# Special Relativity - Regarding the Simultaneity of Events During the Train Paradox

My questions concerns that classic train paradox, wherein there is a train and a tunnel of equal length, and the train is traveling and some fraction of the speed of light towards the tunnel.

According to the Special Theory of Relativity, an observer outside the tunnel will see the train length contracted (Lorentz Contraction), whereas an observer inside will see the tunnel contracted.

Additionally, suppose that there were doors at the ends of the tunnel and that the observer outside the tunnel closed both doors instantaneously when he/she saw that the train was completely inside the tunnel.

The classic resolution of this paradox invokes the non-simultaneity of events, explaining that the observer in the train sees the far door close first, and then, once the train has begun to exit the tunnel, looks back to see the door at the beginning close. Thus, both observers agree that the train does not touch the door when they are closed for an instant.

Now my questions.

Why is it that the observer on the train sees the far door close first? It seems to me that the information coming from the far door would reach the observer on the train only after the information from the other door is reach.

Under this interpretation, the observer on the train would observe the train getting hit by the doors. What if, by some means, this event can be explained in terms of a stationary observer too? Everyone always concludes that the train remains untouched by the doors, but really the only condition that needs to be met is that both observers must agree. Why can't they both agree that the train was hit?

So, to summarize.

1. Why is the door that is farther away from the train observed to close first?

2. Why can't the other possible conclusion (both see a hit train) be observed?

• the other solution is certainly plausible with any sort of delay, – shai horowitz Oct 12 '16 at 3:48
• I want to say that the only factor here is whichever light from the doors reaches the observer first. (So unless the observer is equidistant from the doors when the light meets in the middle as well, the observer will always see one before the other). Is relativity even needed here? – Yogi DMT Oct 12 '16 at 12:41

Why is the door that is farther away from the train observed to close first?

By observe, in SR, we don't mean see, we mean essentially to record the time and place of events according to rods and (synchronized) clocks at rest. For example,

• Assume that, at the front and back of the train, there are identical clocks that are synchronized in the frame of the train.
• Further assume that, at the ends of the tunnel, there are identical clocks that are synchronized in the frame of the tunnel.

By the relativity of simultaneity, the clocks on the train are not synchronized in the frame of the tunnel where it is observed that the clock at the front is behind the clock at the back.

Symmetrically, the clocks in the tunnel are not synchronized in the frame of the train where it is observed that the clock at the entrance of the tunnel is behind the clock at the exit.

• Finally, assume that the length contracted train in the frame of the tunnel just fits within the tunnel.

Thus, there is a moment, according to the tunnel clocks, that the contracted train is completely within the tunnel.

But remember, in the frame of the tunnel, the train clocks are not synchronized. In particular, since the clock at the front of the train is observed to be behind the clock at the back, it must be the case that, as recorded by the clocks on the train, the door at the exit of the tunnel closes earlier than the door at the entrance.

That is, in the frame of the train, the front of the train just reaches the exit, the door there closes for an instant without hitting the train and there is still a trailing portion of the train that is yet to enter the tunnel.

When the back of the train just clears the entrance of the tunnel, the door at the entrance closes for an instant without hitting the train and there is a leading portion of the train that has exited the tunnel.

As always, I recommend that you draw a spacetime diagram of this sequence of events to get a better 'picture' of how this works.

• +1 "By observe, in SR, we don't mean see" This, this, and so much this. While the usual special relativity 101 way of explaining things by looking at light rays moving at speed $c$ in every reference frame makes it easier to visualize things, it seems to sometimes make people confuse things and think that relativity of simultaneity is due to observers seeing things at different times because the speed of light is infinite. ... – JiK Oct 12 '16 at 14:49
• ... That's why I'd recommend getting used to the approach of Lorentz transformation in a more abstract way early on, which can be neatly visualized by spacetime diagrams, as also said in this answer. My physics degree contained only about three lectures of SR, in an elementary course, and what I learned in those three lectures would have been so much easier if I had been shown a spacetime diagram. So please OP (and everyone else), draw a spacetime diagram. (end of rant) – JiK Oct 12 '16 at 15:24

Let the systems-frames : $\;\mathrm{S}\;$ the system of the Tunnel and $\;\mathrm{S'}\;$ the system of the Train. The train is moving with speed $\;v\;$ from the negatives to the positives of the $\;x-$axis of $\;\mathrm{S}\;$ as in Figure 1.

Now, in our case we have the following two simultaneous events in $\;\mathrm{S}\;$ : \begin{align} \mathrm{B} = & \;\text{the back end of the train is at the entrance of the tunnel and the back door} \tag{01.B}\\ &\; \text{ of the tunnel is closed instantaneously.} \nonumber\\ \mathrm{F} = & \;\text{the front end of the train is at the exit of the tunnel and the front door} \tag{01.F}\\ &\; \text{of the tunnel is closed instantaneously.} \nonumber \end{align}

So, for the tunnel observer the rest tunnel and the moving train have the same length, let it be $\;L\;$ as in Figure 1. Suppose that the tunnel observer sets his/her space-time origin on event $\;\mathrm{B}$ so that for the coordinates we have : \begin{align} \left(x_\mathrm{B}\,,t_\mathrm{B}\right) & =\left(0\,,0\right) \tag{02.B}\\ \left(x_\mathrm{F}\,,t_\mathrm{F}\right) & =\left(L\,,0\right) \tag{02.F} \end{align} To determine the space-time coordinates of these two events in the system of the Train $\;\mathrm{S'}\;$ we'll use the Lorentz Transformation expressed with differences \begin{align} \Delta x' & =\gamma\left(\Delta x-v\,\Delta t\right) \tag{03.1}\\ \Delta t' & =\gamma\left(\Delta t-\dfrac{\,v\,}{c^2}\Delta x\right) \tag{03.2} \end{align} For convenience, suppose that the train observer sets his/her space-time origin on event $\;\mathrm{B}\;$ also, as in Figure 2, so for the coordinates of the events in the system of the Train $\;\mathrm{S'}\;$ we have : \begin{align} \left(x'_\mathrm{B}\,,t'_\mathrm{B}\right) & =\left(0\,,0\right) \tag{04.B}\\ \left(x'_\mathrm{F}\,,t'_\mathrm{F}\right) & =\left(???\,,???\right) \tag{04.F} \end{align}

So \begin{align} \Delta x'_\mathrm{FB} & =\gamma\left(\Delta x_\mathrm{FB}-v\,\Delta t_\mathrm{FB}\right) \Longrightarrow x'_\mathrm{F}-x'_\mathrm{B}=\gamma\left[\left(x_\mathrm{F}-x_\mathrm{B}\right)-v\,\left(t_\mathrm{F}-t_\mathrm{B}\right)\right] \Longrightarrow \nonumber\\ x'_\mathrm{F} & =\gamma\,L \tag{05.1}\\ \Delta t'_\mathrm{FB} & =\gamma\left(\Delta t_\mathrm{FB}-\dfrac{\,v\,}{c^2}\Delta x_\mathrm{FB}\right)\Longrightarrow t'_\mathrm{F}-t'_\mathrm{B}=\gamma\left[\left(t_\mathrm{F}-t_\mathrm{B}\right)-\dfrac{\,v\,}{c^2}\,\left(x_\mathrm{F}-x_\mathrm{B}\right)\right] \Longrightarrow \nonumber\\ t'_\mathrm{F} & =-\dfrac{\gamma \,v\,}{c^2}\,L \tag{05.2} \end{align} But, since for the speed $\;v >0$ $$t'_\mathrm{F} =-\dfrac{\gamma \,v\,}{c^2}\,L < 0 = t'_\mathrm{B} \tag{06}$$ that is in the system of the Train $\;\mathrm{S'}\;$ the event $\;\mathrm{F}\;$ happens before event $\;\mathrm{B}\;$ by a time interval $$\vert \Delta t'_\mathrm{FB} \vert =\vert t'_\mathrm{F}-t'_\mathrm{B} \vert=\dfrac{\gamma \,v\,}{c^2}\,L \tag{07}$$ Since in the system of the Train $\;\mathrm{S'}\;$ the front end and the back end of the train stand at rest on the coordinates $\;x'_\mathrm{F} =\gamma\,L\;$ and $\;x'_\mathrm{B} =0\;$ respectively, for the length of the train in its rest frame we have as expected
$$\text{Length of the train in its rest frame}=\vert \Delta x'_\mathrm{FB} \vert =\vert x'_\mathrm{F}-x'_\mathrm{B} \vert=\gamma\,L \tag{08}$$ Now, a train observer on the back end of the train at the time moment that the event $\;\mathrm{F}\;$ is happening, that is on $\;x'_\mathrm{B} =0\;$ at $\;t'_\mathrm{F}=-\gamma\,v\,L/c^2 \;$, will meet the back entrance of the tunnel at time moment $\;t'_\mathrm{B}=0\;$, that is after the time interval $\;\vert \Delta t'_\mathrm{FB}\vert \;$ of equation (07). But, since the tunnel is moving with speed $\;v\;$ from the positives to the negatives of the $\;x'-$axis, the back end of the train and the back entrance of the tunnel at time moment $\;t'_\mathrm{F}\;$ are apart at a distance
$$v\, \vert \Delta t'_\mathrm{FB} \vert =\dfrac{\gamma \,v^2\,}{c^2}\,L= \gamma \left(1-\dfrac{1}{\gamma^2}\right)L= \left(\gamma-\dfrac{1}{\gamma}\right)L \tag{09}$$ This is the length of the back portion of the train outside the tunnel at the time moment $\;t'_\mathrm{F}$. This is also the length of the front portion of the train outside the tunnel at the time moment $\;t'_\mathrm{B}$, see Figure 2. On the other hand from this same Figure we have
$$\text{Length of the tunnel in the Train frame }\mathrm{S'} =\gamma\,L-\left(\gamma-\dfrac{1}{\gamma}\right)L=\dfrac{L}{\gamma} \tag{10}$$ as expected.

For a space-time Diagram see Figure 3.

Numerical Example

Let \begin{align} \dfrac{\,v\;}{c} & =0.60 \tag{NE.1}\\ L & =100\,m \tag{NE.2} \end{align} then $$\gamma = 1.25\,,\quad \dfrac{\,1\;}{\gamma}=0.80 \tag{NE.3}$$

The length of the train in its rest frame is $\;\gamma\,L = 1.25\times 100\,m= 125\,m\;$ contracted to $\;L = 100\,m\;$ in the tunnel frame.

The length of the tunnel in its rest frame is $\;L = 100\,m\;$ contracted to $\;L/\gamma = 80\,m\;$ in the train frame.

The length of the portion of the train in its rest frame outside the tunnel is $\;(\gamma-\gamma^{-1})L=(1.25-0.80)\times 100\,m= 45 \,m$.

The two simultaneous events in $\;\mathrm{S}\;$ are in $\;\mathrm{S'}\;$ apart by a time interval $\;\gamma (v / c)(L/c)=1.25\times 0.60 \times 10^2\,m/(3\cdot 10^8\,m/sec) = 0.25\cdot 10^{-6}sec =0.25\mu s$.

The paradox becomes absurdity if the train is longer than the tunnel. Actually special relativity predicts that unlimitedly long objects can be trapped inside unlimitedly short containers:

http://math.ucr.edu/home/baez/physics/Relativity/SR/barn_pole.html "These are the props. You own a barn, 40m long, with automatic doors at either end, that can be opened and closed simultaneously by a switch. You also have a pole, 80m long, which of course won't fit in the barn. [...] So, as the pole passes through the barn, there is an instant when it is completely within the barn. At that instant, you close both doors simultaneously, with your switch. [...] If it does not explode under the strain and it is sufficiently elastic it will come to rest and start to spring back to its natural shape but since it is too big for the barn the other end is now going to crash into the back door and the rod will be trapped in a compressed state inside the barn."

See, at 7:12 in the video below, how the train is trapped "in a compressed state" inside the tunnel: