I just watched "Absence of Malice" on TV yesterday, with Paul Newman and Sally Field. What a great story that was. I just don't understand how the writers blew the happy ending.

He was in his boat, getting ready to leave on a trip up the coast of indefinite duration. She had come to see him for the last time. As he said goodbye and pulled away for the dock, I was dying for her to shout after him: "Mike!"...and then, when he looked at her, she would say: "Can I come?" Then he would look at her in a funny way, and an unspoken understanding would pass between them. She would come running, jump off the dock, and swim to the boat where he would help her on board.

That's the ending I wanted.

## Tuesday, January 29, 2013

## Sunday, January 27, 2013

### How Much Suction in a Vacuum Cleaner?

I sometimes mention what a good resource Wikipedia is, especially in the sciences; in fact, I contributed $20 to their recent fundraising campaign because I have benefited so much from them over the years. But today I noticed a glaring error in their article about vacuum cleaners. Oh, I know I'm supposed to correct it myself...that's the whole idea of wikipedia. But I have no idea how to do that, so I'm just going to tell you about it here.

(EDIT: What follows is a very good calculation based on a fundamental misconception of the type of motor used in a vacuum cleaner. Eventually I figured out the mistake and posted a link to the corrected calculation at the bottom of the discussions. But since then, this article continues to get more hits than any other blog page of mine, while the correction gets little attention. So I'm posting the link to the revised calculation here. But you should still enjoy the discussion below.)

They say the suction of a typical household vacuum cleaner is around 20 kPa, or 80 inches of water. That's an outrageous amount of suction! It comes to around twelve pounds on a typical vacuum hose. Like you could pick a gallon jug of milk with a vacuum cleaner? That's a lot of suction.

Here's how you do the calculation. First of all, every vacuum cleaner I've seen uses a centrifugal (squirrel-cage) blower. I think that's correct. Now, the pressure you get from such a blower is limited by the maximum speed of the blade tip, and that in turn is limited by centrifugal forces. It is a fascinating and little-known fact of material science that a solid disc of steel has a maximum peripheral speed of around 400 feet per second, beyond which the internal stresses exceed the tensile strength of steel. I actually know enough about these things to do the calculation, but I'm not going to do it today. I'm just going to use that 400 feet per second for my blower calculation.

(First a short digression: at 3600 rpm, you hit 400 feet per second with a rotor diameter of around 2 feet. That's a fundamental size limitation for things like big electric motors or generators, and in fact that limit is observed in practise, with some margin to spare.)

How do you convert this to air pressure? The governing equation is I think due to Bernoulli, and it is

where rho is the gas pressure. I normally like to derive these things but today I just want to get the answer. The density of air is 1.3 kg/meter^3, and using 120 meters/sec for velocity, I get a pressure of 10 000 Pascals, or 10 kPa, which is just about half of what Wikipedia gives for the "typical" home vacuum cleaner. So what's my problem...isn't that pretty much in the ballpark?

The problem is: do you have a 2-foot diameter blower in your vacuum cleaner? I don't think so. Do you have four six-inch diameter blowers in cascaded in series? I've never seen cascaded blowers in a home vacuum cleaner. Do you have a 4-inch blower and a 20 000 rpm motor? I don't think so.

With a six-inch diameter blower at 3600 rpm (that's a huge blower), I make it around 2500 pascals, or about 10 inches of water. You can do some pretty cool stuff with 10 inches of water pressure, but 80 inches would be downright scary.

(EDIT: What follows is a very good calculation based on a fundamental misconception of the type of motor used in a vacuum cleaner. Eventually I figured out the mistake and posted a link to the corrected calculation at the bottom of the discussions. But since then, this article continues to get more hits than any other blog page of mine, while the correction gets little attention. So I'm posting the link to the revised calculation here. But you should still enjoy the discussion below.)

They say the suction of a typical household vacuum cleaner is around 20 kPa, or 80 inches of water. That's an outrageous amount of suction! It comes to around twelve pounds on a typical vacuum hose. Like you could pick a gallon jug of milk with a vacuum cleaner? That's a lot of suction.

Here's how you do the calculation. First of all, every vacuum cleaner I've seen uses a centrifugal (squirrel-cage) blower. I think that's correct. Now, the pressure you get from such a blower is limited by the maximum speed of the blade tip, and that in turn is limited by centrifugal forces. It is a fascinating and little-known fact of material science that a solid disc of steel has a maximum peripheral speed of around 400 feet per second, beyond which the internal stresses exceed the tensile strength of steel. I actually know enough about these things to do the calculation, but I'm not going to do it today. I'm just going to use that 400 feet per second for my blower calculation.

(First a short digression: at 3600 rpm, you hit 400 feet per second with a rotor diameter of around 2 feet. That's a fundamental size limitation for things like big electric motors or generators, and in fact that limit is observed in practise, with some margin to spare.)

How do you convert this to air pressure? The governing equation is I think due to Bernoulli, and it is

Pressure = 1/2*rho*v^2

where rho is the gas pressure. I normally like to derive these things but today I just want to get the answer. The density of air is 1.3 kg/meter^3, and using 120 meters/sec for velocity, I get a pressure of 10 000 Pascals, or 10 kPa, which is just about half of what Wikipedia gives for the "typical" home vacuum cleaner. So what's my problem...isn't that pretty much in the ballpark?

The problem is: do you have a 2-foot diameter blower in your vacuum cleaner? I don't think so. Do you have four six-inch diameter blowers in cascaded in series? I've never seen cascaded blowers in a home vacuum cleaner. Do you have a 4-inch blower and a 20 000 rpm motor? I don't think so.

With a six-inch diameter blower at 3600 rpm (that's a huge blower), I make it around 2500 pascals, or about 10 inches of water. You can do some pretty cool stuff with 10 inches of water pressure, but 80 inches would be downright scary.

## Thursday, January 24, 2013

### Hot in Cleveland

When they started promoting this new TV show "Hot in Cleveland" a few years ago, I was really looking forward to it. I'm a sucker for a good gal-pal comedy, and I like all the stars, especially Valerie Bertinelli, who made an indelible imprint on my psyche as the 16-year old daughter on that 70's show "One Day At A Time". So I was a little disappointed the first time I caught an episode to find that the writing was a little juvenile. But today I caught an episode that I thought was brilliant. It was the one where Wendy Malick's daughter comes back and Valerie and the British girl end up in a Bachelor-parody competition run by her ex's nine-year old son. I thought it was great...good on you, girls.

Anyhow, there are a couple of important sociological observations that need to be made in connection with this show, and now is as good a time as any to make them.

First there is the Betty White connection. Betty of course was part of the original cast of "The Golden Girls". The Golden Girls was a very different show. Those women were

What is shocking is to realize that the cast of Hot in Cleveland is

My second observation is not so positive. In checking out Valerie Bertinelli's background, I noticed something I had forgotten: she was married to heavy metal guitarist Eddie Van Halen. Why do the most beautiful girls fall for those idiot rockers? Pamela Anderson married the drummer from Motley Crue. A

I can't take it any more. I'm going to have to stop here.

EDIT: I remembered one more thing I wanted to mention, on the subject of so-so sitcoms with one brilliant episode. I was flipping through channels one day a few years ago when I caught the tail end of a Drew Carey episode that appeared to be a full-blown musical parody of "How To Succede In Business (Without Really Trying)". I'd

Anyhow, there are a couple of important sociological observations that need to be made in connection with this show, and now is as good a time as any to make them.

First there is the Betty White connection. Betty of course was part of the original cast of "The Golden Girls". The Golden Girls was a very different show. Those women were

*old*, not hot. Well, Rue McLahan was pretty hot, but it was "old-lady hot", not Valerie Bertinelli hot. You couldn't write a plausible Golden Girl episode where any of them took up romantically with a forty-year old stud, while that would be totally believable for any of the Cleveland cast.What is shocking is to realize that the cast of Hot in Cleveland is

*almost the same age*as the cast of Golden Girls when it went on the air! There has been an enormous change in the way society perceives older women over the last thirty years, and I think it is just great. We used to consider a woman old at forty. Those days are thankfully gone. I remember being shocked when Joan Collins on Dynasty looked sexy at*fifty*. No one is shocked by that any more, and I say it's about time.My second observation is not so positive. In checking out Valerie Bertinelli's background, I noticed something I had forgotten: she was married to heavy metal guitarist Eddie Van Halen. Why do the most beautiful girls fall for those idiot rockers? Pamela Anderson married the drummer from Motley Crue. A

*drummer!*That's not even a*musician*. But nothing hurt more than seeing Heather Locklear....oh my god....I just had to look it up....I knew she married a drummer but*I didn't know it was the same drummer as Pamela Anderson.*I can't take it any more. I'm going to have to stop here.

EDIT: I remembered one more thing I wanted to mention, on the subject of so-so sitcoms with one brilliant episode. I was flipping through channels one day a few years ago when I caught the tail end of a Drew Carey episode that appeared to be a full-blown musical parody of "How To Succede In Business (Without Really Trying)". I'd

*really*like to see that whole episode.### In which I figure out how to count modes

Two years ago I did some very cool stuff about the thermodynamic equilibrium of a radiation field with a harmonic oscillator. One of the pieces of the puzzle was to figure out how many electromagnetic standing wave modes you could put in a rectangular cavity. I had some trouble with this: I could get it in the ballpark, but the way I counted modes they came in groups of three; and the actual formula that I looked up didn't have that expected factor of 3 in it. You can read the original article here.

As a result of a recent discussion on stackexchange.com, I think I've found my mistake, and it's quite interesting. First of all, here is the picture I had of a typical mode:

The green lines are the electric field and the purple lines are the magnetic. You can see both fields to to zero on one of the faces. There are three choices for the blank face, so every unit cell gets three standing wave modes. Right?

The guy who posted the question on stackexchange (he goes by the awkward moniker of e-journal_clasica) started off by giving these formulas for the electric field of a standing wave:

My problem was to reconcile these formulas with the picture I had of the physical modes. I started off by noticing that his condition on E1, E2, and E3 was satisfiable infinitely many ways; but in particular, it was easily satisfiable for the choice of either E1, E2, or E3 =0. And weren't those my three mode configuarations?

Ah...but are they

As a result of a recent discussion on stackexchange.com, I think I've found my mistake, and it's quite interesting. First of all, here is the picture I had of a typical mode:

The green lines are the electric field and the purple lines are the magnetic. You can see both fields to to zero on one of the faces. There are three choices for the blank face, so every unit cell gets three standing wave modes. Right?

The guy who posted the question on stackexchange (he goes by the awkward moniker of e-journal_clasica) started off by giving these formulas for the electric field of a standing wave:

My problem was to reconcile these formulas with the picture I had of the physical modes. I started off by noticing that his condition on E1, E2, and E3 was satisfiable infinitely many ways; but in particular, it was easily satisfiable for the choice of either E1, E2, or E3 =0. And weren't those my three mode configuarations?

Ah...but are they

*orthogonal*? That's when I noticed something very interesting, and I wonder if you can see what it is.## Wednesday, January 23, 2013

### How Does the Brain Work

The mind is a funny thing. I just read a very annoying book that my nephew gave me for Christmas called "The Plastic Brain". The writer claims that we have gained huge new insights into the workings of the mind through recent experiments where stimulus from the outside world is mapped to specific cell regions in the brain.

I found the book to be overwrought and tedious at the same time. With all the hyping of the "groundbreaking new research", in the end I found nothing that really enlightened me about how the mind works. Science has a long long way to go in this area.

In the meantime, I made an observation about my own mind the other day which I found surprising, and I'm going to share it with you. I play a lot of Tetris. Quite a lot actually. Well, I guess I'm obsessed with it.

We used to have an old Mac in the house with what I think of as original Arcade Tetris. The music was phenomenal. "DAAA...da da Dumm........Da-da-da-da DAH. (repeat). Duh-DAH...duh-DAH..duh-DAH, duh-DAH, da-da-DAH!..etc. Remember that one? Well, that was a long time ago and I don't have it any more. My online Tetris only beeps, which of course I mute while playing, because that would be really annoying. I have a special way I like to play. I start at Level Three, and I use only the shift keys...left or right, no rotations. I like to score about 10 lines before I allow myself to start using rotations. That's it...it's just how I like to do it.

Anyhow, the other day I was thinking about it, and I realized: you know the L-pieces, there is a left and a right. And it occured to me that at this moment, if I had to tell you what color they were, I wouldn't be able to do it. Oh, I know they're blue and purple...what I

And then I thought some more. There are seven pieces. The square is red. The bar is orange. The "T" is yellow. (That color sticks in my mind more than any of the others.) But what about the zig-zags? I realized...I don't know! Not like the L's...where I knew the colors but I just couldn't say which was which. When I picture the zig-zags in my mind, I can't even remember what colors

I believe in the theory of different types of intelligence. Obviously I'm at the high end of the scale in math intelligence, but I've never believed that made me better than other people. I've always been much too aware of so many others, much "dumber" than me on paper, yet who were much more capable at such tasks as negotiating with a used-car dealer, or picking up girls at a bar. Either you've got it or you don't.

But can't you

So what about Tetris? I'm saying that there are all kinds of people out there who's "intelligence" in relating shapes and colors is far more advanced than mine. And I could sit down with booklets full of shapes and colors, and memorize them, and pass a test in them...but it still wouldn't make me smart in shapes and colors. Either you are or you aren't.

I found the book to be overwrought and tedious at the same time. With all the hyping of the "groundbreaking new research", in the end I found nothing that really enlightened me about how the mind works. Science has a long long way to go in this area.

In the meantime, I made an observation about my own mind the other day which I found surprising, and I'm going to share it with you. I play a lot of Tetris. Quite a lot actually. Well, I guess I'm obsessed with it.

We used to have an old Mac in the house with what I think of as original Arcade Tetris. The music was phenomenal. "DAAA...da da Dumm........Da-da-da-da DAH. (repeat). Duh-DAH...duh-DAH..duh-DAH, duh-DAH, da-da-DAH!..etc. Remember that one? Well, that was a long time ago and I don't have it any more. My online Tetris only beeps, which of course I mute while playing, because that would be really annoying. I have a special way I like to play. I start at Level Three, and I use only the shift keys...left or right, no rotations. I like to score about 10 lines before I allow myself to start using rotations. That's it...it's just how I like to do it.

Anyhow, the other day I was thinking about it, and I realized: you know the L-pieces, there is a left and a right. And it occured to me that at this moment, if I had to tell you what color they were, I wouldn't be able to do it. Oh, I know they're blue and purple...what I

*don't*know is which is which. You'd think that would have sunk in to my brain by now considering all the hours I've spent at it.And then I thought some more. There are seven pieces. The square is red. The bar is orange. The "T" is yellow. (That color sticks in my mind more than any of the others.) But what about the zig-zags? I realized...I don't know! Not like the L's...where I knew the colors but I just couldn't say which was which. When I picture the zig-zags in my mind, I can't even remember what colors

*either*of them are! Now that's really weird.I believe in the theory of different types of intelligence. Obviously I'm at the high end of the scale in math intelligence, but I've never believed that made me better than other people. I've always been much too aware of so many others, much "dumber" than me on paper, yet who were much more capable at such tasks as negotiating with a used-car dealer, or picking up girls at a bar. Either you've got it or you don't.

But can't you

*develop*those intelligences? I suppose you can to a degree. But what I see in the school system, where teachers drill students to despair in the*mechanics*of mathematical calculations...that is ridiculous. The "skill" those kids are developing has nothing to do with what my brain is doing when it solves a math problem. It's strictly behavioral conditioning designed to*mimic*the actual process of mathematical thinking. And even at that, it creates a very poor facsimile that the system justifies by administering tests which only measure the step-by-step mimicry of thinking, rather than trying to measure actual mathematical thinking.So what about Tetris? I'm saying that there are all kinds of people out there who's "intelligence" in relating shapes and colors is far more advanced than mine. And I could sit down with booklets full of shapes and colors, and memorize them, and pass a test in them...but it still wouldn't make me smart in shapes and colors. Either you are or you aren't.

## Friday, January 18, 2013

### Lockshin Kigel

I've been trying to make Lokshin Kigel from about two years now with varying degrees of success and I think yesterday I finally nailed it. I'm posting it here because I need to write down my recipe while I can still remember what I did.

It turns out you can't find a recipe for lokshin kigel on the internet. If you google it, all you get is results for "lockshin

My problem with cooking is that I'm always improvising. My theory is that how far wrong can you go if you stick to good wholesome ingredients? So let me try and remember what I did...

They had cottage cheese on sale for 99 cents at the Food Fare last week. I wouldn't have normally bought it because it was diet (1%) but I figured...what the hell, for 99 cents how bad could it be? I can always make up the fat content by using lots of butter.

And that's what I did...about a third of a pound, and half again as much Crisco (melted together in the microwave). I had two apples that I chopped up and threw in. Salt and pepper (don't skimp on the pepper) and just over a quarter cup of sugar. (That was one scoopful, and I just tossed another scoop into the measuring cup so you readers would know how big my scoop is.) Four eggs, and that was my mix.

I had a half-used bag of fettucini noodles...it was a 900 gram bag to start with, and I'm guessing there must have been 5 or 6 hundred grams left over. So that's what I used.

I think that's it. I baked it for an hour at 350, then turned down the heat and left it in the oven. Some of the noodles were a bit crunchy so you might want to cut that back 10 minutes or so, but I kind of like the jawbreakers. So unless I've forgotten something, there you have it.

I think we'll get back to the physics when we return.

EDIT: I

Oh yeah, one more thing. About the cottage cheese. The reason it was 99 cents is that it was expired...what my Chinese wife calls "past the duty date". But what's five or six days to a tub of cottage cheese?

It turns out you can't find a recipe for lokshin kigel on the internet. If you google it, all you get is results for "lockshin

*kugel*". Are there no other Galitzianers out there posting recipes? Is the Internet run by Litvaks? We'll see if I register in the Google search after I post this...My problem with cooking is that I'm always improvising. My theory is that how far wrong can you go if you stick to good wholesome ingredients? So let me try and remember what I did...

They had cottage cheese on sale for 99 cents at the Food Fare last week. I wouldn't have normally bought it because it was diet (1%) but I figured...what the hell, for 99 cents how bad could it be? I can always make up the fat content by using lots of butter.

And that's what I did...about a third of a pound, and half again as much Crisco (melted together in the microwave). I had two apples that I chopped up and threw in. Salt and pepper (don't skimp on the pepper) and just over a quarter cup of sugar. (That was one scoopful, and I just tossed another scoop into the measuring cup so you readers would know how big my scoop is.) Four eggs, and that was my mix.

I had a half-used bag of fettucini noodles...it was a 900 gram bag to start with, and I'm guessing there must have been 5 or 6 hundred grams left over. So that's what I used.

I think that's it. I baked it for an hour at 350, then turned down the heat and left it in the oven. Some of the noodles were a bit crunchy so you might want to cut that back 10 minutes or so, but I kind of like the jawbreakers. So unless I've forgotten something, there you have it.

I think we'll get back to the physics when we return.

EDIT: I

*knew*I'd forget something! The mable cheddar. I had about 150, maybe 200 grams of marble cheddar in the fridge and I grated it into the mix.Oh yeah, one more thing. About the cottage cheese. The reason it was 99 cents is that it was expired...what my Chinese wife calls "past the duty date". But what's five or six days to a tub of cottage cheese?

## Saturday, January 12, 2013

### Again with the rotor bars

I had a big thing going last year about the rotor bars in an induction motor, and I still can't figure it out. But once in a while someone comes along and posts a comment which purports to set everything straight, as one of my readers did yesterday on this post. John wants me to analyze the motor as magnet-on-magnet action instead of rotor bars in a magnetic field.

It's a good theory but I don't buy it. John doesn't deal with the essential problem: how do you

how do you get torque if the field lines curve

I was tempted to answer this question by suggesting that the field lines exert a torque not on the copper bars, but on the atomic-level circulating currents in the magnetic domains of the iron. We know that iron can amplify a magnetic field up to a thousand times; this can only mean that the atomic-level circulating currents on the surface of the iron are a thousand times greater than the macroscopic copper-wound currents that we can see. Does this explain the torque of a motor?

I don't think so. My problem with this explanation is all of a sudden, there's way

I want to show what I mean when I say the standard calculation gives the right answer (assuming the field passes through the copper bars, so I put some dimensions on my picture. With a 3-cm radius, the rotor shown here would look OK in an ordinary 1/2HP motor. It's not hard to see that there are about 10 square centimeters of rotor bar cross section, which would typically sustain a current of around 2000 Amps. A good strong magnetic field achievable with laminated iron would be around 1 Tesla (that's volt-seconds-per-meter-squared). Is this enough to calculate the torque? Just about...I still need to give you the length of the rotor bars. Let's call it 12 centimeters:

(2000 Amps)(0.12 meters)(1 volt-sec/square meter) = 240 Newtons

To get the torque, we multiply by the radius (3 cm) which gives a torque of about 7 N-m. Now horsepower is torque times frequency (in radians per sec.) Remember that 3600 RPM is 377 radians per second...except the typical motor is four-pole, so you only get half of that...and also, the utilization factor of the copper bars is certainly under 50% with the sinusoidally varying currents and the phase shifts. Lumping it all together I get:

(7.2 Newton-meters)(1/2 * 377 radians/sec)(1/2) = 650 Watts

which is one horsepower...just about where we want it.

Except for one problem: the magnetic field

The irony is that I think I'm going to explain the great paradoxes in quantum mechanics but I still can't seem to figure out how a motor works.

It's a good theory but I don't buy it. John doesn't deal with the essential problem: how do you

*calculate*the torque? The problem is that**I**Lx**B**actually gives you pretty much the right answer, but fails to answer the question I asked with this picture:how do you get torque if the field lines curve

*away*from the rotor bars, avoiding the copper almost entirely?I was tempted to answer this question by suggesting that the field lines exert a torque not on the copper bars, but on the atomic-level circulating currents in the magnetic domains of the iron. We know that iron can amplify a magnetic field up to a thousand times; this can only mean that the atomic-level circulating currents on the surface of the iron are a thousand times greater than the macroscopic copper-wound currents that we can see. Does this explain the torque of a motor?

I don't think so. My problem with this explanation is all of a sudden, there's way

*too much*torque. Because the standard macroscopic analysis gets me in the right ballpark. I don't want a correction that makes the torque a thousand times bigger. The problem with the standard analysis isn't the result...it's that nagging question: don't the magnetic field lines curve away from the copper to stay inside the iron?I want to show what I mean when I say the standard calculation gives the right answer (assuming the field passes through the copper bars, so I put some dimensions on my picture. With a 3-cm radius, the rotor shown here would look OK in an ordinary 1/2HP motor. It's not hard to see that there are about 10 square centimeters of rotor bar cross section, which would typically sustain a current of around 2000 Amps. A good strong magnetic field achievable with laminated iron would be around 1 Tesla (that's volt-seconds-per-meter-squared). Is this enough to calculate the torque? Just about...I still need to give you the length of the rotor bars. Let's call it 12 centimeters:

(2000 Amps)(0.12 meters)(1 volt-sec/square meter) = 240 Newtons

To get the torque, we multiply by the radius (3 cm) which gives a torque of about 7 N-m. Now horsepower is torque times frequency (in radians per sec.) Remember that 3600 RPM is 377 radians per second...except the typical motor is four-pole, so you only get half of that...and also, the utilization factor of the copper bars is certainly under 50% with the sinusoidally varying currents and the phase shifts. Lumping it all together I get:

(7.2 Newton-meters)(1/2 * 377 radians/sec)(1/2) = 650 Watts

which is one horsepower...just about where we want it.

Except for one problem: the magnetic field

*inside the copper bars*is very low. Look at the picture again. The field goes into the iron. I could try and fix this by analyzing the atomic-level currents in the magnetic domanis (1000x greater than the copper bar current) that would really mess up my calculation. So I still really don't know what to do.The irony is that I think I'm going to explain the great paradoxes in quantum mechanics but I still can't seem to figure out how a motor works.

## Tuesday, January 8, 2013

### Group Velocity vs Phase Velocity

When we left off, I told you I had come up against a paradox when applying the principle of least action to the trajectory of an electron. I started by doing it for baseballs. But I didn't do what Feynman does, where he minimizes the action between the starting and ending point. I assumed an initial trajectory, and claimed that the subsequent trajectory had to remain perpendicular to surfaces of equal action. By comparing trajectories of corresponding action at different elevations, I showed the only way to get this to work out was to have the trajectories curve. And I actually got the correct curvature for a baseball.

Something nasty happens when we try to apply this to electrons. Instead of a baseball under the influence of gravity, we send an electron beam between the plates of a charged capacitor. The "surfaces of equal action" naturally become the surfaces of constant phase for the electron wave function. They look like this:

So the upper part of the wave is going faster, making the whole thing bend. But wait...in quantum mechanics, wave number corresponds to momentum...so according to wavelength, it's the

The paradox is resolved when we examine the group velocity of the wave. Electron waves are not like light waves, where all frequencies travel at the same speed. For light waves, you can easily calculate the velocity as the frequency times the wavelength, and this

The formula for group velocity is a little different from the obvious formula for phase velocity. This is how it goes:

Let's see if we can calculate the group velocities for the electron wave along the two trajectories shown here, and see if the faster wave isn't actually going slower!

We can start with the equation for the wave along the lower trajectory. Using the complex exponential, it is just:

To get the group velocity, we need to figure out what is w (omega) as a function of k. Well, omega is the frequency, which lines up with the energy, and k is the wave number, which lines up with the momentum. We can arbitrarily declare the potential energy along the flat trajectory to be zero, which makes the energy simply proportional to the square of the momentum. We can arbitrarily choose our units so k=1, and then the proportionality becomes exact:

So the group velocity it 2, which is twice the phase velocity.

How about the upper path? Well, the total energy of the upper electron is the same, because the two paths started out as a simple wave in free space. But when they enter the capacitor, the one closer to the negative plate (the upper path) loses kinetic energy and gains potential energy. Since we show (in our diagram) that the wavelength is one percent greater, we interpret this as a one percent decrease in momentum. Remembering that total energy is given by w and kinetic energy by k-squared, we must have:

But remember that k for the upper path is numerically 0.99 which means the P.E. along that path must be 0.02 to keep w constant=1. You can see this is correct because for k=0.99 we then get w=1, which are the correct wave parameters for the function as sketched.

Again, if we differentiate w with respect to k, the P.E. term disappears (since it's a constant) and we get once again dw/dk = 2k.

But on the upper path, k = 0.99, so the group velocity is one percent

Something nasty happens when we try to apply this to electrons. Instead of a baseball under the influence of gravity, we send an electron beam between the plates of a charged capacitor. The "surfaces of equal action" naturally become the surfaces of constant phase for the electron wave function. They look like this:

So the upper part of the wave is going faster, making the whole thing bend. But wait...in quantum mechanics, wave number corresponds to momentum...so according to wavelength, it's the

*lower*part of the wave that's going faster. Which one is right?The paradox is resolved when we examine the group velocity of the wave. Electron waves are not like light waves, where all frequencies travel at the same speed. For light waves, you can easily calculate the velocity as the frequency times the wavelength, and this

*looks*like it should work for electrons too. But you may have heard people talk about group velocity and phase velocity. I'm not about to explain why it works this way. That's a long story. But whether or not you understand the math of it, you may have heard that for electron waves you need to look at the group velocity to see how fast the electron is actually moving.The formula for group velocity is a little different from the obvious formula for phase velocity. This is how it goes:

Let's see if we can calculate the group velocities for the electron wave along the two trajectories shown here, and see if the faster wave isn't actually going slower!

We can start with the equation for the wave along the lower trajectory. Using the complex exponential, it is just:

To get the group velocity, we need to figure out what is w (omega) as a function of k. Well, omega is the frequency, which lines up with the energy, and k is the wave number, which lines up with the momentum. We can arbitrarily declare the potential energy along the flat trajectory to be zero, which makes the energy simply proportional to the square of the momentum. We can arbitrarily choose our units so k=1, and then the proportionality becomes exact:

So the group velocity it 2, which is twice the phase velocity.

How about the upper path? Well, the total energy of the upper electron is the same, because the two paths started out as a simple wave in free space. But when they enter the capacitor, the one closer to the negative plate (the upper path) loses kinetic energy and gains potential energy. Since we show (in our diagram) that the wavelength is one percent greater, we interpret this as a one percent decrease in momentum. Remembering that total energy is given by w and kinetic energy by k-squared, we must have:

But remember that k for the upper path is numerically 0.99 which means the P.E. along that path must be 0.02 to keep w constant=1. You can see this is correct because for k=0.99 we then get w=1, which are the correct wave parameters for the function as sketched.

Again, if we differentiate w with respect to k, the P.E. term disappears (since it's a constant) and we get once again dw/dk = 2k.

But on the upper path, k = 0.99, so the group velocity is one percent

*less*than on the lower path. Even though the phase velocity was one percent greater. So now we can understand why the electron wave bends*away*from the direction of slower electrons, while a light wave in a refractive medium bends*towards*the direction of slower "photons".## Sunday, January 6, 2013

### How Electrons are Different from Baseballs

The other day I did a calculation using the Principle of Least Action to calculate the trajectory of a baseball. I didn't do it exactly the way Feynman does it. He takes the initial coordinates (in time and space) and the final coordiantes, and figures out the path which minimizes the action using the Calculus of Variations.

I did it a little differently. I started with the initial position and velocity, and calculated the rate of change of action along that path. Then I looked for parallel paths which had the same rate of change, so the action was growing at the same speed. It turned out that for a slightly higher path, the ball had to travel faster. (The potential energy was higher, so the kinetic energy also had to be higher to maintain a constant difference). I therefore had to bend both paths so that the direction of propagation was always perpendicular to the surfaces of constant action.

This gave me the correct radius of curvature for the initial trajectory. But there is a little problem with the second baseball. The hypothetical baseball on the higher path ends up with the wrong radius of curvature. In my example, it was going one percent faster than the lower baseball:

You can easily see from the geometry that the radius of curvature is 10 meters. If I apply v-squared-over-r to the lower baseball, I get an acceleration of 10 m/sec, which is right for gravity. But if I apply the same formula to the upper baseball, I get an acceleration of 10.1. It's not quite right. So I have to wonder what's going to happen when I apply this analysis to electrons.

We know the electrons are governed by a wave function. Now, the

If we're dealing with electrons, this picture might represent a beam passing through a parallel-plate capacitor, with the positive charge below so the electrons are pulled down. So the potential energy is greater for the upper path, which makes the kinetic energy lower, which makes the momentum lower. So far so good...the wavefronts are farther apart, and we know that wave number is momentum in quantum mechanics, so that means the upper electrons are travelling slower.

But if an electron beam is passing through a capacitor, the electrons with higher potential energy are the ones travelling

I have the answer to this little paradox, and I'll explain it when we return.

I did it a little differently. I started with the initial position and velocity, and calculated the rate of change of action along that path. Then I looked for parallel paths which had the same rate of change, so the action was growing at the same speed. It turned out that for a slightly higher path, the ball had to travel faster. (The potential energy was higher, so the kinetic energy also had to be higher to maintain a constant difference). I therefore had to bend both paths so that the direction of propagation was always perpendicular to the surfaces of constant action.

This gave me the correct radius of curvature for the initial trajectory. But there is a little problem with the second baseball. The hypothetical baseball on the higher path ends up with the wrong radius of curvature. In my example, it was going one percent faster than the lower baseball:

You can easily see from the geometry that the radius of curvature is 10 meters. If I apply v-squared-over-r to the lower baseball, I get an acceleration of 10 m/sec, which is right for gravity. But if I apply the same formula to the upper baseball, I get an acceleration of 10.1. It's not quite right. So I have to wonder what's going to happen when I apply this analysis to electrons.

We know the electrons are governed by a wave function. Now, the

*only*wave a wave can curve is if it looks something like the drawing here, with the parallel wavefronts lock-step with each other. That's how the wavefronts look if you have, for example, light curving through a medium of varying refractive index. If electrons are waves then they have to do something almost the same.If we're dealing with electrons, this picture might represent a beam passing through a parallel-plate capacitor, with the positive charge below so the electrons are pulled down. So the potential energy is greater for the upper path, which makes the kinetic energy lower, which makes the momentum lower. So far so good...the wavefronts are farther apart, and we know that wave number is momentum in quantum mechanics, so that means the upper electrons are travelling slower.

*But if the waves are in lock-step, the upper waves must be traveling faster!*That doesn't make sense. Let's think about it again. Suppose it's light travelling through a variable medium. If its denser on the bottom, it travels faster on top. So it curves downwards. Just like the baseballs. That's what the picture shows. The beam curves downwards because it's travelling faster on top.But if an electron beam is passing through a capacitor, the electrons with higher potential energy are the ones travelling

*slower*. So the beam should curve towards the negative plate, because that's where the electrons have the higher potential. But that's just wrong...we know they beam has to curve towards the positive plate.I have the answer to this little paradox, and I'll explain it when we return.

## Tuesday, January 1, 2013

### The Principle of Least Action

Richard Feynmann tells the story of how his high school science teacher introduced him to the Principle of Least Action, and then he goes on to show how your can use it to calculate a trajectory. (Along the way he shows you how to do Calculus of Variations.) It's a strange method because it makes much of the

The path taken by a projectile in a natural trajectory, we are told, is the path which minimizes the

When we learn about Snell's Law, we see another example of how a path gets chosen which turns out to minimize a certain quantity. In the case of light passing from one medium to another, we find that when we calculate the time to get from point A to point B, that light chooses the path which gets there in the least possible time.

This seems a bit miraculous, but then when we analyze the path in terms of the wave theory of light, we see that it makes a lot of sense. Light travels in a direction perpendicular to the wavefronts at all times, and this basically keeps parallel rays in lock-step with each other. It's not so much that the light chooses the path of least time; rather, that all nearby paths are synchronous, so there is no benefit to shunting onto a side path and then cutting back in at the last minute. That's how it works for light.

That doesn't explain how it works for a ball thrown into the air, but it gives us a hint as to how we might do the calculation. The action integral, which is cumulative over time, should be the same for all nearby paths; and in particular, it should be the same for a

Here's how it works. You take a ball travelling horizontally at 10 meters per second. Then you take a second ball, slightly higher up, and assume that the two balls are, like the case of the light rays, riding the same wavefront. The direction of travel must be at all times perpendicular to the wavefront, and the wavefronts must stay in synch with each other. And the propagation of the wavefront is caluclated by using the action integral. Let's see what happens.

I wonder if you can see that I've chosen these two paths so that they have the same action? The top path has a little more potential energy, so I've given it just enough additional kinetic energy to keep the difference to be a constant.

There's only one problem with my little picture. The "wavefronts" do not remain perpendicular to the direction of propagation. But that's because projectiles don't maintain a flat trajectory. We can fix up the picture by curving the trajectories so the two paths stay in lock-step with each other:

Now the trajectories line up with each other, and the "action" calculated along each line stays in step. We just had to curve the paths. It shouldn't be too hard for you to calculate the radius of curvature; it comes to 10 meters in this case.

And it's not too hard to verify that 10 meters is actually the correct radius of curvature for the path of a baseball at 10 m/sec. (You can see it most easily by calculating v-squared-over-r). So the Principle of Least Action actually gives us a correct result for a real physics problem. And we didn't even have to use Calculus of Variations.

I left us off a few days ago with a discussion about cloud chamber trajectories, and I had promised to make some kind of an argument as to how the wave theory of matter could justify those straight-line paths of ionization. The point of today's digression was to lay the groundwork for the quantum mechanical case. In quantum mechanics, we have the wave function, and the idea is that the wave function guides the particle to find the path of least action. When we return, I'm going to try and apply the same type of analysis as I used today on the baseball, to see what happens if we use electrons instead.

*difference*between potential and kinetic energies. We know that energy is conserved, so the*sum*of potential and kinetic must be a constant. But what could be the physical significance of the*difference*of those two energies?The path taken by a projectile in a natural trajectory, we are told, is the path which minimizes the

*action*, which is the time integral of the difference of energies. How does the projectile choose the correct path out of all possible paths?When we learn about Snell's Law, we see another example of how a path gets chosen which turns out to minimize a certain quantity. In the case of light passing from one medium to another, we find that when we calculate the time to get from point A to point B, that light chooses the path which gets there in the least possible time.

This seems a bit miraculous, but then when we analyze the path in terms of the wave theory of light, we see that it makes a lot of sense. Light travels in a direction perpendicular to the wavefronts at all times, and this basically keeps parallel rays in lock-step with each other. It's not so much that the light chooses the path of least time; rather, that all nearby paths are synchronous, so there is no benefit to shunting onto a side path and then cutting back in at the last minute. That's how it works for light.

That doesn't explain how it works for a ball thrown into the air, but it gives us a hint as to how we might do the calculation. The action integral, which is cumulative over time, should be the same for all nearby paths; and in particular, it should be the same for a

*parallel*path. This is a little different from what Feynmann does in the Lecutres; he takes the starting and ending points as given constraints, and lets the path vary in-between. What I'm going to do is take a*parallel*path and force the action to be equal between the two paths. It turns out you can get a useful physical result this way.Here's how it works. You take a ball travelling horizontally at 10 meters per second. Then you take a second ball, slightly higher up, and assume that the two balls are, like the case of the light rays, riding the same wavefront. The direction of travel must be at all times perpendicular to the wavefront, and the wavefronts must stay in synch with each other. And the propagation of the wavefront is caluclated by using the action integral. Let's see what happens.

I wonder if you can see that I've chosen these two paths so that they have the same action? The top path has a little more potential energy, so I've given it just enough additional kinetic energy to keep the difference to be a constant.

There's only one problem with my little picture. The "wavefronts" do not remain perpendicular to the direction of propagation. But that's because projectiles don't maintain a flat trajectory. We can fix up the picture by curving the trajectories so the two paths stay in lock-step with each other:

Now the trajectories line up with each other, and the "action" calculated along each line stays in step. We just had to curve the paths. It shouldn't be too hard for you to calculate the radius of curvature; it comes to 10 meters in this case.

And it's not too hard to verify that 10 meters is actually the correct radius of curvature for the path of a baseball at 10 m/sec. (You can see it most easily by calculating v-squared-over-r). So the Principle of Least Action actually gives us a correct result for a real physics problem. And we didn't even have to use Calculus of Variations.

I left us off a few days ago with a discussion about cloud chamber trajectories, and I had promised to make some kind of an argument as to how the wave theory of matter could justify those straight-line paths of ionization. The point of today's digression was to lay the groundwork for the quantum mechanical case. In quantum mechanics, we have the wave function, and the idea is that the wave function guides the particle to find the path of least action. When we return, I'm going to try and apply the same type of analysis as I used today on the baseball, to see what happens if we use electrons instead.

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