Fields other than the photon field in the Casimir effect?

1. Dec 13, 2009

bcrowell

Staff Emeritus
Can anyone tell me if the following argument correctly explains why the Casimir effect, as observed in experiments to date, can be explained by arguments that consider only the zero-point energy associated with the photon field, while ignoring other fields?

Between two parallel, conducting plates separated by a distance L, the longest wavelength you can have is essentially L. (Let's not worry about factors of 2.) Then the low-energy cutoff on the modes of oscillation of the EM field is $hc/\lambda$. This cut-off energy sets the scale for the zero-point energy, which determines the strength of the Casimir attraction between the plates.

On other other hand, suppose you have a particle of mass m. It seems like the experiments have been done with L in the micrometer range, and then if we're talking about the electron-positron field, the low-energy cutoff is obtained when the particles are nonrelativistic, so it occurs at $E=p^2/2m$, which (again ignoring factors of 2) comes out to be $(hc/\lambda)(v/c)$. Therefore it's down by a factor of v/c compared to the energy cutoff for photons.

To get the field of massive particles to contribute comparably to the photon field, you could try looking for less massive particles, which might be relativistic even when their wavelengths were comparable to L. But the only particles we have that fit this description are neutrinos, I think, and since they're electrically neutral, the existence of the conducting parallel plates doesn't impose any boundary condition on them.

Is this right?

2. Dec 13, 2009

Count Iblis

Yes, you need to impose boudary conditions on the relevant field. But for massive fields, the effect will fall of exponentially like
exp(-L/lambda) where Lambda is the Compton wavelength.

Suppose you have just one plate. Then that modifies the boundary conditions of the field there and a distance L away where you wan to put the other plate, the effect is already exponentially suppressed by exp(-L/lambda). So, putting the other plate there will always yield an exponentially small effect if L >> lambda.

3. Dec 13, 2009

bcrowell

Staff Emeritus
Thanks, Count Iblis, that's very helpful!