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If an object moves at velocity $v$ relative to my frame its length $l_0$ will be Lorentz contracted to the length in my frame $l$: $$ l = l_0 \sqrt{1-\frac{v^2}{c^2}} $$ If I rotate around myself with frequency $\omega$, objects at distance $r$ will have the relative velocity $v = r \omega$. So the moon at a distance of $r=300000$ km should be Lorentz contracted to length $l= 0$, if I rotate around myself with $\omega =1$ Hz (according to the formula $l$ would even get complex, when I would rotate faster).

I haven't done the experiment, but I believe if the moon would be Lorentz contracted somebody would have already noticed that effect. So something must be wrong here.

The funny thing is, I believe that if the moon or a spaceship would rotate around me with a speed of $v < c$, when I must see it Lorentz contracted according to the formula above. So it seems to me the two cases, I rotate around myself and something rotates around me, are not equivalent. In the first case I see no Lorentz contraction, but in the second case I do. Is my thinking correct? If so, can someone explain why these two cases cannot be equivalent and why using the formula for Lorentz contraction must be wrong when I rotate around myself?

knzhou
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asmaier
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  • I heard that lorentz contraction appears in one dimension, typically parallel to the axis of motion. In this case the direction of motion would be constantly changing and you appear to lose some density as a function of radius from the center of mass. –  Aug 23 '16 at 20:45

2 Answers2

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When you rotate, you are no longer in an inertial frame, so it's tricky to apply special relativity. In particular, the coordinates have a nontrivial connection, which means you cannot directly compare vectors at two different points. In particular, you cannot compute the relative velocity between you and the moon directly.

As an analogy, consider two people on the Earth, each holding compasses. It's impossible to compare if their compasses are both working unless they both meet up in the same place to compare. This process is called parallel transport.

In this case, (supposing for simplicity that the moon's velocity is zero in the nonrotating frame), the physically correct way to compute the relative velocity of the moon is to parallel transport its velocity to you, at the origin. When you do this, you'll find a velocity of zero, as expected.

In the latter case, when the moon is moving and you're not rotating, you are in an inertial frame, so the connection is trivial. So you can directly compare your velocity to the moon's, getting a nonzero result and hence a Lorentz contraction.

This problem also is common in general relativity. For example, naively calculating the coordinate velocity of distant galaxies suggests that they're moving away faster than the speed of light. And naively calculating the coordinate velocity of light falling into a black hole in the Schwarzschild metric says that the light's velocity slows down to zero as it approaches the event horizon. (The problems are worse in GR, since the nontrivial curvature of the spaces means that the result of parallel transport is nonunique.)

knzhou
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  • It is 100% false that non inertial frames have curvature or that general relativity is needed. – Javier Aug 23 '16 at 21:00
  • @Javier We must be disagreeing on what curvature is. In a rotating frame, geodesics aren't straight lines. Does that not count? – knzhou Aug 23 '16 at 21:05
  • I know that you can talk about noninertial frames in special relativity, but it's kludgy, and the real conceptual issues that arise there are all the same as the ones that arise in GR. – knzhou Aug 23 '16 at 21:07
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    @knzhou -- No. Curvature is invariant, so a flat spacetime is still flat from the perspective of an accelerating or rotating frame of reference. – David Hammen Aug 23 '16 at 21:08
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    For example, see http://physics.stackexchange.com/a/13697/52112 . – David Hammen Aug 23 '16 at 21:09
  • @DavidHammen Ah, I see, you're right. – knzhou Aug 23 '16 at 21:11
  • @DavidHammen Actually, I think my answer is independent of that. The curvature is zero, but the connection in a rotating frame is nontrivial. So you do need the concept of, e.g. parallel transport to talk about the velocity of a distant object. You don't need the rest of GR, though. – knzhou Aug 23 '16 at 21:14
  • @Javier (@DavidHammen) Okay, thanks for setting me straight. I think my revised answer is now correct, can you have a look? – knzhou Aug 23 '16 at 21:19
  • Much better. You need parallel transport even in Newtonian mechanics to deal with rotating frames; i.e., $v_\text{rot} = v_\text{inertial} + \omega\times r$ is sometimes called the transport theorem. Even Goldstein uses handwave derivation. A proper derivation requires differential geometry. – David Hammen Aug 23 '16 at 21:24
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One important proviso: According to A. P. French, Special Relativity, MIT Introductory Physics Series (1968) pp 149-152 (still a gem as far as I'm concerned) who noted that there is an important difference between "observe" and "see or look at", the words see and look involve the finite time for the transit of the light from different parts of the body.

French gives references to the following papers J Terrell, Phys Rev 116, pp 1041-45 (1959), V F Weisskopf, Physics Today 13 pp 24-7 (1960) and G. D. Scott and M.R Viner, Am J Phys 33 p534 (1965) and finishes with the comment "This misconception, which must have made every physicist blush a little ....".

jim
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