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There seems to be a consensus that the one - way speed of light is anisotropic in a rotating frame of reference (Sagnac Effect).

According to this article Einstein synchronization "looks this natural only in inertial frames. One can easily forget that it is only a convention. In rotating frames, even in special relativity, the non-transitivity of Einstein synchronization diminishes its usefulness. If clock 1 and clock 2 are not synchronized directly, but by using a chain of intermediate clocks, the synchronization depends on the path chosen. Synchronization around the circumference of a rotating disk gives a non vanishing time difference that depends on the direction used.

Imagine a rotating ring of arbitrarily large diameter. In accordance with the foregoing the one - way speed of light along the ring clockwise and counterclockwise will be different, because simultaneously emitted in opposite directions beams of light that go along the ring will return to the starting point at different times. Hence, it is reasonable to assume that it is anisotropic on any segment of a ring, large or small, say on a segment AB.

Of course, taking into account the Lorentz contraction, the measured round - trip speed of light on any segment of the ring will be exactly equal to c.

Suppose that, a purely inertial laboratory S’ for a very long time moves tangentially to the circumference on which the ring lies, very near to the AB segment.

How does the anisotropic one – way speed of light on the AB segment can magically turn into isotropic one - way speed of light in the co-moving inertial laboratory S’, as the Einstein’s relativity teaches us?

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    $\begingroup$ I'll do you one better: how can rotation magically turn into translation just by using an infinitely large ring? $\endgroup$
    – John Dvorak
    May 20, 2019 at 11:16
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    $\begingroup$ @JohnDvorak, I would ask a different question, namely whether rotation around an infinitely large ring has any true meaning? If you accept the existence of such a ring, then rotation around its axis (which is at an infinitely large radius from the ring) would perhaps be indistinguishable from translation. $\endgroup$
    – Steve
    May 20, 2019 at 11:32
  • $\begingroup$ For any rotating ring $\omega=v/r$ and since $v<c$ then $\lim_{r\to\infty} \omega = 0$ so the rotating reference frame is no longer rotating. $\endgroup$
    – Dale
    May 22, 2019 at 15:57

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There is nothing "magical" about this. For a rotating ring $\omega=v/R$ where $v$ is the tangential velocity of the ring and $R$ is the radius of the ring. Since $v<c$ then $\lim_{R\to\infty}\omega=0$. So then the rotation is 0 and the speed of light is isotropic for any $v$.

This should not be surprising at all. The whole reason that you can approximate a large rotating ring as nearly inertial (to first order) is precisely because as the ring becomes large the angular velocity becomes small. This eliminates both the centrifugal force and the Coriolis force, as well as the Sagnac effect and any other first order non-inertial effects.

EDIT: Furthermore, it is fundamentally incorrect to think of the Sagnac effect as detecting the anisotropic speed of light in the rotating reference frame. For convenience I will use cylindrical coordinates $(r,\theta,z)$.

The anisotropy in the one way speed of light in the rotating frame depends on $r$, but the Sagnac effect does not. The anisotropy in the one way speed of light is maximum in the $\hat \theta$ direction and zero in the $\hat z$ and $\hat r$ directions, while the Sagnac effect is maximum in the $\hat z$ direction and zero in the $\hat r$ and $\hat \theta$ direction. The anisotropy in the speed of light disappears in a local inertial frame, but the Sagnac effect does not.

Basically, although both do occur in a rotating frame they are not the same. About the only thing that they share is that they both depend on a rotation rate.

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    $\begingroup$ @Albert, can you point out what specifically is "very incomprehensible and unusual, completely contradicting elementary common sense, hence absolutely wrong". Are you unaware that $\omega = v/R$? Are you unaware that $v<c$? Do you not understand how to evaluate the limit of $\omega$ as $R$ approaches infinity? $\endgroup$
    – Dale
    Jun 2, 2019 at 12:37
  • $\begingroup$ It is clear that if the rotation stops, the laboratory at the edge will stop and the speed of light will become isotropic. And why have you decided that it would stop? Infinitely tend to zero doesn’t mean equal zero. We can keep linear velocity of the edge infinitely close to c and yes, $\omega$ will infinitely tend to zero if the radius tends to infinity. Is it not clear? If this is hard for you to understand, you can reduce the radius slightly. $\endgroup$
    – Albert
    Jun 3, 2019 at 19:46
  • $\begingroup$ “if the rotation stops, the laboratory at the edge will stop”, nope, this is incorrect. You can fix v at any desired value <c. As R increases $\omega$ decreases even if v is constant. $\endgroup$
    – Dale
    Jun 4, 2019 at 1:37
  • $\begingroup$ Yes, if $v$ is constant and $R$ increases, $\omega$ decreases, that is exactly what I have said. If rotation stops, the edge will stop. What is according to you "incorrect"? Do you want to say, that if there is no rotation of the ring (the ring is completely at rest) , it's edge is still rotating? Or vice versa? $\endgroup$
    – Albert
    Jun 4, 2019 at 4:27
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    $\begingroup$ $\lim_{R\to \infty} v/R = \omega = 0$ does not imply $lim_{R\to\infty}v=0$. Don't blame SR for your inability to correctly take a limit. $\endgroup$
    – Dale
    Jun 5, 2019 at 13:10
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Short answer: it is anisotropic, but that isn't an inertial frame of reference.

Long answer:

"Frame of reference" in special relativity means "coordinate system". When you talk about a frame of reference you need to be clear about what coordinates you're talking about, including (especially!) the time coordinate.

We keep time on earth by means of a bunch of clocks that are located in various places on the planet, and that we keep synchronized somehow. Let's only worry about clocks at the equator because that's enough for this problem.

Take two adjacent clocks A and B and set them so that the light travel time from A to B and B to A is equal as measured by those clocks. This is called Einstein synchronization and it's how clocks in inertial frames are synchronized. Then adjust C, the clock on the other side of B, so that it's synchronized with B in the same sense. Keep doing this and you'll eventually circumnavigate the globe and get to Z, which is next to A. You'll find that Z and A are not synchronized, and if you try to fix that they'll end up out of sync with Y and/or B. There's a lump under the carpet, and you can push it around but you can't get rid of it.

Let's try a totally different approach: broadcast a light speed signal from the north pole and have each clock set itself to a certain time (say t=0) when it receives that signal. This is essentially how we synchronize most clocks in real life (though the signal often travels over the internet, and the details are much more complicated). It works pretty well. If you look at adjacent pairs of clocks on the equator after this procedure, they'll all be more or less in sync. But none of them will be exactly Einstein synchronized. The lump is still there, it's just evenly distributed around the equator now.

Now, let's take two clocks on the surface, say A and B, and extend them to a line of synchronized clocks extending out into space (tangent to the surface). This is your rocket-ship coordinate system. If you synchronize those clocks pairwise to match the way A and B are synchronized, and A and B were synchronized by the signal from the north pole, then every pair of clocks will be not-quite-Einstein synchronized in the same way as A and B. The speed of light will be anoisotropic as measured by these clocks.

This doesn't contradict special relativity because SR only says that the speed of light is $c$ in inertial reference frames, and this isn't one. Inertial reference frames synchronize their clocks by the Einstein procedure by definition. Your reference frame is a perfectly valid coordinate system, and it might be useful in analyzing the motion of the rocket ship relative to clocks on the ground; it just isn't one of the specific family of coordinate systems that special relativity says the speed of light is $c$ in.

(You might be thinking that it's circular to say that the speed of light is $c$ in inertial reference frames when they're synchronized to make it $c$ in the first place. The interesting thing about them, though, is that they exist at all. In a non-special-relativistic world, either you can't define Einstein synchronization (because the speed of light varies) or the laws of physics don't take the same form with respect to all of those reference frames.)

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  • $\begingroup$ Imagine that in a center of circumference there is a source of monochromatic light (a clock). Could you please let me know, what frequency (or clock rate) of the source would measure: 1) rotating around the source observer ; 2) 100% inertial observer, who momentarily coincides with the rotating one? Which clock synchronization would be in accordance with the outcome of the measurement? $\endgroup$
    – Albert
    Oct 6, 2019 at 9:53
  • $\begingroup$ @Albert Synchronization is irrelevant to that problem as you only need time differences on a single clock to determine the frequency. Assuming the clocks measure proper time, the light is blueshifted by a factor $\gamma$ whether measured by (1) or (2). Light emitted by (1) will be seen by the center as redshifted, and light emitted by (2) when coincident also will be. Transform to (2)'s rest frame and you'll see that the center is moving toward (2) when it emits the blueshifted light seen by (2), and moving away from (2) when it sees the redshifted light emitted by (2). $\endgroup$
    – benrg
    Oct 6, 2019 at 15:49
  • $\begingroup$ Einstein 1905, &7, ''From the equation for $\omega$ it follows that if an observer is moving with velocity $v$ relatively to an infinitely distant source of light of frequency $\nu$, in such a way that the connecting line “source-observer” makes the angle $\phi$ with the velocity of the observer referred to a system of co-ordinates which is at rest relatively to the source of light, the frequency $\nu'$ of the light perceived by the observer is given by the equation' $\nu'=\nu \frac {1-cos \phi \cdot v/c}{\sqrt {1-v^2/c^2}}$ $\endgroup$
    – Albert
    Oct 6, 2019 at 16:36
  • $\begingroup$ @Albert Anyone can pick any coordinates, which means anyone can consider themself to be moving or not as they like. If you're trying to suggest that some phenomenon can only be understood if you assume something to be moving (resp. at rest), that's not true since all coordinate systems describe the same world, including those coordinate systems in which that thing is at rest (resp. moving). $\endgroup$
    – benrg
    Oct 7, 2019 at 1:48
  • $\begingroup$ @Albert Incidentally, Einstein (1905) didn't use the modern SR "observer" that's really a reference frame. His observers were just scientists recording data. They didn't have invisible infinite lattices of clocks and metersticks attached to them, but if there was a convenient object like a train or a lattice of clocks and metersticks near a phenomenon of interest, they could use it as a reference body (whether they were moving with it or not). $\endgroup$
    – benrg
    Oct 7, 2019 at 1:50

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