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.

## 2 comments:

Interesting post! Glad I found it.

Thanks, Scott. I only wish I knew what I was doing.

Post a Comment