Time Dilation - measured vs observed time interval

[astrocytosis]

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 ∆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 ∆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) = ∆t₀ [(1 − β)/(1 + β)]^1/2

(which is clearly less than ∆t₀.) Be sure to prove both equalities.

Homework Equations

∆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,

∆t_see = ∆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.

 

[PeroK]

Hint: imagine a light signal leaving A at time $t_1$ and a light signal leaving B at time $t_2$. When do these signals reach the origin?

 

[TSny]

If we start at t = 0, then Bob sees Alice pass A at A/βc,

Why did you divide by βc here?

Then Alice reaches point B at ∆t = γ ∆t₀ = ^1/2∆t₀/[1-β],

Did you use the correct expression for γ?

So the total time between Alice passing A and Bob seeing Alice pass B is ∆t₀/[1-β]^1/2 + B/βc.

Again, why βc?

A>B so if A-B = ∆x

Can you express ∆x in terms of ∆t₀ and βc = V_S'S?

Overall, I think your approach will get you the answer if you address the above questions.

 

[astrocytosis]

Why did you divide by βc here?

A/βc is the time it takes for the light to reach Bob from point A (x=vt -> t = x/v). I'm using γ = [1-β²]^-1/2 but made some typos.

∆t_see = ∆t₀/[1-β²]^1/2 - ∆x/βc

I can use a Lorentz transformation get ∆x: ∆x = γ[∆x' + V_S'S∆t]. I think ∆x' is zero because Alice is measuring a proper time interval, so, plugging in for ∆x with βc = V_S'S:

∆t_see = ∆t₀/[1-β²]^1/2 - ∆t/[1-β²]^1/2

Which still doesn't seem quite right.

 

[PeroK]

A/βc is the time it takes for the light to reach Bob from point A ...

This is light that's not traveling at $c$, then?

 

[TSny]

A/βc is the time it takes for the light to reach Bob from point A

Does βc represent the speed of light or the speed that Alice moves relative to Bob?

∆t_see = ∆t₀/[1-β²]^1/2 - ∆x/βc

This would be OK if you fix the βc part.

You should be able to get an expression for ∆x by thinking about the time it takes Alice to travel the distance ∆x according to Bob.

 

[astrocytosis]

This is light that's not traveling at $c$, then?

Oops... got caught up thinking about Alice's velocity and forgot that has nothing to do with how fast the light will reach Bob.

∆t_see = ∆t0/[1-β²]^1/2 - ∆x/c

∆x = V_S'S∆t = βc∆t according to Bob

∆t_see = ∆t0/[1-β²]^1/2 - β∆t, which is just ∆t = (1-β)∆t.

I was able to get the other equality easily from there. Thank you!