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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

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  • $\begingroup$ Read how Lorentz contraction is derived in theory and you will know. $\endgroup$
    – mastrok
    Commented Jun 30, 2014 at 13:24
  • $\begingroup$ @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. $\endgroup$
    – Ruslan
    Commented Jun 30, 2014 at 13:36
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    $\begingroup$ @Ruslan: This maybe useful. itis.volta.alessandria.it/episteme/ep6/ep6-lars1.htm $\endgroup$
    – user7757
    Commented Jun 30, 2014 at 15:12
  • $\begingroup$ @Ruslan Note that "Rotation" shouldn't be taken on face value. Take for example this: en.wikipedia.org/wiki/… 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). $\endgroup$
    – user12029
    Commented Jun 30, 2014 at 19:28
  • $\begingroup$ 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. $\endgroup$
    – mastrok
    Commented Jul 2, 2014 at 4:29

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Ruslan
    Commented Jul 2, 2014 at 11:24
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    $\begingroup$ @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. $\endgroup$
    – John Rennie
    Commented Jul 2, 2014 at 11:31
  • $\begingroup$ 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. $\endgroup$
    – user4552
    Commented Oct 23, 2014 at 1:05
  • $\begingroup$ 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". $\endgroup$
    – JEB
    Commented Feb 1, 2018 at 17:28
  • $\begingroup$ @JEB why not expand that a bit and post it as an answer? $\endgroup$
    – John Rennie
    Commented Feb 1, 2018 at 17:32
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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.

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  • $\begingroup$ 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. $\endgroup$
    – Ruslan
    Commented Oct 23, 2014 at 4:20

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