# Puzzled by an equation for relativistic time difference...

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1. Feb 22, 2017

### ElPimiento

1. The problem statement, all variables and given/known data
Suppose that A', B', and C' are at rest in frame S', which moves with respect to S at speed v in the +x direction. Let B' be located exactly midway between A' and C'. At t' = 0, a light flash occurs at B' and expands outward as a spherical wave. (A', B', and C' are all on the +x axis, with A' having the smallest x coordinate and C' having the largest x coordinate. Assume A'B' = B'C' = Lp.)

What is the difference between the time it takes the wave front to reach A' and the time it takes to reach B' (Use the following as necessary: v, c and Lp.

2. Relevant equations
\begin{align} L & = \frac{L_\text{p}}{\gamma} \\ \Delta t & = \gamma \big( t' + \frac{v}{c^2}x' \big) \\ \end{align}

3. The attempt at a solution
I think an observer in S' would see the events simultaneously. So the time interval should be 0. but this is not the correct answer. So I will present an alternate attempt at rationalizing the situation.

An observer in S sees the light travel a distance AB = BC which is contracted from the proper length A'B' = B'C' according to:

$$\text{AB} = \text{BC} = \frac{\text{A'B'}}{\gamma} = \frac{L_\text{p}}{\gamma}$$
Since the wave is an electromagnetic one, the time take to traverse these distances will be

$$\Delta t_\text{AB} = \Bigg( \frac{L_\text{p}}{\gamma} \Bigg) \Bigg(\frac{1}{c + v} \Bigg) = \frac{L_\text{p}}{\gamma (c + v)} \\ \Delta t_\text{BC} = \Bigg( \frac{L_\text{p}}{\gamma} \Bigg) \Bigg(\frac{1}{c - v} \Bigg) = \frac{L_\text{p}}{\gamma (c - v)} \\ \implies \Delta t = \Delta t_\text{AB} - \Delta t_\text{BC} = \frac{L_\text{p}}{\gamma}\Bigg(\frac{2v}{c^2 - v^2} \Bigg)$$

Now, I think the time interval being asked for is one from S' (which would not be the proper time right?). So I'll apply the inverted Lorentz transformation:

\begin{align} \Delta t' & = \gamma \big(\Delta t + \frac{v}{c^2} \Delta x \big) \\ & = \gamma \Bigg(\Delta t + \frac{v}{c^2} v \Delta t \Bigg) \\ & = \gamma \Delta t \Bigg( 1 + \frac{v^2}{c^2} \Bigg) \\ & = \gamma \frac{L_\text{p}}{\gamma}\Bigg(\frac{2v}{c^2 - v^2} \Bigg) \Bigg( 1 + \frac{v^2}{c^2} \Bigg) \\ & = L_\text{p} \Bigg(\frac{2v}{c^2 - v^2} \Bigg) \Bigg( 1 + \frac{v^2}{c^2} \Bigg) \end{align}

But this is incorrect. The correct answer is:

$$\Delta t = L_\text{p} \Bigg(\frac{2v}{c^2 - v^2} \Bigg)$$

Which I would get numerically by treating $\Delta x$, above, as 0; but I feel like this doesn't make physical sense since the $\Delta x$ is distance B travels in the time between wavefronts.

2. Feb 22, 2017

### Staff: Mentor

Can you please provide the exact problem statement?

3. Feb 23, 2017

There ya go.