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A relativistic train with a rest length of 500.0 m takes 780 ns to pass a stationary observer standing on the train platform, as measured by the stationary observer.

(a) What is the speed of the train? (Hint: Remember to account for the Lorentz contraction of the spaceship.)

I've been at it for a couple hours and I just don't see how I can use the Lorentz formulas.

If S is the inertial frame with respect to the observer and S' is the inertial frame with respect to the train, then the front of the train and an observer line up at t = t' = 0 and x = x' = 0,

Then, in the S' frame, the front of the train sees itself at $x' = 500 m$ when $t = 780 \times 10^{-9} seconds$

The correct answer is: $$2.72 \times 10^8 \frac{m}{s} = 0.907c$$

But I don't see where to head from where I set up. Any help is much appreciated at this point.

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up vote 0 down vote accepted

You should know that Lorentz-FitzGerald contraction is (in units of $c = 1$): $$L = L_0\sqrt{1-v^2}\text{,}$$ where you are given that $L_0 = 500\,\text{m} = 1.6678\,\text{$\mu$s}$ and that in the inertial frame the contracted length $L$ passes by the observer in $\Delta t = 780\,\text{ns}$. Therefore $v = L/\Delta t$ and: $$L^2 = L_0^2\left(1-\frac{L^2}{\Delta t^2}\right)\text{.}$$ If you solve this for $L$ algebraically and plug in the values that you are given, you should then get $v = 0.906$ from the above relationship to $v$. Close enough up to roundoff error.

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Thank you so much! You have no idea how helpful this was. Have a great day! – Dear Watson Apr 26 '13 at 11:59

Although I'll end up with the same answer as Stan I think it's nice to see how to do this using the Lorentz transformations. We'll take the observer to be moving left to right at velocity $v$, and the front of the train is at $x$ = 0 and the end at $x$ = 500m. I'll call the length of the train $L_0$ for consistency with Stan's answer. The events marking when the observer passes the two ends of the train are (0, 0) and ($L_0/v$, $L_0$). As usual we'll assume the two frames coincide at (0, 0) so we only have to transform the second point ($L_0/v$, $L_0$).


The Lorentz transformations are:

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

$$ x' = \gamma \left( x - vt \right) $$

So transforming into the observer's frame we get:

$$ t' = \gamma \left( \frac{L_0}{v} - \frac{vL_0}{c^2} \right ) $$

$$ x' = \gamma \left( L_0 - v \frac{L_0}{v} \right) = 0 $$

So $x'$ is zero, but we knew this already, because in the observer's frame they are stationary at the origin. The solution to calculating the speed $v$ is going to come from the expression for $t'$ (where the problem tells us $t'$ = 780ns). Rearranging the expression for $t'$ gives:

$$\begin{align} t' &= \gamma \frac{L_0}{v} \left( 1 - \frac{v^2}{c^2} \right ) \\ &= \frac{L_0}{v} \frac{1 - \frac{v^2}{c^2}}{\sqrt{1 - \frac{v^2}{c^2}}} \\ &= \frac{L_0}{v} \sqrt{1 - \frac{v^2}{c^2}} \\ \end{align}$$

Incidentally $vt'$ is the length of the train in the observer's frame, $L$, so:

$$ L = L_0 \sqrt{1 - \frac{v^2}{c^2}} $$

as Stan mentioned in his answer, though I'll stick with my expression for $t'$ above, and rearrange it to get:

$$ \frac{1}{v^2} = \frac{t'^2}{L_0^2} + \frac{1}{c^2} $$

Plug in $t'$ = 780ns and $L_0$ = 500m and we get $v$ = 2.717 $\times$ 10$^8$m/sec or 0.906$c$.

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