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When I first learned about the depletion interaction, my initial reaction was that it looks very similar to the Casimir effect. On making this remark to the professor, he replied somewhat mystically: "It is the Casimir effect." No further detail was supplied, however.

Nevertheless, it really looks like both problems can be understood in terms of degrees of freedom maximising their positional entropy by reducing the volume of some "forbidden region" between two objects. For the depletion interaction, the fluctuating degrees of freedom are small particles, while in the Casimir effect, these degrees of freedom are long-wavelength modes of the free radiation field. However, a key difference is that in the first case the fluctuations arise thermally, while in the second case they are unavoidable quantum fluctuations of the vacuum.

Is it possible to derive the Casimir force between two conducting plates from entropic considerations alone?

Qmechanic
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Mark Mitchison
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2 Answers2

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Your analogy is quite right in the sense that the original derivation from Casimir between two infinite and ideal conducting plates in vacuum leads to an attraction owing to a greater number of modes "outside" than "inside" the slab they form as it is shown at the beginning of this review.

Note that this pseudo counting method only holds because an ideal modelling of the plates has been used here where the electric field is considered to be exactly zero inside the plates.

If you change a little bit these assumptions, you end up with an interaction whose physics is further from a depletion interaction but closer to an actual macroscopic van der Waals interaction which would also be right in the Casimir case (as shown by Lifshitz in the 70s' I think).

gatsu
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Partial answer:

Because the distance between the plates is finite, you have a natural correspondence between the quantum Casimir problem, and a thermodynamic problem.

In fact, in a $d$-dimensional space, if you calculate the Helmholtz free energy, in a grand-canonical formalism, for a perfect photon gaz (so with zero chemical potential), with the correspondence $\beta = \frac{2a}{\hbar c}, V = aL^{d-1}$, for plates with characteristic length $L$ separated by a distance $a$ , you will find exactly the Casimir energy for a photon field.

And it works for fermions too, where you have to consider a perfect gaz of relativist fermions (with zero chemical potential).

Trimok
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