# Time Dilation - measured vs observed time interval

## 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 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.

## 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,

∆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.

PeroK
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Gold Member
2020 Award
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
Homework Helper
Gold Member
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 = VS'S?

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.

∆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.

PeroK
Homework Helper
Gold Member
2020 Award
A/βc is the time it takes for the light to reach Bob from point A ...

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

TSny
Homework Helper
Gold Member
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?
∆tsee = ∆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.

This is light that's not travelling 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.

∆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!

PeroK