Re: Challenging exercises in relativity theory

Spoonfed wrote:
Ben Rudiak-Gould wrote:
I don't see why it would suggest anything at all. This is a very strange
calculation you're asking people to perform.

It's challenging, but it has a definite answer.

Well, so does Feynman's problem, but that doesn't save it from physical irrelevance. But I've thought about it and realized that I was wrong to criticize your problem. My intuition was that the observed earth-ship angles would change with time. I think your rephrasing is better, since it makes it clearer that they don't:

Or you could just calculate the angle between
the two rockets as viewed from earth + the angle between the earth and
rocket from each of the two rockets.

I can tell you immediately that the sum will be less than 180, but I'll work through the problem anyway. I'll solve a more general problem where the angle between the ships from the earth is theta, and the speed of each ship is beta. I'll let one ship move along the +x axis, and the other will have a four-velocity V given by

V_t = gamma
V_x = gamma beta cos theta
V_y = gamma beta sin theta

Boosting into a primed coordinate system where the first ship is at rest, we get

V_t' = gamma (V_t - beta V_x) = gamma^2 (1 - beta^2 cos theta)
V_x' = gamma (V_x - beta V_t) = gamma^2 beta (cos theta - 1)
V_y' = V_y = gamma beta sin theta

The earth is moving along the -x axis, so the apparent angle between the earth and the second ship is

phi = tan^-1 (-V_y' / V_x')
= tan^-1 (sin theta / (gamma (1 - cos theta)))

Hmm, not very reasonable-looking, so maybe I made a mistake. For the special case of theta = 90 we get phi = tan^-1 (1/gamma), which does look reasonable. It always gives a sum of less than 180 degrees for gamma > 1. For gamma=2 I get 143 degrees. Is that what you got?

How did I know the answer would be less than 180? Because the geometry of special relativity is the same as the FRW geometry with Omega=Lambda=0, and you problem is equivalent to adding the vertices of a triangle in that geometry, and I know that it's hyperbolic for Omega < Omega_c.

Don't you want some units on that?
Good enough.

:-)

what point should you use to determine the center of
that coordinate system?
Er, it had better not matter, else your theory is inconsistent.
Shouldn't you use the point P to calculate the force at point P?

The choice of origin is dictated by convenience. If I'm solving the gravitational two-body problem, the most convenient origin is the center of mass of the system, even though neither body is ever located there. I'm free to choose the center of mass because the answer is the same no matter what origin I choose. You can't claim that someone's calculation is wrong just because the point they picked isn't the one you would have picked.

F(r) = -4/3 pi G rho (r - r0)

where r0 is an arbitrary point in space. r0 is the center of a radial
gravitational field which increases without bound as you move farther from
the center.

Correct. This is essentially what Einstein calculated. To get
infinite force at point P, then, he used r=infinity.

Maybe you're right. I don't really understand what he was arguing at the end of the paragraph.

So what's the significance of this? You could argue that the gravitational
field is not well-defined, since it depends on this arbitrary point, and you
could use this to make the case for a modification of Newton's theory
(perhaps an added exponential falloff).

... which is essentially what Einstein did?

Maybe.

But there's another way out. The
difference between any two of these fields (with different values of r0) is
a constant field, i.e. you can move r0 around by adding a constant field.
What's the added effect of the constant field? Nothing!

That is similar to my idea of taking the average field which I
mentioned to Sue.

I assume you're thinking of something like the following: to calculate the field at P, it makes sense to take r0=P; so the field everywhere works out to zero. Or, it makes sense to take an average over all the solutions for all values of r0, and that average is zero everywhere.

These arguments look reasonable, but they must be wrong, for the simple reason that F(r)=0 does not, in fact, satisfy the field equation that it was our original goal to solve! You can concoct a solution to a differential equation by any method you like, no matter how wacky (think image charges), but only if it's actually a solution. If you're interested I can explain the flaw in each argument, but I only have to look at the conclusion to know that there is a flaw.

-- Ben
.

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