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Let's suppose we have a one dimensional rod made of elementary point particles, in contact with each other placed along the x-axis.If the rod is moving along x-axis then we know (because of relativity) that its length will reduce.

But how is this possible? Since the rod is made of elementary particles which are in contact with one another, and elementary particles can't be deformed, wouldn't this mean they can't come any closer?

Where am I wrong?

auden
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Paras Koundal
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    Possible duplicates : What is the real meaning of length contraction? http://physics.stackexchange.com/q/95618 "Reality" of length contraction in SR http://physics.stackexchange.com/q/148216 – sammy gerbil Jul 25 '16 at 12:31
  • @sammygerbil: I do not believe the alleged duplicates address this question, or anything like it. – WillO Jul 25 '16 at 13:54
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    To the OP: What does "in contact with one another" mean? – WillO Jul 25 '16 at 13:54
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    @WillO : I disagree. The question is asking how the idea of a rigid rod can be compatible with the prediction of SR that observers in relative motion measure different lengths for the rod. ie Does the rod really deform? This is the same as asking, What does length contraction mean? – sammy gerbil Jul 25 '16 at 14:30
  • @sammygerbil: But the difference (it seems to me) is that this question posits a specific reason why it appears to the poster that the rod can't deform, and asks what's wrong with that reason. That specific reason seems to me to be the focus of this question, and absent from the others. – WillO Jul 25 '16 at 15:48
  • @sammygerbil: Likewise, there's a big difference between "Are there really no perpetual motion machines?" and "Here is a machine that appears to generate perpetual motion; what am I overlooking?". I think the questions you linked to are more like the former and this question is more like the latter. – WillO Jul 25 '16 at 15:54
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    Elementary particles have zero dimension so you rod would have zero length. – Peter R Jul 25 '16 at 16:24
  • @WillO : re Perpetual Motion Machine argument : exactly. That is the power of generalisation. Once you have shown that Perpetual Motion machines cannot exist, you no longer have to examine every new candidate to find out what is wrong with it. If the OP suggested that the rod is made out of titanium or diamond (or kryptonite!), and therefore should not contract because it is ultra-hard, so SR must be wrong, - what difference would that make? – sammy gerbil Jul 25 '16 at 19:25
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    @sammygerbil: I heartily disagree with you. If someone designs a machine that appears to demonstrate perpetual motion, you can, depending on how much time you've got to spare, either ignore him or try to figure out what's wrong with it. There is a potentially a lot to be learned from the second option. Learning opportunities should not be lightly discarded. – WillO Jul 25 '16 at 19:36
  • @sammygerbil: And notice that the OP never said "SR must be wrong"; those are words you put in his mouth. What he actually said was "Where am I wrong?", which is exactly the sort of question a serious student should be in the habit of asking. Note that John Rennie gave a helpful answer. The existence of helpful answers is often a hallmark of good questions. – WillO Jul 25 '16 at 19:38
  • There are no rigid bodies in relativity.. – ACuriousMind Jul 26 '16 at 12:59
  • I don't think relativity says that the object gets smaller. Such a concept can only make sense in the rest frame of the object. I think that relativity says that if two different observers measure the length, they will get different answers. That's a very different thing. – garyp Jul 26 '16 at 20:31
  • I agree with peterr, "a one dimensional rod" ? -- What is the "one dimension", are you saying it has no diameter or no length. You can have a "one dimensional sphere" with only a diameter, but a rod requires 2 dimensions; much as a cubic requires 3 dimensions. Please edit your question. – Rob Dec 25 '17 at 06:41
  • @Rob, we don't need 2D for a rod - a 1D rod is a segment of line, and that's fine. The problem I see here is the "point particles, in contact with each other", which makes this rod constituted of an infinite number of particles (a continuum really), which is not how it's in nature. But this has already been addressed in the answers. – stafusa Dec 25 '17 at 12:58

2 Answers2

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Your question contains a contradiction because you propose:

elementary point particles, in contact with each other

but by definition point particles have zero size and cannot be placed in contact with each other. As far as we know elementary particles are indeed pointlike, so you can't have a rod of a finite length made up from elementary particles in contact.

Your rod would have to be made up from extended particles like atoms or neutrons, and these have a non-zero size because they are a bound state of elementary particles. However the size of an extended particle is Lorentz contracted in exactly the same way as the rod is, so your rod would shrink at the same rate as the (extended) particles that make it up shrink.

For completeness I should point out that the Lorentz contraction is not really a contraction but a rotation. More precisely it's a hyperbolic rotation in 4D spacetime. In the moving coordinate system the ends of the rod rotate so they are at different times. The rod looks contracted because the observers in the rod rest frame and moving frame disagree about the times they are observing the ends of the rod. For a detailed discussion of this see "Reality" of length contraction in SR.

Community
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John Rennie
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  1. The elementary particles in a piece of normal matter are not in contact with one another; they are in fact very distant from one another compared to their size.
  2. But let's imagine some hypothetical matter made of rigid spheres in contact. The length contraction will affect the spheres as well, so they will deform into spheroids (shortened along the direction of motion).
  3. But what if the spheres are "truly rigid"? This is not possible in relativity theory; all matter must be deformable ultimately to enforce causality.
AGML
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