Proof of length contraction

Question

I am trying to prove relativistic length contraction using the following thought experiment however it does not work out. Please explain where I have gone wrong rather than directing me towards a different proof as I'd like to understand why this one in particular doesn't work. Also note that b is $\frac{v}{c}$ and $g$ is the lorentz factor.

A Light ray travels along a train of length $L$ from the point of view of the passenger. The train travels at velocity $v$ relative to an outside observer in the same direction as the light ray.

Consider the point of view of the passenger:
Distance traveled by the ray: $L$
Time the ray takes to travel: $t$ Speed of the ray: $c= \frac{L}{t}$

Consider the point of view of the observer:
Distance traveled by the ray: $L' + t'v$
Time the ray takes to travel: $t'$
Speed of the ray: $c = \frac{L'+t'v}{t'} = \frac{L'}{t'}+ v$

Equation summary:
(1) $ c = \frac{L}{t}→ t=\frac{L}{c}$
(2) $c = \frac{L'}{t'} + v$
(3) $t' = gt$

Math: equate (1) and (2):
$ \frac{L}{t}= \frac{L'}{t'} + v $
substituting in (3):
$\frac{L}{t}= \frac{L'}{gt}+ v$
simplify:
$L = \frac{L'}{g}+ tv$
sub in (1)
$ L = \frac{L'}{g} + \frac{Lv}{c}$
$ L = \frac{L'}{g} + Lb$
$ L -Lb = \frac{L'}{g}$
$ L (1-b) =\frac{ L'}{g}$ (This is wrong)

Answer 1

Let's make it a bit clear what we are measuring. Let's assume that at time zero the rear of the train is just passing the observer standing by the track, so the situation looks like this:

$S$ is the frame of the observer sitting in the train. In this frame the train is stationary and the light is emitted at the point $(0,0)$ so it starts at the back of the train, travels a distance $\ell$ in a time $t$ and arrives at the front of the train. The time $t$ is just $t=\ell/c$ so we get your first equation:

$$ c = \frac{\ell}{t} $$

$S'$ is the frame of the observer standing by the track. In this frame the light ray is also emitted at the point $(0,0)$ but this time is has to travel the (contracted) length of the train $t'$ plus the extra distance the front of the train has moved $vt'$. This gives us your second equation:

$$ c = \frac{\ell' + vt'}{t'} $$

And you are quite correct that the speed of light is the same in both frames so we can equate these to get:

$$ \frac{l}{t} = \frac{\ell' + vt'}{t'} $$

Your problem is that you cannot assume $t' = \gamma t$. To get the value of $t'$ you need to use the Lorentz transformations:

$$ t' = \gamma \left(t - \frac{vx}{c^2}\right) $$

The equation $t' = \gamma t$ applies only when $x=0$, which is not the case here.

Beginners in SR are prone to casually throwing factors of $\gamma$ around, but this is a dangerous strategy unless you know what you are doing. The only safe way to analyse problems like this is to first identify the spacetime points of interest then use the Lorentz transformations to transform them into the other frame. Trying to shortcut this is likely to lead to problems, as indeed you've found in this case.

Incidentally Lorentz contraction is not as simple as beginners usually believe, and deriving the (simplified) equation for it is a bit involved. Have a look at my answer to “Reality” of length contraction in SR for more on this.