Ladder paradox and reversal of time order

Question

I have a very specific question regarding the ladder paradox, that is not answered in any of the other threads regarding the topic.

Assume a car and a garage. In the garage's intertial frame of reference (IFR) the car moves towards it with velocity $v$ in positive $x$-direction. The garage has proper length $L$.

The car also has proper length $L$ and in its IFR it has an initial distance $d$ to the garage.

Let E1 be the event: The front of the car crosses the back of the garage.

Let E2 be the event: The back of the car crosses the front of the garage.

My assignment is to show that in the car's IFR E1 happens before E2 but in the garage's IFR this order is revered.

Well, I have found out that the order is never reversed, if both car and garage have the same proper length, i.e. in their respective rest frames they each have the length $L$.

I have solved this both algebraically as well as geometrically and now I wonder, if I have it fundamentally wrong somewhere or my professor has made a mistake somewhere.

Let me explain how I calculated. First I write down the world lines of car and garage in their rest frames. Then I transform them into the respective other IFR using Lorentz transformations, which are inverse to each other.

In the car frame, the car's front and back have the respective world lines: $$a_f(t)=\begin{pmatrix} ct\\-d\\ \end{pmatrix}, \quad a_b(t) =\begin{pmatrix} ct\\-d-L\\ \end{pmatrix}.$$

In the garage frame, the garage's front and back have the respective world lines: $$\tilde{g}_f(\tilde{t})=\begin{pmatrix} c\tilde{t}\\ 0\\ \end{pmatrix}, \quad \tilde{g}_b(\tilde{t})=\begin{pmatrix} c\tilde{t}\\ L\\ \end{pmatrix}.$$

Now I transform them into the respective other IFR:

$$g_f=\begin{pmatrix} \gamma & -\beta\gamma \\ -\beta\gamma & \gamma \\\end{pmatrix}\tilde{g} _f =\begin{pmatrix} \gamma c\tilde{t}\\ -\beta\gamma c\tilde{t}\\ \end{pmatrix},$$

$$g_b=\begin{pmatrix} \gamma& -\beta\gamma\\ -\beta\gamma& \gamma\end{pmatrix} \tilde{g} _b =\begin{pmatrix} \gamma c\tilde{t}-\beta\gamma L\\ -\beta\gamma c\tilde{t}+\gamma L\end{pmatrix},$$

$$\tilde{a} _f=\begin{pmatrix} \gamma& \beta\gamma\\ \beta\gamma& \gamma\end{pmatrix} \begin{pmatrix} ct\\-d\\ \end{pmatrix} =\begin{pmatrix} \gamma ct-\beta\gamma d\\ -\beta\gamma ct-\gamma d\end{pmatrix},$$

$$\tilde{a} _b=\begin{pmatrix} \gamma& \beta\gamma\\ \beta\gamma& \gamma\end{pmatrix} \begin{pmatrix} ct\\-d-L\\ \end{pmatrix} =\begin{pmatrix} \gamma ct-\beta\gamma(d+L)\\ -\beta\gamma ct-\gamma (d+L) \end{pmatrix}.$$

In the garage's IFR E2 happens when $$0=v\tilde{t}_2-\gamma(d+L)\leftrightarrow\tilde{t}_2=\frac{\gamma}{v}L+\frac{\gamma}{v}d.$$ E1 happens when $$L=v\tilde{t}_1-\gamma d \leftrightarrow\tilde{t}_1=\frac{1}{v}L+\frac{\gamma}{v}d\geq\tilde{t}_2.$$

In the car's IFR the situation is the same. E2 gives $$t_2=\frac{1}{v}L+\frac{1}{v}d,$$ and E1 gives $$t_1=\frac{\gamma}{v}L+\frac{1}{v}d\geq t_2.$$

So, in both IFRs the order of events is the same. The same can be seen in the Minkowski diagram I have made with GeoGebra for this scenario. It shows the rest frame of the garage with its world lines being the y-axis and the red dotted line. The car's IFR axes are the solid blue lines and it's world lines are dashed in blue. The proper time periods are the red and black arrows for the garage IFR and car IFR respectively. One can see that, albeit the time period between E1,2 differs for each observer, the order in which they happen stays the same.

Either I have made a mistake somewhere or my professor has made one. But I had better be sure.

Answer 1

That the time ordering must be opposite seems pretty straightforward to me. Let the garage frame have the unprimed coordinate system while the car frame has the primed coordinate system.

Since $l_{car} = \frac{L}{\gamma} \lt L$, it must be that E2 occurs before E1 in the unprimed system.

Since the car fits inside the garage in these coordinates, the back of the car must be inside the garage before the front of the car crosses the back of the garage.

On the other hand, since $l'_{garage} = \frac{L}{\gamma} \lt L$, it must be that E1 occurs before E2 in the primed system.

Since the garage is shorter than the car in these coordinates, the front of the car must cross the back of the garage before the back of the car crosses the front of the garage.

By the way, kudos for drawing a spacetime diagram! I wish I had the time to go over your work closely. But, in this case, it's easy enough to show this result with the Lorentz transformation.

Stipulate that, for event E0, the coordinates are

$$x_0 = 0,\quad t_0 = 0$$

$$x'_0 = 0,\quad t'_0 = 0$$

where E0 is the event the front of the car crosses the front of the garage. It follows that

$$x_1 = L,\quad t_1 = L / v$$

$$x_2 = 0,\quad t_2 = L / \gamma v $$

thus

$$t_2 \lt t_1$$

Now, let's find the time coordinates of these events in the primed system:

$$t'_1 = \gamma\left(t_1 - \frac{v}{c^2}x_1\right) = \frac{L}{\gamma v}$$

$$t'_2 = \gamma\left(t_2 - \frac{v}{c^2}x_2\right) = \frac{L}{v}$$

thus

$$t'_1 \lt t'_2$$

and so the time ordering is reversed in the primed system. You should check that the two events of interest have space-like interval (time ordering of events with time-like interval is invariant).

$$\Delta s^2 = (c\Delta t)^2 - (\Delta x)^2 < 0$$