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If one is watching a relativistic object of e.g. spherical shape, which emits enough light to be detectable, it will, despite being Lorentz contracted, appear of its natural shape, although rotated. This phenomenon is called Terrel rotation$^\dagger$.

Citing wikipedia on Lorentz contraction, "length contraction is the phenomenon of a decrease in length measured by the observer of an object which is traveling at any non-zero velocity relative to the observer". So, how can the observer actually measure this decrease in length? Can it be somehow done in a non-relativistic regime of a measurement apparatus?

$^\dagger$Russian version of the page gives more detail with some pictures

Ruslan
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  • Read how Lorentz contraction is derived in theory and you will know. – mastrok Jun 30 '14 at 13:24
  • @mastrok it is derived by just applying Lorentz transformation to positions of both sides of a rod and finding the difference. Fine, but I still don't know. – Ruslan Jun 30 '14 at 13:36
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    @Ruslan: This maybe useful. http://itis.volta.alessandria.it/episteme/ep6/ep6-lars1.htm –  Jun 30 '14 at 15:12
  • @Ruslan Note that "Rotation" shouldn't be taken on face value. Take for example this: http://en.wikipedia.org/wiki/File:XYCoordinates.gif#How_these_images_were_made In diagram 2 a dimple actually forms in the sphere! (Hint: you're actually looking at the projection on the spatial axes of the intersection of a tilted cylinder with a cone). –  Jun 30 '14 at 19:28
  • you are right, lorentz transformation of both sides of a rod and find the difference at "the same time" in your reference frame. This is how you measure it. simply if one can accelerate a long rod to high speed, just take a snap shot of the rod when the center of the rod just passes through the camera (to make sure that photon emitted at the same time arrive the camera at the same moment). Then measure it from the photo.

    The other indirect evidence is that a large number of muon produce in the sky can reach the ground.

     – mastrok Jul 02 '14 at 04:29
  • Since we know that lifetime of muon is short. From the known speed of muon reaching the ground, you can calculate how long the muon can travel before it decay using classical kinematics, it says that they have decayed before they reach the ground. However, we found lots of muon reaching the ground means that time is delayed for the muon in the ground's frame. If you are in the frame of muon, you actually see that the distance between the sky and the ground is contracted! This is a simplest indirect measurement of length contraction. – mastrok Jul 02 '14 at 04:35
  • @mastrok have you read the link I gave in the question? The one about Terrel rotation? It's not that trivial to "just snapshot" the object so that Lorentz contraction would be captured. Muon lifetime is, as you indeed say, an indirect measurement. I ask about direct one. – Ruslan Jul 02 '14 at 06:34
  • You are right, the terrel rotation also counted the time lag in photon reaching the detector. That's why I said, you can snap shot it when some time after the center passes through the detector such that the time lag for both photons are the same. – mastrok Jul 02 '14 at 07:01
  • I guess the other way to do it is to emit photon along a ruler at the same time in your rest frame and then see how the rod blocks the photon in that moment. In this way, there is no effect coming from the terrel rotation – mastrok Jul 02 '14 at 07:06
  • @mastrok The problem would then be analogous to ladder paradox. We won't see Lorentz contraction of the shadow. – Ruslan Jul 02 '14 at 11:17
  • We "do" see lorentz contraction of the shadow! – mastrok Jul 02 '14 at 11:23
  • Nope. Because of relativity of simultaneity the flying object will be rotated in the frame of the observer. So, while the bunch of photons go to the object, they will get collected (i.e. blocked) by it, but not at the same time. This will finally give us the uncontracted shadow. See this part of the article. – Ruslan Jul 02 '14 at 11:26
  • Your link provide a nice explanation of the ladder paradox. In the resolution section, the shadow is contracted. – mastrok Jul 02 '14 at 11:32
  • In the bar and ring paradox, their "rotation" is due to contraction along the direction of motion. "Rotation" here is different from usual rotation. Think of a rod tilted by 45 degree in its rest frame with respect to the x-axis and moves in the x-direction, observer looking at the moving rod will report a different angle. But this rotation is not length preserving. – mastrok Jul 02 '14 at 11:39
  • @mastrok OK, you were right. Now I've simulated this scenario of shooting things at flying object and understood that I understand too little yet :) . – Ruslan Jul 02 '14 at 12:24
  • Related https://physics.stackexchange.com/q/229282/226902 – Quillo Feb 08 '23 at 22:59

2 Answers2

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The problem with experimental measurement of Lorentz contraction is that the only objects we've managed to accelerate to near light speeds are elementary particles, and they're pointlike so they can't contract.

Well, not quite. The RHIC accelerator collides heavy nuclei, and they do have a non-zero radius. The trouble is that it's hard to measure the size of a nucleus. However what you can do is calculate the dynamics of the collision, and if you do that you find it matches the results expected if the nuclei are Lorentz contracted into disks. I would certainly regard this as experimental confirmation of Lorentz contraction, but since it's an indirect measurement I guess it does leave the door open for the sceptics.

John Rennie
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  • Well, this is a technical problem. But it seems there's a problem to in principle measure the Lorentz contraction directly. For example, shooting photons at a relativistic object will give the shadow without Lorentz contraction because of how ladder paradox is resolved. Observing the light emitted by the object will give us Terrel rotation. I can't think of any experiment which would definitely directly show Lorentz contraction. So, this is the point of my question. – Ruslan Jul 02 '14 at 11:24
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    @Ruslan: in principle the measurement is easy. The key thing is to do all measurements locally i.e. at your position. Take a rod of length $\ell$ travelling at a speed $v$ towards you and record the time when the front edge passes you and the time the back edge passes you. Multiply by $v$ to get the length and you'll get the result $\ell/\gamma$. I say in principle because there are obvious formidable practical problems. – John Rennie Jul 02 '14 at 11:31
  • There are plenty of high-precision experiments that show relativistic effects without the necessity of accelerating anything to a significant fraction of $c$. A good example is the famous Hafele-Keating experiment. –  Oct 23 '14 at 01:05
  • In addition to RHIC, there is also DIS (Deep Inelastic Scattering): here an electron beam shines individual (virtual) photons into a relativistically flattened nucleon, with the scattered electron detected. Results are consistent with Lorentz contraction of the target, and one could are that at these energies, exchanging a virtual photon is "seeing". – JEB Feb 01 '18 at 17:28
  • @JEB why not expand that a bit and post it as an answer? – John Rennie Feb 01 '18 at 17:32
  • @JohnRennie Because I haven't scattered an electron in 20 years? But I'll try. – JEB Feb 01 '18 at 17:35
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This depends quite a bit on what you're willing to accept as "direct."

The magnetic force between two parallel current-carrying wires can be interpreted as being due to Lorentz contraction, and that's quite easy to measure -- you can do it with a battery and some strips of aluminum foil.

Some measurements can be interpreted as showing time dilation, but in a different frame they show length contraction. For example, in the earth's frame of reference, we explain the anomalously high flux of cosmic-ray muons at the earth's surface as being due to time dilation: their half-lives are lengthened because of their motion. But in the frame of the muons, their survival is due to the length contraction of the earth's atmosphere.

  • By "direct" I mean measurements of length itself which would show that it is different from that in the rest frame of the object. The examples you provide are quite indirect because they need some interpretation which puts Lorentz contraction/time dilation to be the cause, while these experimental facts might have been caused by something completely unrelated to Lorentz transformations if SR weren't true. What I was after is doing (in principle) the measurement of contracted length with the knowledge of 19th century and getting the correct result not needing special interpretation. – Ruslan Oct 23 '14 at 04:20