It was my understanding that the speed of light is supposed to be constant for every observer, as in the classical mental experiment of the guy in the train with the flashlights which explain the relativity of simultaneity.
(i.e. observer A outside of the train, will see one beam hit the back wall before the other beam hit the front wall, because for him the “front beam” has to go for a longer way.
Observer B in the moving train, instead, will see the beams hit the two walls simultaneously.
But both the observers see the two beams move at c).
Now: with the Sagnac experiment, I would expect the same thing: I would expect observer A (still) to see the two beams arriving at the detector at different times because one beam has gone for a longer way.
I would expect Observer B, rotating with the device (including source and detector), to see the beams arrive simultaneously.
However, that is not the case, as observer B actually see one beam arrive after the other: how is that? What am I missing?
Best Answer
@JohnDuffield is indeed correct. Let me try to quantitatively show the anisotropy of the speed of light for an observer on the rotating platform. I think the following reasoning is to be credited to Langevin (I don't remember the reference, sorry, and anyway it would be in French!).
So let's denote by $\omega$ the angular speed of the platform with respect to the still observer. Clocks and rods on the rotating platform will be affected by their motion with respect to the still observer. What order in $\omega$ shall we expect? If there is a term proportional to $\omega$, then the clocks and rods will behave differently depending on whether the platform rotate in one or the other direction. This is silly, as this amount to simply having the still observer look from the top instead of from the bottom for example. So the first term has to be proportional to $\omega^2$. But then, that means that if we limit ourselves to an approximation at order $\omega$, we can completely neglect the fact that clocks and rods on the rotating platform will not measure time and lengths as for the still observer. What that means is that we can use a good old Galilean transform!
So let's denote by $(x, y, z, t)$ and $(x',y',z',t')$ the spacetime coordinates on the platform and of the still observer respectively, then
$$\begin{align} x'&=x\cos\omega t-y\sin\omega t\\ y'&=x\sin\omega t + y\cos\omega t\\ z'&=z\\ t'&=t \end{align}$$
In differential form, this reads
$$\begin{align} dx'&=dx\cos\omega t-dy\sin\omega t - \omega (x\sin\omega t+y\cos\omega t)dt\\ dy'&=dx\sin\omega t + dy\cos\omega t + \omega(x\cos\omega t-y\sin\omega t)dt\\ dz'&=dz\\ dt'&=dt \end{align}$$
Then we look at the spacetime interval $ds$. The still observer being inertial,
$$ds^2 = dt'^2-dx'^2-dy'^2-dz'^2.$$
I have used units where the speed of light $c=1$. Then substituting the above change of coordinates, and a bit of trigonometry, gives
$$ds^2 = dt^2 -2\omega(xdy-ydx)dt\underbrace{-dx^2-dy^2-dz^2}_{-dl^2}.$$
The propagation of a light signal corresponds to $ds^2=0$. The presence of cross terms $dxdt$ and $dydt$ results in an anisotropic speed of light. Let's investigate that precisely.
We note that $dA=\frac{1}{2}(xdy-ydx)$ is the area of the infinitesimal triangle whose vertices are the origin of the coordinates, the point $(x,y,z)$ and the point $(x+dx, y+dy, z+dz)$, that is to say the area swept by the vector from the origin to the position of the light signal during the duration $dt$. This is an area with a sign: positive if the sweeping is anti-clockwise and negative otherwise. So we get
$$dt^2 - 4\omega dAdt -dl^2=0.$$
We can solve this 2nd order equation in $dt$:
$$dt = 2\omega dA + \sqrt{dl^2+4(\omega dA)^2}.$$
But since we neglect terms of order $\omega^2$,
$$dt = dl + 2\omega dA,$$
i.e.
$$\frac{dl}{dt} = 1 - 2\omega \frac{dA}{dt}.$$
This is the speed of light for an observer on the platform: not only it is not equal to 1 but it depends on the direction of propagation since it depends on the sign of $dA/dt$.