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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?

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In both quantum and thermal cases, as the particles get closer than the size of their screening cloud (Debye radius or e-h Compton wavelength respectively) the effective charge becomes larger. In the sense I agree with your prof. –  Slaviks Oct 1 '13 at 14:05
    
@Slaviks Interesting comment. I hadn't thought about it in terms of screening since there does not necessarily need to be a bare long-ranged interaction between the particles for depletion flocculation to occur. –  Mark Mitchison Oct 1 '13 at 16:20
    
Yes, you're right - pure depletion interaction needs a lot of entropy around and I see no obious way to connect it to the explainable-by-ground-state-energy Casimir force... –  Slaviks Oct 1 '13 at 16:59

2 Answers 2

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).

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Yes, although in the ground state the Helmholtz free energy is actually just the energy, since the temperature of the pure vacuum state is zero. This is the key difference that I am wondering about: is it possible to treat the uncertainty in the vacuum fluctuations as some kind of statistical uncertainty to which we can associate an entropy. I suspect not, but it would be very interesting to know what precisely is wrong with this kind of argument. Other than the obvious one that pure quantum states are states of maximal knowledge/zero entropy. –  Mark Mitchison Oct 1 '13 at 17:57
    
@MarkMitchison: I am not sure this answers your question...., but the entropy is (minus) the derivate of the Helmholtz free energy relatively to the temperature, so we are able to get an entropy $S(a,L)$. If I am not mistaken, the dependence is $S(a,L) \sim (\frac{L}{a})^{d-1}$ –  Trimok Oct 1 '13 at 18:36
    
But then you must set the temperature to 0. Surely then the entropy vanishes? –  Mark Mitchison Oct 1 '13 at 18:40
    
No, as indicated in my answer, $\beta = \frac{2a}{\hbar c}$ –  Trimok Oct 1 '13 at 18:40
    
Oh no I see, sorry –  Mark Mitchison Oct 1 '13 at 18:40

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).

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Good point about the conductors not being perfect mirrors. This aspect is certainly different from the depletion interaction, where the blocking objects can be assumed to be infinitely hard. –  Mark Mitchison Oct 1 '13 at 17:54
    
I think it goes a bit further than that. Zero point electrodynamic interactions between neutral objects are in general different from depletion interactions. However, in some specific conditions they share similar features such as the fact that to a bigger volume corresponds more allowed states for a (non existing anyway because we are in vacuum) photon. –  gatsu Oct 1 '13 at 20:00
    
Just to mention to which extent the point about "counting" is subtle. If we trust the "number of modes argument", on the one hand we have a slab of size $L$ corresponding to an infinite but countable set of modes and on the other hand we have two semi-infinite spaces corresponding to an infinite and uncountable set of modes. The difference between the two should be infinite and that's about it end of the story.... –  gatsu Oct 1 '13 at 20:09
    
...what saves the story is that these modes are used as sampling points to estimate the area of a function via $\sum'_n f(n)$ for the slab and $\int dn\: f(n)$ for the outside. Overall the discrepency that remains is of the type that a rectangle of base unity and height $f(0)$ say does not accurately represent the area bellow $f$ between $n=0$ and $n=1$. Wonderfully this mathematical problem leads us to Casimir's prediction. –  gatsu Oct 1 '13 at 20:24
    
That's interesting, do you have a reference that expounds the sampling interpretation? –  Mark Mitchison Oct 1 '13 at 20:28

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