Is this a fundamentally relativistic phenomenon?

This question was inspired by some silliness in other threads but is independent of that silliness.

Say that a train car sitting on a track is accelerated uniformly along its length if each point on the train car experiences the same positive acceleration $a(t)$ at each time $t$ as measured from the track's frame. (The track is not accelerated --- it remains in the same inertial frame.)

Clearly such an acceleration cannot change the length of the car in the track's frame, so its proper length (which has to be greater than its length in any other frame) must increase. That is, an observer on the moving car must say that the car has stretched. But there is a limit to how much you can stretch a train car, so beyond some velocity the train must snap. The snap should be observable to anyone, including an observer stationary with respect to the track.

Therefore we have what I will call the Curious Phenomenon:

If a train car reaches a sufficiently high velocity as a result of being accelerated uniformly along its length, then the train car must snap.

Note that the statement of the Curious Phenomenon (as opposed to the derivation of that phenomenon) has nothing to do with relativity. Note too that the phenomenon is in principle (though perhaps not in practice?) directly observable.

This leads me to two questions, which might or might not be the same question in disguise:

Question 1: Is there a clear conceptual explanation of the Curious Phenomenon based on classical mechanics without invoking relativity? Or does one really need relativity to explain this?
Question 2: Suppose we knew nothing about relativity, but had observed the Curious Phenomenon. Would the search for an explanation naturally lead to relativity in the same sense that say, a search for an explanation of the Michelson-Morley phenomenon could naturally lead to relativity?
• I do not fathom your logic. If the acceleration is uniform, why would the carriage snap, as long as it can move on the tracks? – anna v Mar 19 '16 at 6:36
• @annav: The velocity of the carriage is increasing. Its length in the track frame is staying constant. Therefore its proper length must be increasing. There is a limit to how far its proper length can increase. Another way to see the same thing: The front and back of the carriage begin accelerating at the same time in the track frame; therefore an observer on the carriage must say that the front began accelerating before the back. Therefore that observer will see the carriage as stretched. – WillO Mar 19 '16 at 6:38
• But uniform acceleration means at the same time too. In addition in special relativity there is length contraction, not increase, for the stationary observer. The movint one sees nothing. – anna v Mar 19 '16 at 6:44
• @annav: The assumption of uniform acceleration tells us that $L$ (the length in the track frame) does not change. If $L'$ is the proper length, we have $L'=L/\gamma$, where $\gamma$ is a Lorentz factor. The velocity is increasing, so the Lorentz factor must be decreasing. Therefore $L'$ is a constant divided by something that is decreasing. Therefore $L'$ is increasing. – WillO Mar 19 '16 at 6:46
• sorry, I cannot follow but I have to go on a trip soon. I am sure you are invertin the gamma relation. the proper length is the length of the moving carriage at its rest frame. bye – anna v Mar 19 '16 at 6:51

There is a clear conceptual explanation that happens all the the frame of one small train car.

The idea is that the operator of each car is handed instructions telling them when according to their clock to fire rockets on which parts of their car.

And they follow the instructions. And the instructions are handed to every single car. Clocks are synchronized and then the instructions get followed. When the instructions are labelled "Einstein simultaneous acceleration profile a(t)=blah, car #508" the person in the car is astounded to find out the rockets first fire at the same time, but their instructions say they are instructed to fire rockets harder on the front of the car before they fire the rockets harder on the back end of the car. These instructions, when followed, stretch out the car as well as generating an acceleration. And the mismatched timing of the increased thrusts rips the car apart.

They will think it was strange the instructions were labelled with the name "simultaneous acceleration" when they had to accelerate the front part before the back part. But the labels your boss sticks on your instructions isn't a physical cause. The physical cause is the rockets ripped the car apart.

The only place relativity came up is when you decided you wanted to have each car fire thrusters so that the whole thing accelerated in a way that was simultaneous to the inertial observer. But without relativity, no one would hand you those instructions to follow. So you wouldn't do the experiment, so you wouldn't observe the phenomena.

And if it's so vague to just say that there is some velocity at which cars break and it doesn't predict the velocity then it's not falsifiable.

• @WillO I edited. But I think you could have answered this just as well yourself, so maybe my answer was pointless. – Timaeus Mar 19 '16 at 12:19
• Thank you for your answer and for the pointer to the Bell Spaceship Paradox. I had not seen this before and I agree that it's the same phenomenon. Re your answer: What if we just order all rockets to fire for one instant (all simultaneous in the track frame)? This does not seem vulnerable to the objection that nobody would think of it without relativity. – WillO Mar 19 '16 at 13:20
• @WillO That's not a different scenario. An instant is just a small interval and three nearest cars are the same as one car with rockets in different parts. I'm saying that if you give relativistic instructions it rips the car apart. And if you give unrelativistic instructions then in a universe with SR the acceleration won't be simultaneous in the track frame. – Timaeus Mar 19 '16 at 13:24
• Thanks again. I hope I'm not being dense but let me try one more variant. Following Bell, the train consists of an engine and a caboose, each with a rocket, attached by a rubber band. The rockets begin firing simultaneously in the track frame and continue to fire, applying a force that is constant over time. This (I think) requires no relativistic instructions. Still the band snaps. Is there a nonrelativistic explanation? – WillO Mar 19 '16 at 14:04
• @WillO Doesn't work: objects in relativity aren't rigid. Extended bodies need force applied in a time interval in a region. If you want the same force applied for each track-time then the back and the front of the same car need to see different accelerations in their two different MCIRFs. The whole point is that the instructions so that the SR track IRF sees simultaneous acceleration requires the parts of each car to experience different accelerations in each part's MCIRF. The front and back disagree about the other one firing uniformly in time. So instructions tell them to do it differently. – Timaeus Mar 19 '16 at 23:38

As Timaeus says this is another version of the Bell's spaceship paradox and as such has been discussed many times over the years. Let me suggest a way that seems to me to clarify what is going on.

Consider two observers on the train, $A$ who is at the origin at time zero and $B$ who is some distance $d$ along the train at time zero. If the train is accelerating with constant proper acceleration $a$ then the positions of the observers in the track frame as a function of track frame time are given by:

\begin{align} x_A(t) &= \frac{c^2}{a}\left(\sqrt{1 + \left(\frac{at}{c}\right)^2} -1 \right) \\ x_B(t) &= x_A(t) + d \end{align}

This is a standard result that you'll find in e.g. chapter 6 of Gravitation. As you say in the question the spacing between the observers is constant in the track frame.

Now let's switch to the rest frame of observer $A$. The key thing you need to know is that for an observer with constant proper acceleration $a$ their spacetime geometry is described by the Rindler metric:

$$ds^2 = -\left(1 + \frac{ax}{c^2}\right)^2c^2dt^2 + dx^2 \tag{1}$$

Proving this is straightforward but tedious so rather than do it here I'll just refer you to the first hit that came up when I Googled it.

For our purposes the key feature of this metric is that it predicts there is time dilation comparable to that you'd find in a gravitational field. If we take $dx=0$ and use the fact that $ds^2 = -c^2d\tau^2$ equation (1) becomes:

$$\frac{d\tau}{dt} = 1 + \frac{ax}{c^2}$$

where $t$ is the time measured by observer $A$ and $\tau$ is the time measured by an observer at the position $x$. So in our scenario $A$ observes $B$'s time to be dilated by a factor of:

$$\frac{dt_B}{dt_A} = 1 + \frac{ad}{c^2}$$

I use the conventional term dilated, but actually $B$'s time runs faster than $A$'s. This matters because that means the acceleration of $B$ measured in $A$'s frame, call this $a_B$, is greater than the proper acceleration $a$ by a factor of $(dt_B/dt_A)^2$:

$$a_B = a\left(1 + \frac{ad}{c^2}\right)^2$$

So even though $A$ and $B$ have the same proper acceleration, $A$ observes $B$ to be accelerating away at $a^2d/c^2$.

And that's why the train stretches.

• Thanks very much for this. It is enlightening, though I'm not sure we need to worry about the Rindler metric, because it's enough to let the accceleration last for a finite time, and to look at the train before and after the acceleration ends. And I'm afraid I still don't know the answer to my question, which is whether it's possible to account for the snapping without relativity. – WillO Mar 19 '16 at 12:14
• @WillO: the fact that in $A$'s rest frame $B$ is accelerating away is a purely relativistic effect. I thought that was obvious! You should be worrying about the Rindler metric. Understand it is key to understanding accelerated motion in SR. – John Rennie Mar 19 '16 at 12:16
• Yes, I agree that the fact that in A's rest frame, B is accelerating away is a purely relativistic effect. I was asking whether it's a purely relativistic effect that in the track's frame, the train snaps. You could argue, I guess that these are not different things, so the question is answered --- but I'm left with a vague uncertainty about whether there might be some explanation for the snapping that doesn't go through the relativity. – WillO Mar 19 '16 at 12:22
• Also --- I do not think you need the Rindler metric to convince me that B is accelerating away from A. (Though this will not stop me from reading about it!) Unless I am mistaken, you only need the Lorentz contraction. If we halt the acceleration at any moment, then the part of the train that stretches from A to B has retained its length in the track frame, and must be Lorentz contracted in the track frame relative to its proper frame, therefore must have grown in proper length. No? – WillO Mar 19 '16 at 12:27
• @WillO: waving the Lorentz contraction about is an exceedingly dangerous tactic as it can so easily be applied where it isn't relevant. My point here is that I can show the stretching of the train explicitly starting from the metric. Perhaps you don't think that level of rigour is worth the effort, but I do. – John Rennie Mar 19 '16 at 12:30

Say that a train car sitting on a track is accelerated uniformly along its length if each point on the train car experiences the same positive acceleration a(t) at each time t as measured from the track's frame. (The track is not accelerated --- it remains in the same inertial frame.)

If I understand the above correctly, you're specifying that each point of the train car has the same coordinate acceleration as observed from the inertial reference frame (IRF) of the track and so, the world-lines of the points of the train car are congruent.

Now, in SR, an object cannot have uniform coordinate acceleration since, in that case, the speed of the object would eventually reach and then exceed $c$. So, let's further stipulate that the coordinate acceleration of the points of the train is of the form

$$a(t) = \alpha \left(1 - \frac{v^2(t)}{c^2} \right)^{3/2},\qquad t \ge 0$$

where $v(0) = 0$. That is to say, the acceleration begins at $t = 0$ and each point of the train car has constant proper acceleration $\alpha$ in the rest frame of the track.

It's easy to see with a spacetime diagram that, for $t \ge 0$, and in a momentarily comoving reference frame (MCRF) of any point of the train car, other points along the length of the train car have different velocity and proper acceleration - the points more forward are moving faster and their accelerometers read larger numbers while the points more rearward are moving slower and their accelerometers read smaller numbers.

In other words, there is no MCRF for the train car as a whole; from a MCRF of a point, the train car is expanding along the direction of the acceleration.

Note that the situation is vastly different in Galilean relativity where there is a MCRF for the train car as a whole.

But there is a limit to how much you can stretch a train car, so beyond some velocity the train must snap.

Not according to your setup. You've stipulated that "each point on the train car experiences the same positive acceleration a(t) at each time t".

Since this is the case, the train car does not snap by stipulation. Yet we know that there are MCRFs in which the ends of the train are moving with vastly different velocities and have vastly different accelerations which is clearly not physically reasonable.

So, it is this stipulation that you must closely examine. That is to say, one cannot observe the points of the train to have the same constant proper acceleration.

As another answer has pointed out, for the points of the train to maintain constant local distance in a co-moving frame, the points in back must have greater proper acceleration than the points in front.

• Thanks for this. I will study it. I agree that during the acceleration there is no MCRF for the train as a whole. My intention was to finesse this issue by having the acceleration become zero after some finite time, so that the entire train occupies one inertial reference frame before the acceleration starts and a different inertial reference frame after the acceleration ends. – WillO Mar 19 '16 at 14:56
• It does snap. Read the wiki on Bell's spaceship paradox. The idea is that if something starts at rest in an IRF and then in that IRF each part accelerates equally at every time in that IRF, then they each had the same velocity profile so each part displace the same so the distance between parts in that IRF is the same as when it was at rest. So the proper length increased because it must be longer than it was at rest in so that its length contracted version now is equal to the original IRF length. – Timaeus Mar 19 '16 at 18:47
• @Timaeus, my point is clearly made in the first sentence of the last paragraph - it is the stipulation the must be closely examined. The points of the train car cannot have the same constant proper acceleration because that leads to the contradictory state of affairs where the rest frame observes, by stipulation, that each point of the train has uniform coordinate acceleration (no snapping), while some MCRFs observe parts of the train to have vastly different velocities and accelerations (which is clearly unphysical). – Alfred Centauri Mar 19 '16 at 19:33
• @AlfredCentauri In order to have an IRF simultaneous acceleration you have to stretch the cars. So if that is your stipulation, then you have to stretch your cars. There isn't a MCIRF during the acceleration (which you argued well), so your last paragraph isn't clear either. And you make it sound like the acceleration doesn't cause it to snap, so it sounds like you are wrong. Simultaneous acceleration in the IRF leads to snapping, just like with Bell's spaceship. There isn't any contradiction because the OP knows it snaps and stipulated things that make it snap. – Timaeus Mar 19 '16 at 20:04
• @AlfredCentauri The OP clearly indicated that the acceleration is the coordinate acceleration in the track IRF, and hence the coordinate distances in that IRF are constant in that IRF and hence the cars must be physically expanded (larger proper length) in order to achieve that acceleration. Just like Bell's Spaceship. The OP didn't suggest a constant proper acceleration. The OP suggested a time varying acceleration that has the same coordinate acceleration for the same coordinate time, all measured in the track IRF. I went back and forth with the OP to get them to clarify this. – Timaeus Mar 19 '16 at 23:43

Here is my answer in words attempting to translate the math in the other answers.

First, the Curious Phenomenon:

If a train car reaches a sufficiently high velocity as a result of being accelerated uniformly along its length, then the train car must snap.

should be more accurately stated as

The Curious (But Not So Much Because Difficult To Set Up) Phenomenon:

If one manages to accelerate each car (using a rocket probably) in a train car such that at very high velocity those cars are accelerated uniformly as seen from the train station, then the train car must snap.

It will snap because the rocket in a given car has to be more powerful than the one behind, else we cannot observe the uniform acceleration of cars from the train station.

To say it otherwise: if the train is simply accelerating like a proper train, all cars following the lead engine, then when relativistic velocities are reached each car will not have the same acceleration from the point of view of the train station.

Question 1: Is there a clear conceptual explanation of the Curious Phenomenon based on classical mechanics without invoking relativity? Or does one really need relativity to explain this?

It is relativistic because it is only at relativistic velocities that one cannot accelarate a train car and get uniform acceleration along the cars (as seen from the train station) from the train engine alone.

Question 2: Suppose we knew nothing about relativity, but had observed the Curious Phenomenon. Would the search for an explanation naturally lead to relativity in the same sense that say, a search for an explanation of the Michelson-Morley phenomenon could naturally lead to relativity?

We would probably never observe something similar in nature: how and why would a extended compound natural system arrange the behaviors of its longitudinal components in such a way that those components mutual spacings appear longitudinally stable in a referential relatively to which they are permanently accelerating?

• Thank you. I hate to sound stubborn, but although I agree that we would probably never observe this in nature, I still wonder this: If we were (improbably enough) to observe this in nature, and if we didn't know about relativity, is there a natural train of thought that would lead us from this observation to relativity? I continue to feel like I don't have a good answer to that, and I continue to wish I had one. – WillO Mar 19 '16 at 18:08
• @WillO. It would not lead me to discover relativity. But more clever people, surely :) – Stéphane Rollandin Mar 19 '16 at 18:11

This isn't an answer, but is too long for a comment.

To answer @anna-v's concerns:

I think the problem here is that there's no such thing as instant acceleration, and you can derive contradictory results by assuming there is.

Instead, let's break up a single instantaneous acceleration into two smaller instantaneous accelerations. This is still incorrect, of course, but gives an idea of what happens as we apply smaller accelerations more frequently, as we would do in the real-life limiting case of applying continuous non-instant acceleration over a period of time.

Suppose two points $A$ and $B$ are already moving at $0.5 c$ in the track reference frame, and are some non-zero distance $x$ apart in that frame. Since we're debating the value of $x$, I'm not assigning it a fixed number, just pointing out that it's greater than 0.

In the track frame, you now accelerate/push both points at the same time. So, the time and distance between the two pushes in the track frame is:

• Time difference: 0 (you pushed them both at the same time)

• Distance difference: x (they were some distance apart when you did this, but we don't know what this distance was)

We now use the matrix for $0.5 c$ to translate this vector $\{x,0\}$ to point A's reference frame:

$\left( \begin{array}{cc} \frac{1}{\sqrt{1-v^2}} & \frac{v}{\sqrt{1-v^2}} \\ \frac{v}{\sqrt{1-v^2}} & \frac{1}{\sqrt{1-v^2}} \\ \end{array} \right).\{x,0\}$

The result (for $v=0.5 c$) is:
$\left\{\frac{2x}{\sqrt{3}},\frac{x}{\sqrt{3}}\right\}$

Note that the time difference is non-zero, since $x$ is non-zero.

In other words, as far as the points are concerned, they are being pushed at different times, one more frequently than the other.

Thus, in their own reference frame, their distance is not constant.

Of course, if the points were coupled by the electromagnetic force (as the the two ends of a train car would be), that force might be sufficient to hold the points at an equal distance, at least for a while.

Why this contradicts our relativistic "intuition":

A meter stick is shorter when it's moving at $0.5 c$ than when it's not moving at all. Thus, if we take a meter stick and instantly change its speed to $0.5 c$, we would expect it to shrink.

However, we can't really change its speed instantly. We have two options for realistic continuous acceleration:

• Have the meter stick accelerate at a fixed rate in its own reference frame. This is the normal definition of constant acceleration and yields the expected result of the meter stick shrinking.

• Have the meter stick accelerate at a fixed rate in its original reference frame (ie, the "fixed" frame). This is what we're doing here, and it yields very different results:

I would argue the phrase "instantaneous acceleration" is ambigious since it could mean either of the above. I would suggest one of the following:

• Acceleration over time where the object's speed in its own reference frame increases uniformly. This is the "standard" definition of uniform acceleration, and an object can theoretically accelerate indefinitely by this definition.

• Acceleration over time where the object's speed in a "fixed" reference frame increases uniformly. This is the definition being used here. In this case, the object must stop accelerating at some point since it's velocity can never exceed 'c'.