Consider an observer Bob in S standing close beside the x axis as Alice, who is holding a clock,
approaches him at speed VS'S along the axis. As Alice and her clock move from position A to
B, Alice’s clock will measure a (proper) time interval ∆t0, but as
measured by Bob’s “latticework of clocks”, the time between the two events (Alice and her clock
pass point A, Alice and her clock pass point B) is ∆t = γ ∆t0. However, since B is closer to Bob
than A is, light from Alice’s clock as it passes B will reach Bob in a shorter time than will the
light from A. Therefore, the time ∆tsee between when Bob sees Alice’s clock passing A and when he sees it passing B is less than ∆t.
∆tsee = ∆t(1 − β) = ∆t0 [(1 − β)/(1 + β)]1/2
(which is clearly less than ∆t0.) Be sure to prove both equalities.
∆t = γ ∆t0
β = v/c
Velocity of Alice: βc
The Attempt at a Solution
If we start at t = 0, then Bob sees Alice pass A at A/βc, where A is the distance between point A and Bob, because that is how long the light takes to reach him.
Then Alice reaches point B at ∆t = γ ∆t0 = 1/2∆t0/[1-β], but the light takes another B/βc to reach Bob.
So the total time between Alice passing A and Bob seeing Alice pass B is ∆t0/[1-β]1/2 + B/βc. But since Bob observes Alice passing A at A/βc, the time difference between Bob observing Alice passing A and Bob observing Alice passing B is
∆t0/[1-β]1/2 + (B-A)/βc
A>B so if A-B = ∆x,
∆tsee = ∆t0/[1-β]1/2 - ∆x/βc
This is as far as I was able to get. I'm not sure how to relate this equation to the equalities I'm supposed to prove. Am I approaching this wrong and/or have I made an incorrect assumption in my reasoning?
Thanks for the help.