Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A few years ago, when i studied the casimir effect interpretation as the filtering out of vacuum modes with appropiate boundary conditions, i had the following dilemma; supposedly the derivation of the force between the walls was entirely equivalent to just calculating Van der Waals forces. This, as it was argued, compelled us to not take too seriously the idea of a negative energy density of the space between the walls.

However, i was left wondering what would happen if one would take the interpretation based on vacuum modes a bit further, and we could somehow switch on and off the permitivity $\epsilon$ of the conductor; what sort of long range radiation we would expect to see? A possible experimental implementation would be if the layers where superconductive, and then a magnetic field would be switched on just to break the superconductive phase.

We probably need to use the Dirac-Heisenberg-Wigner formalism for these sort of dynamic quantum systems, but what would be intuitively expected to be detected?

share|cite|improve this question
to be clear, this is NOT the sort of dynamic casimir effect i'm after in this question, but definitely worth a read: in particular, it is interesting that they are modulating the inductance with a SQUID, which is not the same but it is quite related to the permittivity – lurscher May 26 '11 at 16:17
up vote 3 down vote accepted

A friend recently brought to my attention that this experiment was actually performed 6 months after i posted the question in this site:

Christopher Wilson from Chalmers (and his team) used the same mechanism that i've proposed in here: using a superconducting magnet to oscillate the mirror surface.

I'm glad to see that the idea really works!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.