Marty Does Relativity

I said last time that I was going to try and explain why kinetic energy is ½*mv^2. I’m going to resort to relativity to make my case. I don’t like doing this because relativity really isn’t my territory. I find the arguments really mathematical and hard to follow. But that’s what I’m stuck with.

We start by introducing the concept of four-vectors. I hope you know what an ordinary vector is but if you don’t you don’t. That’s my starting point. (Ordinary vectors, that is.)

The distance to a point (x,y,z) is of course given by the three-dimensional Law of Pythagoras:

D^2= x^2+ y^2+z^2

In relativity, we don’t stop here. We say the true relativistic “distance” must include a time component in addition to the three space components, so the relativistic equation looks like this:

D^2= x^2 + y^2 + z^2 - t^2

Why the minus sign in front of the t^2 term? That’s a long story, and it has to do with the detailed mathematical structure of the space-time continuum. It seems that time and space, while interchangeable up to a point, are not really on a completely equal footing in relativity. Simply put, time is like an imaginary number. So when you square it, it goes negative. Also, to get consistent units, we really have to multiply by the speed of light. Otherwise we can’t have time and distance in the same equation. We can rewrite our formula for relativistic distance, for convenience letting the y and z components go to zero, and here is what it looks like:

D^2= x^2 + (ict)^2


I should mention that the intuitive meaning of relativistic distance is very different from what ordinarily think of. For example, the “distance” between two points in space-time is “zero” when they are at opposite ends of a beam of light! But that’s another question. The point is that in relativity, everything that we usually think of as an ordinary vector, or a “three-vector”, is actually a four-vector. And momentum is a perfect case in point. When we track the momentum of something, we do it as a three-dimensional vector. What then is the fourth piece of the vector in relativity, the so-called “time component” of momentum?

It seems that the answer is: mass. Something has momentum because it is careening through space in three dimensions. It is also careening through time, and that is the fourth component. Letting momentum be denoted by the vector p as is customary, and following the exact same rules as when we converted ordinary distance to a four-vector, we can formally write the relativistic equation for momentum as follows (with the y and z components set to zero):

p^2= (mv)^2+ (icm)^2

Now how are we going to make sense of this? You can see that the final term is starting to look a lot like Einstein’s formula for energy, except there’s an extra factor of m thrown in. But the important thing to remember is something that doesn’t show up explicitly in Einsteins formula, but you can’t understand anything if you don’t know about it. It’s the relativistic change in mass.

In relativity, the mass increases as you go faster, and it increases by a factor of the square root of (c^2 – v^2)/c^2). In our equation, wherever the mass appears it happens to be squared, which is nice because we won’t be carrying around square root signs. Putting in the correction for change in mass, what we get for momentum squared is:

p^2 = m^2*v^2 – m^2*c^2 – m^2*v^2

It turns out that the increase in mass exactly cancels out the “ordinary” momentum term, and all that is left is the mc^2 term. Dividing both sides of the equation by m, we get

(p^2)/m = m*c^2

which is just Einstein’s famous formula. It seems that Einstein’s mc^2 is a relativistic invariant, a property of a body which remains constant regardless of its apparent momentum. Because of the negative sign in the t component of the four-vector, the apparent gain in p^2 is compensated by the increase of mass in the t component.

Of course I cheated in this analysis. When I evaluated p^2, I used the relativistic change of mass for the t term, but I ignored it for the x term. I did it because it made my answer come out cleanly. I don’t know why it works.

That's the first and last time I'm going to try and do relativity calculations in this blog.

Why not E=mc^3???

Peter, my brother-in-law, tells me that his son Charlie asked him a physics question the other day that he couldn’t answer: why is it E=mc-squared, and not mc-cubed or some other power? I told him I’d think about it and maybe post something on my blog. Well here goes.

For starters, this question isn’t really about relativity. It’s about energy, and E=mc^2 only makes sense if energy has the units of (mass)*(velocity)^2. So we might as well ask: why is kinetic energy defined as KE = ½*mv^2?

The crazy thing is I don’t have a really good answer to this question. I did once upon a time, but I don’t any more. That time I knew the answer was almost forty years ago when I was in high school and we had just learned the formula for kinetic energy. I remember asking why is that the kinetic energy, when all of a sudden another student declared with conviction: “Can’t you see? It’s just the indefinite integral of momentum with respect to veloctiy.” (My brother-in-law will not have a hard time guessing that the “other student” in question was none other than Randy Ellis, currently a professor of computer science in Kingston, Ontario.)

This was an amazing explanation which blew my mind. We were all (including Peter) taking calculus together, and it struck me as obvious that if you just wrote

∫mvdv

that the solution was indeed 1/2mv^2, the formula for kinetic energy.

It took me about three days to realize that this didn’t even mean anything! Integrating what for what purpose? It was just a bunch of letters on the page, and it didn’t explain anything. I still don’t know what it means. I’m not going to say you absolutely can’t make any sense of this. I’m just saying that at the high school level, it couldn’t have meant anything to any of us.

Still, there are pieces of the truth in all these things. We know that work is (force)*(distance), and we know that force is (mass)*(acceleration). Putting these together we know that the units of work, and hence energy, must at the very least be equal to kilograms*(meters)^2/(seconds)^2, which are indeed the units found in the formula for kinetic energy, and also in Einstein’s formula. So it really can’t be E=mc^3 or something else.

But just having the units line up still doesn’t completely satisfy me. Why is kinetic energy proportional to the square of the velocity? Why do we keep track of 1/2mv^2 and not some other power of v? In fact, we do keep track of another power of v, namely the first power! We keep track of mv and call it momentum, and we say that momentum is conserved. Then we keep track of 1/2mv^2 and call it energy, and we say that energy is conserved! Why don’t we keep track of 1/6mv^3 and call it something else, and look for a new conservation law? Where does it stop?

We can look for some guidance to other forms of energy, because of course it doesn’t start and stop with kinetic energy. There is, for example spring energy, with the formula ½*kx^2 where x is the displacement, and there is also the energy in a capacitor, ½*CV^2 where V is the voltage. These formulas look very much like the formula for kinetic energy. If we can explain one, perhaps we can explain all the formulas?

It turns out to be not so easy. The spring formula and the capacitor formula are indeed easy to explain, and their explanations turn out to be very similar to Randy’s calculus-style explanation of the kinetic energy formula: the only difference is that unlike kinetic energy, the calculus explanation makes perfect sense in these cases! In both these cases the energy is realy the product of two different causes: for the spring, it is the force and the displacement, and for the capacitor it is the voltage and the charge. In both cases the two variables are proportional to each other, so instead of writing (voltage)*(charge), you can just write (voltage)^2 and multiply by “capacitance”, which just hides the fact that capacitance is simply the proportionality factor which relates voltage and charge. Same with the spring constant.

I don’t know any way to apply this same argument to the formula for kinetic energy. It’s not clear to me in any way that kinetic energy must be the product of momentum and velocity in the same way that the energy in a capacitor is the product of charge and voltage.

So how do we explain it? I suppose the easiest thing is to just say that that’s how it is, it works and gives us useful results, and so we accept it. But I can’t just give up that easily. I’ve been turning this over and over and I’ve come up with an explanation. I’m not that happy with it but it's going to be the topic of my next post.

Shout out to Slovenia

I have been blogging for going on two years now, but only last month I realized that Google Blogger tracks all kinds of statistics for me. I had no idea people were actually reading me, but they were! I have had going on 4000 hits in the 20 months I’ve been blogging, and the last two months I’ve averaged over 500 per month. The greatest number of hits are from the United States, Canada, England, and Germany. Those are the top four countries by a fairly wide margin. However, the race for fifth place is quite hotly contested by Russia, India, the Netherlands and…Slovenia?

Just today, for the first time since I’ve been monitoring my stats, little Slovenia edged past Russia for sole possesion of fifth place, with 85 all-time visits to my blog. Congratulations, Slovenia, and keep up the good work. We’re thrilled to have you here.

Why Not Write Hebrew with Arabic Script?

I follow the news from Israel on the internet, usually with some dismay. Today, however I had a surprise. Moshe Arens, one of the more stubborn right-wing commentators, had an article in Haaretz where he supported the retention of Arabic as an official language in Israel. It’s hard to believe that there is a move afoot to downgrade the status of Arabic, but it’s even more of a surprise to find Arens speaking out against such a move.

For a long time, I have been advocating that Israel and Israelis should be more proactive in identifying themselves as belonging to the Middle East rather than as a European outpost. This means showing more interest and respect towards Arabic culture. One of the most conspicuous aspects of that culture is their incredibly beautiful written script, recognized instantly wordlwide. I therefore came up with the idea that we should adopt the Arabic script for use in Hebrew.

Of course I am not talking about replacing the holy letters of the Torah. These cannot in any circumstances be tampered with. It’s our second alphabet I’m talking about, our written script. What do we need it for? It’s totally replaceable.

The beauty of my proposal is that since Hebrew and Arabic are so closely related, many words would become instantly recognizable to speakers of both languages, just as English and French words are mutually recognizable in print even when they are pronounced differently.

More importantly it would be a huge gesture towards the Arab world that we respect their culture and with to be a part of the Middle East. Such a gesture is long overdue on our part. Sadly, my proposal has been ignored since I first raised it five years ago. With this posting, maybe I can give it a small bump.

Does Radioactive Decay Cause Lightning?

Next week I am supposed to teach a unit of Grade 9 Science: static electricity. I am going crazy trying to make those damned experiments work! You rub this or that with a piece of fur and little pieces of paper are supposed to jump up and down: but try and explain it? Oh, I know, it’s all very simple: there is the triboelectric effect, and there is induced charge, but do you think it actually works the way the lab manual says it does? Try it yourself.

I can hardly begin to count the number of contradictions I find when I actually try to resolve the theory with what I am actually seeingin my basement. Actually, I’m exaggerating: I can resolve quite a few of them,but when I step into the classroom tomorrow I’m supposed to explain this stuffat the Grade 9 level: no Coulombs law, no concept of voltage or capacitance,nothing but “like repels like” and “electricity is made of particles”. It’s a scary prospect.

I’m not going to get into detail today on the experiments, but there is one glaring fact that everybody knows. It’s a lot easier to put a negative charge on something than a positive charge. Why is this?? When I rub a balloon with a piece of wool, doesn’t the wool get just as much positivie charge as the balloon gets a negative charge? But if I walk over to the wall, I can stick the balloon on the wall and it stays right there. If I put the piece of wool on the wall, it just falls down. Yes, I know, the wool is heavier than the balloon. But I have not been able over the last three days of effort to find one indication that there is positive charge on the wool. What am I doing wrong?

I came up with a theory that the air is actually full with a surplus of negative charges. So when you put a negative charge on the balloon, it stays there because the negative air charges are repelled. But the positive charge on the wool is rapidly neutralized by the ambient negative charge.

This sounded OK, but then I read about lighting on the internet. It seems that there is no real theory of how lightning works! There are bits and pieces of a theory, but it just doesn’t all add up. And the first thing that doesn’t make sense is that on some level, the whole planet seems to have a negative charge and the atomsphere is positively charged. So the potential rises as you gain altitude, at a rate of about 100 volts per meter. This trashes my theory of negative charge.

But it got me thinking about lightning, and I asked my self the question: could the charge of the planet be caused by radioactive decay? Alpha particles carry a positive charge and they are being given off by uranium atoms. How many of them make it out of the earth’s crust unscathed to the atmosphere? Are they enough to account for the quantity of atmospheric electricity?

From Wikipedia I got a ballpark figure of 10^20 grams of uranium in the top 25 km of the earth’s crust. This is on the order of 10^18 moles. Since there are 10^5 coulombs in a mole, it is 10^23 coulombs of available electricity. These atoms are decaying at a rate of one decay per atom per four billion years. I ran the figures and it comes to about one million amperes.

Some of these alpha particles will be neutralized before they escape the earth’s crust. But how many? Also from Wikipedia, I learn that there are, worldwide, about 50 lightning strikes per second, each delivering an average of 15 coulombs. In other words, planetary lightning represents an average current of about 1000 amps.

It seems that if less than 1% of the alpha particles generated by uranium decay actually make it into the atmosphere, it would be enough to explain lightning. I guess that’s my theory for today. I recently wrote about how I got ripped off for the Nobel Prize because I was just 78 years late in explaing the Compton Effect by the wave theory of light. Well, maybe I’ll have more luck with this theory.

How I got Cheated Out of the Nobel Prize

Two weeks ago I wrote about how I almost won the Nobel Prize. Now I have to come clean and tell you what went wrong. You may remember that I had just come up with a way to explain the Compton Effect without photons. It’s actually a really straightforward application of wave-on-wave interactions. I’d been trying to think of an explanation for about ten years without success. I think the reason I couldn’t do it was I was trying to build on my earlier success with explaining the photo-electric effect. In that case, the key was to look at the superposition of the two electron wave functions, before and after, and match the frequency of that superposition with the frequency of the driving e-m wave.

The Compton effect is entirely different! You don’t match up frequencies, you match up wavelengths. The “before” and “after” electron wave functions are simply a plane wave moving to the right, and a plane wave moving to the left. There is no frequency involved because the energies of the incoming and outgoing electrons are the same. But the superposition sets up a standing wave of charge, and it’s the wavelength that interacts strongly with an e-m wave. It works because in quantum mechanics, wavelength is momentum: so an electron interacts with a “photon” whent they have the same wavelength. The only thing to remember is you don’t need to parcel your light up into “photons”: it works because of the wavelength relationship, and that’s all you need.

You can imagine that people told me I was a quack when I tried to tell them about this. It was obviously wrong, way out in left field, nonsense, you name it. I didn’t mind because I knew my day would come. The analysis was correct, and some day the world would recognize it.

This went on for about six months until disaster struck. I was browsing articles on the internet one day and I stumbled across a paper by a fellow named Strnad. His paper was about Schroedinger’s 1927 explanation of the Compton Effect. I started reading with disbelief: Schroedinger’s explanation was my explanation.

Did the naysayers change their tune? Yes they did, without missing a stride. One day it was “nonsense, garbage, quackery” and the next day is “been done before, nothing new, everybody knows that already”. Either way, the bottom line was clear. I’m not getting the Nobel Prize after all.

By the way, Strnad had a peculiar take on the whole question. While he agreed with Schroedinger’s description, he didn’t think it should be taught to students. “It’s important not to confuse them about the existence of photons.” (I’m paraphrasing.) It seems if the students were exposed to Schroedinger’s ideas, it might shake their faith in what their professors are telling them.

I have to wonder why someone would say that’s a bad thing.

Transmission Lines and Charateristic Impedance

I once did a calculation on an electromagnetic wave propagating between two parallel plates to see which was greater: the electrostatic attraction between the opposite charges induced on the plates, or the magnetic repulsion between the parallel currents induced in the same plates. You can almost see where this is going if you have any intuition for these things. It turns out that in Nature, these two forces are perfectly balanced. It’s just one of those things.

I was actually trying to answer the question of “why do like currents attract?” instead of repelling, the way like charges do. I wanted to draw some cosmic conclousion about the fact that if they repelled, then there would be a net repulsion between the two parallel plates; and that the only way the universe made sense was that if there was no net force of any kind between those plates. I was able to verify by direct calculation that the net force indeed went to zero for a freely propagating e-m wave, but I fell short of my ultimate goal. I never came up with a reason why the net force had to go to zero for reasons that were self-evident.

Someone posted a problem the other day on a website I go to called stackeckchange.com . The problem was a parallel-wire transmission line connected to a resistor. The poster had recognized that when you connect a battery, opposite charges appear on the two wires, so they attract; and also, when current flows, the currents are opposite so they repel. The amount of charge hardly depends on the resistance, so it (and the attraction) are virtually constant. The amount of repulsion depends on the amount of current, or the resistance. So the poster askes: for what value of resistance does the net force go to zero?

You can see that this is similar to the problem I analyzed with the parallel plate. So I posted an answer in the form of a conjecture: that if you treat this as a transmission line, the condition for zero net force ought to be a freely propagating wave; and the way to get this was to have your load impedance matched to your line impedance, so there is no reflected wave.

This answer generated a lot of ridicule from the experts in the group, and we’re still arguing about it. But it looks to me very much like I’m turning out to be right. Of course no one will admit it, but what else is new? In the meantime, the process of arguing has solidified some ideas I’ve had floating around for a while about how to calculate the impedance of different transmission lines, and I’ve come up with some pretty cool tricks that I think I’m going to post one of these days.

You can check out the discussion at http://physics.stackexchange.com/questions/3306/when-is-the-force-null-between-parallel-conducting-wires