# Mistake in simple calculation in special relativity [closed]

I am just learning a little bit of special relativity on my own and am obviously still quite confused. I would appreciate it if someone could point out the error in the following calculation:

Let A be a stationary coordinate frame with coordinates x',y',z',t' and B a moving coordinate frame (moving along the x axis with velocity $v$ with respect to A) with coordinates x,y,z,t. Let us think of B as attached to a spaceship moving at velocity v.

At time 0 (in both frames), the origins of A and B are concurrent.

The spaceship is of length $L$ (in frame B) and is aligned along the x axis and a light ray begins moving from the origin at time t = t' = 0. This is event $E_1$ and event $E_2$ is when the light ray hits the other end of the spaceship.

Now in frame B (the spaceship one), $E_1$ happens at $t = 0$ and $E_2$ happens at $t = \tau$, say. Since the spaceship is of length $L$ in this frame, we have the relation: $$L = c\tau$$

On the other hand, in frame A, we see a light ray leave the origin and hit the other end of the spaceship after some time $t' = \tau'$. The light ray has to travel a distance the length of the spaceship (in frame A)+ the distance the spaceship moved in this time.

This should be equal to $\gamma L + v\tau'$ where $\gamma = 1/(\sqrt{1-v^2/c^2})$. On the other hand, this should also be equal to $\tau'c$ since the light ray traveled for time $\tau'$ in between the events.

Equating the two and substituting in $L = c\tau$, I obtain: $$\gamma c\tau + v\tau' = c\tau' \implies \tau' = \tau/(1 - v/c).$$

On the other hand, by time dilation, I think the right answer should be: $$\tau' = \tau/\sqrt{1-v^2/c^2}.$$

Where did I go wrong?

## closed as off-topic by Kyle Kanos, sammy gerbil, ZeroTheHero, ACuriousMind♦Jun 4 '18 at 16:42

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• Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions over those just asking for a specific computation. – ACuriousMind Jun 4 '18 at 16:42
• Thanks for the comment but 1) this is not a homework question because I am not a part of any institution currently. I am learning the material on my own and have no one to ask questions to. 2) This is not quite a check-my-work type question because I knew I was wrong, I just didn't know where. Anyway, I already received several great answers and I figured out my mistake. – Asvin Jun 4 '18 at 17:33

Have to admit I haven't checked your working line by line, but I can see what might be your only mistake. You are mis-applying the time dilation equation (your last equation). It applies only if $\tau$ is the time between two events in the frame in which the event occur at the same place. But in the spaceship frame the events occur at opposite ends of the spaceship! Suggest you rename the time interval.

Alternatively, you could put a mirror at the far end of the spaceship, and consider the time between the event of the light leaving the source and returning to it. This really will be a 'proper time', correctly denoted by $\tau$, in the spaceship's frame.

A small nitpick: it's conventional (and therefore easier to follow) for the dashed frame to be moving in the x direction wrt the undashed frame. In other words, suggest you use your dashes for B frame co-ordinates.

• Ah, okay that is what I was thinking too. Let me try and figure out how to apply the correct time dilation equation and see if that matches what I get. – Asvin Jun 2 '18 at 15:52
• Good. But I do recommend those notation changes. I've added a paragraph to my (partial) answer above. – Philip Wood Jun 2 '18 at 16:01

Main confusion in special relativity is that people think observing object getting small. Actually, we observe as small so light propagates all the $L$ way instead $/gamma L$. Secondly, you transform $x$ distance (you used $L$, it seems confused with $x-x_0$) with Galilean transformation should be Lorentz transformation.

so $x=c \delta t$ (you find correctly)

and $x'= c \delta t' + v t'$

then use the Lorentz transformation that

$x= \gamma (x'-vt')$ will give the time dilation.

I think Mirror example better to understand this phenomenon. Such as https://users.sussex.ac.uk/~waa22/relativity/Complete_Derivation_files/derivation.pdf

or even better Mechanics: Berkeley Physics Course. vol. 1. Charles Kittel chapter 12

I hope it will be helpful.

best regards