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  How to calculate the Casimir force on a closed hollow frustoconic resonant cavity in vacuum?

+ 2 like - 0 dislike

It is well known from the explanation of Casimir effect that the attraction between 2 conducting plates in vacuum is due to the fact that between the 2 plates, due to boundary conditions, there are less electromagnetic modes than in the rest of space (that is, in the space outside the space between the 2 plates). Therefore, there is a pressure differential of the zero point vacuum oscillations between the space between the plates and the space outside the plates. The long wavelength modes that are excluded (by the boundary conditions) from the space in between the plates act in the space outside the plates and their pressure makes the plates approach each other.

Similarly, in closed resonant cavity, due to boundary conditions there are fewer modes in which the zero point field can oscillate than in the exterior of the cavity. Therefore a closed cavity can be regarded as a rigid body in a perfect incompressible fluid. Now the pressure due to the more numerous zero point oscillating modes that are in free space as compared to the interior of the cavity, acts on a non-symmetrical body and it makes it move.

It is well known that an asymmetrical body immersed in an ideal incompressible fluid moves by itself in an accelerated way. See, e.g., http://www.sciencedirect.com/science/article/pii/S0022123610002727

In my case $$\vec{F} = \int_{cone \ surface} p\vec{n} dS $$ where $p$ is the difference between the pressure of the internal modes of the zero point field allowed inside the cavity and the external pressure of the zero point field outside the cavity (in vacuum), where are more oscillation modes available.

Question: How to calculate $p$?

Thank you!

asked Jan 4, 2017 in Theoretical Physics by mhrt (55 points) [ no revision ]

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