SR time dilation in a centrifuge dilema Picture a giant centrifuge gyrating close to the speed of light (indulge me).
Person A is sitting in it close to the axis, person B further away on the radius and they are facing each other. To someone outside of the centrifuge person B is going faster than A, and according to special relativity B would experience more time dilation.
However, relative to A, B would just be sitting still, and therefore shouldn't experience any dilation at all. Can someone please explain this to a poor physics enthousiast?
Just a note, even though the force human centrifuges produce is mesured in G, I don't believe general relativity applies here since there is no mass generating actual gravity.
 A: First of all, note that if A shines a beam of light at B, the beam will not hit B but will instead seem to curve away. This tells A that they are in a "non-inertial" frame (in inertial frames light travels in straight lines) and that something funny is going on. In particular, the second postulate of relativity (that the speed of light is constant) needs to be adapted for non-inertial frames.
Special relativity can handle this situation. Minkowski developed an elegant formulation of SR in which the important thing is a kind of "distance" between events in spacetime, defined as:
$$
ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2
$$
For convenience we'll use units where the speed of light $c$ is 1, so we'll omit that in future equations. Also note that unlike distances in space, the spacetime "distance" doesn't need to be positive. In fact there are 3 possibilities for the sign: a positive value is called a "timelike" distance, a negative value is a "spacelike" distance, and a 0 value is a "null" or "lightlike" distance. The choice of signs is of course arbitrary. But a very useful thing is that for timelike distances, the square root is the elapsed time for an observer "at rest". To figure out how much time has elapsed (according to A) between ticks of B's clock, we'll need to figure out the interval between the events of the start of one B tick and the end of one B tick.
Doing this requires figuring out the equation of the path followed by light (remember this needs to work out to 0, a "lightlike" distance) according to A. It's not actually too bad, and if you grind out the math (google "Born coordinates" for the details) you'll get:
$$
ds^2 = (1 - \omega ^2 r^2) dt^2 - 2 \omega r^2dtd\phi + dz^2 + dr^2 + r^2d\phi^2
$$
where $r$ is radial distance from the hub, $\phi$ is the angle away from the direction A is looking in, $z$ is height, and $\omega$ is the angular rotation speed, which A will have to determine by experiment to match the path of beams of light.
Note that the $dt^2$ value decreases as $r$ increases... so since B has a positive $r$, A can conclude that B's clock is slow relative to their own.
The situation is not symmetric, because if B wants to use a frame in which they are at rest they'll have to figure out the appropriate expression for $ds^2$ for their chosen coordinates, and it will be different (and will show that A's clock is ticking faster).
A: 
However, relative to A, B would just be sitting still,

No. Their velocities are the same as the velocities of their cats that they let go off of simultaneously.
We know that those cats move linearly, so we are not so much confused, hopefully.
If A and B have some difficulty figuring out the momentary velocity of each other, they can ask those aforementioned cats.
When a cat measures the velocity of the other cat, it should do it the right way. It should  use a meter stick and two clocks. The stick should not rotate.
