrogen atoms, there all the atoms are in a pure state: some of them in the ground state, and some of them in the higher excited states. Radiation occurs by means of something called a quantum leap, whereby an atom instantly jumps from one state to another, and in the process either absorbs or emits a photon.

This picture dates back to the Bohr atom of 1915 with its planetary model. The Bohr atom became obsolete in 1926, with the advent of the Schroedinger equation. Schroedinger's atom had clouds of charge density instead of little orbiting charged particles. The planetary orbits of the Bohr atom were replaced by the so-called "pure states" of the Schroedinger atom.

What made the "pure states" special? Absoblutely nothing. Not in the Schroedinger atom at least. They were arguably special back in the Bohr atom because they were the only states. More importantly, they were special because they received a special exemption from Maxwell's Equations, the laws of electromagnetics. Contrary to those well-established laws, an electron in the Bohr atom was allowed to orbit forever without radiating, as long as it stayed in one of its allowed states. Bohr gave no explanation of how it could do this: that's just the way it was.

The Schroedinger atom was very different. If an atom didn't radiate it was simply because there was no oscillating charge to drive the radiation. No special exemption from Maxwell's Equations was needed. If the charges began to oscillate, Schroedinger's atom would do just what it was supposed to do according to Maxwell's Equations.

Then what were the so-called "pure states" all about? It's like this. Schroedinger wrote a differential equation to describe the hydrogen atom. There is a way to solve differential equations called "Separation of Variables". It's a funny method because you declare at the start that you are only going to look for solutions which can be broken down into the product of two functions, where one of them is a function only of space, and the other a function only of time. When you solve the equation this way, you get the so-called "pure states", also called eigenstates. There is really nothing special about them in a physical sense.

Are they the only states in which the atom can exist? Of course not. They came about only because you chose to restrict your mathematical solution to only those types of states. There is absolutely nothing in the physics which says that the wave function of hydrogen must be the mathematical product of a space function and a time function. That came from using the technique of Separation of Variables.

Why do people use this technique if it only gives you a tiny fraction of all possible solutions? There is method in the madness after all. Under the right circumstances, it turns out that all real physical solutions of the equation can be expressed as the mathematical sum of these special eigenstates. It may be a sum of just two or three of them, or it may be an infinite sum. It is especially useful to do this for things like the hydrogen atom because the time-dependence of the eigenstates is especially simple: they are all states of pure frequency. These kinds of summations are easy to handle mathematically, and they tend to have clear physical interpretations.

The physical picture for the hydrogen atom is especially satisfying. The Schroedinger equation ultimately tells us where the charge is distributed in the atom. It turns out that for the eigenstates, the charge distribution is stationary. That's why they are stable: because stationary charges don't radiate. Once you mix two states together, however, you find that there is an oscillating charge distribution. The frequencies of oscillation are exactly those that you measure in a heated gas by means of spectoscopy. It's obvious that the oscillating charges are responsible for the thermal radiation of a hot gas.

So the question becomes: why do the adherents of the Copenhagen interpretation insist that they are not interested in the mixed states but only in the pure states? It's hard to say. Perhaps in the eleven short years when the Bohr atom held sway, those scientists got so used to thinking about special states and quantum leaps that they just couldn't bear to give it up, even when that point of view became totally unnecessary.
So they cobbled together their own interpretation of Schroedinger's equation, whereby his eigenstates were accorded special status. Unlike any other differential equation, they weren't going to allow you to construct new states by adding together eigenstates. Every atom was going to be in a pure eignestate, and instead of allowing the atoms to make a smooth transition from one eigenstate to another, they would calculate the probability of the atom jumping between states, just as they used to do with the Bohr atom.

So which picture is right, the Copenhagen interpretation or the Schroedinger picture. Incredibly, the Copenhagen people are able to show that their interpretation gives all the correct answers when applied to experiments. Doesn't this mean that they are right and Schroeginer is wrong?

No it doesn't. To do that they would have to show that you get the wrong answer if you treat the atoms as oscillating charge distributions and apply Maxwell's equations to calculate the resulting radiation. That's easier said than done. I know, I know, people are jumping up and down and shouting: "Ultraviolet catastrophe! Ultraviolet catastrophe!" There is no ultraviolet catastrophe with the Schroedinger atom. There is no ultraviolet catastrophe because the vibrations which would cause it are suppressed at the mechanical level.

Think about it: why wouldn't you get the correct black-body spectrum just by analyzing the classical radiation you would calculate from the charge oscillations as given by the Schroedinger equation? This is a straighforward question with a well-defined answer: either it's right or wrong. What is astonishing is that you could probably ask this question to 100 physicists, and not one of them would have ever thought of asking it. Oh, most of them would "know" the answer in the sense that they would "know" that you obviously get the wrong spectrum. But how do they know? They've never done the calculation, and it's not in any book. Everyone knows you're not allowed to use Maxwell's equations to calculate the radiation from quantum systems. That leads to nonsense. Quantum systems operate according to completely different laws.

Or do they?

Evolution or Intelligent Design?

2012-02-18T07:21:00.000-08:00

The other day I said I had a problem with the theory of evolution. I broke the theory into two parts, which I called the Weak Theory and the Strong Theory. I said I was OK with the Weak Theory. That's the part that says we can trace our ancestry all the way back to the primordial slime. Who but an ignorant hillbilly could doubt that, in the light of the massive scientific evidence?

No, I'm not going to argue with the Weak Theory. It's the Strong Theory that gets me. I said I was going to hold off for a day or two and see if anyone could guess what I mean when I talk about the Strong Theory. Well, time's up.

The Weak Theory says,"this is exactly what happened". The Strong Theory says we know how and why it happened. The intricate development of millions of different life forms occurred through the mechanism of random errors in the transmission of the genetic code. The overwhelming majority of such transmission errors would naturally be harmful if not fatal; but here and there, it results in a characteristic which is useful. The bearer of this characteristic is more likely to procreate than his "normal" siblings, and the new characteristic is thereby incorporated into the gene pool of the population, where over time it ultimately dominates and becomes the new norm.

Is there one single scrap of scientific evidence to support the Strong Theory? In school we learned about the grey moths and the white moths. It seems that 99% of the moths used to be white, but during the Industrial Revolution smog coated everything in London with a layer of grey; the white moths became easy prey for birds, and as a result now all the moths are grey. Evolution at work.

But there were already grey moths in existence. The characteristic of interest was already present in the gene pool. Where was the random mutation that supposedly created it? There was none, because the gene already existed.

How does a brand new trait come into existence? That's a tough one. The much-maligned proponents of Intelligent Design like to talk about the problematic case of the human eye, and rightly so. There was once a time when not a single organism was sensitive to the presence or absence of light. They were all just blobs floating around in the ocean, and when one of them bumped into another, presumably the bigger one would wrap itself around the smaller one and mulch it down into digestible bits. A case of the blind eating the blind.

Now, what does the Strong Theory tell us? That somehow, one of these blobs laid its blob-eggs somewhere, and a cosmic particle came along and messed up the DNA in one of the eggs. The result was that a baby blob was born who was different from all the other blobs: somewhere on the surface of its body, there was a little patch of skin that contained photo-sensitive chemicals: compounds that would change state when exposed to light, and change back again when the light disappeared. Quite a trick for a random cosmic particle.

But that's not nearly enough. Somehow this patch of skin had to be hooked up to whatever rudimentary brain this blob must have possessed: the blob had to know that it was sensing light. The blog had to know that this tingling on its skin meant light, and it had to want to swim towards the source of that tingling, because perhaps there was more food to be had in that direction. By eating more food, the blob got bigger, and that made him more likely to survive in his blob-eat-blob world to the age of procreation. And his photo-sensitive offspring, over time, came to rule the world of the blobs through their inherent superiority.

So let's recap all the things that had to happen in a single generation for this genetic mutation to have taken hold. First, you needed the appearance of photo-sensitive compounds. This doesn't just happen by shooting a cosmic particle at a piece of DNA: the DNA doesn't become light sensitive. The DNA is a kind of a blueprint that creates an enzyme, and it is this new enzyme that manufactures the photo-sensitive chemical. But that's not enough. This new chemical, which is presumably just floating around in the protoplasm, has to somehow stimulate whatever locomotive mechanism this blob uses to propel itself around. It doesn't do any good to have a mutation which does one without the other.

That doesn't begin to address the question of how the blob is supposed to know whether it wants to swim towards the light or away from it. But never mind. Let's suppose it all works: that a single cosmic particle disrupts the genetic code in just the right way so that these three simultaneous, highly improbable developments occur (and without accidentally killing the blob in the bargain). The hard part is behind us; we've initiated the evolutionary chain, and now it's only a matter of time before this primitive photo-sensitive blob, the ancestor of us all, evolves into the myriad of life forms with the sophisticated organs of vision which we are all familiar with. Right?

The Creationists like to talk about a hurricane blowing through a junk yard, and when the dust has settled you find that by randomly flinging one piece of metal against another, it has managed put together a Boeing 747. I'd say this is a pretty close to how the Strong Theory explains the evolution of the human eye. I know that saying this brands me as an ignorant hillbilly, but if the shoe fits I might as well wear it. I'd just like to throw one more wrinkle into the story of vision. It's the question of color.

Not all mammals have color vision. Presumably we humans have some ancestor who saw only black and white. What does the strong theory tell us? That somewhere along the line, a random mutation produced an individual with color vision, and because of his superior skills, his progeny came to dominate the gene pool over time.

A random mutation produced an individual with color vision? That's some cosmic particle we're talking about. Because anything less than a fully developed sense of color vision couldn't have been all that useful. Do you have any idea how much more complex a color TV is as compared to black and white? There are three different receptors, and there's all kinds of signal processing you've got to handle all of a sudden, and its a huge complication. And you can't argue that it came about gradually, because what kind of benefit is there to the individual who can see just a little red?

Come to think of it, how much of an advantage is color vision even when fully developed? Yes, it's nice to have, but it's not clear to me how an individual born out of the blue with the miraculous gift of color vision would have had all that much advantage over his black-and-white fellows. There were no color-coded electrical wires to worry about back then; yes, you could more easily tell the difference between a blackberry and a raspberry, but aren't they both delicious after all? I'm not saying it wouldn't have come in handy now and then, but to the extent that one such individual would ultimately dominate the whole gene pool? I just don't buy it.

The thing that bothers me the most about the whole evolution thing is that it's taken as a litmus test for admission into decent society. Question any aspect of the theory and you're branded as an ignorant hillbilly. Why is this?

I think it's because the proponents of evolution really want to use it as a weapon against religious faith. That's why the Strong Theory is so important to them. The Weak Theory is, after all, fully compatible with the idea that God has guided the process along with the intent of creating us in his image. But that's not good enough for the evolutionists. They demand that we acknowledge the random nature of the process, so that there is clearly no role for a Creator in their world-view. And if that means they have to argue for a Strong Theory that by its very nature cannot possibly have any scientific evidence to back it up, then so be it.

Why people don't understand quantum mechanics

2012-02-17T11:30:00.000-08:00

People don't understand quantum mechanics because they think that energy can only come in descrete lumps. This was an early interpretation of the old quantum mechanics, that has become solidly entrenched in the collective psyche even at the very highest levels. It is wrong and misleading in almost every instance where it is applied. The most imporant example of its misuse is in the case of emission and absorption of radiation between atoms.

Everything that atoms do in the context of thermal radiation can be understood by analyzing them as tiny antennas that work exactly the same as classical radio antennas. This is not some kind of crackpot idea: I know exactly what crackpot ideas are because I've taken a crackpot test and scored 155 on it. Seriously, I know exactly what I'm saying and I can back it up. If you understand this about atoms and radiation then all of quantum mechanics starts to make a little more sense.

The first thing you need to do is throw away the old planetary model of the atom. That was invented in 1915 by Bohr, and it became obsolete in 1926 when Schroedinger came along. Unfortunately, the image of the Bohr atom is burned into everyone's brain, thanks in no small measure to the efforts of the school system. The electron cloud of Schroedinger's atom is a vague and fuzzy concept for most people. Fortunately, there are some good animations of it available. I had a really nice one from the University of Saskatchewan website that I linked to last year in my article on the crystal radio, but as I check it out right now it's giving me a hard time over my browser. So I'm not sure if it works. Here's another guy selling applets that make the same point; even if you don't buy it, you can see some nice animations on the web page. The point is that in quantum mechanics there is such a thing as a charge density, and if you take an atom in any kind of superposition of pure states, then you get an oscillating charge density. That's what you see in the applets, and there's nothing controversial about it.

The stumbling block for many people is the idea that you have an atom in a superposition of pure states. I can't count the number of times people have told me this is nonsense...people who should know better. They say a hydrogen atom, for example, is either in the ground state or the first excited state...never half and half. If it is in the presence of thermal radiation, it may absorb a photon and instantly jump from the lower state to the higher state, or it may emit a photon and jump the other way. That's the old "quantum leap" made popular by Neils Bohr, and made obsolete by the Schroedinger Equation. Or it should have been made obsolete, except that the proponents of the Copenhagen School adapted it to the new Schroedinger atom. Schroedinger developed his theory to show that there was a sensible way for the atom to make a smooth transition from one state to the other, and he was appalled when his theory was hijacked to support the quantum leap.

It's true that the Copenhagen school cobbled together a workable interpretation whereby only pure states exist, but the point is it's not a necessary interpretation. What people don't realize is that there is no experimental way to distinguish between the Copenhagen model of discreet states and quantum leaps, versus the Schroedinger model of mixed states and continuous transitions. This point was made really well by people who know all about these things in a stackexchange.com discussion which I link to in this article. The fact that you can analyze everything in terms of pure states and quantum leaps doesn't mean that you have to analyze it that way.

So what does the universe look like in Schroedinger's picture? Every atom is in some mixture of pure eigenstates, and if you look at the applets I mentioned earlier, it's easy to see that the charges are oscillating. As oscillating charges, they ought to radiate according to the classical laws of antenna theory; and that's where I hit the second wall. "Even if you're right about the oscillating charges..." (notice how they start with "even if", not even admitting what is clearly a mathematical consequence of the theory), "even if you're right about that, you're not allowed to use Maxwell's Equations to calculate the resulting radiation."

Not allowed? Why the hell not? There are two possibilities. Either you get the wrong answer for the black body radiation field, or you get the right answer. If you can show me that you get the wrong answer, then the game is over and I have to pack up my marbles and go home. But what if you get the right answer? Then what becomes of your quantization of energy into discrete lumps? Either way, it has nothing to do with what I'm supposedly allowed to do.

Let's leave it there for now. I'll come back to this before long. But I think I've also left some unfinished business about the theory of evolution...

What's Wrong with the Theory of Evolution?

2012-02-17T05:18:00.001-08:00

There are two parts of the theory of evolution, which I'll call the Weak Theory and the Strong Theory. There's probably a pretty good case for the Weak Theory. It's the Strong Theory I have a problem with, and today I'm going to tell you why.

The Weak Theory tells us that we're all descended from lower animals. It's not too hard for us to believe we share an ancestor with monkeys. I don't know how good the fossil record really is on this. They used to talk about the quest for the "missing link"...the fossil specimen that would bridge the gap between them and us. I don't know how well that continuum has been filled in over the last hundred or so years, but let's give the theory the benefit of the doubt.

If you buy into the idea that we are related from the monkeys, then you have to allow that it can go back even farther. Were monkeys once tiny like groundhogs, or chipmunks? I suppose maybe they were. And aren't the chipmunks just overgrown rodents? Let's agree that there is a chain of descent.

If we accept that our ancestors were mice, then all the mammals have an equally good claim on that line. Which makes us cousins to lions and tigers, cows and horses, whatever. That's the gist of the Weak Theory. It actually explains a lot of things: we all have similar skeletal structures, we all have two eyes and two ears, fingers and toes, teeth, hearts that pump blood, etc.

But of course it doesn't stop there. The mice from which we are descended were in turn descended from lower forms. Once upon a time all animals were reptiles; the mouse and the frog and the snake are therefore distant cousins; the birds are another branch of the same family; and so on. The frog was once a fish; the fish was once a jellyfish; and the jellyfish was once a festering slime. And so we are told that there is a continuous line of descent, whereby we can trace our own ancestry to that same primordial slime.

This is the Theory of Evolution, and who but an ignorant hillbilly could possibly doubt it? Haven't we all been shown the evidence in the form of the fossil record, which clearly shows the continuous transformation of all these various forms of life from one to another? What, you haven't seen the fossils yourself? No matter; the scientists assure us that the case is well established, and that's good enough for me. Because all that I have described so far is the Weak Theory, and I don't have an issue with that. It's the Strong Theory that bothers me.

What do I mean by the Strong Theory? I'd like you to think about that, because it's a distinction that people seem to gloss over. I've never seen it explicitly discussed in quite the way I'm going to tackle it, and I even wonder if it's obvious where I'm going with this. So I think I'll let you mull it over for a day or so, and then continue where I left off.

Marty does some real thermodynamics

2012-02-16T11:55:00.000-08:00

I've been promising to post this for about two years now and I guess now is as good a time as any. You go on the internet and get into physics arguments with people and sometimes you lose. But sometimes you win pretty big. This was undoubtedly my finest hour, and now I'm going to tell you about it.

It all goes back to my long-standing gripe with the idea of photons. One of the major arguments as to why light needs to be in the form of particles has to do with the photographic process. If light is a wave, then how come you can get a developable photographic image from a very distant star, when any reasonable calculation shows that there is no way to concentrate enough energy on a single atom of silver to reduce it from silver bromide to the metallic form.

I had been arguing that you didn't need photons because the appearance of silver flecks was thermodynamically spontaneous. It's an important point because people argue that you need the energy of a whole photon to drive a detection event, whereas I was arguing that you didn't. It's actually a little more complicated than that. There are two thresholds people worry about. One is the amount of energy to drive the overall process, and the other is the amount of energy to push an electron into the intermediate state. The latter is theoretically recoverable, but the former definitely needs to be accounted for. Even if the reduction of metallic silver is thermodynamically favored, the proponents of the photon theory still claim you need photons just for the activation energy, because no one has ever proposed a mechanism whereby you could recover this energy: that is, until I came up with the idea of Quantum Siphoning.

The problem is I turned out to be completely wrong. I had been arguing this point for about a week with a guy named SpectraCat when he finally came up with the numbers. The enthalpy of the reaction was 99 kilojoules per mole, and it was the wrong way. It was a crushing defeat for my argument. It wasn't just the activation energy: the complete process required a definite input of energy.

Incredibly, I was able to recover from this huge setback by means of an ingenious thermodynamic argument, which I am presenting below. It's odd that I happened to know enough thermodynamics to be able to make this argument. It never would have happened if not for a project I had been involved in as a junior engineer fresh out of university, thirty years previously. I talk about that in my series Karma and carbon monoxide.

Here is how I answered Spectracat's point about the enthalpy of the reaction.

You have presented the following equation insupport of your position:

AgBr ----> Ag + ½ Br2 ΔH = +99 kJ/mol

You point out that the reaction is endothermic and therfore cannot procedewithout the input of energy from a photon.

It’s a good argument, and it should be a good argument because it’s myargument. I was the one who suggested, in the face of some initial ridicule,that we could settle the question by looking at the thermodynamics of theprocess. It turns out that I was right to look at the thermodynamics but wrongto think that we would settle the question in this way.

The Gibbs Free Energy is, of course, the parameter which normally tells us if areaction procedes to the right or to the left. You have used the enthalpyinstead in presenting your numbers; but no matter. The correction for theentropy is in any event rather small; only a little more than 3 kJ/mol in thiscase. Nothing decisive.

The important factor you have neglected is concentration. The Gibbs Free Energyequation gives us the change in free energy only when the reactants and theirproducts are present in stoichiometric ratios. In the photographic process, atypical crystal may have, after being exposed to light, only a literal handfulof silver atoms out of trillions. It is apparent that the silver and silverhalide species are very far from their stoichiometric proportions, andtherefore a more careful analysis is required.

The themodynamically correct method must be to treat the crystal as a solidsolution of silver and silver bromide. In its initial state, the crystal is100% AgBr. Is the formation of a single silver atom thermodynamicallyspontaneous, or does it require a net input of free energy?

For convenience, we will double the reaction to clear fractions:

2AgBr -------------> 2Ag + Br2 ΔG = +192 kJ

We can calculate the familiar equilibrium constant from first-year chemistrywith the formula

K = exp(-ΔG/RT)

With RT = 2.2 kJ (approx) we get

With this information we can write the chemical equilibrium equation:

This equation easily solves for the equilibrium concentration of Ag beingapproximately one part in ten trillion (10^-13). This is within an order ofmagnitude or so of typical concentrations in an actual exposed crystal. It isapparent that for an initially pure crystal of AgBr, the spontaneous creationof trace amounts of silver is thermodynamically favored. The stimulation of theincident light wave (I’m going to stop calling it a photon) merely speeds upthe process, but is not strictly necessary energetically.

We can do the same calculation a different way. Use the Gibbs Free Energy tocalculate the conversion of one part per trillion (10^-12) of silver. It comesto 192 nanojoules. This is the input of free energy required to drive theprocess. But this assumes that the two species are unmixed. In fact, we oughtto treat the silver as being in solution. Then we can show that, within theaccuracy of this calculation, the needed energy is available from the entropyof mixing.

The entropy of mixing is given by the formula

ΔS = nRm*ln(m)

where n is the number of moles and m is concentration of the mixed species.Multiplication by T gives you the free energy of mixing:

ΔG = T ΔS = nRTm*ln(m)

For two moles at a concentration of one part per trillion, the free energycomes to 121 nanojoules; and make no mistake, it is in the right direction todrive the reaction forward. It is true that with the numbers I have chosen weare just a little short of the 192 nanojoules we said we needed, but rememberwe haven’t yet accounted for the contribution from the mixing of the bromine.In any case, you only need to drop the concentration another factor of ten totilt the reaction decisively to the right.

The case becomes even more convincing (convincing to me, you understand: I knowyou're still not convinced) when we recall that the crystal in its pure form isconsidered to be a poor photodetector. In practise the material must bedoctored by the addition of impurities, dislocations, and what are called“electron traps” to become really effective. It’s not hard to imagine (OK, for meto imagine) that the energy needed at the trace concentrations we are dealingwith comes at least partly if not in large measure from the “doping” of thecrystal. In other words, there is plenty of chemical energy available to drivethe transition from silver bromide to silver without needing the energy of aphoton. This is in line with my original description of the process asproceding from a metastable state to one of lower energy.

And that's why you don't need a photon to supply the energy to create a detection event. The energy needed is already available in the "detector" (the photographic plate)!

This still doesn't explain where you get the activation energy. The way we understand the process, a bound electron first has to get into the conduction band, from whence it is captured by a silver atom. At the time I wrote this analysis, I still didn't have an explanation of how that could happen. That's where Quantum Siphoning comes in, which brings my argument full circle. But I've already said that, haven't I?

How to tell if you’re a “Self-Appointed Defender of the Orthodoxy”

2012-02-15T14:51:00.000-08:00

Last week I took John Baez’s on-line test to find out if I was acrackpot. One item that cost me 20 points was Question 28: have you ever usedthe phrase “self-appointed defender of the orthodoxy”. Then it occurred to methat maybe those people have as much trouble recognizing themselves as thecrackpots have, so there ought to be a test to help identify them. So based onmy own many years of personal encounters with that type, here is my list of tenquestions that will tell you if you are indeed a “self-appointed defender ofthe orthodoxy”.

1. Do yourespond to internet postings by people you consider crackpots?

Why? I go on the internet looking for good discussions with smartpeople who know what they are talking about. I’m not interested in the peoplewho post garbage, and I don’t understand people who make a big deal of it.

2. Haveyou ever said: “Even if you are right about X, how do you explain Y?”

It’s the “even if” that bothers me the most, because you avoid taking astand either way. If you think someone is right about X, you should acknowledgeit. If you think he is wrong about X, then argue him down on his specific case.Don’t change the subject by challenging him on Y, especially if he has made noclaims about Y.

2a. (EDIT: I thought of this one later.) A nasty variation on item 2: "Even if you're right, so what?"

How do people get off pretending that the whole game of physics isn't about trying to be right? As though you're not motivated by petty things like the quest for personal glory, but somehow your opponent is. And the worst thing about is is the "if" clause: you don't even commit one way or another on the physics, you just question the motivational psychology of your adversary.

3. Do younormally refute false arguments by giving a published reference instead ofmaking your own counter-argument?

For me, this is the biggest problem with the world of physics. Peoplelove arguing by reference to authority. That’s a much bigger problem than theexistence of so-called crackpots.

4. Do youargue that a theory is useless unless it makes testable predictions, presumablydifferent from those of the Standard Model?

This is basically John Baez’s clincher, his 50-point final question. Ifyou take it seriously you have to throw away every insight that makes existingknowledge understandable in a different way. Bohmian mechanics? I’m no fan ofit, but I certainly don’t claim it’s useless because it “merely” claims toduplicate the results of the Standard Model. You might as well throw out LagrangianMechanics because it doesn’t predict anything different from F=ma.

5. Haveyou ever said “QED is the most accurate theory known to man, verifed experimentallyto eleven decimal places”?

Or used any other line from Feynman as a debating point, withoutattribution. Unless you happen to be Feynmann himself. Then you get a pass onthis one.

6. Do youchallenge your opponents to refute arguments in published articles availableonly to paid subscribers, which you link to in your posts?

This is another nice tactic from the “What-do-you-say-about-THIS-then?”grab-bag. In fact there is way too much argument by reference to authority. That’snot what physics should be about.

7. Do yousee your participation in discussion groups as a form of public service?

I get into discussions on the internet because it’s fun for me. I also liketo challenge myself by getting into skirmishes sometimes. If you participate indiscussions because you believe someone ought to protect the naïve from beingmisled by false ideas, then you are probably a self-appointing defender of theorthodoxy.

8. Haveyou ever commented on the sanity or otherwise of people you argue with on theinternet?

If they’re so crazy, why are you wasting your time arguing with them?

9. Do youmake fun of people who question the theory of evolution?

There is a good case to be made that we are indeed descended fromearthworms, but it is not so unreasonable to doubt that the mechanism wherebywe got from there to here was based on an accumulation of random errors in the transmissionof the genetic code.

10. Haveyou ever gotten someone banned from a discussion group?

Enough said.

I don’t have an official scoring guide, but if you answered “yes” tofour or more of these, then you are probably a self-apointed defender of theorthodoxy. If you answered “yes” to nine or more, then you are probably eitherZapperZ, Jim Carr of Mati Meron.

Calculating that Beer-Bottle Frequency

2012-02-12T11:57:00.000-08:00

When we left off the other day, I had set up some calculations on theHelmholts Resonator. You had a beer bottle with a volume of 300 cc’s, and aneck 3cm long with a cross-section of 2 square centimeters. The idea was totreat the plug of air in the neck as the mass, and the enclosed volume as thespring. Here’s how the calculation should go.

I suggested allowing for a one percent compression of the volume, so you’dhave a pressure excursion of 1kpA.That’s a little extreme; you can see it leadsto an excursion of 1.5 cm each way for the plug. A little much, but it willwork for purposed of calculation.

I explained that the 1% compression gave you a volume energy of 0.15joules. Of course, that turned out be wrong: I should have gotten 1.5 milliJoules. Let's assume we're all agreed on that. If you use 1.3 kg/m^3 for the density of air, you should be able toequate that to the maximum kinetic energy of the plug, giving you a velocity of just under20 m/s. Are you with me so far?

Now comes the fun part. You have to equate the maximum plug velocity andthe maximum excursion to a sine wave oscillation. It will help to draw a graph:

Without going through too many details, it should be pretty clear from the graph that the only way you can get an excursion of 30 mm with a velocity of 20 m/sec is if it's spread over a time span of 1.5 msec. That pretty much defines the frequency of the sine wave. You can just eyeball it from the graph, but it comes in handy to know that the maximum slope of a sine wave projected from peak to peak gives you a span of exactly two radians. Either way, you get a frequency of about 200 Hz. It's close to Ab below middle C.

Let's check it against the formula from Wikipedia, which I included in the earlier post. Plugging in the known value of 340 m/s for the speed of sound in air, we get a frequency of 250 Hz. Not bad, but why the discrepancy?

It turns out that in my calculation, I didn't account for the effect of adiabatic compression. This actually baffled Isaac Newton, who worked out a formula for the speed of sound in air and couldn't figure out why it was off by 20%. It's because air heats up when it is compressed, and that has the effect of making it stiffer, or harder to compress. It cools off again when it expands, so you don't actually notice a change in temperature, but it definitely affects the frequency and speed of propagation. You're supposed to use what's called the adiabatic compressibility of air to get the calcuation right. It's a bit of thermodynamics that I'm not going to get into any deeper at this point.

And that's how the calculation goes, using the energy method. There's a bit of a post-script to the story. As I mentioned, this all happened in first-year engineering physics. Around the same time, the prof did an in-class demonstration of the effect of breathing helium, and how it makes your voice "go higher". He explained that the reason your voice got higher was because the speed of sound in helium was higher than the speed of sound in air. I'm not making this up, that was his explanation. He actually said something about the waves reaching your ear more quickly. It just didn't make sense.

I went up to him after class and told him it didn't make sense to me, but he scoffed at my objection. He said something about how he wasn't here to spoon-feed me; if I didn't like his explanation, then it was up to me to figure out a better one. It was pretty clear to me that he was just expressing his contempt for a typical undergrad, and an engineer at that. But I took his advice at face value, and it turned out to be pretty good advice.

It was after I figured out how the Helmholtz resonator worked that I figured out what the deal was with the helium. Your lung/mouth/nasal system acts as a resonant cavity, and when you replace normal air with helium, you change all the resonant frequencies. Your voice doesn't so much get higher as the higher frequencies get preferentially amplified. It's like turning up the treble setting on your stereo. (In case you're wondering, a "stereo" is a kind of big giant iPod that people used to have in the 70's).

I went to the prof after I figured this out, and told him what I thought. He said it was nonsense and he could prove it. He took me up to the lab where there was a bottle of Helium, and we both sang a note, then breathed in and tried to sing the same note. "There! You see?" he said. I tried to tell him that we had actually both sung the same pitch each time, but he had had enough. He told me to go away and stop wasting his time.

That's not quite the end of the story. Twenty years later the professors at the University of Manitoba went on strike, and I went down to the physics building and found a classroom of students with no teacher. I started doing problems from the back of the chapter, when this same professor walked by the classroom and saw me. I'm pretty sure he's the one who called the campus police and got me arrested.

It didn't work out all that badly for me. I made all the papers, and I was a bit of a hero for a day or two. The strike ended the next day, and I like to think all that embarrassing publicity had just a little bit to do with it.

I score 155 on John Baez's Crackpot Index

2012-02-12T06:20:00.000-08:00

Yesterday I asked if there was some objective way to tell if you were a crackpot. It turns out that a guy named John Baez, the head moderator of sci.physics.research put something up on his website a few years ago that claims to answer this question objectively. I was pretty interested in trying it out, so I gave myself the test. It turns out that Baez wasn't really serious about devising a useful measurement tool. He just wanted to make fun of crackpots. Or that's how it looks to me. In any case, I went through his test and got a score of 155. Baez doesn't provide a scoring guide, so I don't know if that makes me a crackpot on his scale. But for the record, here it is.

The Crackpot Index

John Baez

A simple method for rating potentially revolutionary contributions to physics:

A -5 point starting credit.
1 point for every statement that is widely agreed on to be false.
2 points for every statement that is clearly vacuous.
3 points for every statement that is logically inconsistent.
5 points for each such statement that is adhered to despite careful correction.
5 points for using a thought experiment that contradicts the results of a widely accepted real experiment. I claim that a beam of atoms going through a Stern-Gerlach magnet spreads into a donut instead of the widely-accepted two dots.
5 points for each word in all capital letters (except for those with defective keyboards).
5 points for each mention of "Einstien", "Hawkins" or "Feynmann". I refer to Feynman as one who supports the two-dot interpretation of Stern-Gerlach (see point 6.)
10 points for each claim that quantum mechanics is fundamentally misguided (without good evidence).
10 points for pointing out that you have gone to school, as if this were evidence of sanity. I point out that I was kicked out of school. Do I get a 10-point credit for this?
10 points for beginning the description of your theory by saying how long you have been working on it. (10 more for emphasizing that you worked on your own.)
10 points for mailing your theory to someone you don't know personally and asking them not to tell anyone else about it, for fear that your ideas will be stolen.
10 points for offering prize money to anyone who proves and/or finds any flaws in your theory. In the physicsforum thread where I got kicked out, I offered to split my Nobel Prize money with anyone who would help me publish my theory.
10 points for each new term you invent and use without properly defining it.
10 points for each statement along the lines of "I'm not good at math, but my theory is conceptually right, so all I need is for someone to express it in terms of equations".
10 points for arguing that a current well-established theory is "only a theory", as if this were somehow a point against it.
10 points for arguing that while a current well-established theory predicts phenomena correctly, it doesn't explain "why" they occur, or fails to provide a "mechanism".Yes, that's what I say about the collapse of the wave function.
10 points for each favorable comparison of yourself to Einstein, or claim that special or general relativity are fundamentally misguided (without good evidence).
10 points for claiming that your work is on the cutting edge of a "paradigm shift".Yes, that's what I say about Quantum Siphoning.
20 points for emailing me and complaining about the crackpot index. (E.g., saying that it "suppresses original thinkers" or saying that I misspelled "Einstein" in item 8.)
20 points for suggesting that you deserve a Nobel prize. Guilty.
20 points for each favorable comparison of yourself to Newton or claim that classical mechanics is fundamentally misguided (without good evidence).
20 points for every use of science fiction works or myths as if they were fact.
20 points for defending yourself by bringing up (real or imagined) ridicule accorded to your past theories. Yes, I complain about the know-it-alls on physicsforums who ridiculed me.
20 points for naming something after yourself. (E.g., talking about the "The Evans Field Equation" when your name happens to be Evans.)I might have dodged this bullet but just last week I said I wanted to use the name "Marty's Function" for my donut distribution for the Stern Gerlach experiment.
20 points for talking about how great your theory is, but never actually explaining it.
20 points for each use of the phrase "hidebound reactionary".
20 points for each use of the phrase "self-appointed defender of the orthodoxy".I think I called ZapperZ and Jim Carr something almost exactly like this.
30 points for suggesting that a famous figure secretly disbelieved in a theory which he or she publicly supported. (E.g., that Feynman was a closet opponent of special relativity, as deduced by reading between the lines in his freshman physics textbooks.)
30 points for suggesting that Einstein, in his later years, was groping his way towards the ideas you now advocate.
30 points for claiming that your theories were developed by an extraterrestrial civilization (without good evidence).
30 points for allusions to a delay in your work while you spent time in an asylum, or references to the psychiatrist who tried to talk you out of your theory.
40 points for comparing those who argue against your ideas to Nazis, stormtroopers, or brownshirts.
40 points for claiming that the "scientific establishment" is engaged in a "conspiracy" to prevent your work from gaining its well-deserved fame, or suchlike.
40 points for comparing yourself to Galileo, suggesting that a modern-day Inquisition is hard at work on your case, and so on.
40 points for claiming that when your theory is finally appreciated, present-day science will be seen for the sham it truly is. (30 more points for fantasizing about show trials in which scientists who mocked your theories will be forced to recant.)
50 points for claiming you have a revolutionary theory but giving no concrete testable predictions.I probably do this with Quantum Siphoning and the collapse of the wave function. I don't claim to predict different results from everyone else, I just claim to have developed a causal mechanism which explains how it works.

How do I know I'm not a crackpot?

2012-02-11T20:26:00.000-08:00

You have to ask yourself that sometimes. No one ones to be a crackpot, but how do you know if you are one or not? I don't think I am, but that doesn't prove anything. Is there some objective measure that would let me decide?

Notice I say "let me decide", not let you decide. I can't worry too much about other people thinking I'm a crackpot. And honestly, I don't worry too much about it either. I have some idea what I am, and I'm OK with it. It's really more of a theoretical question for me: how would I know if I was or I wasn't?

I definitely have show some of the characteristics of crackpottery. I rant about my real or imaginary enemies out there. I've been kicked out of places where respectable people go. And I claim to have a "paradigm-shifting" new point of view that will change the way we think about quantum mechanics. Last but not least, I publish my physics discoveries in a blog. Those are not good signs.

On the other hand, I've been blogging for just over two years now, which means I've got a pretty good paper trail behind me. If I'm crazy, it should be pretty easy to tell by reading my old posts. Let's see what we have.

First, there are my rants about my real or imagined enemies. It's true that I take a couple of potshots at some of my discussion group nemeses like ZapperZ from physicsforums, and I got a little personal when critiquing Scully's argument on the photoelectric effect. But in close to 100 blogposts, that's about it. I think I'm about 2 on a scale of ten when it comes to personal vendettas.

Then there's the business of getting kicked out of places. If you read every single post over the last two years you will know that I was:

charged with assaulting a police officer while picketing a former employer
banned for life from the University of Manitoba for teaching physics during a professor's strike
banned for life from physicsforums.com for indeterminate reasons
expelled from the Teacher Certification program at the University of Winnipeg

And to be honest, there's a few more instances I haven't bothered to mention. It's a pretty scary looking record, and you have to wonder how I justify it to myself. Maybe I just don't believe in backing down when people in authority try to throw their weight around. I'm not saying I would have the guts speak out if I lived in some brutal dictatorship; I'm just saying what's the point of living in a free country like Canada if you're not ready to stand up for what you believe in? The fact is, the day I was arrested at the U of M for teaching physics still counts as a pretty good day for me.

No, as far as being a crackpot the most disturbing indicator is my claim to have come up with an original explanation for the collapse of the wave function that will change our way of looking at the universe. That's the kind of thing that crackpots say. What makes me think I'm different?

That's a tough one. Quantum Siphoning, as I call it, is a pretty good theory if you ask me. I came up with it about a month after I first started blogging, and I've brought it up maybe three or four times over the last two years. As far as I'm concerned, it's out there waiting to be noticed, and if it happens it happens. Meanwhile, I keep doing physics. It's worth noting that when I submitted my essay to that contest at FQXI last year, "There are No Pea-Shooters for Photons", I didn't mention anything about quantum siphoning. I think a true quack would probably have played it up, but I didn't.

The thing about all these things is that none of them really characterize what this blog has been all about. I couldn't have really put my finger on it until just this week when I started thinking about it, but today I went over all my old posts and it's true: my blog is really about the physics, and my real "comfort zone" seems to be right on the edge, where I'm putting myself on the line. Last week I did a couple of posts first-year university topics, like explaining the Helmholtz Resonator, but that's really not typical for me. It's too safe. I generally don't do topics where I'm not sticking my neck out to some degree.

I take chances, and sometimes I get it wrong. More often I get it right, but sometimes I get it wrong. If I get it wrong it's because I stick my neck out by putting numbers and graphs down to back up my conclusions. Sometimes the calculations are a little sketchy, sometimes I'm lucky to get within an order of magnitude, but in every case the physics are laid out and the numbers are there. Real crackpots don't do that.

The Helmholtz Resonator

2012-02-10T10:53:00.000-08:00

I was taking first-year Engineering Physics when we had a lab where we were supposedly going to measure the speed of sound in air. I didn't understand it at the time, but this is a nasty trick they pull on you in school. There is a formula that has the speed of sound in it, and you measure a bunch of parameters and plug them into the formula and the result is you get the speed of sound. It's very wrong because you don't know what you're doing, but that's how they teach it.

This was a good example of the method. You blow across the top of a beer bottle and note the frequency. If you're a musician, you can do it by ear, assuming you know that A is 440 Hz. Well, you also have to know how the ratios go: if A above middle C is 440, the next A down is 220; the E in between would be 330. You get the idea. For accurate work, you're supposed to factor in the twelfth root of two, but we don't need to talk about that here. I think in the lab we had a microphone and we put the waveform into an oscilloscope, where we used the time calibration to work out the frequency.

How do you get the speed of sound from that? From a formula, of course. I looked up the Helmholtz Resonator on Wikipedia, and there's a pretty good article, including this formula:

You just blow across the beer bottle to get the frequency, then measure the length and cross-sectional area of the neck, divide by the volume, take the square root and there's the speed of sound. Simple. My only question is: how are you supposed to learn physics by doing all that?

The Wikipedia article is actually pretty good. They don't give you that formula right off the bat: they actually explain what is going on physically, which involves the pressure and density of air; then they show you how you can collapse that formula by expressing the speed of sound as a function of those two parameters. But that's really a math trick, and it hides the real physics. It's no way to teach students.

I might have only been in first year, but I wasn't going to plug numbers into a formula that someone else pulled out of thin air. I was going to figure out for myself how to calculate the frequency. That's when I came up with the idea of using the energy methods, which I talked about the other day in connection with the tides.

Here's how the physics works: it's your basic mass on a spring. You've got a plug of air in the neck, which we treat as being incompressible. That's the mass. Then you've got the volume of air inside the bottle, which is compressible to accomodate the motion of the plug. That's the spring. When you have a mass on a spring, there is kinetic and potential energy. At the moment of maximum displacement, all the energy is in the form of compressional energy in the spring. One-quarter cycle later, all that energy has been transferred into kinetic energy of the moving mass.

The thing that makes these calculations especially easy to follow is the fact that the frequency of the oscillation normally doesn't depend on the size of the oscillation. So we can take an arbitrary displacement, e.g. 1.0 mm, and work out all our energies on that basis. We just need to know the density of air and the compressibility.

The density of air is just 1.3 kg/m^3, as I'm sure you know right off the top of your head. (Really. You ought to.) The compressibility is a little harder to justify. We know that the pressure of air is 100 kiloPascals, or 100 Newtons per square meter. What does this tell us, about the compressibility of the air inside a bottle? It's a little tricky, and here's how it works.

If you compress the air in the bottle by one percent, it's easy to believe the pressure should go up to 101 kPa. (Actually, we're supposed to worry about something called the isothermal case versus the adiabatic case here, but let's not.) The work you have to do is (pressure)x(volume), but it's not the whole 100 kPa you're working against, it's only the difference of 1kPa, because you've still got 100 kPa atmospheric pressure behind you. So the energy to compress a 300ml beer bottle by 3ml is just (1kPa)x(3ml) which comes to 0.3 joules if you work your units correctly. (EDIT: That's a mistake. It should come to 3 millijoules. Sometimes we get things wrong.) Actually, it's just half of that because you're not working against the whole 1kPa through the whole distance: the resistance starts off at zero and only builds as you compress it. That's why a lot of energy formulas start off with a factor of 1/2, as I discussed in my earlier article about E=mc^2.

Maybe you'd like to try the calculation yourself to see if we get the same answer. Here's a picture of a typical beer bottle:

The idea is to figure out how far the plug had to move to compress the volume by one percent. (It comes out to an absurdly large figure, but that's OK for purposes of calculation). Then figure out the weight of the plug, and how fast it has to travel to have that amount of kinetic energy. Then, relating the maximum speed and the distance it has to travel to complete a cycle, you ought to get a pretty good handle on the frequency.

Knock yourself out.

Let's Think Twice about Bombing Iran

2012-02-09T08:59:00.000-08:00

The Israeli threats to bomb Iran are nothing new. They’vebeen going on for about five years. But this last week or so there’s been adisturbing upsurge. So I feel I have to say something. If there’s a one in athousand chance someone might hear me, I have to try. And if no one listens, atleast I’ve said it. So I’m double-posting this article in both my blogs, forwhat it’s worth.

Can we please think twice about bombing Iran? Let’s say we couldactually stop them from getting the bomb. Even then, I’m not sure we shouldtry. We have a bomb, what makes us think they shouldn’t have one? But that’s amoot point. Everyone agrees that the best case scenario is that we’d slow downtheir bomb program for a year or so. If we’re afraid they might use the bomb onus, doesn’t bombing them first pretty much guarantee it?

I know what’s going on. There’s an atmosphere of panic inIsrael, whipped up by our leaders. “Achmadinejad is the new Hitler, and hewants to wipe out Israel and all the Jews.” Well, I don’t buy it. If they lovekilling Jews so much, why don’t they start with the 20,000 Jews livingpeacefully right in the middle of Iran? And by the way, if they tried to dothat, I’d be the first to say let’s hit them with everything we’ve got, anddamn the consequences. But life goes on pretty much normally for the Jews ofIran. Can’t we just leave it that way?

Iran is a proud country, the home of an ancient culture,that has not launched a war of aggression in some five hundred years. They haveno love for Israel, but that is their business. It is true that they providematerial support to our enemies in Lebanon and Gaza, but that is theirperogative, just as America supplies us with arms. Iran does not threaten tobomb America on that account; why should we threaten to bomb them for the samereason? It is true they might have had a hand in the outrage committed twentyyears ago in Buenos Aires, and that is arguably less forgivable; but equallydespicable acts have been carried out by agents of the American, British, andother secret services. Do I need to mention Pinochet, the Contras, and even theSAVAK? At least support of these kind ofacts does not appear to be an element of current Iranian foreign policy. Can wesay the same for ourselves? Before we climb up on our moral high horse, weought to watch what we’re sitting in.

The leaders of Iran are not shy about expressing theirdislike of Israel; but at the same time, they have clearly stated that theyhave no interest in interfering with any agreement we might reach with thePalestinians. They have a legitimate interest in seeing that any such agreementis not arrived at through undue coercion, and we need to respect their supportof Hizbullah and Hamas in that light. And other than the occasional tastelesscomments questioning the Holocaust, I find their public positions on thesematters are much more reasonable than Israel’s ongoing campaing of threats.

And what about the consequences? I don’t want to argue thatwe should hold back because we’re afraid of Iran’s response. If we need to doit, we need to do it. But let’s be realistic. We got away with this kind ofthing twice before, in Iraq and recently again in Syria. Let’s not push ourluck. Those enemies were not in a position to do us much damage. Iran isanother ball of wax. If we poke them they will hit back with everything they’vegot, and it won’t be pretty. Think World War III, Armageddon, End of Days…youget the idea.

The point is it’s all so unnecessary. The surest way for usto avert the threat of an Iranian attack is to make peace with thePalestinians. The Saudis have done all the diplomatic leg work for us, gettingthe whole Arab League to sign on to their peace plan, and yet we just ignorethem. I just don’t know where we think we’re going with all this.

So how fast was that wave anyways?

2012-02-08T12:14:00.000-08:00

When I left off yesterday, I had a one-kilometer high tidal wave stretching half-way around the earth. Here is the picture we were working with:

Treating this as a standing wave, we figured out that when the potential energy was entirely converted to kinetic energy, so that the surface of the ocean was perfectly flat, we would have a maximum water velocity of 140 meters per second. The question was: how long would it take to flatten the bump and fill in the trough?

We might want to use calculus to handle those tricky shapes, but for what we are doing it will be good enough to just ballpark it. You can see that the bump is 5000 km long by 1 km high, which is roughly 5x10^9 square meters. We are moving water at a speed of 140 m/s through a cross-sectional area 1000 meters in height, for a net throughput of 140 000 square meters per second. It's not to hard to work out that the time required is 35,000 seconds, or pretty close to ten hours.

That's pretty sloppy, on two counts. First, the actual volume is less than we estimated, because the height of one kilometer does not prevail over the entire 5000 kilometer span of the ocean. Secondly, the average speed is less than we estimated, because the peak velocity only prevails at the moment the ocean is perfectly flat. Happily enough, these two errors cancel each other out, and if I'm not mistaken that makes our value of ten hours pretty much exact.

What does that mean physically? Ten hours to go from peak amplitude to flat, ten more hours to peak the other way, and double again to return to our starting configuration. That's forty hours for one full cycle. I already said I'd cheated by looking at Wikipedia, and their answer comes to thirty hours. It's not bad considering that they've done the calculation for the entire spherical planet, and our calculation was only for a rectangular strip.

So how fast would an ocean wave be traveling? Remember what we've just done is a standing wave calculation, which is equivalent to two traveling waves passing through each other. It should be fairly clear that the forty hours we calculated is the time for the traveling wave to traverse the length of the trough, which was 20 000 km, or half the circumference of the planet. So the time to circle the planet at the equator would then be eighty hours, or three and a half days. That gives a speed of close to 300 mph, not too far from the figure of 200 mph that I guessed last week as a ballpark.

As I mentioned previously, the energy method is a great way of avoiding differential equations. To me, this isn't just a matter of trading one mathematical technique for another. The nice thing about tracking the energy is that it lets you keep your analysis firmly grounded in actual physics. You've always got physical pictures to check yourself against if you find yourself getting lost. Of course, the "special-casing" (that is, using actual arbitrary values instead of letters to symbolize your parameters) is a big part of that. It works because in so many situations, the frequency of these oscillating systems is pretty much independent of the amplitude. So pick a nice convenient amplitude and just run with it.

I'm not sure, but I think I remember the first time I used the energy method on an oscillating system. If I recall, it was a first-year engineering physics lab where we were supposedly going to calculate the speed of sound by blowing across the mouth of a beer bottle. I think I'll take up that topic when I return next time.

How Fast is a Tidal Wave?

2012-02-07T06:22:00.000-08:00

When I worked out the effect of the tides on the earth-moon system, I said at one point that the natural frequency of the oceans was going to be a factor. Then I didn't really make much use of it in my analysis. At one point I guessed that a giant tidal wave might travel at something like 200 mph, but that was just a wild guess. Today I'm going to actually calculate it.

You ought to know that what I'm going to show you today are some of my very best tricks. I don't know how the rest of the world does this calculation, but I use tricks: good tricks that really work. One thing I really don't like is differential equations, so I've come up with ways of doing stuff with energy that lets me avoid a lot of hard math. The other thing I don't like is formulas: I like numbers. So instead of letting the height be h and the velocity v, I "special-case" the hell out of everything. Let's start by considering the case of a giant tidal wave one kilometer high.

We're going to assume the oceans are everywhere one kilometer in depth, and that there are no continents to get in the way of our nice calculations. You will recall that the moon creates a tidal bulge on both sides of the planet: so I've sketched a slice of the ocean taken half way around the equator, going from one bulge to the other. The full circumference of the earth is just twice as much as I've shown here:

Below my sketch I've drawn a graph of the potential energy of the system. You can see that I'm showing a maximum for the potential energy where the water is piled the highest, but I'm also showing the same maximum for the deepest troughs. I think that's the right way to do it. The potential energy is zero where the water is at its equilibrium level, shown by the thin red line.

We can consider this sketch as representing the traveling wave pulled along by the moon, racing around the planet; but that's actually not what I have in mind. I want to treat this as a standing wave, a stationary system sloshing up and down. So the peaks and troughs stay where they are, and the neutral points stay where they are. You ought to know that such a system is equivalent to two travel waves passing through each other in opposite directions. We'll come back to that perspective later.

As a standing wave, this system goes through a full cycle like so: first, as shown in the sketch, it is stationary at its maximum displacement. The potential energy is distributed as I've shown in my graph, and there is no kinetic energy because the water is at its moment of reversal. Next it starts to fall, and after a certain time, the bulges have reversed: what was high is now low, and what was then low is now high. Oddly enough, the graph of potential energy will look exactly the same at that moment, because I calculated it based on the displacement from equilibrium, regardless of whether it was positive or negative. But the more interesting question is what happens halfway between those extremes?

It should be clear that at some point the water is completely flat. So what happened to all the potential energy we had calculated? Does it just simply disappear? Obviously not. In fact, it is converted into kinetic energy. When the ocean is flat, the water is rushing to fill in the giant void, and at the moment of flatness the kinetic energy of rushing water is at its greatest. I've drawn a graph to show how the potential energy compares to the kinetic energy at those critical points of the cycle:

All of the energy that was potential is now converted into kinetic form. Most significantly, the concentration of energies at the respective peaks of the two graphs is obviously the same: a given quantity of rushing water has exactly the same kinetic energy as did the same quantity of elevated water, a quarter of the way around the world, in potential form.From this we can calculate how fast the water is rushing. Now, I haven't yet put any numbers on the graph yet, and the nice thing is that I don't have to. The potential energy of that mountain of water at longitude 0 is converted into kinetic energy of the same quantity of water at longitude 45, 5000 kilometers away: and we already know (or we ought to know) how to convert potential to kinetic: the speed is the same speed you get if you just let a stone fall through that distance under gravity. Since the change in elevation is one kilometer (that's an average change, based on the maximum difference of 2 kilometers from peak to trough) we just have to ask: how fast does a stone travel if you let it fall freely through a distance of one kilometer? I make it 140 meters per second, and you can check that if you like.

This is actually such a cool calculation that I don't want to spoil it for you at home. I hope you can see where I'm going with this: I need to figure out how long it will take to drain that giant bulge and fill in the void if the water is travelling with a maximum velocity of 140 meters per second. That will give me a pretty good ballpark of the natural frequency of oscillation. I have to tell you I've already looked up the "answer" on Wikipedia (go to the section on Amplitude and Cycle Time ) and it comes out pretty close if you do it right. So I'm going to leave you in suspense until tomorrow, in case you want to try the calculation for yourself.

How the tides slow down the moon

2012-02-06T16:42:00.000-08:00

I’ve noticed some of my most-read posts aren’t the heavyduty quantum mechanics but the more bread-and-butter physics questions. I gotinvovled with the moon last month because I wanted to make a point about thequadrupole component of the magnetic field in the Stern Gerlach experiment, butthen the moon topic took on a life of its own. First I got messed up with thereflectance, and then I said I was going to explain the tides. I guess now it’stime to pay the piper.

The easy thing to explain about the tides is why there aretwo of them, I’ve already shown the picture that explains it completely. It’sthe distortional component of the moon’s gravity that causes the tides, not thedirect force, and it’s easy to see that the nature of the distortional force iswhat we call the quadrupole field.

The tough question turns out to be: what is the effect onthe earth-moon system of the tidal friction? It’s not too hard to convinceyourself that it ought to slow down the rotational speed of the earth. But whatabout the moon’s orbital speed? If the earth is slowing down, shouldn’t themoon speed up? Otherwise how would angular momentum be conserved?

It turns out this question drove me crazy for about twodays. I thought I had it all figured out: the moon speeds up to conserveangular momentum, and therefore it is dragged down into a lower orbit. Whylower? Because the velocity of a satellite in a low orbit is greater than thevelocity of the same satellite in a high orbit. Then I looked it up onWikipedia, and it was backwards. They had it going into a high orbit toconserve angular momentum!

I have to say I never like to be on the wrong side ofWikipedia when it comes to questions like this. The people who write articleslike this one aren’t just pulling that stuff out oftheir ass. So where was my mistake?

It’s actually hard to believe, as near as I can figure out,this is how it works. If you somehow get up into orbit and push on the moon tospeed it up, it actually slows down. No fooling. Let’s say you set off anatomic bomb on the leeward side of the moon, to boost its orbital speed. Theeffect of that one impulse is to throw the moon into an elliptical orbit. Butif you keep on continuously setting off bombs, until you’ve gone one fullorbit, the effect of the ellipticity (is that a word?) has to cancel itself out.The net result is you’ve got yourself back into a circular orbit, but it’s higher up than the one you started with.But being a higher orbit, it’s also a slowerorbit. (Yes, that’s how it works.) By pushing on the moon to speed it up,you’ve ended up slowing it down. You’ve driven it to a higher orbit, but it’svelocity is less than when you started out.

(It’s velocity is lower, but its angular momentum isactually higher. That’s because angular momentum is the product of velocity andradius, and the increase in radius is more than compensates for the decrease invelocity.)

There’s still one thing missing from this explanation, andthat is a mechanism. It’s fine to say that the moon speeds up to conserveangular momentum, but what makes it speed up? I have to admit I had an awfultime getting this question all straightened out, but to give myself credit, Ihad the mechanism right all along. It’s the shift in the tidal bulge. Thepicture we showed earlier showed how the bulge would look if the earth and moonwere just sitting there. In fact, the earth is spinning fast, dragging thebulge along with it, and so the moon is effectively trailing behind the bulge.

I’ve drawn the picture here showing the bulge at a 45 degree angle to the earth-moon axis: this isn’tthe physical case, it’s the ideal case of a uniform ocean with a naturalfrequency exactly equal to the driving frequency of the earth-moon system. Inother words, it's an ocean at resonance, cresting to the maximum possible powerand dissipation. You can see that the gravitational effect of the bulges is todrag the moon along, tending to speed it up to synchronize it with the earth.The leading bulge, being closer to the moon, is more effective than the trailing bulge. The perverse response where the moon actually slows down in response to beingdragged along is another thing altogether.

Proto-Helium: What is it good for?

2012-02-05T07:24:00.000-08:00

Last week, with great fanfare, I announced the solution of the six-dimensional Schroedinger equation for Proto-Helium, a mythical atom composed of two electrons and most of a proton, such that the species is just marginally stable with respect to loss of the second electron. It's undoubtedly a picturesque idea; even if my partial solution, such as it is, happens to be incorrect, it still illustrates the necessary existence of a mathematical solution fulfilling those particular specifications. But what can we do with proto-helium? Does it hold the key for understanding the solution of the helium atom?

Sadly, after all that, I still don't know where to go from here. I don't even know how to write the radial dependence of the wave function for proto-helium, but that doesn't even bother me so much. The worst of it is I don't know how to take this idea and leverage the solution for helium. I've laid out the isoelectronic series of helium from one end to another, starting with my mythical proto-helium at one end and ending with the limit of infinite nuclear charge; and I've got plausible expressions for the solutions to both those limiting cases in terms of simple product functions or sums thereof. What I still don't know how to do is to blend from limit to the other.

And it's not for lack of trying. I've been playing around with all kinds of superpositions, but none of them are helpful. I know a lot of things that won't work, but I haven't come up with anything that does. That doesn't necessarily mean I've hit a dead end: it just means I don't know where to go yet. I might figure it out tomorrow, or I might never figure it out. That's how it goes in physics.

There is one small advantage I've gleaned from inventing proto-helium. I've shown myself how to write a spherically symmetric charge distribution by starting with an arbitrary axis of separation for the two charges. That's a non-trivial problem: you always have to start that way, but the question is how do you end up symmetrizing it? Do you need an infinite distribution of such solutions taken over all possible orientations? What proto-helium shows me is that you don't: you can satisfy your need for spherical symmetry with a superposition of as few as four simple product functions. I think that's worth something: I just still don't know exactly what to do with it.

How do you get spherical symmetry from all that junk?

2012-02-03T07:43:00.000-08:00

For the last couple of posts, I've been all excited because I say I've written a spherically symmetric wave function for two electrons bound to a central charge. I know you must be thinking: the guy is nuts. Look at the wave function, and it's got all kinds of stuff in theta and phi; it's got one electron leaning this way and another one leaning that way. How can you possibly get spherical symmetry from all that junk?

It's a good question and it actually goes to the heart of some very deep issues in quantum mechanics. When Schroedinger invented the wave function in 1926, the big motivation was to put the mysteries of quantum theory firmly in the realm of causal interactions taking place in real space-time. It was a huge success for the hydrogen atom. But almost immediately, everyone who was anybody understood that the obvious generalization to multi-particle systems took you into 3n-dimensional phase space, where n is the number of particles. So the helium atom has to be solved in six-dimensional phase space, not in "real" space-time. It's a problem.

What people seem to forget is that these n-dimensional solutions, although mathematically correct, seem to overspecify things to a certain extent. The obvious example is the phase of the wave function: clearly, you can multiply a solution by any arbitrary phase and it's still the same physical solution. But the bigger problem is the whole business of the particles. The generalized n-dimensional Schroedinger equation requires you to describe the physics in terms of particles being here and there; but in reality, you are not allowed to distinguish between different particles. So there is something called "symmetrization" you have to do with your solution, which means your saying "A is here and B is there OR B is here and A is there...". Or something like that.

But in our deepest understanding of reality, we can't really believe that electrons A and B have independent existence...otherwise, why would we have this whole structure for interchanging them and getting the same physics? And yet we don't seem to know how to solve for a system of electrons without starting off, in our human perspective, by labelling them A, B, C etc, as though they are independent, distinguishable entites; and then, after the work is all done, we "symmetrize" the solution, effectively robbing the electrons of their independent existence. We just don't know how to get there without without going through that whole song-and-dance.

In the case of proto-helium, I have one electron biased towards the Northern Hemisphere, and the other one biased towards the south. The angular dependences of charge density are cos-squared for the first one, as measured from the North Pole, and sin-squared for the other one: obviously the charge density is uniform over the sphere. Similarly, the spin orientations add up so that the first one is pointing everywhere radially outwards, and the second one everywhere radially inwards: the total spin must therefore be everywhere zero. It's almost obvious that if I had started with one electron in the eastern hemisphere and the other in the west, I would have gotten the same physical solution in the end. The whole north-south orientation was just a convenient fiction that helped us do our bookkeeping.

The big question is: if every physical manifestation of our solution is spherically symmetrical, then why does the equation have so much intricate structure built into it? The answer can only be that there is a huge redundancy built into the n-dimensional Schroedinger equation, and it flows from the premise that we must start off with independent distinguishable particles. If, in the end, we get a physical solution which contains no trace of those independent entities, we have to ask ourselves: did those independent electrons ever really exist?

What does Proto-Helium look like?

2012-02-02T10:05:00.000-08:00

I’m sure you’reall eagerly awaiting the conclusion of the story, where I write out the fullequation for the wave function of proto-helium, that mythical atom consistingof two electrons and a fractional nuclear charge, which is just marginallystable with regard to loss of the second electron. At the same time, I canalmost hear the naysayers scoffing at me, because I’ve heard it before: “So what!! Who cares if you can calculatethe wave function of an atom that doesn’t even exist. Don’t you know there isno such thing as nine-tenths of a proton?”

I know that’swhat people are saying because people have said that kind of thing to me all mylife. Well, I’m going to give the scoffers more ammunition today because I’mnot going to write out any more mathematical equations than I’ve alreadywritten. The fact is, I don’t know what the radial portion of the functionlooks like because I haven’t actually solved for it: to which I’m sure thescoffers will howl out gleefully: “Then you have nothing!!!”. Well, let my readers be the judge of that. I have theangular part of the function, which I think is still more than nothing. Thequestion is: have I got it right?

When I solvedfor the two-electron potential well, I did it by taking the simple productfunction solution and adding a bit of second harmonic. I added it with oppositephases to electrons A and B, which had the net effect of pushing the chargedistributions a little bit apart. This was the graph of the two wave functions:

The problem withthis is that it ignores the whole business of the indistinguishability ofindividual charges, which brings about the need to symmetrize your wavefunction. What I have sketched so far is equivalent to a simple productfunction, and it can’t be right because it’s not antisymmetric with respect toexchange of particles, as must be the case for Fermions. The symmetrization forthis very simple case turns out to be surprisingly complicated especially once you include the effects of spin, and it lookslike this, as I explained in my post from last year, "The Two-Electron Well Revisited".

All thisoverhead is what makes the helium atom so very difficult. The one-dimensionaltwo-electron well must ultimately be solved in two-dimensional phase space; andas I have written it out above, it may be most simply expressed as the sum of no less than foursimple product functions. My proto-helium atom will be very similar: the fullsolution must exist in six-dimensional phase space, but in can be most simplyexpressed as the sum of four simple product functions in ordinarythree-dimensional space. Of course, every wave function in 3-d space canultimately be expressed as the sum of simple product functions, but how manyfunctions does it take? I don’t know how to do it for helium, and I suspect itmight not be so simple. That’s why it’s so useful to work with the limitingcases, such as proto-helium.

The nice thingabout simple product functions is that they retain a certain degree of physicalreality that gets lost when we make the transition intosix-dimensional phase space. If we look back at the one-dimensional potentialwell, we can see that all the real physics, from a descriptive point of view,takes place in the very first iteration of the solution, when we simply allowedthe two charges to push each other a little bit apart. Proto-helium is verymuch the same: we have regular helium, where the electrons are almost on top ofeach other; and as we weaken the central attraction of the nucleus, the chargestend to move apart, until in the limiting case, their angular distribution follows this marvelous function which I would like to name after myself:but since there is already a Green’s Function, I’m going to call this one “Marty’sFunction”:

In proto-helium, the angular distribtuion of the first electron is given by this function, and the second electron gets the same function except turned upside down. So one electron leans towards the northern hemisphere, the other one towards the southern. You then just take the simple product of those two functions. The marvellous thing is that if you carefully examine the distribution of charge and spin, you get a perfectly uniform charge distribution over the surface of the sphere, and the spin vector points everywhere radially outwards. So the function is spherically symmetric.

But we're not done yet. Just like the two-electron well, we haven't properly symmetrized the function with respect to exchange of electrons. And just like the two-electron well, the full solution will be the sum of four simple-product functions identical to the one we've already described, except with spins this way and that. But the essential physics is to be found in the description we've already made so far.

How do you guess the solution of an equation?

2012-02-01T07:13:00.000-08:00

The most important thing in guessing the solution to an equation is to start with the right equation. The problem with the helium atom is that it's not the right place to start. It only dawned on me gradually, after I derived the approximate solution for the isoelectronic series of helium, that the helium atom wasn't the essential case. It's actually the limiting cases we should look at first.

Yesterday I talked about the limit of high Z, or high nuclear charge. For atoms like uranium and beyond, the solution for the wave function is basically the scaled-down version of the hydrogen atom, with the two electrons in a simple product state. This approximation becomes exact as Z approaches infinity.

What about the other limit, the low-Z case? You think the series starts with helium, where Z=2, but that's not right: hydrogen also has a stable negative ion, with Z=1. Is that the lower limit? Not quite. The hydrogen negative ion is just slightly stable with regard to dissociation, by about 3/4 of an electron volt. (Compare this to the binding energy of atomic hydrogen, at 13.6 eV.) No, the lower limit is actually a very peculiar case. It's where you shave off just a little bit from the proton to get an atom with Z=0.92, or something like that. I haven't done the calculation, but obviously at some point the stability goes to zero. You obviously can't hold two electrons with, say, a quarter of a proton, so there's got to be some definite value where the energy of the bound system is exactly equal to the energy of the dissociated system. I'm going to call it "proto-helium", and that's the case we want to look at.

Incredibly, we can guess the form of the solution for this case! And that form will be familiar to readers of this blog, because it's almost exactly the same form I've used to solve not one but two problems in just the last month. I used this function to guess the deposition pattern of a pencil-beam of silver atoms passing through a quadrupole Stern-Gerlach field, and it was also the solution for the effective brightness of a drywall-sheet-cutout moon tilted for maximum illuminating power in the night sky. The function looks something like this:

What I'm showing here is the square of a spinor amplitude function, and in quantum mechanics we have to ask the question: just what was the function we had to square to get this distribution? In the case of the quadrupole Stern-Gerlach experiment, this was the function:

and when we draw it to show which way the spin is pointing, it looks something like this:

This is almost the function I need to solve my "proto-helium" atom, but unlike the quadrupole field, where the spinor orientation rotates retrograde to the pattern, we want standard rotation for proto-helium, so the spinor is always pointing radially outwards. It turns out that's easy to do: we just flip the sign of the spin-down component:

and you get the spins all pointing straight out:

I know it looks a little weird: the positive sign gives you the retrograde spin, and the negative sign gives you the conventional rotation. Shouldn't it be the other way around? The truth is I don't really know which is which. There's a certain amount of human convention in spinor algebra, and up to a point you're free to choose your own convention as long as you stick with it consistently. Since I solved the retrograde case first, that became my convention, so by flipping the sign now I'm just being consistent. I think it could have gone either way.

But we're not quite done yet. This is supposed to be the angular distribution amplitude of an electron in three-dimensional space, but it's just a two dimensional function. We have to extend it to three dimension, and we have to do it so that everywhere on the surface of a sphere, the spinor is pointing radially outwards. I think I've done the algebra correctly when I write the function like so:

And from that angular distribution I am going to construct the wave function for proto-helium. We're not there yet. This represents an electron whose density is biased to the Northern Hemisphere. The other electron is going to have a very similar function, but biased towards the southern hemisphere. When we combine those two electrons, we will find that the spin is pointing everywhere radially outwards so there is no net spin, and the charge density is equally distributed over the spherical surface. These are critical requirements for a physically reasonable solution. These conditions apply to proto-helium, and they also apply to helium and its entire isoelectronic series. We're not done yet. There is still the symmetrization to talk about, so that the wave function changes sign upon exchange of the two electrons; and we haven't even considered the radial distribution yet. But all in good time, my friends; all in good time.

The Wave Function of Helium

2012-01-31T09:16:00.000-08:00

When I started this blog two years ago, I said I didn't know how to handle two electrons at the same time. And then, almost right away, I got into some very cool calculations about the helium atom. Which is, of course, all about two electrons at the same time. The catch is that I cheated: I optimized the wave function of the helium atom assuming the two electrons were in what's called a "product state", where, to use the Copenhagen terminology which I abhor, the probability of finding electron A is independent of the position of electron B.

It's cheating, but it's the same cheat that everyone else uses. (Except that I happened to do an especially nice job of streamlining the math by using some very cool scaling arguments!) You get a pretty good approximation to the ground state energy level, and you can't really improve on it unless you go the whole hog and calculate the 6-dimensional wave function of both electrons.

Then I did something that was very very cool. I generalized the solution method to apply to the whole isoelectronic series of Helium: that is, every possible ion consisting of a nucleus with two electrons. That would be H(-) (yes, there is a stable species of Hydrogen with an extra electron), He(neutral), Li(+) and Be(2+) etc. The funny thing about the series is that the higher up you go, the more accurate it gets. I actually worked out a table of values which I promised to post, where I compared the calculated energies to the experimentally determined values. Then I got distracted and I never did post my table. Well, here it is now:

The calculated values come from a formula I derived for the series. The formula has two parameters, k and Z. Z is just the nuclear charge, and k is what I call a relaxation parameter whereby the wave function spreads itself out to accomodate the mutual repulstion of the two electrons. It works like this: You put the first electron into the atom, and it naturally goes into the ground state which is just a scaled version of the Hydrogen ground state. Then you put in the second electron and assume it goes into exactly the same state. But it turns out there's an advantage to be gained, in terms of lowering the energy of the system, if the wave function spreads out just a little bit. Here is the formula which we have to work with:

You can see there are three terms in the formula: the first is the potential energy, the second is the kinetic energy, and the third is the interaction energy of the two electrons, due to their mutual repulsion. The k factor tells you how much the wave function spreads out to accomodate this repulsion, and you calculate it just by using basic first-year calculus to minimize the energy. For helium you get a relaxation factor of 27/32. (Other people get 27/16 but they are defining their terms a little differently from me.) It turns out the relaxation factor approaches one as z approaches infinity. This means that the interaction energy becomes less significant as the atoms get bigger. For U(90+) the electrons just basically share the scaled-down hydrogen ground-state orbital. (That's uranium stripped down to its last two electrons, in case you didn't figure that out.)

My formula is very accurate at the high end, but starts to diverge at the low end. In fact, for the hydrogen negative ion, it does something quite bad. My calculated value is -12.86 eV, and the actual energy is -14.35 eV, which is more than 10% off. That might not seem so bad, but in fact it makes a huge difference in the physics. The binding energy of ordinary hydrogen is 13.6 eV, or as it is known, one Rydberg. My calculated binding energy is just under one Rydberg, so it is actually preferable for the hydrogen to eject the extra unwanted electron. In fact, the true binding energy is just slightly greater than one Rydberg, and this makes the configuration stable with two bound electrons. In fact, H(-) is a stable species, which you wouldn't have known from my calculations.

Why doesn't my formula work so well for the lightest atoms? Because the electrons have a more intricate way of minimizing their energy, which goes beyond what I can account for with my simple model. You have to treat the system in 6-dimensional phase space in order to do the optimisation. Not about to happen on this blog. (Not without some tricks anyhow).

But that's not exactly where I wanted to go with this. There's something very funny about the binding energies, and I'm not sure anyone has actually solved this problem. As the atoms get lighter, the binding energy gets less and less. What is the lightest atom which is stable with two electrons? Obviously, hydrogen. But what if you could shave a bit of charge off the proton? Wouldn't it still be stable, up to a point? I'm interested in that very special atom consisting of something like 0.92 protons and two electrons, where the binding energy is...exactly zero. Has anyone calculated this atom? Because I think I know how to write down the wave function for it, and I mean the real wave function, not just an approximation.

The Spreading of the Wave Packet

2012-01-30T18:00:00.000-08:00

One of the big problems for the wave interpretation of quantum mechanics is the spreading of the wave packet. Shroedinger did some very clever stuff with his equation to construct cases where a little packet of waves maintained its spatial integrity, most notably in the case of the harmonic oscillator. However, once the electron was propagating in free space, there was nothing he could do to keep his wave packets from spreading out. How was one to reconcile this with the obvious fact that an electron produced at point A was always observed to travel intact to point B?

This is a very puzzling question indeed, but I must ask: just exactly which experiments do we have where an electron is produced at point A and detected at point B? Just as I argued in an earlier essay that we have no pea shooters for photons, I don't believe we really have any pea shooters for electrons either. Oh, there are obviously devices which produce a constant stream of electrons; and the rate of production of these electrons can be slowed down, probably to any arbitrary extent. But isn't the appearance of single electrons still unpredictable? I do not believe we have any reliable method for producing a single electron at point A and detecting it at point B. So what is the problem with the spreading of the wave packet?

People say that you can actually see individual electrons in a cloud chamber. The traces are undoubtedly very compelling: yet they may not be what they seem. In 1927 (or 1929? could it have been that soon after the Schroedinger equation) Nigel Mott published an analysis which showed that for a spherically propagating wave, the most probable observed ionizations would be those lying on a straight line: in other words, the straight-line rays of the cloud chambers were in fact just what you should expect for spherically-propagating waves generated according to Schroedinger's equation.

The common belief that we can actually see individual particles, whether photons or electrons, is with us at every turn. Even Feynmann is guilty of it when he talks about the photomultiplier tube: he says you can actually see individual photons, and they are indicated by the click of the detector. There is a video of him kicking around YouTube where he makes this point, and it strikes me that he is uncharacteristically agitated, for want of a better word, when he makes this argument. Is he showing his frustration because deep down he knows his reasoning is flawed?

Guide to the Perplexed: Part IV

2012-01-28T14:23:00.000-08:00

My series on Bell, entanglement, and the EPR paradox begins with a kind of restrospective essay on more or less how I got to where I am. It's not until my second post that I get into the very interesting history of how we actually got from Einstein to Bell. It's not until my third essay that I get to the crux of the matter: this whole business with the 22.5 degrees is highly overemphasized in the popular narrative. People don't realize that there are huge problems with causality even when the polarizers are aligned, "pre-Bell" so-to-speak. I explain why in this series of essays. Somewhere in the middle of all this I had another one of my Jewish digressions, this one on the fascinating history of double-dipping as originally discussed in the Talmud and later revived in a famous Seinfeld episode.

My next article began with a discussion in StackExchange.com where I guy posed the very interesting question: can you distinguish experimentally between a system where you have atoms in two different states, versus the same group of atoms except they are each in a superposition of those two states? It seems that the people who know how to do these things, using density matrices and such, conclude that there is no difference: and this has deep and far-reaching implications.

After this, I decided to talk about quantisation and the measurement postulate in the context of the
Stern-Gerlach experiment. It seems to me that people who should know better are awfully confused about where exactly the wave function supposedly collapses. Then, in following up on this article, I came across a fascinating Master's Thesis by a fellow from Utah named Jared Rees Stenson, who wants us to analyze the Stern Gerlach experiment in terms of a pure quadrupole field. Stenson does the very interesting analysis for the case of an unpolarized beam, but stops short of the polarized beam; so I set myself the challenge of doing this calculation. I spend the next three essays developing the necessary analytical machinery for my attack on this problem, and then, in my subsequent essay, I abandon all this machinery and simply guess the solution! It's more than a blind guess, of course: it has to satisfy some basic physical parameters, not least of which it has to duplicate Stenson's solution when applied to the unpolarized beam. But the real test would be whether my solution would meet the test of rotational symmetry. The quadrupole field has a four-fold symmetry which would be hard to duplicate unless my solution were just right. It would take some fancy spinor algebra but I should be able to test it against the special case of a 90-degree rotation.

I began girding my loins so to speak to tackle this problem when it slowly began to dawn on me: the quadropole version of the Stern-Gerlach experiment and the so-called "traditional" version were actually one and the same thing! What difference could the addition of a steady-state DC field have on the distorting effect of the quadrupole component? I realized that it was exactly like the way you calculate the tides: it wasn't the direct force of the moon's gravity that caused them, it was purely the distortional or quadrupole component of that field. Why should the Stern-Gerlach experiment be any different? If if that were the case, then the standard description you find everywhere of the beam splitting in two...had to be wrong, because the spatial symmetries of the quadrupole field demanded nothing less than a four-fold symmetry in the detection pattern!

Before going into my final calculation, I have one last brief digression on tides, where I consider the ocean as a driven oscillator, where there are three frequencies that need to be accounted for: the earth's rotation, the moon's period, and the natural frequency of the oceans. Leaving that problem for another day, I proceed to set up the final test of my solution for the polarized beam in the quadrupole field: can I take my solutions for the spin-up and spin-down cases, and add them together to get the correct solution for the spin-sideways case? The answer of course must be the original solution rotated by 90 degrees, and I show in this article that it does indeed work out.

Along the way I had a handful of random blog topics including a link to an awesome gospel harmony song by the Gaither Vocal Band, "There Is A River"; a promo for a physics retreat I held over Christmas at the Maskwa Wilderness Lodge; a link to where a guy from Jordon had been reading my blog in its Arabic translation; and a complaint about getting ripped off by my University of Winnipeg dental insurance plan.

My next topic started with what I thought would be a simple calculation involving the reflectance of the moon, which got a little hairy when I realized that the seemingly flat appearance of the moon in the sky was inconsistent with the theoretical properties of the ideal diffuse or "Lambertian" scatterer. It turns out there are at least three moons which are interesting to calculate: the ideal Lambertian moon, the moon as a polished steel sphere, and the moon as a flat sheet of drywall tilted for maximum nighttime effectiveness. It turns out this last case has some eerie similarities with the mathematics of....the quadrupole Stern Gerlach effect! Check it out if you don't believe me.

Along the way I had a few more random posts: this one, a reprint of an old mail-out I did pointing out the arrogant and dismissive manner in which Israel had been presenting itself towards its neighbors; a topic which unfortunately has not lost its timeliness. Again with the Jews, I wrote up a historical analysis comparing the life of the Palestinians living under Israeli rule with the life of the Jews in the Czarist Pale of Settlement, which was later reprinted in the local Jewish weekly. Then there was something messed up with my blog posts, and it turned out to be a technical problem which a guy from Finland helped me solve over at the Blogger Help Forum.

Most significantly, I was finally, after an acrimonious battle with my professors, expelled from the Teacher Certification program at the U of W. I started a separate blog to talk about that, which you will find if you follow the link.

And that pretty much brings us up to the present four-part series, A Guide to the Perplexed, which takes a retrospective look at two years of blogging to see what I've actually accomplished. On the one had, I look at it and see that I've actually done quite a lot of physics. On the other hand, the problem inspired the name of this blog is basically still with me: I still really don't know how to do quantum mechanics. The stumbling block is, and always was, how to handle a problem with two electrons in it. Oh, I know everyone says you just solve the Schroedinger equation in six-dimensional phase space; and I know there are some people who can actually do just that. I just believe that most of the people who talk about it really have no idea what they're talking about: the only difference with me is that I actually know that I don't know what I'm doing.

At any rate, that's how I felt when I started doing this retrospective series. The funny thing is that during the course of the week that it's taken me to get through all my old topics, an idea of a solution has started taking shape in my head. I'm thinking I might have an angle on the two-electron problem, and I wonder if it's for real. I'm going to leave off for today and come back to this question next time.

Guide to the Perplexed Part III

2012-01-26T10:13:00.000-08:00

It had been five months since my last post when I returned to blogging with an article on the collapse of the wave function. One of the most famous examples of wave function collapse is the appearance of flecks of metallic silver on a photgraphic plate exposed to the light of a distant star. Any reasonable calculation of the energy density of an electromagnetic wave shows that it is impossible to gather enough energy to drive the chemical transition AgBr => Ag + 1/2Br2, and this circumstance is taken as proof that the energy of the light must be concentrated into particles called photons. In this article, I outline the thermodynamic argument which shows that when we treat the whole silver bromide crystal as a solid solution, the reaction actually becomes spontaneous at very low concentrations. In other words, the energy necessary to detect the light is already present in the crystal, and you don't need to invent a "photon" to supply it.

I don't know if it was before or after I wrote that article that I discovered something that would change my life: Google Blogger tracks your statistics! I discovered that people were actually reading my articles, and suddenly everything was different. I don't know if you'd call it an awful lot of hits, but I was getting two or three hundred clicks a month. This changed everything.

The first thing I posted was an old article about the Mid-East conflict that I had first circulated by email back in 2006. Then I got right back to physics. My next article was about something that I figured out over twenty years ago. I had been trying to calculate how much power you could absorb from an AM radio station with a well-designed crystal radio, and I discovered that the theoretical power was independent of the length of your antenna. This seems like an absurd result, but there is a formula for it in the books, and it's true. The catch is that you can't easily build a perfect antenna because of imperfections in real materials, mostly due to the resistivity of copper. But in theory the result is true and it is mathematically derived. What is different in my approach is that I show how you can understand the result pictorially, and working from simple pictures you can get a pretty good ballpark of the exact theoretical result.

What the whole world seems to have missed about this calculation is its enormous implications for quantum mechanics. In Schroedinger's picture, a hydrogen atom is nothing more or less than a tiny crystal radio antenna, and everything that a hydrogen atom does, in terms of its interaction with the electromagnetic field, can be understand in terms of its properties as a classical antenna. In particular, this new perspective makes a mockery of those old textbook calculations where you evaluate the photo-electric effect by looking at the cross-sectional area of an atom. The effective electrical cross-section of an antenna has nothing to do with its physical cross-section, and this is a purely classical effect that you don't need to explain with "photons".

Meanwhile, now that I was posting again my hit count had taken a sudden upswing. Google doesn't just give you the hits, it tells you the country of origin and the search engine keywords; so I was pretty excited one day to notice my first visitor from the Palestinian Territories, so I couldn't resist giving a friendly shout-out to whoever he was. It was only aftertwards I realized that he might have just as easily been an Israeli "settler"; but either way, I'm glad to see him.

Another surprise from the Blogger statistics was the number hits I got from people who googled "perturbation theory" and "ladder operators". So I wrote a follow-up to my earlier musings on the subject. It's a more mathematical topic than I normally ought to bite off, but I still think my pictorial perspective adds something to the big picture.

In my many internet discussions over the years about the photo-electric effect, I had often heard of the semi-classical school of Jaynes and Scully. I had always assumed that my approach was more or less in line with theirs, and when people ridiculed me for my ideas, I would sometimes invoke Jaynes for moral support. I was pretty shocked only last year to learn that I was wrong: Jaynes and Scully take a classical field and apply it to the quantum atom, but then instead of allowing the atom to evolve through time evolution from the excited state to the ground state (which is what I do), they still calculate the quantum leap transition probabilities. In other words, if my approach is "semi-classical", then Jaynes and Scully should actually be considered hemi-semi-classical. Or whatever. You know what I mean. I explain it all in the linked blogpost.

Coming up with a semi-classical ("no-photon") explanation for the photo-electric effect was a defining moment in my life, and for ten years I would go on the internet and try to argue it. The most common way people would shoot me down was to say "maybe you can explain the photo-electric effect, but you can't explain the Compton effect." And they were right: I couldn't. Until one day I did. This was a paradigm breaker! I thought for sure I would win the Nobel Prize for this. Sadly, it was not to be. It turns out my explanation was identical to the explanation that Schroedinger had already published in 1927. It's true that in 1919 Compton had "proved" that you couldn't explain the effect according to the wave theory, but that was because he treated the electron as a little charged ping-pong ball. The wave theory explanation is a totally natural outcome of the Schroedinger equation of 1926, but by then the photon paradigm had been so firmly established that even Schroedinger was ignored and marginalized when he argued against it.

In the meantime I had gotten into a discussion on StackExchange.com about transmission line impedances, and so to get myself back in the game, I re-did some old calculations about a parallel-plate waveguide. You get some very interesting results if you just assume that for a freely propagating wave between two plates, there must be no net attraction or repulsion between the plates. This discussion led to some very cool calculations of transmission-line impedances, which I calculate as usual with very pictorial methods. Next I take a pretty big leap and apply these methods to calculate the radiation resistance of a half-wave dipole. Everybody knows this is supposed to come out to 73 ohms, but for most people that number is pretty mysterious. I don't get it exactly right, but I definitely justify it to within a reasonable accuracy, and all by very graphic methods.

In my next post, I find myself again dragged back to the Middle-East conflict. Here I resurrect an old proposal of mine to adopt the usage of Arabic Script to write Hebrew. It's a fantastic idea on all kinds of levels, and it would do huge things to bring Arabs and Jews together, but nobody in Israel is listening to me. I'll keep trying.

As I mentioned, Google Blogger gives me statistics on country of origin, and naturally the U.S. and Canada lead the list, followed by Germany, Russia, and the United Kingdom. Surprisingly, the next spot on the list is up for grabs, and is hotly contended by such countries as India, the Netherlands, South Korea, and...tiny Slovenia, which for a brief moment edged out the other contenders for sole possession of sixth place. I acknowledge them in this post.

My next series of articles was motivated by a question from my nephew, who asks "why is energy e=mc^2, and not m-c-cubed or whatever? Although this question can be easily answered with dimensional analysis, the actual reason is harder to justify than you might think, and I was led pretty far into uncharted territory (for me anyway) when I tried to justify it via relativity.

My next series of articles deals with perhaps the most baffling and certainly the most talked-about paradox in all of quantum mechanics: the question of entanglement, with Alice and Bob and the crossed polarizers and all that stuff. It gets pretty involved and I think we'll continue with this topic when I return.

What if the Moon were Made of Golfballs?

2012-01-26T07:43:00.000-08:00

I've been talking about the reflectance of the moon in recent posts, and the issue came up of why the appearance of the Moon is so out of whack with what should be called for by the theory of diffuse reflection. The classic Lambertian reflector has the singular property that it looks just as bright whatever your viewing angle: but this property does not extend to the angle of illumination! In other words, it most certainly does not look just as bright no matter what angle you illuminate it from. So why does the moon look uniformly bright all the way across its disk? Why does it look like a flat cut-out?

It occurs to me to ask the question: what if the moon were made of golf-balls? Each golf-ball would be a mini-moon: when the moon was at half-phase, each of the little golf-balls would look like half moons to us here on earth. Except they'd be too small to see. So we'd just see the average of all of them, and the local average would be the same wherever we looked. So maybe the moon would look uniformly bright everywhere.

Except the golf-balls low on the horizon would also be partly in the shadow of other golf-balls, so the fringes should still look darker. In other words, I'm not completely buying my own explanation. But it's a thought.

Meanwhile, I tried to apply some calculus to this question, and I'm not too sure of my technique, but I got an answer and I wonder if anyone out there would like to double-check it. I compared the real moon...that is, the ideal, bright-white Lambertian "real" moon...to a flat cut-out sheet of drywall. According to my calculation, if you compare these two models at midnight on a full moon, the flat drywall cut-out provides 50% more night-time illumination than my "ideal/real" Lambertian moon. I'm not going to try and repeat my calcuations here, but I'm just wondering if anyone out there wants to see if they get the same answer as me.

Guide to the Perplexed: Part II

2012-01-23T07:42:00.000-08:00

In my last post, I catalogued all my blog articles for 2010, up to August of that year. What followed was a long period of inactivity, which followed a visit to Winnipeg from my grad school buddy Richard Epp. I remember it was a really hot summer day, and started telling him about Quantum Siphoning. Richard objected to the interpretation of the wave function as charge density. He pointed out something I already knew: if the charge is distributed, then the calculation for electrostatic energy is messed up. How could I answer that objection? I had no answer, and every time I tried to work on physics, that was all I could think about.

Then in February I stumbled across a website called The Foundational Questions Institute which was apparently sponsoring an essay contest on the question "Is Reality Analog or Digital?". Incredibly, the deadline for submissions was that very day, so I immediately wrote up an article which I gave a title that says it all: There are No Pea-Shooters for Photons. The targets of the essay are the three pillars of the photon theory, as universally recognized in the popular narrative: the Blackbody Spectrum, the Photo-electric effect, and the Compton Effect. In my essay, I show how each of these has an intuitively reasonable wave theory explanation, which could not have been understood by Planck, Einstein, or Compton in their time because the wave theory of the electron was not revealed by Schroedinger until 1926, long after the particle paradigm had taken firm hold. The title of the article refers to the well-known post-modern justification for photons based on experiments where you supposedly fire one photon at a time and track where it goes. I point out, as I did in an earlier blogpost called The Clicking Detectors, that none of these experiments are really quite what they claim to be, for the plain and simple reason that there really are no pea-shooters for photons.

When I submitted my essay, I was in a rush to meet the deadline, and I didn't feel that I had dealt as well as I might have with the question of the blackbody spectrum, so I went back to my blog and that became the subject of my next article, which soon mushroomed into a nine-part series. The basic idea was that instead of attacking the ultraviolet catastrophe at the electromagnetic level, you take it on at the mechanical level. All electromagnetic radiation has to have its source in the mechanical vibration of charges, and if there is no mechanism to set those vibrations in motion, you don't have a problem at the electromagnetic level. The one nagging problem with my argument is that thermodynamics doesn't depend on specific mechanism: there's something called the Equipartition Theorem which supposedly rules no matter what, and I had to somehow explain it away. I struggled with this through my next four blogposts until I made a huge breakthrough.

There's something people do in physics which drives me crazy and that is putting all their arguments in the most abstract, mathematical form. I need to see real examples and real mechanisms, and in particular I wanted to figure out just how an equilibrium is established between a mechanical oscillator and the electromagnetic field. This doesn't look like it should be an insurmountalbe problem: we know how to do the driven harmonic oscillator with damping: just apply that to an atomic oscillator with the electromagnetic field as the driving force. How hard could it be?

The problem is that we don't just have a driving field of known intensity: we have a random, distributed field. There is no simple number we can pick out of the blue and say "we are driving the atom with an oscillating field of so-and-so-many volts-per-meter: the field strength is expressed in volts-per-meter-per-hertz, and how the hell do you interpret that. What happened is after years and years of not knowning how to handle this question, all of a sudden I figured it out! It's a beautiful, very pictorial explanation that starts off by considering that familiar old chestnut of statistical theory, The Drunkard's Walk . Analyzing the driving force on the oscillator as a special case of the Drunkard's Walk, I show that you are allowed to truncate your frequency distribution at any arbitrary limits and you still get the same oscillation regardless. I then carefully count up the cavity modes of the electromagnetic field, and then run a numerical example of a special case to come up with the amazing conclusion, which is the basis of the Rayleigh-Jeans derivation of the ultraviolet catastrophe: the energy per mode of the electromagnetic field is equal to the energy per mode of the mechanical oscillators!

What this means is if you can show that the high-frequency oscillations are suppressed at the mechanical level, then they are automatically suppressed at the electrical level. You don't need to throw out Maxwell's equations to avoid the ultraviolet catastrophe.

My next post was four weeks later. A year previously, I had been very excited to work out the solution for the problem of two electrons sharing the same potential well. It came as a huge surprise to learn that in contrast to the well-known single-electron case, with two electrons the shape of the solution depended on the size of the box. I had a huge argument with a guy named SpectraCat in physicsforums.com over this: he said that the shape of the solution depended on the strength of the interaction, and I said that was exactly the same as saying it depended on the size of the box. Of course I was right: the very suprising thing about it was that the case of the very small box corresponded the case of independent particles, and vice versa: the strong interaction corresponded to the case of the very large box! It may seem counterintutitive, but in my analysis of the iso-electronic series of helium, I show that's exactly how it works. A "helium-like atom" is any atom stripped down to its last two electrons. The series actually begins with hydrogen: it turns out that the negative hydrogen ion, consisting of a proton and two electrons, is marginally stable. As you add more protons to the nucleus, the electrons get more tightly bound: helium, lithium, beryllium, etc: I actually found binding energies for all those atoms and you can clearly see from the values for the energy, that as you go to higher atomic numbers, the electron configuration approaches the simple product state of hydrogen-like orbitals. In other words, as the box gets smaller, the interaction of the electrons becomes insignificant.

What brought me back to this question was I realized I had made a big mistake: I had botched the symmetrization! The wave function was actually quite a bit more complicated than I had drawn it, because my answer did not preserve the correct symmetry for fermions: the function must reverse polarity when you switch particles. You can always do this by taking appropriate sums and differences of whatever function you already had, and that's what I do in this article. It turns out this is the exact same symmetrization method that needs to be applied to the case of two isolated hydrogen atoms. You are not allowed to simply say that the electron here is spin-up, and the electron there is spin-down: try writing it down that way, and then interchange electron A with electron B. You know that according to theory, you must get back the same function with a negative sign, and you'll see that you don't. The wave function isn't right until you symmetrize it the way I've shown in my article, and it turns out that this leads to some very distrubing and surprising consequences.

In the meantime, I wrote a couple of other articles in April about Fourier Transforms, Ladder Operators, and Pertubation theory. My son's friend was taking a course in Mathematical Physics, and I had been helping him with assignments. It drives me crazy the way they suck all the physics out of these things and just give you math question that amount to manipulation of symbols according to a set of rules. That's not physics to me, and it's not even math. For me, it's all about the interpretation, and I got into some very cool stuff in this article about solving differential equations with Fourier transforms. The very last problem on the homework assignment was a weird-looking differential equation that I vaguely recognized as having something to do with the quantum harmonic oscillator: I couldn't quite put my finger on it but finally figured out that it had to do with ladder operators. One thing led to another and I started writing some very cool stuff about Perturbation Theory. This is something that's taught, as usual, as a set of rules for symbolic manipulation of functions, but here I make it into something pictorial, relating it to ladder operators and Taylor expansions. It's a bit half-baked, but it's still good.

And then it stops. I don't know exactly what happened, except I remember I was working all summer on a construction survey crew, and we had a lot of fifty-hour weeks. My next article wasn't until almost six months later, and that's where we'll continue when I come back again.

A Guide to the Perplexed:Part I

2012-01-22T10:14:00.000-08:00

I started this blog almost two years ago with a rant about how I didn't feel like I knew what I was doing any more, and quantum mechanics was to blame. Since then, I've posted eighty-something articles, including some pretty cool stuff, and maybe you're wondering: do I feel any better today? That's a tough question. Let's start by going through my old posts and listing the topics I've covered. That will be a useful thing in itself.

One of the main reasons I started this blog was because I felt I had all kinds of original ideas on how to do physics problems, and I wanted to start recording them. How original were these ideas? I don't know, you can judge for yourself. I just thought I did physics differently from other people, and I wanted to stake out my territory. So it's a bit ironic that my first physics post after the inaugural rant was entitled Something I just Figured Out Yesterday . This is what happened. I complained in my first post that I didn't know how to do quantum mechanics. I actually was quite good at doing most of the undergrad level calculations, like the hydrogen atom. More importantly, I felt I wasn't just good at doing the calculations, but I could actually interpret the results with some insight. The big problem was when I came up against two electrons at the same time. I just couldn't begin to handle multi-electron problems.

So I was thinking about this, and it occurred to me that there was one two-electron problem that I had to be able to solve: the problem of two isolated hydrogen atoms. There are two protons and two electrons. Since I already knew how to solve the single hydrogen atom, how could it be any harder if there were two of them. They key was the shift of perspective: I was going to solve it as a two-electron problem.

I came up with an apparent solution that raised more questions than it answered: it took me quite a while to find my mistake . In the meantime I put up a very important post called The Clicking Detectors . This one wasn't brand new, it was stuff I'd figured out a long time ago but never written up. People have the idea that you can split a beam of photons with a half-silvered mirror, and the statistics of the detector clicks show that each photon must have gone either one way or another. This is supposed to be a proof of the particle theory of light. In "The Clicking Detectors", I show that any reasonable wave theory of light would give the exact same detector statistics.

I didn't post again for about a month. What had happened in the meantime is that I'd got into a very intense discussion thread at physicsforums.com on the nature of wave function collapse, and in the course of that discussion I actually figured out a mechanism whereby you can explain all kinds of collapse phenomena via the ordinary time-evolution of the Schroedinger wave function. I called the mechanism Quantum Siphoning . In my article on the Clicking Detectors, I said that any "reasonable" wave theory would give the right statistics: the catch was, I didn't actually have a reasonable wave theory at that time. Quantum Siphoning turns out to be that theory. If I ever win the Nobel Prize, it will be for Quantum Siphoning.

My next article was a reposting of something I'd actually written about a year earlier in private correspondence with my physics buddy, Richard Epp, who was in grad school with me and now works at the Perimeter Institute. Ramanujan and the Casimir Effect shows how you can use some of the weird seemingly divergent series studied by Ramanujan to calculate the force between parallel plates due to the Casimir Effect.

In the meantime, I was still engaged in the quest for a two-electron problem that I could actually solve, and I was now working on The Double-electron Potential Well , which became a series of two articles. Not the third-year physics problem with fictional "non-interacting electrons", but the real thing. I had some preliminary sketches of two-dimensional wave functions that I had posted on physicsforums.com and I had written that I was having trouble getting a handle on it when a very smart guy named Peter Atcam suggested I "turn off" the interaction and then turn it back on again very gradually. This was a hugely productive suggestion that not only clarified the situation with the two-electron well, but led to a whole new topic, The Iso-electronic Series of Helium .You can find solutions for the helium atom on the internet, but no one does it with quite as much finesse as I do in this four-part series of articles. In the meantime, my article on the two-electron well holds the distinction of being my most widely-read posting
of all time, according to Google Blogger statistics; which is a bit ironic since it contains an error of omission which I didn't find and correct until over a year later.

I was just finishing my series on helium when the tin-pot dictators of physicsforums.com led by the omniprescent ZapperZ kicked me off the forum for the third and final time. I talk about the circumstances in my article Banned for Life . All I can say now is I've been banned from better places than physicsforums.com, including a couple that I mention in my next article, A Tale of Two Strikes . But back to physics. My next article came in response to a question posted in physicsforums wherein I show how the anomalous specific heats of diatomic gasses at very low temperatures can be intuitively understood as a consequence of the wave nature of matter .

When I talked about Quantum Siphoning, I said it represents my best shot at getting a Nobel Prize. I wasn't joking. You might think it's not much of a shot, and maybe it isn't but if it's one in two hundred, it's still a better shot than anyone else who's every read any of my blog posts. It's a shot because if it's right (and it is right) then it's a paradigm breaker just like relativity or the uncertainty principle: it's not just a different calculation but a whole new way of looking at physical reality. What makes me think I'm capable of making a breakthrough of this magnitude when people smarter than me have tried and failed?

I'm a bit of a student of the history of quantum mechanics, and from my readings I know that there was no one who would have been more sympathetic to my viewpoint that Schroedinger, who was himself ridiculed by the "mean physicists" (Born, Heisenberg, Lorentz etc.) for trying to find real-time, causal mechanisms for quantum phenomena: and make no mistake, these would be wave-theory mechanisms, not particles. The particle concept is the source of all evil in physics: the quantum leap, the collapse of the wave function, the Alice-and-Bob correlation of entangled particles. No one hated particles more than Schroedinger: so if quantum siphoning is the answer to these paradoxes, why didn't Schroedinger invent quantum siphoning?

The pathway to the truth is fraught with obstacles; and what makes me different is I have always done things my own way. Over a lifetime of tinkering with ideas, I have put together my own unique bag of tricks. These are mostly things that other people already knew; people much smarter than me, including Schroedinger. But no one single person, not even Schroedinger, ever knew exactly all the little things I knew. It is the strange uniqueness of my own personal bag of tricks that has placed me in the fortuituous position of being the paradigm breaker.

I'm not going to list all of my tricks right now: after all, that's why I started this blog in the first place. But my next series of articles, spread over three months and beginning with the post Karma and Carbon Monoxide , tells the story of one of the most unlikely links in the chain which led to Quantum Siphoning. If you read through the previously-mentioned physicsforums thread where I ended up getting banned for life , you will see that at one point in the argument I was dealt a near-death blow relating to the activation energy needed to drive the chemistry of the photographic process. Incredibly, I recovered from this blow with an ingenious thermodynamic argument where I showed that despite the unfavorable enthalpy, that at the very low concentrations of metallic silver in a developable exposure, we were entitled to treat the silver halide crystal as a solid solution, and taking into account the Gibbs Free Energy of the total system, it could be shown that the thermodynamics of the transition was actually borderline favorable! This argument led directly to Quantum Siphoning, and I never would have been able to make it if not for the story told in this series of posts about a project I was given as a junior engineer in my very first job out of university at a government research station in Pinawa, Manitoba: hence, Karma and Carbon Monoxide.

This brings us up to August 2011, except for one article from the summer of 2011 which I haven't yet commented on: Why do Solids Absorb Light . This turns out to be one of my most-often googled posts, and not without good reason. It is a question which is asked all the time, and people never give a decent answer. The standard answer about electronic transition levels is fine for explaining why light would be scattered, but where is the mechanism for absorption? I have to admit my answer is a bit sketchy, but I'm pretty sure it is a semi-classical mechanism very similar to the one Schroedinger demonstrated for the Compton effect. That in itself is another very important paradigm which is buried in obscurity in the conventional group-think, but more on that later.

After my Carbon Monoxide series ended, I stopped posting for about six months, not to resume again until 2011. I think we'll stop this recap as well for now, and continue next time from where we leave off today.

Explorer Bad, Firefox Good

2012-01-20T04:57:00.000-08:00

I mentioned in some earlier posts that some of my articles were screwed up: not that I had physics mistakes, but that the files seemed to be corrupted. When I clicked on the link, they would show up for a second, and then go blank. I had to take down the articles and repost them.

Yesterday I decided I ought to go through my whole blog, all 84 posts, and make sure they were all working. I was appalled to find about ten of them crashed the site, including some of my best posts with really good search engine exposure. This was very disappointing.

I prepared to delete the bad posts by making backup copies. and then it occured to me to check out discussion groups for possible help. A quick search showed that Google Blogger maintains its own help forum. I posted my problem and within an hour a user named "mspotilas" replied:

Blogger rolled out threaded/two level commenting, and that does not work on Internet Explorer. With Firefox and Chrome you should be able to open the posts with comments.

To make them work on Internet Explorer, too, while waiting for Blogger to fix the bug, you can change in blog's settings your comment form from embedded form to full page or popup. That disables the threaded commenting, and also Internet Explorer will work again.

You can see the exchange here at theBlogger Help Forum . Could it be the browser? I could hardly believe it so I tried the links in Firefox. They all worked! The lesson: Explorer bad, Firefox good.

But there was one puzzling thing: My problem had nothing to do with comment fields. It was just random throughout my posts. Or was it??? I looked through my list of posts again. Son of a gun, if the bad posts weren't all the ones with comments. How could I have missed that? It was 4:00 am when I went through my files, but still...

POSTSCRIPT: I hope Potilas doesn't mind me posting his user profile here. For your information:
(Oh, and he's also got a YouTube channel mostly about classical piano ).

Introduction

Sairastan MS-tautia ja käytän Bedrocan-lääkekannabista taudin kurissa pitämiseen.

I have multiple sclerosis and I use medical cannabis (Bedrocan) to control the disease and symptoms.

Gender

Male

Other names

MS-potilas

Is Israel an Apartheid State?

2012-01-18T16:32:00.000-08:00

I originally posted this article two months ago but something funny has started happening to some of my links, whereby they seem to show up for about a second and then go blank when you click on them. So I'm reposting it here.

* * * * * * * *

These days you read about things happening in Israel that would have been unthinkable in the idealistic days of the first few decades of the beleaguered state. Even then, with its back to the sea and deadly enemies surrounding it on all sides, Israel was widely vilified as a second South Africa; even more outrageously, it was often compared to the Nazis. Sadly, with the passage of time, some of the old slanders have begun to take on elements of truth as the rights of Arabs under Israeli rule have come under increasing attack. Not that Arab life is Israel is remotely comparable to the worst excesses of South African apartheid, let alone the grotesque comparisons with Nazi persecution. Ironically, however, the critics and defenders of Israel alike have missed the most obvious and glaring comparison : there is an excellent case to be made for equating Arab life under Jewish rule with Jewish life in Imperial Russia!

We Jews are supposed to place great importance in knowing our history, but our collective memories of life in Russia are oddly skewed. I happen to have some expertise in these things because I am one of the few people in my generation who is able to read the literature of our people in its original language, and I have done so extensively. So I know something about Jewish life in Russia. I also know that if you ask modern Jews to give a single word that most succintly expresses the nature of that life, they will almost unanimously say "pogroms".

Unfortunately, it seems we Jews, like all other peoples, remember only what it suits us to remember. We choose to remember life in Russia as an unmitigated series of horrors not least because it helps to justify our Zionist mythology. (By the way, because I call it a "mythology" does not mean I don't personally buy into it. I am a proud Zionist, but I am not proud of everything Israel does!) In fact the pogroms were a horrible episode in our history, but they were far from a dominant feature of Jewish life in old Russia. The actual picture is much more rich and nuanced. It is true that there were many episodes of persecution and injustice, but it these were interspersed with periods of great freedom and opportunity. And as we review the history it is surprising how many parallels we will find with lot of the Arabs in Israel.

We can begin with how the Jews came to be Russian subjects. Medieval Poland was a place of refuge for Jews fleeing the persecution of the Crusaders, and once established we fluorished there. Poland was in those days a huge kingdom covering much of present-day Eastern Europe; over the centuries, it was gradually picked apart by the surrounding powers of Austria, Germany, and Russia, until with the final Partition of Poland in 1793 the Russian Tsar awoke one day to find himself the proud ruler of close to a million Jews. Thus the Jews became unwanted subjects of Russia much the same way as the Palestinians became unwanted subjects of the Jews: through military conquest.

These Jews were not the educated intellectuals of North America who typify the modern Jewish stereotype: they were mostly primitive, black-frocked and bearded religious fanatics. The Tsars mistrusted the Jews for their alien beliefs and their close ties to their co-religionists living in hostile states across the border. Sound familiar? Keep reading. Over the course of the nineteenth century the pendulum swung from one extreme to another, the government sometimes trying to integrate the Jews into the modern economy as productive citizens, and sometimes trying to contain them by harsh discrimination. A great concern was the Jewish birthrate, with early marriages and up to a dozen children being the norm.

There is much nonsense written about the actual facts of daily life. People say that Jews weren't allowed to own land. They certainly were: however, there were restrictions on where they were allowed to buy land. Sound familiar? And there were certainly cases where the Jews were cheated out of their lawful property rights by the government. But at the same time we were entitled to go to court and contest such expropriations, and occasionaly we would win these cases. Mendel Bailiss was famously acquitted by a Russian judge and jury in the infamous blood-libel trial of 1912. But on the whole there is little doubt the courts were stacked against us. Sound familiar?

One of our greatest grievances against the Tsar was the "Pale of Settlement". Jews were forbidden to take up residence outside the areas which basically constituted the original Polish kingdom: in other words, they weren't allowed to leave the Occupied Territories to live inside the Green Line where there were greater economic opportunities. Oops, it wasn't called the Green Line...that's what we have in Israel.

Now let's remember a thing or two about the pogroms themselves. There were three significant waves of pogroms. The first was in the 1880's in the aftermath of the assasination of Tsar Alexander. Although the news of these outrages terrified the Jewish community throughout Russia, the total number of fatalities in this period was in fact less less than one hundred. A more serious outbreak began with the Kishinev pogrom in 1903, and over the next few years perhaps a thousand Jews died in the unrest.

There is a lot of nonsense about Cossacks and Russian police officers leading these outrages. In fact, while high officials in the Government undoubtedly knew and approved of what was going on, the fact remains that Russia was a country of law and justice and it was unthinkable for the police to allow these things to go on with their knowledge, let alone to participate in them.

The catch was: if they didn't know, they couldn't very well do anything about it! It's called plausible deniabilty and it's the oldest trick in the book. As long as they could pretend they didn't know, they would let it go on; but after a day or two they would invariably show up and restore order. A few ringleaders might be slapped on the wrist, but that would be the extent of it.

How similar is this to the pogrom which we allowed the Christian Phalangists to carry out under our noses in the Sabra and Chatilla refugee camps in 1982? It is true that Ariel Sharon was eventually found accountable and demoted from his cabinet post, but that didn't stop him from later becoming Prime Minister. How do you think our Arab citizens should have felt about that?

There were some fifty deaths in the Kishinev pogrom, and the world was shocked. It was a time when people believed that freedom and human dignity were marching forward, and the backwardness of Russia was a huge embarrassment both inside and outside the Empire. The plight of the Jews attracted worldwide attention, and an upsurge in the Zionist movement was one of the immediate consequences.

This is not quite the end of the story. In 1914 war broke out and within four years the old order of kings and emperors simply ceased to exist. It was replaced by a new harsh world of nationalisms and ideologies. Civil war raged in Russia and in the Ukraine, a nationalist government took over that virtually declared war on the Jews. A hundred thousand died in the pogroms of 1919-20, and the world scarcely took the time to yawn. Even the Jews hardly remember these martyrs, as their suffering was eclipsed by the much greater disaster of the Holocaust twenty years later. But we ought to at the very least not blame those pogroms on the Tsar, who had already been killed by his Bolshevik captors.

History is a funny thing. We choose to remember whatever suits our purpose, and it suits our purpose to demonize the Tsar and everything he stood for. Yet when the Palestinians do the same to us (and they do), we feel aggrieved. "Why don't they appreciate that living under the Jews, they are far better off than their bretheren living under brutal dictatorships elsewhere in the Middle East?" We ought to remember that their attitude toward us is simply human nature, and it is not so different from our attitude towards the Russian Empire. Perhaps we have more in common than we like to admit.

Again with the Jews...

2012-01-18T05:06:00.000-08:00

A couple of months ago I posted on this blogsite a letter that I had first circulated five years ago, "Ten Things We Jews Believe About the Middle East" . I have always been a staunch supporter of Israel, and until 1992 I supported every military strike and every new settlement. Then Arafat signed the Oslo Agreement, and the conflict was over. We had won! or at least, we had established our right to exist within recognized borders. There was no more need to fight.

But in the aftermath of this great event, nothing seemed to change! For the next eight years it was business as usual. The occupation went on, and the settlements grew and multiplied. Why were we surprised in the year 2000 when the intifada broke out? Did we look at our own behavior? No! We said that the Arabs were finally showing their true colors; that the intifada proved that they had never given up their goal of driving us into the sea. That's when I sent out the letter I referred to above.

As you can guess, the letter was not well received in the Jewish world. The negative reaction was not by any means unanimous, but it was nonetheless...well, it was negative. Among non-Jews, on the other hand, my letter was very well received. I got so used to this pattern that one day, when an acquaintance I thought to be non-Jewish wrote critically, I answered back (totally unselfconsciously, I swear!): "I didn't know you were Jewish, Ralph." Ralph took offense at that comeback: what right did I have to assume he was Jewish? It was only very recently that I learned the rest of the story: Ralph isn't Jewish, but his wife is.

Anyhow, what brings all this to mind is that I was browsing through some old correspondence and I stumbled across a follow-up to my original letter. The facts cited are a bit out of date, but the general picture has hardly changed. Here then is what I wrote five years ago. God help us if we don't change our attitudes.

************

ISRAEL SHOWS ITS CONTEMPT FOR THE ARABS.

Six weeks ago, I sent out my ten-point plan for attitude change alongwith my two-point action plan for peace in the Middle East (see below). Since then, the news from Israelhas only strengthened my conviction of the dire need to change our attitudes towards our Arab neighbors before we bring down disaster on ourselves. Consider these five examples of Israelibehavior:

1. Syrian President Basher Assad proposes peace talks on the Golan issueand we brush it off with contempt.

2. Arab League Secretary Amr Moussa calls for a renewed initiative basedon the 2002 Beirut peace plan and we ignore him.

3. Israeli Prime Minister Ehud Olmert stands on a platform in Moscow beside VladimirPutin and warns Iran to"be very, very afraid" of what Israelwill do unless Iranbacks off its declared plans for peaceful nuclear development. (And just days ago anothercabinet minister mused publicly about a "first strike"!)

4. Foreign Minister Tzipi Livni accepts and then rejects an invitationfrom Qatarto take part in a UN conference in the Gulf Stateswhere she would have an opportunity to present Israel's best diplomatic face to its neighbors. Reason? She refuses to attend the sameevent as representatives of the elected government of the Palestinian people.

5. Prime Minister Olmert appoints ultra-right-wing-nationalist AvigdorLieberman to a senior cabinet post. Remember the fuss we made when Joerg Haider became a member of theAustrian government? Our man Lieberman makes Haider look like a boy scout. And now he's ourpoint man on the Iranissue!

I thought we Jews were supposed to be smart. Maybe we are when it comesto inventing the theory of relativity or finding the cure for polio, but it seems like we've gota few things to learn about civilised relations between nations of different backgrounds. If it's not alreadytoo late.

I get to re-use a graph from another problem

2012-01-17T19:49:00.000-08:00

Last time I said it was interesting to calculate how effective the moon would be if it was a simple hunk of drywall, just a big round cut-out. Actually the interesting calculation is if we hang the drywall up in the sky, and then optimize it by tilting its angle so as it revolves around the earth, it always delivers the maximum amount of illumination.

It's not hard to calculate that the effective illumination is maximized if the angle of tilt simply bisects the angle between the sun and the earth, as measured from the drywall. It's a consequence of Lambertian diffuse reflection and plain old geometry. So, for example, when the moon, sun, and earth form a right angle, the effective illumination is exactly 50% of what it was in "full moon" conditions: you tilt the drywall at 45 degrees so it intercepts 71% of the sunlight, and because of the viewing angle it's apparent size is only 71% of the full disk. Compound these two effects and you get 50% power. We can sketch the function of effective illumination power as the moon revolves around the earth. We imagine the sun is at the bottom of the picture, and the graph of radiance looks like this:

Look familiar? It's the same graph I plotted last month for the atom deposition pattern of a Stern Gerlach beam in a quadrupole field ! Interesting.

But that's nothing. The sheet of drywall does something very cool as it moves around the circle, always tilting itself so that it delivers the maximum power to the surface of the planet. It starts off at top dead center, square to the sun. As it moves around the ring clockwise, it tilts 45 degrees when it is at 3:00 position; then, in the 6:00 position (between the sun and the earth) it tilts at 90 degrees, parallel to the sun's rays, so it intercepts nothing; moving on, at 9:00 it is tilted 135 degrees to its original orientation, until finally when it returns to home, it is....flipped by 180 degrees! You have to make two full revolutions before it is restored to its original orientation. Now, where in all of physics has anyone ever heard of a situation where you need to make two full revolutions to get back to where you started? (HINT: That's a trick question!)

You better paint both sides of the sheet of drywall.

The Polished Steel Moon

2012-01-16T19:35:00.000-08:00

Not everyone knows this, but twenty years ago I used to bethe math guy on community access TV here Winnipeg. “Math With Marty” started in1989 and ran for three years; it quickly attracted what everyone calls a cultfollowing. Just recently I started putting up old video clips on YouTube. Iguess it was one of those old shows that got me started on the topic of themoon.

I originally posed the problem in terms of billiard ballcollisions: If the moon is a giant billiard ball, and you shower it withregular billiard balls at high speed, how would you describe the distributionof billiard balls bouncing off the moon? You can watch me solve this problemhere Actually, in the opening clip, it’s not me solving theproblem, it’s my friend Neil, who was co-host and lead guitar player on theshow with me. I pick it up toward the end of the clip, and…well, you can seefor yourself.

Although I solve the problem in terms of billiard ball collision,it’s obviously exactly the same problem if you consider it as light reflectingoff a polished sphere. And the answer is the same: the scattered light is uniformly distributed through space. Allangles are equally illuminated. If you can’t see the source beam, you have noway of knowing even what direction it came from based on the scattered light,because the scattered light goes equally in all directions.

Despite its deceptive simplicity, this is not a trivialresult. In fact, it is only true in three dimensions, as we see from Neil’sattempt to solve the two-dimensional equivalent. In my way of thinking, whenthings come out this way it seems to illustrate some kind of cosmic property ofthe universe. In other words, I don’t know what it means, but it must meansomething.

What would the moon look like if it were a polished steelball? Evidently it would appear equally bright from whatever angle we viewedit, regardless of the relative angle of illumination. In fact, all we would sewould be a bright glint of the sun, and we wouldn’t even know whereabouts onthe face of the moon it came from. (except that the silhouette off the moon wasblocking whatever stars would have been behind it.)

The interesting question is: would this polished ball be abetter or worse source of illumination, on average, than our ordinary every-daymoon? Bu ordinary and everyday, I mean of course the theoretical moon whichought to be a Lambertian scatterer, an ideal bright-white sheet of paper. Inother words, all the light that comes in must go out, just like the polishedball, except now the scattering is diffuse. Cosine-law and all that.

Which one would provide more illumination to the earth, onaverage? It’s a funny question, and for the longest time I was drawn to thetantalizaing prospect that either moon would be equally good. Actually, inpractical terms, the real moon is more useful because it scatterspreferentially in the backwards direction, so it is a better night-light than adaytime light. This is in contrast to the polished ball, which scatters in alldirections equally, so much of its utility is wasted in brightening the day byan infinitesimal amount. The point is, what goes in must go out, and since overthe course of a whole month the moons are on average located at all possibleangles with repect to the sun, the earth must receive the same totalillumination from either one of them.

Except it’s not quite right. The moon does a circular orbitin the place of the sun, but it is never found, for example, above the northpole. This screws up everything. It’s actually a case where the two-dimensionalcase has a much tidier solution. For cylindrical earth-moon systems, theaverage illumination of polished versus difffuse is of course equal. Not forthe three dimensional case.

Because of the cosine-law for the scattering angle, thediffuse scatterer keeps more of its illumination in the equatorial plane, whichis of course where the earth is. The polished sphere wastes more of itsscattered light outside the plane. So if the purpose of the moon is toilluminate the night, the actual moon is actually a better moon than thepolished sphere after all. Partly because it keeps more of its light in theplane of the solar system, and more importantly because it is a more effectiveback-scatterer, so it wastes less power on the daylit skies. Did God maybe knowwhat he was doing when he put it up there?

The unfortunate thing about the “real” moon, the diffusescatterer, is that I haven’t found any neat and tidy way to do the calculation.I wanted to use its equivalence “on average” with the polished moon to drawsome nifty conclusions, but I still don’t know how. The polished moon, at worstcase, reduces to a Grade 11 science problem in focal lengths, so I think I cando it. I just don’t know exactly what I’ll do with it.

A couple of funny things about the real moon. First, interms of total reflected radiance as an illuminator of the nighttime sky, itought to have an equivalent polished version. Not the polished sphere: that is clearlydifferent. I’m saying that for some distorted, squashed-down version of thesphere, we should be able to generate a polished surface that has the sameilluminating effect as the real, diffuse moon. That would be an interestingthing to calculate. I’m going to guess that it might be a cycloid ofrevolution, but that’s just a wild guess.

The other point I still wanted to come back to was the one Italked about last time, the departure from Lambertian diffusion. As Wikipediapoints out, if the moon were a Lambertian scatterer, it should look darkeraround the edges and brighter in the center: in other words, it should looksmore spherical. One of the most obvious facts of the moon is that it just doesn’t:it looks more like a flat dish than a round ball. Wikipedia explains this as adeparture from Lambertian scattering: since the outer edges are just as brightas the inside, the scattered power must be greater at lower scattering angles.

I said last time that I didn’t buy it, and now I have a goodreason to back this up. The Wikipedia theory would seem to explain theflat-dish appearance of the full moon, but then it’s a total contradiction withthe half-moon, which also appearsuniformly bright. If the flatness of the full moon is indeed due to enhancedlow-angle scattering around the outside, the the half moon should show evenmore drastic darkening toward the diametral line, which is the zone of veryoblique illumination. In other words, if the scattering is enhanced at the lowangles, it must be depleted at the high angles. But the half-moon looks just asuniform as the full moon. You can’t have it both ways.

I still say it’s a psycho-visual effect having to do withthe saturation of the eye receptors, the rods or the cones or whatever theyare. Some of those photo-shots of the full moon look pretty dramaticallyspherical. I guess the effect shows up when you balance your light levelsproperly.

There was one more type of moon configuration which turnsout to interesting to analyze, and that’s the big flat drywall cutout moon. Afterall, when we look at the moon, it looks flat…so why now analyze how it wouldbehave if it were just a big round hunk of drywall stuck up there in the sky. Ifigured out some cool stuff about it, but I think we’ll leave that for out nextpost.

Teacher Certification: Let's try this again

2012-01-15T10:00:00.000-08:00

I posted this article a few days ago and something really weird happened to the page. Whenever I click on a link to this page, it shows up for about a second, and then goes blank. I'm going to try reposting it here to see if it happens again.
------------------------------------------------------------------------------------------------------

For those of you who don’t know me, I recently went back to university after many years to certify as a physics teacher. For years I had considered doing this, but was reluctant for two reasons: first, I always thought that I should be involved in things at the university level, because it was so important what was going on there. I just didn’t think high shool mattered that much. Second, the idea of going back and doing two years of university to “learn how to teach” seemed like something of an idignity to me.
Actually, I said there were two reasons, but there’s probably a third: the idea of choosing one single career seems to close the door on all kinds of other possibilities of where life might lead. Maybe I got to a point in life where it seemed like there weren’t all that many doors left regardless, so I bit the bullet and put in my application. I was accepted last year at the University of Winnipeg.

It was the best decision I ever made! The great thing about the U of W post-degree program is that they put you into the schools right away, one day a week in practicum. I loved working with the kids, and my co-op teachers were great about letting me go up to the board and do random topics. But the biggest impression I got from being in the practicum was that I had something different to offer that the kids were desperately hungry for but just weren’t getting from the system. They needed me, and I wanted to be there. All of a sudden, it was important.

The university courses were a bit annoying, but I still enjoyed them. The profs were not very smart, and they really had nothing to tell me about teaching that I didn’t already know. But it was actually fun doing assignments, and even educational sometime. At the U of W they give you a lot of assignments to write up lesson plans, so you have to go through the Provincial Curriculum and see what’s required. I really got to learn the science curriculum inside and out. At the other university across town, they say they had a lot of essays to write about the so-called great philosophers of education: what did Dewey say about this, and what did Piaget say about that? That would have been unbearable for me.

But the profs couldn’t stand me. I was older than all the other students, and I obviously knew a lot more about my subject areas (math and physics) than anyone on the faculty. It seems they all took it into their heads to cut me down to size, to prove to me that I wasn’t as smart as I thought I was. One thing led to another, and the end result was, as of yesterday, I am kicked out of the program.

How did it happen? It’s funny how a negative vibe gets picked up on and amplified. One by one people started complaining about little things that happened in class, until one day the dean called me up and told me I had to report for a meeting. “What is the subject of the meeting?” I was told that there had been numerous complaints about my conduct in class. I replied with what was to become my mantra over the next eight weeks:

“Please put the allegations in writing and I will respond to them”.

Just what were those allegations? That’s a long story, and I’m going to save it for another day. For my former fellow classmates who might be wondering why I didn’t show up for the test this morning, if you happen to be reading this, now you know.

-------------------------------------------------------
I think I might want to stick to physics on this blog, so I've set up another blogsite to talk about my situation with the university. Check it out at Due Process, Natural Justice, and the University of Winnipeg

The moon was a ghostly galleon

2012-01-14T07:03:00.000-08:00

I started working on this problem the other day and Ithought it would be a piece of cake, but now it’s driving me crazy. Thequestion is: how bright would the Moon be if it was a flat cardboard cutoutinstead of a round ball? Would the full moon look just the same?

It’s a funny question because to me, the moon always did look like a cardboard cutout. Inever get a strong impression of roundness, or sphericity. When I look at themoon, it just looks like a flat disk to me. In my mind, I had already assumedthat this was some general property of reflectivity, emissivity, and angles:for a uniformly illuminated object, the angle of viewing does not affect theapparent brightness. That was the general principle, roughly speaking, and Ipretty much assumed you could go to the bank on that. Was I right?

Here’s a thought experiment. Let’s say you take a big sheetof painted drywall and hang it in outer space directly overhead. When it getsdark at night, have someone (!?) slowly rotate the drywall this way and thatway. What can you conclude about the drywall? All you see is a whiterectangular shape. No, that’s not even true; it won’t even be necessarillyrectangular. A parallelogram at best. The question is: what can you tell meabout the drywall? Can you tell how far away it is? Can you tell how big it is?Can you tell what angle it is tilted at? Can you tell its true shape? Or is itjust a featureless white quadrilateral occupying a measurable angular fractionof the sky?

Like I said already, I had this general principle in my mindwhereby you’d see exaclty the same white color no matter which way you tiltedthe drywall, so it wouldn’t make any difference how far away or what angle. That’swhy the disk of the moon is uniformly white. It doesn’t matter what angle thesun hits it or how you view it, the sand which covers the moon is allilluminated to the idential brightness, and that’s all you can see from theearth.

That’s what I thought, and I had a little math problem Iwanted to solve (I’ll tell you what the problem was eventually) so I tried toapply this principle. To my great annoyance, things just wouldn’t add up. Thetotal available radiant power surely depended on the angle at which thesunlight hit the drywall, so obviously an obliquely oriented sheet couldn’tcapture as much sunlight as one tilted straight on. So therefore the differencemust be compensated for by the angle of viewing: the sheet which intersectsless radiant energy must at the same time present a larger profile to theviewer, and vice-versa. So everything balances out.

But this was crazy! There are two angles in the problem andthey are completely independent: the angle of illumination and the angle ofviewing. There is something called “Lambertian emission”, or ideal scattering,and it is indeed the principle that something in uniform illumination looks thesame no matter what angle you view it from. It works like this: an illuminatedsheet puts out, let’s say, 100 watts per square meter per solid radian (figureit out!) when viewed head on, but only 71 watts per square meter per solidradian when viewed from an angle of 45degrees. The result is that you see exactly the same brightness from whicheverangle you view it: you don’t need toget 100 watts when you view it at a 45 degree angle, because the apparent sizeof the sheet is is only 71% of it’s true size. On account of the cosine of 45degrees. It all adds up.

(I’m going to come back later to that business of the watts persquare meter per solid radian. It might be important…)

So the illuminated sheet looks the same from all angles. But not if you tilt the sheet! You canmove around all you want, and you’ll see the same apparent brightness from anyviewing angle. But if you tilt the sheet with respect to the sun, everythingchanges! You’re not intercepting the same amount of radiant energy, so you can’texpect to have the same brightness. That’s the difference. I got those twothings mixed up.

But if that’s how it goes, why does the moon look uniformlybright? On a full moon, the center is illuminated directly by the sun, and thefringes only obliquely. A given square meter of lunar surface near the edges isintercepting less illumination than a square meter in the middle of the disk,so how can it look just as bright? Something doesn’t add up.

I checked Wikipedia, and they actually comment on this veryquestion, and I quote:

“…if the moon were a Lambertianscatterer, one would expect to see its scattered brightness appreciablydiminish towards the terminator due to the increased angle at whichsunlight hit the surface. The fact that it does not diminish illustrates thatthe moon is not a Lambertian scatterer, and in fact tends to scatter more lightinto the oblique angles than would a Lambertian scatterer.”

So Wikipedia agrees with my perception of uniformbrightness, and they agree with me that this conflicts with Lambertianscattering. But is their explanation correct? Now, I think Wikipedia is aphemonenal resource, and the quality of science and math is generallyfirst-rate. But this explanation is a little to easy. There is a natural wayfor objects to randomly scatter light, and it is called Lambertian. The moon,according to Wikipedia, deviates from this natural scattering pattern in some randomarbitrary way, and as a result, appears…randomly blotchy?...no!...it appearsperfectly uniform! How can this be? It’s too perfect.

Lambertian scattering predicts a global effect whereby theobject appears uniformly bright no matter the viewing angle. What then is the conciseprincible whereby an object might appear uniformly bright regardless of illumination angle??? It’s totallyfarfetched to think you might get such a neat tidy result simply on account of “deviationfrom Lambertian scattering”, without some bigger principle at work. In fact, Ijust don’t buy it.

So I googled images of the moon, and found some beautifulshots. Check out this article buy a guy named Kash Farooq at The Thought Stash . Now look at the moon. It looks round! Yes, we all know it’s round, but I mean it looks spherical!The camera doesn’t lie. Could the flatness we are all used to be the result ofpsycho-visual effects rather than pure physics?

I think what’s going on is that in the usualviewing conditions, the brightness of the moon saturates the eye’s receptors.Once you hit white, it’s white, and it doesn’t get brighter by adding morewhite. I have to admit I’m not completely comfortable with this theory, butthat’s the best I can come up with.

In Grade Seven we had a poem on the curriculumcalled “The Highwayman” that begins with the very memorable line:

“The moon was a ghostly galleon tossed on stomyskies…”

I’m sorry if my moon turns out to be just asheet of drywall, but in physics if you want to get to the bottom of things yousometimes have to reduce things to the lowest common denominator. We’ve gotsome interesting calculations coming up, and I’m not yet sure if they’re goingto work out. I told you there was a problem that got me going on this topic,and it has to do with the surface area of a sphere. The question is: is thereanything in the physics of emissivity that relates to the fact that the truearea of an illuminated spherical surface just happens to be exactly twice thecross-sectional area? We’ll return to this quesiton on another day.

Let's do the math

2012-01-04T08:56:00.000-08:00

I promised I would do this calculation, andhere goes. That is, I’m going to walk through the steps of the calculation andlet you fill in the blanks.These spinor affairs can be really confusing, and ultimatelyno one can lead you through it unless you’re really motivated to do it yourself.So what I’m going to give you really amounts to a series of signposts along theway to check yourself against. Good luck.

First lets remember what we’re doing: we havea pencil beam of silver atoms going through a Stern Gerlach magnet, and we’veproposed…proposed?...no, I’d say we’ve erstablishedthat it spreads itself out into a donut pattern, not the two dots that everyonetalks about. I’ve presented what I said is the only logical form that theamplitude function can take for a polarized beam going through the fieldspin-up. You’ll see that although the silver atoms started out with their spinsall supposedly pointing up, when they hit the screen they have all differentspins, depending on where they hit the screen. I’ve already shown the picture,so for now let’s just remember what the function looks like. Of course it’s aspinor function:

Now it’s a question of geometry as to what thedistribution on the screen should look like if you start with your beam ofatoms polarized the other way, spin down. This is the first “signpost” I’mgiving you to check that you’re doing the calculation right. If you transformyour geometry correctly, your function should look like this:

I’m not going to pretend to explain why itshould look like this: that’s the kind of thing you can only do for yourself byworking out the geometry with pencil and paper, according to whatever methodsyou are comfortable with. I’m just saying, this is the answer you should get.Like I said: it’s a signpost.

Again, let’s remember where we’re going withthis. We now have the ring patterns for the two special cases, spin-up andspin-down. In theory, any arbitrary case can now be analyzed as a superpositionof these two cases. What we want to consider is the special case of thespin-sideways beam. From the geometry of the quadrupole field, we know that forthe sideways beam, we should get the same deposition pattern of silver atoms asour initial case, except it must be rotated 90 degrees. So let’s do the math.

We know how to make spin-left out of our basisstates. Its just (neglecting the scale factor) spin-up plus spin-down. Or maybe its spin-up minus spin-down. It doesn’treally matter, because either way the pattern is rotated by 90 degrees.

Except I’ve worked out this case already andit gives me a uniform distributionpattern around the ring. It turns out that with respect to the axis ofpropagation, what we’ve actually done here is the case of spin-forward and spin-backward. There’s always that element of ambiguity in spinoralgebra, where you don’t quite know where your phase is calibrated to. It turnsout that in this case our spin-left/right cases should have actually been the complex combinations:

These are the spinors we want to analyze. Infact, we’ll just do the first one, because it comes to the same thing. As gothe spinors, so go their deposition patterns. So to find the composite sidewaysdeposition pattern, we just have to put phi(up) for the up-arrows and phi(dn) for the down arrows,and then add them together component-wise. The answer should look like this:

We’re really almost there, but how to verify thatthis function is just a rotated version of our initial function? It doesn’tlook exactly right, but you know how it is with complex phases and things.Looks can be deceptive. But that’s because we’re dealing with amplitudefunctions. There is no confusion with phases if we convert to a denstityfunction. So let’s see what happens.

Last time we did this, we worked out the intensitypattern for the case of the unpolarized beam…remember how it went?...square first,then sum. This time it’s the opposite. For the coherent superposition, we’ve alreadysummed first, and now we square! It’s not that hard to see that the intensity goes as2 + 2sin(theta). That’s actually double our initial function, but remember, wehaven’t been bothering to keep track of our scale factors. The main thing tosee is that the pattern is rotated 90 degrees…just as the doctor ordered!

Now you know what the Stern-Gerlach apparatus really does to a beam of silver atom,and don’t let anyone tell you differently.

Spinors and More Spinors

2012-01-03T10:18:00.000-08:00

The topics are piling up for me faster than I can get rid of them. I was doing this really intricate spinor stuff and then got sidetracked on a discussion of the tides, whereupon I realized what those two problems have in common: everybody gets distracted by the steady-state component of the field, but all the interesting physics really takes at the level of the variation in the field: namely, the quadrupole component.

I threw up a quick sketch of the tidal forces the other day to show how the oceans really follow the quadrupole field. The sketch looked like this:

But actually, the tides are more interesting than that. They don't track the moon directly, as seen in this picture...they are actually displaced. I'd like to say they lag behind the moon, trying to keep up, but that may or may not be right. It's also quite possible that they run ahead of the moon. It's not a question you can settle without some careful thought.

There are three frequencies you have to account for in this calculation. There is of course the 24-hour rotation of the earth, and the 29-day period of revolution of the moon. But there is also the very important natural frequency of the oceans. If you look at the bulging oceans in the sketch, you can imagine that if the moon suddenly disappeared, the oceans wouldn't just instantly return to their spherical equilibrium. They would slosh back and forth for awhile, and they would do so at a certain characteristic frequency. Just what is this frequency?

That promises to be not so easy to calculate. But anybody who does any work with waves knows that a standing wave can be analyzed as the superposition of two travelling waves. The present example is a case in point. We can analyze the sloshing of the oceans as the simultaneous rushing, east-to-west and west-to-east, of two massive travelling waves. We can then ask: how fast does the wave travel?

It's not that we're about to do the actual calculation, it's just that you can actually visualize a massive travelling mile-high monster wave on the oceans, and the point is it's got to be moving really fast...say 100, maybe 200 miles per hour. The significant thing to notice is that it's travelling slower than the equatorial velocity of the earths rotation, namely 1000 mph. So the sloshing of the oceans takes place on a time scale of something like five days; in other words, somewhere in between the rotational period of the earth and the orbital period of the moon. This ordering of frequencies will be critical in evaluating the true physics of the tides, including the lead/lag phase relationship with the moon, and the fascinating question of whether the effect of tidal friction might eventually slow down the moon and bring it crashing down to earth?

But I started off today's post by noting that I was getting way behind, and before we get around to doing the tides, we still have to tidy up some work on that other quadrupole field problem, the Stern Gerlach effect. Now, where were we?

As I recall, I had just come up with the solution for the problem of the polarized beam in the quadrupole field. When I say "came up with", I really mean guessed, because that's what I actually did. Or more accurately, what I did was to keep guessing and reguessing until I came up with something that made sense. This was the distribution I came up with:

and this is the equation that describes the distribution of spin about the ring:

Having "guessed" this solution, the question becomes: does it make sense? And the first thing to verify is that if we have a random beam, composed of equal parts spin-up and spin-down, we get a uniform donut pattern. And since the spin-down pattern has to look just the same except upside-down, the uniformity of the resulting pattern is easy to verify by summing the squares of the amplitude. (For random beams, square first, then sum!).

But the real test will be the case of the coherent beams, where you have up and down spins in coherent combinations. This time, of course, it's just the opposite: sum first, and then square! It is well known that in spinor algebra, the coherent combination gives you sideways polarization. And because of the four-fold symmetry of the quadrupole field, we know that we ought to then get a resultant distribution which is just the picture we already have, rotated by 90 degrees. (Here is the quadrupole field in case you've forgotten it:)

So the question is: what happens when we analyze a sideways polarized beam, using our up and down basis states? Will we get the same pattern we've already derived, except this time rotated by 90 degrees?

Stay tuned.

GreenShield Dental Insurance a Rip-Off

2012-01-02T16:41:00.001-08:00

As a student at the University of Winnipeg, I'm required to buy into their Dental Plan. What a ripoff! I broke a tooth last month and it cost $400 to fix it. I put in the claim and last week they sent me a check for...$38! How can a dental plan not insure you for fixing a broken tooth? Sure, they have "reasons", but what a crock! The insurance company is called GreenShield Canada out of Windsor, Ontario, and they must be ripping everybody off. If anyone else has a story like this,
let's get together and launch a class action suit. Don't let them get away with it.

The Quadrupole Field: Tidal Forces and Particle Beams

2012-01-01T08:51:00.000-08:00

People have an awful time eplaining the tides. Everyoneknows the moon attracts the oceans, making them bulge out on one side: but howdo you explain the second bulge on the wrong side of the earth? The tides go inand out twice a day, not once. How doyou explain it?

The mistake people make is to think that it’s the pull ofthe moon that causes the tides. It is and it isn’t. There is absolutley notidal force in a steady gravitational field. The tidal force occurs because themoon’s gravity is not a constant over the volume of the earth. It get’s weakeras you get farther. It’s the variational component of the steady field which isresponsible for the tides. You can draw a picture of it easily:

The big grey arrow represents the average magnetic field inthe vicinity of the earth, and it has absolutely no influence on the tides. Thefour little black arrows, representing the deviation from the steady field, areentirely responsible for the bulges in the ocean. You can see in the picturethat the distortion of the ocean is exactly in line with the four small arrows.These arrows represent what is called the “quadrupole component” of the field,and they cause the tides.

It is very significant to notice that the up/down arrows arethe same strength as the left-right arrows…at least I’ve drawn them that way.In real life, they are exactly half strength…because there are two moreoff-axis arrows in the actual field, representing the component in and out ofthe page. The TOTAL strength of the four off-axis arrows is equal to thestrenght of the on-axis arrows. This is not an accident, and it’s not somethingthat applies only to gravitational fields. It’s a basic property of allelectric and magnetic fields that is also known as Gauss’s Law: flux in isequal to flux out.

I bring this up because it applies in a huge way to theStern Gerlach experiment. Everyone, and I mean everyone explains Stern-Gerlach by saying the field is strongernear the pointy magnet and weaker near the flat magnet. Yes it is, but whatabout the off-axis component? It’s just like the moon’s gravity! In fact, it’seven more so, because for the moon, we have three-dimensional geometry, so theoff-axis field lines are weaker by a factor of 2; but in the two-dimensionalgeometry of the long pointed magnets, the off-axis component (which everyone ignores!) is exactly as strong as the on-axiscomponent. Just as I’ve drawn in the picture.

And just as the steady-state component of the moon’sgravitational field has absolutely no effect on the tides, why should the steady-statecomponent of the magnetic field have any effect on the silver atoms? Everythingthat happens in Stern-Gerlach has to be explainable on the basis of thequadrupole component of the field. Why should it be different from the tides?

But if this is true, the implications are huge. Everybodysays that the beam splits in two. Even Feynman talks about splitting the beam,and then taking one component of the split beam and putting it through a secondStern-Gerlach magnet rotated through 90 degrees; he says then it will split intwo again. He even goes so far as to acknowledge that the experiment has neverbeen done quite this way, but he doesn’t doubt that it is theoreticallypossible.

And why shouldn’t it be possible? Everyone knows the beamsplits in two. But does it??? In theoriginal Stern-Gerlach experiment, there was no pencil-shaped beam: the beamcame out of an oven through a slit, not a pinhole, and the orientation of theslit was perpendicular to the axis of the magnet. The silver atoms came out ina fan-shape, and spread into two lineson the detection plate…not two dots, but two lines. If I am correct, and I can’tsee why I wouldn’t be, there is no way to split a pencil beam into two dots.Because the quadrupole field has just as much splitting force in the left-rightdirection as it does in the up-down direction.

That's why you can't expect the beam to split into two dots. If you haven't been following my blog regularly, you might have missed my post from last month when I calculated the spreading of an actcual beam through the quadrupole field. Instead of two dots, you get a donut that looks like this:

That's the pattern for a beam that goes in with it's spins polarized in the up direction. For an unpolarized (random) beam, you just get a simple donut pattern.

There is no physical way to construct a magnetic field thatgets stronger from top to bottom without at the same time have it show just asmuch variation in the left-right direction. So it is impossible to split apencil-shaped beam into two dots. Yes, you can split a fan-shaped beam into twolines, but that is not at all the same thing. In particular, all of Feynmann’sthought-experiments that he works through in Vol. 3 Chapter 5…are based on anon-physical misinterpretation of the apparatus. Feynmann got it wrong. Oh, it’sall right for the point he’s making about adding amplitudes and transformationof basis states…all that is immaculate. But the actual experiment does notexist.