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## Homework Statement

Consider an observer Bob in S standing close beside the x axis as Alice, who is holding a clock,

approaches him at speed V

_{S'S}along the axis. As Alice and her clock move from position A to

B, Alice’s clock will measure a (proper) time interval ∆t

_{0}, 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 ∆t

_{0}. 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 ∆t

_{see}between when Bob sees Alice’s clock passing A and when he sees it passing B is less than ∆t.

Prove that:

∆t

_{see}= ∆t(1 − β) = ∆t

_{0}[(1 − β)/(1 + β)]

^{1/2}

(which is clearly less than ∆t

_{0}.) Be sure to prove both equalities.

## Homework Equations

∆t = γ ∆t

_{0}

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 ∆t

_{0}=

^{1/2}∆t

_{0}/[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 ∆t

_{0}/[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

∆t

_{0}/[1-β]

^{1/2}+ (B-A)/βc

A>B so if A-B = ∆x,

∆t

_{see}= ∆t

_{0}/[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.