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This is more like a conceptual question. We define rigid bodies as solid bodies with zero or almost zero deformation (meaning the deformation should be negligible). So no distance between two points should change in time.

Yet, if I have an object with proper length $L_{0}$ and I move this body at relativistic speeds, I will see the length of the body contracted as

$$L = L_{0}\sqrt{1-\frac{v^2}{c^2}}$$

So does that mean there are some exceptions with special relativity, or it simply means those rigid bodies we assumed rigid are not actually perfectly rigid, or something else?

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  • $\begingroup$ Does physics.stackexchange.com/questions/148216/… help? $\endgroup$ – Andy Newman Dec 22 '20 at 22:38
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    $\begingroup$ Because it's all relative to their POV (point of view). $\endgroup$ – RBarryYoung Dec 23 '20 at 15:05
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    $\begingroup$ @Fattie I think the issue with a lot of what happens on this forum is that a lot of people think that knowing a formula and being able to calculate a value using it is the same thing as understanding physics. Unfortunately the people who really know their stuff (and no, I don't count myself in their number) rarely bother to challenge these confused answers. $\endgroup$ – JimmyJames Dec 23 '20 at 18:42
  • $\begingroup$ More a question than an answer, but is the rigidity of an object not defined by the space it finds itself in? Whether or not that space itself then warps is then irrelevant as far as rigidity is concerned. Think of it this way, if I tell you I'm walking from the head to the tail of a northbound train (inside it, of course), I am walking southward. "Southward" is my direction relative to the train, not the ground. If you look at me relative to the ground, since the train moves faster than I do, I actually move northward. Similarly, rigidity is (afaik) defined relative to the space. $\endgroup$ – Flater Dec 24 '20 at 0:58
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    $\begingroup$ What I mean by that is: if you had a row of light bulbs along the train and had a switch that turns them on all at once, I (ground obs.) would see a wave going from one end of the train to the other; by the time I see the last light bulb light up, in your frame that event is already in your past. So it's not even that the train gets "squashed" in the traditional sense, it's that I see a different 3D slice of this 4D spacetime object that is the train (think of a stack of movie frames sliced at an angle), and that this slice happens to be shorter according to the Lorentz transformation. 2/2 $\endgroup$ – Filip Milovanović Dec 24 '20 at 12:57
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Perfectly rigid bodies are not possible in relativity, although this is not directly related the Lorentz contraction mentioned in the question. One immediate consequences of relativity is that no signal can travel faster than the speed of light; and this actually rules out perfectly rigid bodies.

The reason, although it may not be instantly obvious, is actually fairly simple. If we had a long (length $L$), perfectly rigid rod and apply a force to it, it would need to accelerate uniformly. Perfect rigidity would mean that both ends need to be moving exactly in synchronization; as soon as a force is applied at $x=0$, the other end at $x=L$ has to start to move. (If the don't move together, then the length of the rod has changed.) However, it is impossible in relativity for the far end to start moving at the same time, because that would require a signal to travel instantly down the length of the rod. In actuality, when the force is applied at one end, the rod will deform slightly, and the deformation will propagate at speed $v$ ($v$ is the sound speed in the material, and $v<c$) down the length of the rod. Only after a time $L/v$, when the signal reaches the other end, will the far end start to move.

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    $\begingroup$ The big conceptual problem is not length contraction, but what happens to a rotating "rigid" body where different points on the body are moving with different speeds. $\endgroup$ – alephzero Dec 22 '20 at 3:53
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    $\begingroup$ More generally, requiring that the speed of sound in a body be less than the speed of light sets constraints on its elastic properties. The speed of sound is given by $1/v^2 = \partial \rho/\partial P$, and this must be greater than $1/c^2$. A truly rigid body would have $\rho$ = constant and therefore $1/v^2 \to 0$. $\endgroup$ – Michael Seifert Dec 22 '20 at 13:03
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    $\begingroup$ Sorry, this doesn't answer the OP question. Looking at any body (be it rigid or not) in uniform motion, we see length contraction, but body (in its coordinate system) doesn't actually deform. See other answers. $\endgroup$ – Arvo Dec 22 '20 at 15:39
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    $\begingroup$ Just wanted to double down on what Arvo said. This answer does not answer the specific question asked. It merely circumvents providing an answer by explaining a tangential topic, which has probably just further confused the OP because they now feel like they understand the answer to their question when they really don't. The OP appears to be under the misapprehension that length contraction causes a physical (instead of observed) deformation of a (rigid) rod and is trying to reconcile that. The correct answer is one which corrects that misapprehension. $\endgroup$ – zephyr Dec 22 '20 at 16:50
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    $\begingroup$ @zephyr, length contraction is a physical deformation. All frames are equally valid, including the ones in which an object is contracted. $\endgroup$ – Harry Johnston Dec 22 '20 at 21:43
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Buzz's answer is correct in that there's no such thing as a perfectly rigid body in relativity. But even more importantly for your question, a body in uniform motion does not feel any kind of squeezing force, even if it's moving very quickly.

Consider two spaceships moving past each other at high speed. Ship A will see ship B compressed, and by the symmetry of the situation, ship B will see ship A compressed. But for the people on each ship, things will appear to be normal length, time will appear to be going at the same rate as it always does, and there will be no squeezing force.

In the end, this effect is a result of coordinate systems. The people on the two different ships use different coordinate systems, which is what causes them to disagree on the length of things.

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    $\begingroup$ This is misleading in that it suggests that length contraction is some kind of illusion. In a frame of reference in which an object is moving, there is no external force squeezing the object, but the internal forces holding the object together change in such a way that the object contracts. This frame of reference is just as valid as the one in which the object is stationary and uncontracted. $\endgroup$ – Harry Johnston Dec 22 '20 at 21:52
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    $\begingroup$ @Harry Johnston: It's not misleading at all, it is an accurate description of what observers will experience. If you are the observer on one of the ships, you will perceive the rest of the universe as being contracted, but will see yourself and your ship as normal. There's no physical contraction: in your reference frame, all the forces &c are exactly as usual. To you, the rest of the universe appears to have changed. $\endgroup$ – jamesqf Dec 23 '20 at 4:30
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    $\begingroup$ @jamesqf, in one reference frame the object is not contracted, but in the other reference frame it is. Both of these reference frames are equally "physical" and the length contraction (in the appropriate frame) is a real effect, not an illusion. The object doesn't just look shorter (in the appropriate frame) but actually is shorter. $\endgroup$ – Harry Johnston Dec 23 '20 at 12:11
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    $\begingroup$ @JimmyJames, I disagree. The OP is talking about a measurement made in the lab frame, and in that frame the object is contracted. Besides, it is very important to understand that rigid objects don't exist in SR; you can't always weasel your way around the problem by saying "well, there's one particular frame in which the object still appears to be rigid, so let's pretend that's the only frame that counts." :-) $\endgroup$ – Harry Johnston Dec 23 '20 at 22:05
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    $\begingroup$ @JimmyJames, we're going around in circles now. In the lab frame, the egg can deform without breaking because of the way the electromagnetic forces that hold the atoms together change. This has to be true, because the laws of physics are frame-independent. $\endgroup$ – Harry Johnston Dec 23 '20 at 22:49
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The issue really isn't deformation. According to special relativity if you are at rest with a rigid rod of length L, and I am moving at speed v relative to you, then I will measure a shorter length. I've done nothing to the rod, but in my reference frame the rod simply has a shorter length. But regardless, a rigid body is only an idealization anyway. Your very rigid rod can still be deformed, in your reference frame, but this may have limited practical effect because it is very rigid. So we say it is absolutely rigid, to simplify analysis. But back to relativity, if you shorten the rod, I will measure an even shorter length.

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    $\begingroup$ In other words, the body doesn't contract, the ruler we use to measure it changes. $\endgroup$ – Barmar Dec 22 '20 at 15:59
  • $\begingroup$ This is misleading in that it suggests that length contraction is some kind of illusion. In a frame of reference in which an object is moving, there is no external force squeezing the object, but the internal forces holding the object together change in such a way that the object contracts. This frame of reference is just as valid as the one in which the object is stationary and uncontracted. $\endgroup$ – Harry Johnston Dec 22 '20 at 21:52
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    $\begingroup$ @Barmar, yes it does. And there is no such thing as accelerating reference frame (in SR or GR). See hyperbolic motion, Rindler coordinates and early chapters of Misner-Thorne-Wheeler (the ones on SR). You need a better grasp of differential geometry to work with accelerated objects, but spacetime does not have to become curved. So special relativity applies, unless you are using gravity to do the acceleration. Also, en.wikipedia.org/wiki/Acceleration_(special_relativity) $\endgroup$ – Cryo Dec 23 '20 at 0:01
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    $\begingroup$ @JimmyJames. This is straying off topic now. I have dealt with this before physics.stackexchange.com/questions/598473/…. This is why one should not jump between reference frames - it leads to confusion. From the point of view in the lab frame objects are squeezed as they are accelerated, this is different in the rest frame of the object, once it is inertial. Please create a separate post for dicussion $\endgroup$ – Cryo Dec 23 '20 at 22:15
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    $\begingroup$ @Cryo I think the problem is that you have strayed off topic. The question never mentions lab frames. It specifically mentioned 'deformation'. The object is not deformed. in it's frame which is the only one where its rigidity is relevant. $\endgroup$ – JimmyJames Dec 23 '20 at 22:20
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Consider an observer that sees an arrow fly by. The observer measures the length of the arrow as the spatial distance between the head and the tail at the same time. However, “at the same time” is not a relativistic invariant. From a traveller’s perspective (travelling with the arrow) the observer did not observe the two points at the same time: it observed the tail slightly later than the head, and by that time the tail moved forward a bit.

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  • $\begingroup$ Right. And this even happens if the observer makes their measurements when the midpoint of the arrow is directly in front of them, so that (in their frame) the times light takes to travel to them from the head & tail of the arrow are identical. That is, the relativity of simultaneity isn't just an artifact of the finite travel time of light, it's due to rotation of the spacetime axes. $\endgroup$ – PM 2Ring Dec 24 '20 at 20:59

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