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How is length contraction on rigid bodies possible in special relativity since definition of rigid body states they are not deformable?

Physics Asked by Monopole on February 11, 2021

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?

4 Answers

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.

Answered by Buzz on February 11, 2021

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.

Answered by doug_h on February 11, 2021

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.

Answered by Ricky Tensor on February 11, 2021

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.

Answered by WimC on February 11, 2021

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