# Casimir effect for spinning Casimir plates

I recently thought of the following experiment. Let's say I have two plates in vacuum facing each other. Now, due to the Casimir effect, there will be some internal attraction between the plates. Now let's say we spin the plates while facing each other about their combined centre of mass.

The shortest path between the plates is no longer a straight line as the geometry changes when undergoing acceleration. What is the new shortest path between the plates and force between them due to the change in geometry?

• Why is spacetime curved? As far as I know this can be handled with special relativity. – Javier Aug 3 '15 at 19:27
• @Javier I massively edited the thought experiment hopefully now it does have elements of general relativity :) – drewdles Aug 6 '15 at 13:25
• I don't understand what the Casimir energy has to do with this. The problem can be formulated independently of the Casimir effect by demanding a spinning ring of radius R. Are the two points under consideration on the ring or are they exterior to the ring but in close proximity to the spinning ring? – user58089 Aug 6 '15 at 13:45
• So my understanding of the Casimir effect is as follows: If I have two plates with a gap .. They are attracted to each other ... Now similarly the ring should also experience this force which is dependent on the distance between two points (on the ring) ... I hope this explains what I'm thinking (please help me rephrase the experiment if you can) – drewdles Aug 6 '15 at 13:51
• I tried to solve it in Born coordinates, but I am afraid it is a bit too involved. I am getting like $r^2 (1 - 2\omega^2 r^2) R'' + r (1 - 4\omega^2 r^2) R' - k_z^2 + [r^2( \alpha_t^2 - \omega^2 \alpha_\phi^2- \omega \alpha_{\phi t}) + \alpha_\phi^2]R = 0$ just for one part of the solution. Maybe you could try to apply a coordinate transform to the stress energy tensor using Born coordinates and see what happens? I suspect it may not be enough, since the transformation applies to $\langle 0 \vert \hat{T}_{ab} \vert 0 \rangle$ as a whole, but it might be worth a try, perhaps – Slereah Aug 10 '15 at 11:18

## 1 Answer

In layman's terms, the Casimir effect is an outside pressure pushing the plates together. It comes from modes of quantum fields that have longer wavelengths than the separation of the plates. Therefore, these modes can no longer be excited by vacuum fluctuations.

Since special relativity is a basic ingredient in QFT, these fields are homogeneous and isotropic. Thefore a constant Lorentz transformation will not alter the measurement of the Casimir effect.

In general relativity, we allow for non-constant Lorentz transformations and even for fully general diffeomorphisms. Then, there are new effects to consider, most prominently the Uhruh effect.

The derivations of the Unruh effect assumes constant acceleration for simplicity. A rotating frame implies non-constant acceleration for parts of your plates. A naive consideration in your thought experiment could go like

• Let us consider only the outermost edge of each plate
• These edges are accelerated towards the center of mass
• If we are allowed to consider infinitesimal time slices, the acceleration is approximately constant for any given time slice
• The acceleration towards the c.o.m. implies (in the rest frame of the edge) Unruh radiation, reducing the magnitude of the acceleration
• Therefore the measured attraction between the plates should be reduced

The above argument assumes that we are allowed to approximate the acceleration to be constant at each instant, which might not hold. Also, the Unruh radiation only exists in the rest frame of the edges, which are different frames for each edge, so concluding that there is less attraction from just considering one edge is quite bold. Finally, I did not carefully consider the transition from the rest frames to the lab frame.

All these caveats could be solved by careful consideration, but that is beyond the scope of my interest for this thought experiment :-P

• I suppose 50 bounty points aren't enough to tempt you ... :( – drewdles Aug 7 '15 at 17:43
• @AnantSaxena No, I'm afraid they are not. Looking into these questions in detail easily involves hours of work and from my experience QFT in GR backgrounds can be so confusing that even then I'm not sure I'd get the answer right... – Neuneck Aug 10 '15 at 12:58
• I'm glad that you atleast told me the considerations ... Cheers and thank you! – drewdles Aug 10 '15 at 14:29