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In another question evaluating the reality of length contraction, the circular motion was involved and some answers argued that centrifugal force would negate any possible length contraction. A famous paradox called "The Ehrenfest Paradox" analyses a similar situation with relativistic motion in a circle and it suffers from the same criticism.

The effects due to centrifugal motion in a circle (as far as length contraction is concerned) can be removed by considering the following thought experiment. A large circular wall is constructed and a railway track is built on the inside vertical surface of the wall. In fact, the wall can be built in a pit dug in solid granite, to support the wall and the wall is made of the toughest materials known to man. The idea is that the radius of the track is not allowed to increase to any significant extent.

Initially, trains are placed on the track one at a time, each linked to the next by a short spring, until the track is completely filled with connected trains.

Now the trains are accelerated to relativistic speed. SR seems to predict the trains will length contract according to the observers at rest with the track and the connecting springs (made of lower tensile strength material) will be stretched.

One of the answers in the other thread contradicts this conclusion, saying from the point of view of the passengers on the trains the track is length contracting. He implies that since both cannot happen at the same time, neither the track nor the train length contracts and the gaps between the trains will not increase.

How is this paradox resolved? What is the explanation for what is going on and what will actually be observed? How do the observers on the trains explain why the gaps between the trains increased and the springs were stretched if that is what happens?

Please don't say SR cannot analyse this and GR is required. The train is operating in essentially flat spacetime and SR can cope with circular motion and acceleration.

In summary, what actually is measured to happen and how is it explained?

Additional note: For clarity, I intended the trains to be accelerated to a final constant velocity in a Born rigid manner, such that the trains maintain a constant proper length as measured by observers at rest in the trains.

Edit: This part is transferred and paraphrased from the comments to avoid an extended discussion there: Consider a slight modification, this time without any connectors. Lets say the trains are accelerated to a velocity such that they are measured to contract by about 50% and then maintain that constant speed. On the roofs of the trains are poles that are the same length as train.

When the train is at constant speed, the passengers can feed these poles into the gaps between the trains so that they fit snugly without any (parallel to the track) stress or strain on the poles or the trains. Now we could have a collection of objects with a total rest length equal to twice the rest length of the track, yet fitting into a circumference equal to the rest circumference of the track. This would seem to indicate that something really has contracted relative to the other. (Let me know if this second part should be moved to its own separate thread.)

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    $\begingroup$ Please link to the other question and answer. I think you are misrepresenting my answer, but it could be simply a different question than I am recalling. $\endgroup$
    – Dale
    Commented Feb 9 at 1:49
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    $\begingroup$ I don't think you made an answer in that thread. @Dale P.S> Are you Dale or Dalespam in PhysicsForums? $\endgroup$
    – KDP
    Commented Feb 9 at 1:56
  • $\begingroup$ Yes, although I have been just Dale for many years over there. $\endgroup$
    – Dale
    Commented Feb 9 at 2:23
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    $\begingroup$ Also, since length contraction applies to lengths between events, not just to physical objects, the gaps between the trains will contract as much as the trains do. $\endgroup$ Commented Feb 9 at 7:55
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    $\begingroup$ @Dale said "You cannot avoid the physical stretching" The whole point of the question is demonstrate that the trains will be physically stretched. Thanks for confirming that. $\endgroup$
    – KDP
    Commented Aug 12 at 12:22

4 Answers 4

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To gain insight, replace your track with a long thin rectangle with slightly rounded corners. That way every car is inertial most of the time.

Now the trains are accelerated to relativistic speed. You don't specify exactly how the acceleration takes place, but let's suppose for concreteness that every part of every train car is accelerated in the same way, as measured in the track frame. Therefore no car's length changes in the track frame.

Now suppose you're in a car on one of the long sides of the track. Here is what you will say:

  1. The front of my car began to accelerate before the back of my car. Therefore my car got stretched. The same thing happened to all the other cars on my side of the track, and to all the springs connecting us. Therefore either the cars or the springs or both are now longer than they used to be.

  2. As for those cars on the opposite side of the track, exactly the opposite thing happened. Their backs began to accelerate before their fronts did. Therefore either those cars or the springs between them or both are now shorter than they used to be.

  3. There are now more cars on the far side of the track than there are on my side (because the two sides of the track are equally long but the cars [and/or the springs] have different lengths).

Now once you've digested that, go back to the case of a circular track. Riding on a given car, and describing the train in your instantaneous inertial frame, you'll say that each car accelerated differently, and you'll assign different lengths to different cars. Those near you (including yours) will have gotten longer; those more-or-less diametrically opposite will have gotten shorter, and each by a different amount. But they'll still all fit on the track.

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  • $\begingroup$ For the long thin rectangle case I did the detailed calculations here physics.stackexchange.com/questions/321074/… and here physics.stackexchange.com/questions/798472/… $\endgroup$
    – KDP
    Commented Feb 10 at 0:55
  • $\begingroup$ By applying greater proper acceleration at the front than at the back you are not accelerating the cars in a Born rigid manner and are using energy to stretch the cars in their own rest frame. As soon as the engines are turned off to cruise at a constant tangential velocity, the cars will contract from everyone's point of view. Your scheme only works if there is constant proper acceleration for ever. $\endgroup$
    – KDP
    Commented Feb 10 at 1:04
  • $\begingroup$ You are forgetting the the track itself contracts from the point of view of the observers on the trains. Let say we have a long rectangular track as you suggest and put flags at the ends of the longest axis. The train observers will measure the separation of those flags to be shorter than measured by the inertial trackside observers. $\endgroup$
    – KDP
    Commented Feb 10 at 1:08
  • $\begingroup$ @KDP : Your only assumption is "Now the trains are accelerated to relativistic speed". Neither the words nor the concept of "Born rigid acceleration" appears anywhere in your question. Your question is impossible to answer without some assumption on how the acceleration takes place. I made a simple assumption and answered the question, via the same sort of general reasoning that you can re-apply to any kind of accceleration you want. As for the cars re-contracting, you are adding even more assumptions after the fact; plenty of metal bars will stay stretched after you stretch them. $\endgroup$
    – WillO
    Commented Feb 10 at 3:23
  • $\begingroup$ And no, I did not forget that (part of) the track contracts from the viewpoint of the on-train observers; I just didn't explicitly mention it. $\endgroup$
    – WillO
    Commented Feb 10 at 3:23
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The other thread is a more complicated scenario than this one, so I wouldn’t say that you should expect the answers of that one to be exactly similar to this one.

SR seams to predict the trains will length contract according to the observers at rest with the track and the connecting springs (made of lower tensile strength material) will be stretched.

You are correct that the springs will stretch, but it is incorrect to call this length-contraction. Length contraction is a disagreement between two inertial frames regarding the distance between two points which are at rest in one frame. There is no material strain involved in length contraction.

In this case the trains are not inertial. And angular acceleration is not a rigid motion so it is always associated with material strain. So yes, what you describe does happen, but no it is not length contraction.

from the point of view of the passengers on the trains the track is length contracting

This is a very “fraught” statement. The tricky issue is that the passenger’s frame is non-inertial. So there is no standard meaning to their point of view.

Because they are non-inertial they are not symmetric with non-rotating observers. Only the congruence of observers that underwent angular acceleration experience a measurable material strain, and both the train and wall observers agree that is the train observers.

What is the explanation for what is going on and what will actually be observed. How do the observers on the trains explain why the gaps between the trains increased and the springs are stretched if that is what happens?

If strain gauges are placed on the springs then all observers will see that there is a measurable strain in the springs. The observers on the train attribute the measured stretching of the springs to the forces that produces the angular acceleration. These forces include the external forces on the train cars as well as the internal forces keeping the train cars intact and the forces exerted on the springs. This is a mechanical strain and it is caused by mechanical forces.

This would seem to indicate that something really has contracted relative to the other.

I would interpret “really has contracted” as referring to a measurable strain. So yes, there is real measurable strain in this scenario. The contraction is therefore not “length contraction” which is strain-free.

Consider an inertially moving train car on a straight track adjacent to the curved track. As it passes the curved track the two cars (one on each track) are momentarily at rest and immediately adjacent to each other.

In the ground frame the inertial car is shortened. There is no material strain: this is length contraction.

The non-inertial car is strained, so its shape is distorted compared to the inertial car. This material strain is unavoidable, physically measurable, asymmetric, caused by real forces, and is not length contraction.

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  • $\begingroup$ Would you agree that even though it is not called length contraction, the length of each individual train as measured by the trackside observers (on a very large track with millions of trains) will be to a reasonable accuracy in agreement with the length contraction predicted by SR for the tangential velocity? In this case assume the trains are connected by very thin rubber bands and the forces exerted by them is negligible or remove the rubber bands. $\endgroup$
    – KDP
    Commented Feb 9 at 2:59
  • $\begingroup$ @KDP I would not agree with that as a blanket statement. The train cars are also undergoing angular acceleration, so there must unavoidably be some material strain within the cars themselves. Those material strains will lead to deviations from length contraction. Whether those deviations are negligible will depend on the details of the stresses and the material. I don’t know the relativistic version of Hooke’s law and I am not sure I could compute this even if I did. $\endgroup$
    – Dale
    Commented Feb 9 at 3:07
  • $\begingroup$ @KDP I already answered as much as I can and explained why I cannot answer more. There is still angular acceleration so there is necessarily material strain $\endgroup$
    – Dale
    Commented Feb 9 at 3:17
  • $\begingroup$ A rocket flying in circles in space far from any gravitational bodies has angular acceleration, but the strain on the rocket is orthogonal to the motion and not tangential to the motion. In the train example, the angular acceleration is crushing the 'height' of the train so the ceiling is nearer the track, but largely it is not stretching the length of the train. It is trying to stretch the radius of the track and that is the reason for the reinforced concrete walls buried in hundreds of kilometres of solid granite. $\endgroup$
    – KDP
    Commented Feb 9 at 3:45
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Uniform circular motion along the circumference and motion along the diameter while it is spinning are two different reference frames:

  1. Motion along the circumference can be treated as an inertial frame.

The tangential velocity along the circumference is constant and there is no tangential acceleration. All the acceleration is centripetal and is uniform at all points of equal radius. If half a circumference C is stretched into a straight line then the centripetal acceleration will resemble a constant force along the the straight length (like gravity) If the stretched line exerts a constant normal force that counters the centripetal acceleration, then the sum of the acceration is zero.

Now all the requirements for an inertial frame have been met so for relativistic tangential velocity $$C' = \frac{C_o}{\gamma}$$

  1. Motion along the diameter of a circle exhibiting uniform circular motion is a non-inertial frame of reference.

Any motion along the diameter has a constantly changing centripetal acceleration parallel to the direction of motion which is maximum at R and approaches a minimum (0) as one approaches the center. The motion along the diameter is identical to the motion of the amplitude of a sine wave (Simple harmonic motion) and can be modeled as rotational motion by looking at the shadow of a peg on a wheel rotating at a constant angular frequency. Figure  15.4.1
: SHM can be modeled as rotational motion by looking at the shadow of a peg on a wheel rotating at a constant angular frequency.

The motion along the radius can be expressed as $x(t)=Acos(ωt)$

If there is a full special relativity solution to the Ehrenfest paradox, it is in the proper treatment of the measurement of length or motion along the diameter such that the problem can be expressed in an inertial frame of reference.

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Special relativity can be used to analyze simple rotations from the point of view of the non-rotating frame. However, the rotating frame is not equivalent to the non-rotating frame. While, there is no absolute inertial frame, there is an absolute frame of rotation, it is the one in which the universe is not rotating. In the rotating frame, space is non-Euclidean, time is dilated, extra forces exist, and light does not travel in straight lines. Using just SR is not enough. It is best, if one is using special relativity, to examine what things look like from the non-rotating frame. Notice that all of these paradoxes are paradoxes in special relativity and involve accelerations which special relativity was not meant for.

For example, one might, from using just SR, conclude that passengers on the train would see the track contract. But since they are in a rotating frame, they measure the circumference of the circular track to be increased by $\gamma$ because space has become non-Euclidean and the circumference of such circles is $\gamma 2 \pi r$.

In the non-rotating frame, the centrifugal forces on the train will, for real materials, squeeze it down and lengthen it. The springs would also be crushed and lengthened. So, presumably, no further spring extension would be needed.

But note, if we could ignore gravity (easy, just do this in space) and centrifugal force (it is a thought experiment), then it is reasonable to think that the train would lift off the tracks and make a smaller circle. The springs would be contracted but not stretched, so just static forces are involved (ie., no energy). The passengers on the train would see the radius magically shrink. Only a physicist on the train could explain what is happening to them, with a little differential geometry. Now, invoke centrifugal force and the springs expand. For a moment you have the simple configuration described in the question. But, after the train hits the wall, it get smashed like a bug on the windshield.

We can, also for the sake of the thought experiment, suspend reality by imagining that the train is composed of impossibly rigid materials. If one alters the order of events, one can better understand where the forces and energy are coming from. Disconnect one spring and then accelerate the train to relativistic speed. One gets a shortened train and a large gap. Now slow down the back car and speed up the front car until you can connect the two by a spring. Now, one has the same final configuration. However, one finds that it required extra energy and force on the part of the train engines to stretch the springs and bring the total length of springs and cars back to their inertial total length. The order of coming to this configuration should not matter, so the same must be true if the springs had all been connected from the beginning.

As in the Bell's Spaceship Paradox, it can require force, and hence energy, to undo some of the effects of a Lorentz contraction; for instance, returning the length of an object now moving to its length when it was at rest. In essence, the Bell's Spaceship Paradox and this Ehrenfest Paradox are the same. The rotating band and rotating disk problems are more interesting because they look like there is potential to extract energy from length contraction. But, in reality, there is always a net expansion and no way to extract energy.

So why do I focus on where the energy for the forces comes from? Since relativistic length contraction is not one of the forces of nature, it follows that one can not extract energy from it. You can only expend energy to make it appear that it is not happening.

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  • $\begingroup$ I never claimed there was potential to extract energy from the length contraction. In the Bell's paradox and the chain drive paradox the increase in tension comes from the input of energy of the accelerating rockets or from the motor driving the chain. There is no free lunch here. However +1 for the correct conclusion "One gets a shortened train and a large gap" $\endgroup$
    – KDP
    Commented Feb 9 at 21:52
  • $\begingroup$ If there is nothing stopping the gear on the left moving to the right as the belt contracts it takes x amount of energy to accelerate the belt. If there is a device on the left extracting y amount of energy, then it takes (x+y) amount of energy to accelerate the belt. It's the same in the Bell's paradox. If the rockets try to keep the string length the same as measured by inertial observers in the launch frame, then the string is stretched in its instantaneous rest frame and this requires additional energy to be input by the rockets. $\endgroup$
    – KDP
    Commented Feb 10 at 0:48
  • $\begingroup$ Why does it require x+y to accelerate the belt? There is no force acting contrary to the motion of the gear that is doing the acceleration. In Bell's paradox, the energy is spent by rockets to exert a force to directly break the string. In the belt paradox, there is a force extracting energy by stretching, but no force requiring more energy to accelerate a stretched belt. That is the essence of this paradox! The only possible explanation I see is that the mass of the belt is increased by the stress energy $\Delta E/c^2$. Is there another explanation? And, is it enough? $\endgroup$
    – eshaya
    Commented Feb 13 at 19:20
  • $\begingroup$ I almost forgot. There is another answer. The belt never contracts because centrifugal forces at the gear stretch any real material. $\endgroup$
    – eshaya
    Commented Feb 13 at 19:29
  • $\begingroup$ This is not the question about the belt. This is about a set up with trains riding a wall that is explicitly designed to prevent any tangential stretching due to centrifugal force. Centrifugal force $ = \gamma m v^2/R$. We can choose R that is arbitrarily large such that the centrifugal force is arbitrarily small, so forget about centrifugal force as an explanation. $\endgroup$
    – KDP
    Commented Feb 13 at 19:57

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