# Casimir effect - silly question, but

1. Oct 25, 2009

### nomadreid

Casimir effect -- silly question, but...

The summation of the standing waves of virtual particles is calculated between the plates, so that you only have the harmonic series to sum over. Fine. However, what about non-virtual background radiation? Wouldn't only standing waves survive between the plates, thereby again giving you only the harmonic series to calculate over, as opposed to outside? I perceive this as being wrong, because otherwise one could explain the effect classically, which one cannot. So where is the difference?

2. Oct 25, 2009

### Born2bwire

Re: Casimir effect -- silly question, but...

That is not part of the Casimir force effect. That would be a radiative pressure problem independent of the Casimir force, it would be hard to say how to treat it in the first place. Do we consider the background radiation traveling waves or standing wave? I don't actually but if we have a net traveling wave then the radiative pressure from the background radiation would be direction dependent. More likely it's just a random wave I guess.

Still, it's pretty much a moot point because it would probably be drowned out, or replicated, by temperature effects with the Casimir force. The simplest case is zero temperature but there are plenty of extensions to the theory that allow for non-zero temperature, finite conductivity, and dielectric constants.

The Casimir effect is unique because it is a direct result of the vacuum fluctuations of the electromagnetic fields. Any background radiation would be a non-vacuum state since it carries photons. There are a number of ways you can calculate the Casimir effect though. You can directly calculate the change in the vacuum energy by finding all possible modes that exist in your system. You can find the energy change in the fluctuating fields by the introduction of your system. You can find the energy change in the equivalent sources of the fluctuating fields (no photons assumed here, real or virtual).

3. Oct 26, 2009

### conway

Re: Casimir effect -- silly question, but...

As best I can do the calculation, it looks to me like there ought to be a classical Casimir effect which should be linear with temperature. Therefore it should disappear at absolute zero. The leftover effect would be the quantum phenomenon that everyone talks about.

4. Oct 26, 2009

### Bob_for_short

Re: Casimir effect -- silly question, but...

The Casimir effect is akin to the long-distance interaction of atoms (Van de Waals force) and its strict calculation can and should be done in this way. Concentrating too much on "vacuum filed fluctuations" is quite misleading since one thinks of them as of some separate object whereas it is an interaction of distant neutralized charges including the quantized electromagnetic field. Interaction of non-neutralized charges also contains this effect (the Lamb shift).

Now it is clear that finite temperature effects come in addition to that.

Last edited: Oct 26, 2009
5. Oct 26, 2009

### Bob S

Re: Casimir effect -- silly question, but...

The long distance Casmir effect was first seen by 18th century sailors:
http://infao5501.ag5.mpi-sb.mpg.de:8080/topx/archive?link=Wikipedia-Lip6-2/7555.xml&style [Broken]
A macroscopic effect analogous to the Casimir effect was observed by 18th century French sailors. Where two ships are rocking from side to side in conditions with a strong swell but light wind, and the ships come closer together than roughly 40Â m, destructive interference eliminates the swell between the ships. The calm sea between the ships has a lower energy density than the swell to either side of the ships, creating a pressure that can push the ships closer together. If they get too close together, the ships' rigging can become entangled. As a countermeasure, a handbook from the early 1800s recommends that each ship should send out a boat rowed by 10 to 20 sailors to physically pull the ships apart.
Bob S

Last edited by a moderator: May 4, 2017
6. Oct 26, 2009

### Bob_for_short

Re: Casimir effect -- silly question, but...

No, it is an external wave pressure effect. We know that the light exerces a pressure on a surface but it is not the Casimir effect.

7. Oct 26, 2009

### Born2bwire

Re: Casimir effect -- silly question, but...

The problem with using the retarded Van der Waals force is that the Van der Waals force is not an additive force except in what I think are called rarefied materials. If one were to calculate the Casimir force effect using simple inter-particle retarded Van der Waals force you come up with a decidedly incorrect answer. The much more efficient and accurate method is to work with the vacuum fluctuations since then it is just a matter of finding the change in the energy of the fluctuation modes by the introduction of the scatterers.

8. Oct 27, 2009

### Bob_for_short

Re: Casimir effect -- silly question, but...

Really? The plates will not attract or what?
Saying "The energy of fluctuation modes" disconnects the quantized EMF from charges. Why plates should attract if it is the vacuum properties that change? Can vacuum make pressure? Why the vacuum energy density does not fill the gap from outside?
In fact, the quantized EMF comes always with charges, it is a charge property, if you like, and it is the charge interaction energy which is responsible for the forces. This understanding in the only correct one and it makes it possible to calculate the forces between any materials and at finite temperatures in a natural way.

Last edited: Oct 27, 2009
9. Oct 27, 2009

### Born2bwire

Re: Casimir effect -- silly question, but...

With the retarded van der Waals, the calculated forces are around an order of magnitude less than experiment. This was found out in the early 50's and motivated Lifgarbagez to derive his theory for dielectrics in '56. The microscopic method of using the corrected van der Waals is not a good means of calculating the Casimir force because van der Waals is not an additive force. In fact, I have not seen such a calculation or derivation, instead, we use macroscopic methods for calculation. Under these conditions, we must work with the macroscopic fields or equivalent sources of the fluctuating fields.

There are no charges of the like assumed to be explicitly involved in the vacuum explaination of the Casimir force. The vacuum state, while absent of photons, still has electric and magnetic fields. These fields fluctuate about a mean value of zero (hence why there are no observed fields in macroscopic ensemble of measurements). However, the energy density of these fields is infinite, only bounded by the frequency limit specified on electromagnetic waves. And, just like in many other examples in physics, if we disturb the energy density of the vacuum, the spatial dependency of this disturbance will cause a force.

The Casimir energy and force can be directly calculated using such assumptions. The introduction of scatterers into the vacuum forces the fluctuating fields to conform to the boundary conditions of the scatterers. In doing so, we restrict the number of modes that can exist in the vacuum state. Thus, the modes of the system of interest are now a subset of the original vacuum state. By quantifying the change in the energy between the original and desired systems we get the Casimir energy. To get the Casimir force, we can find the gradient of the Casimir energy (or by some equivalent method). How we quantify the energy content can be done by calculating the modes of the system, like in Casimir's original parallel plate or Lifgarbagez' method, finding the energy density of the fluctuating fields, this can be done by using the Maxwell Stress Tensor to find the force, or by finding the energy contained in the equivalent sources that, when impressed upon and inside the scatterers, perfectly reproduce the fluctuating fields. The latter method is based off of Schwinger's source theory model that he applied to the problem in a series of papers.

As for temperature and dielectric dependence, it is just a simple extension to these theories. Lifgarbagez' and Schwinger both present ways of dealing with arbitrary materials and temperatures.

Another way that gives equivalent results is to calculate the radiation pressure of the vacuum's virtual photons. This is to be expected of course since it would be equivalent to dealing with the energy density of the fluctuating fields directly.

10. Oct 27, 2009

### Bob_for_short

Re: Casimir effect -- silly question, but...

Thank you, Born, for your popular explanation. Tell me, please, why the additional energy density does not flow into the slit from outside?

11. Oct 27, 2009

### Born2bwire

Re: Casimir effect -- silly question, but...

It's a static system, so the Hamiltonian is time invariant, and more importantly since there aren't any real photons there isn't any energy flow associated with the fields. But the virtual photon perspective can help provide insight. In the simplest case, we have two infinite parallel perfectly conducting plates, then the plates will only allow specific modes, field solutions, to exist between the plates. Even we allowed the energy of the fields to flow about the volume of our system, not all of the energy could go in between the plates for the same reason why there is a countable number of modes that can propagate down a parallel plate waveguide. The boundary conditions enforced by the scatterers restrict the field modes that can exist. Outside our plates though, a far greater number of modes can still exist. The difference in the energy densities outside and in between the plates can be manifested in virtual photons. But there are more virtual photons outside the plates, so the pressure they exert will be larger than the opposing pressure from in between the plates.

12. Oct 27, 2009

### Bob_for_short

Re: Casimir effect -- silly question, but...

The electric field boundary conditions is zero inside and outside, so how the zero electric field can influence the plates?

13. Oct 27, 2009

### conway

Re: Casimir effect -- silly question, but...

People keep bringing up the restricted number of modes between the plates as though it explains the force. It doesn't. Yes, there are fewer modes between the plates but they are much stronger. The photon in a given mode hits each plate many times per second, while the corresponding photon in the same mode outside the plates is randomly off somewhere near infinity and only rarely hits the plates. So merely counting modes doesn't tell you anything about the pressure.

I have written up a simple calculation of the force which I can post if anyone is interested.

14. Oct 27, 2009

### Born2bwire

Re: Casimir effect -- silly question, but...

It's just like the difference between free-space propagation and a parallel plate wave guide. In free-space, classically, there is a continuum of modes that can propagate. But when you introduce a parallel plate waveguide, the boundary conditions only allows for a countable number of modes with a lower limit the continuum of frequencies that can propagate. So in comparison with the free-space case, the parallel plate waveguide reduces the number of modes that can exist.

With the Casimir force, we are not talking about propagating modes of electromagnetic waves, but the frequencies of the quantum field oscillators that represent the electric and magnetic fields. The presence of the scatterers enforces boundary conditions which change the modes that can be present. Since the physical arrangement of the scatterers alters these boundary conditions, the energy of the system is also dependent on their position. So a slight change in the position of a scatterer changes the energy of the system. This change is what gives rise to the force. An alternative way to look at is with the Maxwell Stress Tensor. The tensor describes the momentum flux of the fields, the gradient of which gives the force volume density.

15. Oct 27, 2009

### Bob_for_short

Re: Casimir effect -- silly question, but...

You are nearly here!

You recognize the importance of charges. It is they who "makes" the boundary conditions. It is only possible in case of strong filed-charge interaction or coupling, isn't it? So the quantized EMF comes with charges, not with vacuum. The only step to make is to recognize that it is the potential energy of interacting charges that is actually calculated and that includes corrections from the quantized EMF which are essential in case of neutralized charges.

16. Oct 27, 2009

### Halcyon-on

Re: Casimir effect -- silly question, but...

Could you give me references of what you are saying? Thanks.

17. Oct 27, 2009

### Born2bwire

Re: Casimir effect -- silly question, but...

We aren't counting the number of modes, we are finding the modes that exist and add up the energy of these modes, the energy of each mode is different, dependent on the mode's frequency. More correctly, we are finding the change in energy in the vacuum state if we perturb the position of our scatterers. To do this, we have a two step problem. The first step is to characterize the energy of our given system, and the second step is to characterize the energy with respect to a reference system, usually the empty vacuum state. We need the reference to account for the divergence of the vacuum energy. Even in the restricted case of the system with scatterers the energy is still divergent. Otherwise, we wouldn't care about the change from the empty vacuum state since this reference energy is independent of the scatterer's displacement. Though the invariance of the Casimir force to the empty vacuum energy is probably another good reason to use it as a reference.

You can find this in just about any treatment of the Casimir force because it is the typical method for finding the force between two parallel plates. Peter Milonni's "The Quantum Vacuum" is a good reference text and he goes into great detail about the Casimir force. A quicker read would be Casimir's second paper on the subject. Technically, his first was the Casimir-Polder paper where they found that including the speed limit of light that they can derive a retarded van der Waals potential. This is the same effect as the Casimir force that he derives using the vacuum modes in his second paper, http://www.historyofscience.nl/search/detail.cfm?pubid=2642&view=image&startrow=1 [Broken] .

Oh fer Christ's sake. Look man, I ain't 12 years old. If you want to waste my friggin time so that you can somehow try to massage some little gotcha out of me you could at least give me the common courtesy of knowing ahead of time so that I don't sit here and waste my evening trying to explain this to you. The quantization of the electromagnetic fields is independent of charges, the vacuum state is devoid of all sources and photons but it still predicts the quantum field fluctuations. This can be explained in one way using the fluctuation-dissipation theorem as is described in Milonni's text. And as I explained before, we can separately describe the Casimir force using source theory, which impresses the equivalent currents and charges of the vacuum fluctuation fields on and inside the scatterers. But all of these various techniques, whether or not you include the vacuum, whether or not you include sources, all give equivalent results and thus raises ambiguities about any physical interpretation, as discussed in a paper by R. L. Jaffe which I have posted several times previously. I am not going to waste any more time on this with you. Honestly.

Last edited by a moderator: May 4, 2017
18. Oct 27, 2009

### Bob_for_short

Re: Casimir effect -- silly question, but...

If we charge the plates with additional electrons, it does not change the boundary conditions for the vacuum fluctuations, right? So the force will not change from attractive to repulsive, will it?

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