# Determining the time elapsed between two events in

I want to determine the time a photon needs in order to cover a distance, say $$l_0$$, where $$l_0$$ is the length of a spaceship (reference system S'). So, the photon is going from one end of the spaceship to another. But, what I am after is the time elapsed between the two events in the system of reference of an observer that is located outside the spaceship (reference system S). My attempt is as follows: we know that the speed of light $$c$$ is constant in all frames of rerefence, so

$$c=\frac{l_0}{t'}=\frac{l}{t}$$

Therefore, the time needed to cover the distance is $$t$$ and

$$t=\frac{l}{c}=\gamma\frac{l_0}{c}$$

Is this line of reasoning correct, or should the velocity of light in the reference frame S be something else? If it should be something else, then why?

Thanks in advance.

• Voting to reopen. Clearly a conceptual question rather than a question about specific computations. Commented Feb 21 at 10:12
• Agreed.${}{}{}$ Commented Feb 21 at 12:37

## 1 Answer

You are overlooking the fact that the observer in the rocket uses two different synchronised clocks to measure how long the photon takes to traverse the rocket and these clocks will not be synchronised according to an observer with motion relative to the rocket. This is known as the "relativity of simultaneity" and is the thing most likely to catch people out in relativity calculations.

It is best in general to use the Lorentz transformations that take care of all these issues.

Unfortunately, you have defined the primed and unprimed frames differently to how Wikipedia does, so you will have to swap those values when using the Wikipedia Lorentz transformation equations.

For your example, start with $$t = \gamma (t' -v l_0 /c^2)$$ where v is the velocity of the observer relative to the initial reference frame.

As you have stated, $$t' = l_0/c$$ and we can substitute this into the transformation and get: $$t = \gamma (l_0/c -v \ l_0 /c^2) = \gamma \ \frac{l_0} c (1 -v/c)$$ which differs from your result by a factor of $$(1-v/c)$$.

Your correctly stated that the speed of light is the same in all reference frames, so let's check how that works out here. The Lorentz transformation for the spatial coordinate is $$l = \gamma (l_0 -v t')$$ and since $$t' = l_0/c$$ we get: $$l = \gamma (l_0 -v l_0 /c) = \gamma \ l_0(1 -v/c)$$

$$c = \frac l t = \frac{\gamma \ l_0(1 -v/c)} {\gamma \ \frac{l_0} c (1 -v/c)} =c$$ confirming c is the same in both reference frames.