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Simple Relativity

by Tedjn
Tags: relativity, simple
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Tedjn
#1
Dec17-07, 01:03 AM
P: 738
1. The problem statement, all variables and given/known data

I'm having some trouble understanding the fudge factor (gamma) from Professor Shankar's relativity notes (http://open.yale.edu/courses/physics...notes_2006.pdf). It's at the bottom of the first page and top of the second.

I don't understand why the fudge factor is the same for a stationary [A] and moving [B] observer (relative to Earth).

2. Relevant equations

To simplify this in my head, I assume there is no time dilation. I do this because Shankar leaves this option open when he writes down his equations. That is t = t' always--there is an absolute time.

3. The attempt at a solution

Now, I make this concrete by fixing [B]'s velocity at u = 0.75c to the right. [A] and [B] synchronize their clocks, and after some time t = t', they watch a photon to their right. To [A], the photon travels with speed c. [A] will think, to [B], the speed of the photon should be 0.25c. This makes sense from the equation x' = x - ut, because then speed x'/t' = (x - ut)/t' = x/t - u (because t = t') = c - 0.75c = 0.25c. But in this case, because [B] sees the photon traveling with speed c, the x' [A] calculates must be too small. So, then [B] will need to take this x' and increase it by a factor gamma.

This, from what I understand, is the reason for the equation x' = gamma * (x - ut) -- from my reasoning, this gamma > 1

Now, take the opposite case. To [B], the photon is traveling at speed c. Then, [B] thinks, [A] will see the photon travel at 1.75c. Again, this makes sense from x/t = (x' + ut')/t = x'/t' + u (because t = t') = c + 0.75c = 1.75c. But [A] sees the photon traveling at speed c. In this case, the x reported by [B] must be greater than what x really is. So [A] must decrease it by some factor.

This, from what I understand, is the reason for the equation x = gamma * (x' + ut') -- from my reasoning, this gamma < 1.

I'm probably missing something simple, because I don't see how the two gamma's are equal. Maybe I'm getting my observers mixed up?

Thanks for any help!
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Hootenanny
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Dec17-07, 04:12 AM
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You can't assume that t=t', because in general it isn't true. If you were to factor in time dilation, you would obtain the expected result.
Tedjn
#3
Dec17-07, 04:26 AM
P: 738
The way I understood it was that as long as gamma is not equal to 1, there will be some changing of the lengths. I just don't see how they can be the same gamma. The reasoning in the notes is that both observers are equal. But taking the simple case of two people in different trains traveling at relative velocities to each other, for person A to think he isn't moving, he'd think that person B is moving forward, and for person B to think he isn't moving, he'd think person A was moving backward. There's a sign switch in that case. I guess I don't understand why the gammas need to be exactly the same.

Hootenanny
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Dec17-07, 07:00 AM
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Simple Relativity

Quote Quote by Tedjn View Post
But taking the simple case of two people in different trains traveling at relative velocities to each other, for person A to think he isn't moving, he'd think that person B is moving forward, and for person B to think he isn't moving, he'd think person A was moving backward.
Here is the crux of confusion. Both observers A and B will say that the other observer is moving towards them, in their reference frame; there is no sign change here. Both the velocity of A measured by B and the velocity of B measured by A will be negative.

For a similar thought experiment imagine your driving down a road with some cars traveling in the other direction. To you the other cars look as if their driving towards you (which of course they are). And to the other drivers coming the other way, you appear to by driving towards them. Both observed velocities are negative. This isn't special relativity, this is plain old Galilean relativity, does that make more sense?
Tedjn
#5
Dec17-07, 03:07 PM
P: 738
Thanks, I understand that now. What about the original case of the light photon. It seems that for observer A, B is moving in the same direction as the photon, so he would have to assume that B sees the photon moving at less than the speed of light. From B's perspective, A is moving away from him, the photon is moving away from him the opposite way, so he would have to conclude A's velocity is greater than the speed of light. If that is the case, then if length changes by the factor gamma, but in opposite directions for each observer.

I'm sorry if I'm missing something fundamental.
Hootenanny
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Dec18-07, 05:08 AM
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Quote Quote by Tedjn View Post
Thanks, I understand that now. What about the original case of the light photon. It seems that for observer A, B is moving in the same direction as the photon, so he would have to assume that B sees the photon moving at less than the speed of light. From B's perspective, A is moving away from him, the photon is moving away from him the opposite way, so he would have to conclude A's velocity is greater than the speed of light. If that is the case, then if length changes by the factor gamma, but in opposite directions for each observer.

I'm sorry if I'm missing something fundamental.
When formulating special relativity, Einstein postulated that the speed of light in a vacuum is constant and equal for all observers irrespective of their relative motions.
Tedjn
#7
Dec18-07, 03:49 PM
P: 738
Sorry about that misunderstanding. I was starting with Newton and trying to follow the relativity notes to build up to relativity starting from the Galilean transformation. From the notes, it seems that each observer will take the value given by the other observer and scale it by the same factor gamma, but I don't see why that factor is the same for both observers.
Hootenanny
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Dec19-07, 03:47 AM
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See this FAQ http://math.ucr.edu/home/baez/physic.../velocity.html


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