Let a clock A be at rest in an inertial frame and let a clock B rotate

[Yayness]

Let a clock A be at rest in an inertial frame and let a clock B rotate around it with constant velocity and constant distance from A. Let v be the velocity of B relative to the inertial frame.
Both clocks count how long it takes for B to rotate once.
According to clock A, the time it takes is t, and according to clock B, the time it takes is τ.

My book used the formula from SR about time difference: t=γτ
I tried to find out if that was right and ended up with t=γ²τ. Is the book wrong?

 

[PAllen]

Let a clock A be at rest in an inertial frame and let a clock B rotate around it with constant velocity and constant distance from A. Let v be the velocity of B relative to the inertial frame.
Both clocks count how long it takes for B to rotate once.
According to clock A, the time it takes is t, and according to clock B, the time it takes is τ.

My book used the formula from SR about time difference: t=γτ
I tried to find out if that was right and ended up with t=γ²τ. Is the book wrong?

How did you arrive at your answer? I think the book is right. This is speed dependent effecat.

 

[Yayness]

I'll try to explain it shortly.

The path of the clock B forms a circle. Let this circle be at rest relative to the inertial frame and let the radius be r.

I added a non-inertial frame of reference where B is at rest. This reference frame does not rotate around its own axis, so left will always be left, and right will always be right etc.
In this reference frame the circle will look like an ellipse (because of length contradiction, as the whole circle has the same velocity relative to the reference frame), and B will always be located where the curvature of the ellipse is largest.

I added an imaginary circle inside the ellipse, and the perimeter goes through the clock B. The imaginary circle and the ellipse have the same curvature in this point. (See attachment.)

Because the curvature is the same, the clock B will feel an acceleration equal to v² divided by the radius of the imaginary circle (which is r/γ²). So the acceleration B will feel, is γ²v²/r, like if the radius of the real circle were r/γ².

I let t=2πr/v and τ=2πr/(γ²v) (because the radius of the imaginary circle is r/γ²). That's basically how I ended up with t=γ²τ.

 

[JesseM]

I'll try to explain it shortly.

The path of the clock B forms a circle. Let this circle be at rest relative to the inertial frame and let the radius be r.

I added a non-inertial frame of reference where B is at rest. This reference frame does not rotate around its own axis, so left will always be left, and right will always be right etc.
In this reference frame the circle will look like an ellipse (because of length contradiction, as the whole circle has the same velocity relative to the reference frame)

The length contraction rule is only defined relative to inertial frames, you can't assume it would work the same way in a non-inertial frame. In the most common type of rotating frame, the path would still be a circle. If you want to use a different type of frame, you need to at least give a specific definition of the coordinate transformation from the inertial frame to your non-inertial frame, only then can you figure out how clocks will behave in this frame (just like with the length contraction formula, the standard time dilation formula also doesn't generally work in non-inertial frames).

In any case, why bring non-inertial frames into this? You can figure out the time on the rotating clock using an inertial frame, it's much easier.