Suppose a car is approaching a railroad crossing at a high relativistic speed, while on the tracks, a square rail car is also approaching the crossing at the same speed. (The road and the railroad meet at a right angle.)

In the rest frame of the car, the rail car is approaching diagonally, and is contracted diagonally along its line of approach, making a lozenge shape. However, the tracks are approaching the automobile from directly ahead, so the separation between the tracks is decreased in the auto's rest frame, but the angle of the track is not.

This seems like a contradiction, because all of the train car's wheels are on the track. The sides of the train car have to be parallel to the track, but, in the rest frame of the automobile, the train car and the tracks are contracted along different axes. A lozenge-shaped train car won't have sides parallel to the track.

So how is this contradiction resolved? It is clear that there is no problem in the rest frame of the tracks or the rail car, but events also have to be consistent in the frame of the automobile.

I thought up this question while thinking about an example in the Epstein book Relativity Visualized about a box falling relativistically onto a conveyor belt also moving at relativistic speeds. It seems like it might be a textbook question, but I don't recall seeing it anywhere, and I'm having trouble figuring out the answer. My suspicion is that I should be thinking of moving frames of reference, rather than moving objects, but I am unclear on how that plays out where the rail car meets the tracks.

Diagram of track, auto, and rail car.

In this picture, the automobile is supposed to be the same size in both drawings. A drawing program would have been tidier but slower.

  • $\begingroup$ Have you accounted for aberration of the tracks at varying angles? $\endgroup$
    – Asher
    Commented Oct 12, 2015 at 6:00
  • $\begingroup$ @asher I'm not sure what you mean, but if you are talking about the aberration of the light coming from the tracks, then, no, that is not what this question is about. I am only talking about Lorenz effects. $\endgroup$ Commented Oct 12, 2015 at 12:18
  • $\begingroup$ Maybe later I'll try to come up with an answer, but for now here's some advice that might work. In this type of problem the best solution is usually to draw a spacetime diagram, but since this is 2D that won't be easy. Instead, write the explicit trajectories of all the objects in the tracks' frame and transform them to the train's frame using the appropriate Lorentz transformation. $\endgroup$
    – Javier
    Commented Oct 12, 2015 at 16:35

2 Answers 2


Your problem is that you're assuming that the train moves at $45º$ in the car's frame, but this is not true! One could probably conjure up an "intuitive" reason based on length contraction and so forth, but since I was never very good with those I'll just show you some of the math.

Like I said in the comments, we'll want to write the position of everything as a function of time. In the ground's rest frame, place an $x$ axis on the bottom rail and an $y$ axis pointing upwards, parallel to the car's velocity. The train is then moving in the negative $x$ direction with speed $v$. Let the proper separation between the tracks be $L$. I will work in units with $c=1$ to simplify a bit. Since there's only one speed involved in this problem, define $\gamma = 1/\sqrt{1-v^2}$.

The bottom track has $y=0$ and the top track has $y=L$. The train has four corners; call the top-left one corner $1$, and proceed with $2$, $3$, $4$ clockwise (I'm sorry, I can't make a drawing right now). Their trajectories in the ground frame are

$$\begin{align} (x_1,y_1) &= (-vt, L) & (x_2, y_2) &= (L/\gamma -vt, L) \\ (x_4, y_4) &= (-vt, 0) & (x_3, y_3) &= (L/\gamma -vt, 0) \end{align}$$

I have chosen $t$ so that at $t=0$ the left side of the train touches the $y$ axis. Also note that I've already taken into account length contraction; this one is easy because it's just 1D movement.

Now let's take a primed frame moving in the positive $y$ direction with speed $v$. The Lorentz transformations and their inverses are

$$\begin{align} t' &= \gamma(t-vy) & t &= \gamma(t'+vy') \\ x' &= x & x &= x' \\ y' &= \gamma(y-vt) & y &= \gamma(y'+vt')\end{align}$$

Now we need to transform everything to the primed frame. I'll skip the calculations, but remember that you need to use the transformations and their inverses. When finding the $(x',y')$ coordinates of the train's corners you want to use the transformations that give the primed coordinates as a function of the unprimed ones (left column), but then you'll have them written as a function of $t$; use the right column to write $t$ in terms of the primed coordinates.

We want to know the train's shape in this frame, so we need to evaluate the positions of its corners at some fixed time $t'$; obviously using $t'=0$ is the easiest choice. We get the following result:

$$\begin{align} (x'_1,y'_1) &= (-v^2 L, L/\gamma) & (x'_2, y'_2) &= (L/\gamma -v^2 L, L/\gamma) \\ (x'_4, y'_4) &= (0, 0) & (x'_3, y'_3) &= (L/\gamma, 0) \end{align}$$

This should be reassuring, because the $1-2$ and $3-4$ sides are still horizontal, and they're separated by $L/\gamma$; precisely the separation of the tracks in this frame! The train's shape has been distorted into a parallelogram instead of a rhombus, as you can check by making up some value for $v$ and plotting these four points. Why does this happen?

I anticipated this at the beginning of the answer; the train is not moving at $45º$. I haven't shown you the full expression for its trajectory in the primed frame, but what happens is that the horizontal velocity is smaller than the vertical velocity by a factor of $\gamma$. Therefore, length contraction happens at some weird angle instead of along its diagonal, so it becomes a parallelogram. Looking at the formulas, we see that the $x'$ velocity is smaller because of time dilation. While the $x$ coordinate doesn't change, time does, so unless the Lorentz transformation provides an appropiate $\gamma$ factor (as it does for $y$), the speed will become smaller.

  • 1
    $\begingroup$ Excellent answer. The first sentence was eye-opening, and of course you are right about the benefits of actually sitting down and doing the calculations. I do now think it is possible to visualize what happens in a 3D space-time diagram, but I'm not sure I can draw it. I may try a bit later. $\endgroup$ Commented Oct 17, 2015 at 3:41

Javier's answer is nearly right and his parallelogram is the correct shape, but the key to understanding this is rotation. The rhombus in the original diagram will naturally become a parallelogram with two of its sides parallel to the track simply by rotating it a little while the exact same degree of length contraction is applied to it at the exact same angle as before. But how can there be any rotation involved in this? Well, if you're considering the automobile to be stationary and the rails are moving towards it, the square rail car would have started out with two of its sides parallel to the rails and the other two perpendicular to them before it was accelerated up to high speed along the rails. In the process of that acceleration the part of the rail car over the further-away rail (meaning further away from the automobile) will move off first with the side over the nearer rail only following later, and this is what turns it into a parallelogram, crucially with two sides remaining parallel to the track. That act of acceleration provides a hidden rotation, as you'll find if you rotate your rhombus, because just by rotating it a bit you will make it take up the exact same shape as the parallelogram. The more you accelerate the rail car, the greater the amount of hidden rotation you're applying to it.


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