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

Question

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?

Answer 1

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.

Answer 2

The George Gamow "Mr Tompkins" description of SR is wrong

As other posters have already said, SR does NOT predict that an observer sees moving objects to be Lorentz contracted. This belief appears to originate with a mistake in George Gamow's popular "Mr Tompkins" books, which spread into the education sector like an infection. It's incorrect. Gamow was a respected physicist, but any decent physics teacher would have given him a fail grade in special relativity.

To understand how this really works, you have to know a little bit about signal propagation timelags.

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Suppose that the speed of light was "really" fixed wrt the tunnel. Once the train was in the tunnel, a tunnel observer at the first door would see a Doppler recession redshift of E'/E = c/(c+v) , and also a Lorentz redshift. Multiplied together this gives the physical SR prediction of

E'/E = c/(c+v) × gammaShrink = SQRT[(c-v)/(c+v)]

But the observation timelag on the more distant front end of the train makes its apparent position out-of-date – it seems closer than its real position, and as a result, the train seems shorter, by Len'/Len = c/(c+v). But SR also says that the train is moving and Lorentz-contracted, so the total length that appears in a photograph taken by this observer is

Len'/Len = c/(c+v) × gammaShrink = SQRT[(c-v)/(c+v)]

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Notice that the proportional change in apparent photographed length is identical to the proportional change in apparent photographed frequency. This equivalence is pretty general across different theories, and might even be a universal law.

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Suppose now instead that the speed of light was "really" fixed wrt the train. Once the train was in the tunnel, a tunnel observer at the first door would see a different Doppler recession redshift due to signal propagation timelags, of E'/E = (c-v)/c , and also a Lorentz blueshift (because we're now saying that the train is stationary and the tunnel is moving, and time-dilated). Multiplied together this gives the physical SR prediction of

E'/E = (c-v)/c × gammaStretch = SQRT[(c-v)/(c+v)]

Again, the observation timelag on the more distant front end of the train makes its apparent position out-of-date, and it seems closer than its real position, making it seem shorter. But the new propagation-based prediction is now Len'/Len = (c-v)/c. But SR also says that the tunnel observer is moving and their reference-rulers are length-contracted, so they see the train to be Lorentz-elongated, giving a total photographed length of:

Len'/Len = (c-v)/c × gammaStretch = SQRT[(c-v)/(c+v)]

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Regardless of which frame the speed of light is supposed to be fixed in, train, tunnel, or Something Else, and regardless of whether the tunnel observer's photograph is supposed to include a Lorentz contraction or Lorentz expansion, the resulting photograph is identical.

The SR photo cannot tell you whether you are seeing a contraction or an elongation, or no effect at all. The choice of supposed frame for the propagation of light makes zero difference to the actual physics. That's the genius of special relativity – it declares that light can be treated as travelling at fixed speed in a flat fixed frame, but it makes it absolutely impossible for anyone to work out from what they see, which frame that is.

As to whether the train is seen to be longer than the tunnel or shorter ... it depends only on where the trackside observer is standing. If the trackside observer is standing by the first tunnelmouth, the receding train appears shortened, and is photographed as fitting entirely inside the tunnel with room to spare. If the trackside observer is near the second tunnelmouth, then because the signal timelag effect is stronger than the Lorentz effect, the approaching train appears longer, and appears in a photo to be longer than the tunnel, and sticking out from both ends.

The physics community didn't seem to get this right until the 1960s, when James Terrell's paper triggered a flurry of papers in the American Journal Of Physics, all agreeing the the Lorentz contraction wasn't directly visible, and that some people had been misinterpreting the linguistic descriptions of SR rather than treating it as a 100% "raw" geometrical exercise.

Answer 3

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 $\;v >0$ \begin{equation} t'_\mathrm{F} =-\dfrac{\gamma \,v\,}{c^2}\,L < 0 = t'_\mathrm{B} \tag{06} \end{equation} that is in the system of the Train $\;\mathrm{S'}\;$ the event $\;\mathrm{F}\;$ happens before event $\;\mathrm{B}\;$ by a time interval \begin{equation} \vert \Delta t'_\mathrm{FB} \vert =\vert t'_\mathrm{F}-t'_\mathrm{B} \vert=\dfrac{\gamma \,v\,}{c^2}\,L \tag{07} \end{equation} 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
\begin{equation} \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} \end{equation} 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
\begin{equation} 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} \end{equation} 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
\begin{equation} \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} \end{equation} 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 \begin{equation} \gamma = 1.25\,,\quad \dfrac{\,1\;}{\gamma}=0.80 \tag{NE.3} \end{equation}

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 $\;\gamma (v / c)(L/c)=1.25\times 0.60 \times 10^2\,m/(3\cdot 10^8\,m/sec) = 0.25\cdot 10^{-6}sec =0.25\mu s$.

Answer 4

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:

http://www.youtube.com/watch?v=Xrqj88zQZJg "Einstein's Relativistic Train in a Tunnel Paradox: Special Relativity"

It is not difficult to realize that trapping long objects inside short containers drastically violates the law of conservation of energy. The trapped object, in trying to restore its original volume ("spring back to its natural shape"), would produce an enormous amount of work the energy for which comes from nowhere.

At 9:01 in the above video Sarah sees the train falling through the hole, and in order to save Einstein's relativity, the authors of the video inform the world that Adam as well sees the train falling through the hole. However Adam can only see this if the train undergoes an absurd disintegration first, as shown at 9:53.

Clearly we have reductio ad absurdum: An absurd disintegration is required – Adam sees it, Sarah doesn't. Conclusion: The underlying premise, Einstein's 1905 constant-speed-of-light postulate, is false.