Re: Challenging exercises in relativity theory
- From: "Spoonfed" <good4usoul@xxxxxxxxx>
- Date: 28 Jun 2006 16:48:03 -0700
Ben Rudiak-Gould wrote:
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?
Oh, I made a trig mistake. I was calculating ArcCos[1/2] when I should
have been calculating ArcTan[1/2]. Anyway, after looking at your
method, I re-did the problem cosidering sets of events.
For instance, events on earth
E0[n]={n,0,0}
Events where rocket 1 passes distances n along the x axis:
E1[n]={n,ß*n,0}
and events where rocket 2 passes distances n along its path
E2[n]={n,ß*n*cos[θ],beta*n*sin[θ]}
I wanted to use sets of events as a check to make sure all the events
remained in a line at the end.
E0' = {n/Sqrt[1 - ß^2], -((n*ß)/Sqrt[1 - ß^2]), 0}; Appropriate,
because all these events are in the -x direction.
E1' = {n/Sqrt[1 - ß^2] - (n*ß^2)/Sqrt[1 - ß^2], 0, 0}; Appropriate,
because all these events are at x'=y'=0.
E2'={n/Sqrt[1 - ß^2] - (n*ß^2*Cos[θ])/Sqrt[1 - ß^2],
-((n*ß)/Sqrt[1 - ß^2]) + (n*ß*Cos[θ])/Sqrt[1 - ß^2], n*ß*Sin[θ]}
Plugging in beta=.866 and theta=Pi/2,
E2[n]'= {1.99982 n, -1.73185 n, 0.866 n}
So the angle towards E0' is 180 degrees and the angle toward E2' is
153.433
2*26.5670 + 90=143.134
and so yes, you had the correct answer. Sorry about that. Again, I
just made a silly mistake. Taking ArcCos instead of ArcTan.
.
- References:
- Challenging exercises in relativity theory
- From: Spoonfed
- Re: Challenging exercises in relativity theory
- From: Ben Rudiak-Gould
- Re: Challenging exercises in relativity theory
- From: Spoonfed
- Re: Challenging exercises in relativity theory
- From: Ben Rudiak-Gould
- Challenging exercises in relativity theory
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