Time Dilation - measured vs observed time interval

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1. Mar 18, 2017

astrocytosis

1. The problem statement, all variables and given/known data
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.

Prove that:

∆tsee = ∆t(1 − β) = ∆t0 [(1 − β)/(1 + β)]1/2

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

2. Relevant equations

∆t = γ ∆t0

β = v/c

Velocity of Alice: βc

3. 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.

2. Mar 18, 2017

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?

3. Mar 18, 2017

TSny

Why did you divide by βc here?

Did you use the correct expression for γ?

Again, why βc?

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

4. Mar 18, 2017

astrocytosis

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.

∆tsee = ∆t0/[1-β2]1/2 - ∆x/βc

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

∆tsee = ∆t0/[1-β2]1/2 - ∆t/[1-β2]1/2

Which still doesn't seem quite right.

5. Mar 18, 2017

PeroK

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

6. Mar 18, 2017

TSny

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

7. Mar 18, 2017

astrocytosis

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

∆tsee = ∆t0/[1-β2]1/2 - ∆x/c

∆x = VS'S∆t = βc∆t according to Bob

∆tsee = ∆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!