Question regarding Lorentz Transformation and Space Contraction- Contradiction I stumbled upon this question regarding Special relativity- and have seemed to reach a contradiction.
I am trying to find the distance that the ball travels
I am obviously not looking for the numerical answer, but I'm trying to understand what should be my intuition when looking at these types of problems.

The train is moving at a velocity of $c/2$ relative to earth, while the ball is moving at a velocity of $c/3$ relative to the train.
The train has a proper length of $L_0$.
Now, when trying to find the distance that the ball travels regarding earth, I see two approaches:

*

*If we set $t=t'=0$ as the time where the ball is at the back if the train ($x=x'=0$), we find that the time he travels in the train's frame of reference is $3L_0/c$ and the distance is $L_0$. Using Lorentz transformation we find the distance traveled is $5L_0/\sqrt{3}$ in the earth frame of reference.


*In the train's frame, the ball is moving at a speed of $c/3$ a distance of $L_0$. Using the proper length we are able to calculate the train's length in the earth's frame as $\sqrt{3}L_0/2$. Seeing as the ball starts at the back of the train (in both frames) and reaches the front of the train, this is the balls distance traveled.
What am I missing here? Which approach should I use and where does this contradiction come from?
Thanks a lot in advance!
 A: When dealing with problems in Special Relativity its best to deal with individual events. So let's consider two events here: the ball leaves the "back end" of the train, and the ball arrives at the "front end" of the train.
Now let's see what we know: The ball leaves the back end ($x=0=x'$) at $t=0=t'$, and the front end of the train is at $x'=L_0$, the proper length. Using just this information, we can figure out everything else. For starters, in the train's rest frame, the ball covers a distance of $L_0$ with a speed of $c/3$ so, as you point out, the ball will hit the front end at $$t' = \frac{3L_0}{c}.$$ To make this clearer, let's make a table:
\begin{array} {|c|c|}\hline \textbf{Event} & \text{Train Frame} & \text{Earth Frame} \\ \hline \text{Ball leaves the back} &\,\,\quad t'=0,\quad\quad\,\, x'=0 \quad\quad & t=0, \quad x=0 \\ 
\hline \text{Ball arrives at the front} & t'=3L_0/c, \quad x'= L_0 & t = {\color{red}?}, \quad x = {\color{red}?} \\ \hline  \end{array}
We can now use the information we have to find both $t$ and $x$, the coordinates of the second event ("Ball arrives at the front of the train") in the Earth Frame. Let's write the Lorentz Transformations in terms of the difference of the events: \begin{aligned}\Delta x &= \gamma (\Delta x' + v \Delta t')\\ ~\\\Delta t &= \gamma \left( \Delta t' + \frac{v}{c^2}\Delta x'\right)\end{aligned}
From the table, you should be able to see that $\Delta t' = 3L_0/c$ and $\Delta x' = L_0$. Given that $v=c/2$, you can show that \begin{aligned}\Delta x &= \frac{5}{\sqrt{3}}L_0\\ \Delta t &= \frac{7}{\sqrt{3}}\frac{L_0}{c} \end{aligned}
So your first approach is correct, the distance between the events in the Earth frame is indeed $5L_0/\sqrt{3}$. However, we can now see why the second approach is wrong: the ball doesn't just cover the length of the train. The front wall of the train is also moving! Therefore, the distance covered by the ball in the Earth frame is given by $$\text{Length of the train (in Earth frame)} + \text{Distance covered by the front wall (in Earth frame)}$$
Clearly (as you pointed out) $$\text{Length of the train (in Earth frame)} = \frac{L_0}{\gamma} = \frac{\sqrt{3}}{2}L_0,$$ and the distance the front wall covers in the time $\Delta t$ is $$\text{Distance covered by the front wall (in Earth frame)} = v\Delta t = \frac{7}{2\sqrt{3}}L_0.$$
Adding them up, you can see that $$\text{Distance covered by ball in Earth frame} = \frac{L_0}{2}\left(\sqrt{3} + \frac{7}{\sqrt{3}}\right) = \frac{5}{\sqrt{3}}L_0,$$ as you'd imagine, since it's exactly what we calculated above.
A: You would end up in the same contradiction (wich is not) in Newtonian mechanics. Position is relative in both newtonian mechanics and Special Relativity. If i am moving relative to you you would see me covering a distance, but i would see myself standing still covering no distance.
