- #1
Tedjn
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Homework Statement
I'm having some trouble understanding the fudge factor (gamma) from Professor Shankar's relativity notes (http://open.yale.edu/courses/physics/fundamentals-of-physics/resources/relativity_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 observer (relative to Earth).
Homework 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.
The Attempt at a Solution
Now, I make this concrete by fixing 's velocity at u = 0.75c to the right. [A] and 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 , 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 sees the photon traveling with speed c, the x' [A] calculates must be too small. So, then 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 , the photon is traveling at speed c. Then, 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 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!