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i was a little confused about the casimir effect. my understanding is with the 2 plate experiment there was a force pushing the plates together because there were less virtual particles in between the plates than outside them. Why are there less particles in between the plates?

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2 Answers 2

The Casimir effect (plate attraction) is better understood as a Van der Waals force due to mutual polarization of neutral but extended quantum mechanical systems. Consider (find out about) two distant neutral atoms, for example. The case of plates is physically not too different from that. It is funny, but this attraction can be attributed to the virtual particles between atoms, contrary to the OP statement.

Force is a (minus) gradient of the electric potential of interaction. Calculation of possible modes inside plates helps calculate this potential energy.

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The objects whose number is lower in between the plates are not really particles per se but the different modes - different possible values of the wavelength or frequency, in particular - in which the particles may be created.

If the distance of the parallel plates is $L$, then the electric field has to vanish at the boundary between the vacuum and the metals - at the plates - and this implies that the electromagnetic waves may be decomposed to standing waves. The wavelength of such photons in the $x$ direction (the direction in which the plates are separated) has to be an integer multiple of $2L$. So only frequencies $$f_n = n\times c / 2L$$ are possible ($c$ is the speed of light). That's needed, once again, for the electric field to vanish at the plates. Note that I am talking about the transverse electric field and $-\nabla \Phi$ has to vanish in the directions $y,z$ because the potential $\Phi$ is constant throughout the metal.

If there were no plates, the frequency $f$ could be any real number - which is equivalent to the $L\to\infty$ limit of the formula above. In this sense, the plates restrict the allowed frequency - they reduce the number of possible values of the frequency.

(For the sake of simplicity, I was assuming that the photons don't have any motion in the directions $y,z$ along the plates - in general, they do have this motion. This fact would be correctly and easily accounted for by switching from the wavelength to the wavenumbers $k$.)

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Just for the record (and not for high-energy theorists): although you can 'easily' take the extra directions into account, the calculation itself is non-trivial, and does require some form of regularization to make the $k$-integrals meaningful. –  Gerben May 3 '11 at 8:23
    
Could you explain, please, why the plates should attract each other if 1) they are neutral and 2) the electric field (force) is zero at the plates anyway? –  Vladimir Kalitvianski May 3 '11 at 20:16

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