Saturday, July 27, 2013

Compound Velocities in Relativity

There's one more class of relativity problems that you encounter in first-year physics that I haven't done yet with my pictorial method: that's the question of compound velocities. You know: the man in the train shoots a bullet inside the train...how fast does the observer on the ground see the bullet?

I already worked out the coordinate transformation for the train going at 80% of the speed of light, so it will be convenient to use that as our starting point. And let's have the man inside the train shoot a bullet at 80% of the speed of light. Obviously in Gallilean relativity the observer on the ground will see the bullet moving at 1.6 times the speed of light. But how does it work in relativity?

We already worked out that the coordinates of the point (0,1) moved to (4/3, 5/3) in the coordinate system of the moving train. And similarly for the point (1,0) which moves to (5/3, 4/3). (Remember, that any Lorentz transformation maps points along the trajectory of a hyperbola.) From these pieces of information, we can actually re-draw the entire coordinate system like so:

You should be able to verify that the equations I've written do indeed map points from the (x,t) coordinate system (the stationary system) to the (x', t') system (the moving system).

You should know that coordinate transformations of this kind can be written in matrix notation. We're going to do that, and the look at what happens to a point that is transformed twice: first, a stationary point that is transformed to the moving train, and then the same transformation applied within the moving train. On the graph, it looks like so:

Applying the transformation once moves the vector from a to b; and applying it again moves the vector from b to c. We can do this by matrix multiplication:
So the point (1,0) (that's "a") gets mapped to the point (41/9, 40/9) (that's "c"). If I know what I'm doing, that means the bullet inside the train is travelling at 40/41 times the speed of light. You might want to check me on that.

I. J. Kennedy said...

40/41, check!

In general you can add velocities relativistically by

v⊕w = (v+w)/(1+vw)

Furthermore, if you let V = [-1,+1] then V is a vector space over ℝ using the above as vector addition. Of course a vector space needs a compatible scalar multiplication a⊙v as well. You might enjoy working out what it is in this case.

Once you have it, you can easily compute the ground speed of Russian doll that has been fired at ⅘ the speed of light from a Russian doll that's been fired at ⅘ the speed of light from a Russian doll that's been fired at ⅘ the speed of light from a Russian doll that's been fired at ⅘ the speed of light from a Russian doll that's been fired at ⅘ the speed of light from a Russian doll that's been fired at ⅘ the speed of light from a MiG-29 that's moving at a ground speed of ⅘ the speed of light!

Simply compute 7⊙⅘ (if I counted right).

It's also interesting to contemplate what it means to multiply by a non-integer scalar.

Marty Green said...

Sorry to be responding so late (if you're still around). This is a very thought-provoking perspective which I'm not sure if I'm smart enough to follow. I took up you challenge on scalar multiplication and got a bit lost...will try again. But am I correct in suspecting that in your vector space, v + w - v is not equal to w? Or did I just muck up the algebra?

I. J. Kennedy said...

Hi Marty,

I think you must've mucked up the algebra. v⊕w⊖v, aka v⊕w⊕(−v), is definitely equal to w.

Looking at the definition of ⊕,
v⊕w = (v+w)/(1+vw)
you can easily see, because of its symmetry, that it's commutative (v⊕w = w⊕v) and that v⊖v = v⊕(−v) = 0. Thus
v⊕w⊖v = v⊖v⊕w = 0⊕w = w.
(In truth, I've not addressed associativity, which is also needed to legitimize this argument.)

Or, you can calculate v⊕w⊕(−v) by applying the definition of ⊕ twice. The algebra is a little messy, but it works out. I suspect this is where you made a mistake.

To figure out the velocity of the final Russian doll, you can apply the ⊕ operator 7 times, or, if you had the magic formula for scalar multiplication, simply compute 7⊙⅘. ☺

Marty Green said...

Okay, you win. I played your game and it wasn't exactly fun. I computed 2v, then 2v, then 4v, and 5v, and I still couln't make sense of it. Then I noticed what it is: you take (1+v)^n, and get the binomial expansion...then you take the ODD powers and divide them by the EVEN powers. And that quotient is your scalar product. It's crazy and I have no idea why it works.

Except that it reminds me vaguely of something we did in third-year filter theory (Elec. Eng.) where you calculate a transfer function and then do something with odd powers divided by even powers. Never made sense back then either.

So what gives? Can you explain it?

I. J. Kennedy said...

Interesting! I hadn't noticed the odd and even powers business. It seems your formula only works for integer (positive integer, in fact) scalars, though. How would you use your result to compute ½⊙v for example? Our scalar multiplication should work for all real scalars (I claim this is a full-fledged vector space.)

As a hint to what's going on, which I won't pretend to understand hardly at all, note that our relativistic addition operator

v⊕w = (v+w)/(1+vw)

mimics the the hyperbolic trig identity

tanh(a+b) = (tanh a + tanh b) / (1 + tanh a tanh b)

Like the strange measurements of the ancients indicated we were residing on a sphere instead of a plane, this is probably a hint that the universe is hyperbolic! (only half kidding)

[SPOILER]
Here is a scalar multiplication defined for any ℝ scalar. It agrees with yours on the positive integers.

a⊙v = [(1+v)^a - (1-v)^a] / [(1+v)^a + (1-v)^a)]

Oh how I wish I could insert TeX into Blogger comments! Or at least an image.

I. J. Kennedy said...

crickets still a chirpin