I have great trouble in understanding simultaneity in special relativity. Let me illustrate it with a concrete example.
Assuming there is a train, its two end points are $A$ and $B$, the length of the train is $x$. The train moves at speed $v$. Assuming the train is moving in the direction from $A$ to $B$.
For a ground observer observing the train movement, he notices that two lightnings strike simultaneously at $A$ and $B$ when the middle part of the train $O'$ passes through right in front of him. In other words, the ground observer is located at the middle part of the train($O'$) when the lightnings strike simultaneously at $A$ and $B$.
Now there is another moving observer sitting inside the train and he sits right in the middle of the train ($O$, equidistant from $A$ and $B$). Does this moving observer think that the lighting happens at the same time? If no, how much time has passed before he notices lightning at $B$, after he had observed lightning at $A$?
Best Answer
The only safe way for beginners to answer questions in special relativity is to sit down with a large sheet of paper and work through the Lorentz tranformations:
$$\begin{align} t' &= \gamma (t - \frac{vx}{c^2}) \\ x' &= \gamma (x - vt) \end{align}$$
Let's be absolutely clear what the tranformations tell us. If we use a coordinate system $(t, x)$ to label spacetime points, and another observer moving at constant velocity $v$ relative to us uses another coordinate system $(t', x')$, the transformations convert our labels $(t, x)$ to the other observer's labels $(t' x')$.
So to answer your question we take the two spacetime points labelling the ends of the train and apply the transformations. This tells us where those two points are in the moving observer's coordinates.
In our frame at $t = 0$ the middle of train is at $(0, 0)$, so the front of the train is at $(0, d/2)$ and the rear of the train is at $(0, -d/2)$ (I've called the length of the train $d$ to avoid confusion with the $x$ coordinate):
To find the potition of the front of the train in the primed frame we just feed $t = 0$ and $x = d/2$ into the Lorentz transformations:
$$\begin{align} t' &= \gamma (- \frac{vd}{2c^2}) \\ x' &= \gamma \frac{d}{2} \end{align}$$
So in the moving frame the lightning strike at front of the train is at $(-\gamma\tfrac{vd}{2c^2}, \gamma\tfrac{d}{2})$. I won't go through the details, but same calculation puts the lightning strike at the end of the train is at $(\gamma\tfrac{vd}{2c^2}, -\gamma\tfrac{d}{2})$.
So the answer is that the observer on the train sees the lightning strike the front of the train at $t' = -\gamma\tfrac{vd}{2c^2}$ and the rear of the train at $t' = \gamma\tfrac{vd}{2c^2}$. The time between the lightning strikes is $\gamma\tfrac{vd}{c^2}$.
Quick footnote
Rereading my answer it's just occurred to me that I've called the length of the train $d$ in the rest frame of the track. The length of the train for the observers on it will be greater - you can use the Lorentz transformations to calculate this too.
Length of the train
Re Graviton's comment, the easiest way to calculate the length of the train in the train's rest frame is to work backwards. Let's call the length of the train in its rest frame $\ell$, and we'll choose our zero time so that the rear of the train is at $(0, 0)$ and the front is at $(0, \ell)$. To transform from the train frame to the track frame we just use the Lorentz transformations as before, but in this case the velocity is $-v$ because if the train is moving at $v$ wrt to the track then the track is moving at $-v$ wrt the train.
When we do the transformation $(0, 0)$ just goes to $(0, 0)$ so we just need to work out where $(0, \ell)$ is in the track frame. Plugging in $t = 0$ and $x = \ell$ we find the point in the track frame is:
$$\begin{align} t &= \gamma (t' - \frac{(-v)x'}{c^2}) = \gamma\frac{v\ell}{c^2} \\ x &= \gamma (x' - (-v)t) = \gamma\ell \end{align}$$
So in the track frame the front of the train is at $(\gamma\tfrac{v\ell}{c^2}, \gamma\ell)$. But we don't want to know where the front of the train is at time $t = \gamma\tfrac{v\ell}{c^2}$, we want to know where is was at $t = 0$. So we take our value for $x$ at time $\gamma\frac{v\ell}{c^2}$ and subtract off the distance moved in time $\gamma\frac{v\ell}{c^2}$, which is just the time multiplied by the velocity. This gives us the value for $x_0$:
$$ x_0 = \gamma\ell - \gamma\frac{v\ell}{c^2} v $$
The rest is just algebra. We write the expression out in full to get:
$$\begin{align} x_0 &= \ell \left( \frac{1 - \frac{v^2}{c^2}} {\sqrt{1 - \frac{v^2}{c^2}}} \right) \\ &= \ell \sqrt{1 - \frac{v^2}{c^2}} \\ &= \frac{\ell}{\gamma} \end{align}$$
And since the rear of the train is at $x = 0$ at time zero and the front of the train is at $x = x_0$ at time zero the length of the train is just $x_0$ so:
$$ d = \frac{\ell}{\gamma} $$
At all speeds $> 0$ the value of $\gamma > 1$, so the length of the train as observed from the track is less than the length of the train in its rest frame i.e. the train is shortened. This is the Lorentz contraction.