# I Casimir effect and vacuum energy and a bit of relativity....

1. Aug 24, 2016

### asimov42

Hi all,

Silly question perhaps: I had understood that the energy density of the vacuum is constant throughout spacetime. But, with the Casimir effect, for example, the geometry of (real) matter (i.e., parallel plates), changes the vacuum energy density in between the plates - is this correct? (since only certain modes of the EM field are allowed between the plates)

So, if the vacuum energy density of various regions of space is influenced in various ways by the configuration of matter in that space, doesn't this wreck Lorentz invariance? I.e., you need to have a very specific vacuum energy spectrum to maintain Lorentz invariance, and if the spectrum between e.g., the plates is different, wouldn't this have implications for relativity if the plates were, e.g., accelerated?

Thanks.

J.

2. Aug 24, 2016

### Mentz114

There is no vacuum energy. The vacuum expectation of all fields is zero, nihil, rien, zut, nothing.

There is no vacuum in the plates experiment because the plates are there.

3. Aug 24, 2016

### asimov42

Um, then what produces the Casimir force?

See, e.g., the article here: http://www.scientificamerican.com/article/what-is-the-casimir-effec/

From the article, "the result being that the total amount of energy in the vacuum between the plates will be a bit less than the amount elsewhere in the vacuum."

4. Aug 24, 2016

### Mentz114

That language has sacrified accuracy for a simple explanation. If you scroll to the end of this page you will see links to other threads about the Casimir effect.

I should add that the vacuum in general relativity, transformed to a local frame is always Lorentz invariant. See 'Lambda Vacuum' .

5. Aug 24, 2016

### asimov42

Hi Mentz114,

Thanks for your responses - as a novice / layperson (in physics, anyway), I'm somewhat unclear on what you've said above. A number of the other threads mention vacuum energy.

I can understand that the expectation value (mean) for the EM field is zero - but I'm not sure how this plays into the idea that the vacuum energy density (see e.g., wikipedia article on vacuum energy, which also mentions the Casimir effect) should be non-zero (i.e., should in fact be enormous, but isn't)... why should the the ground state for the EM field have non-zero energy? I think I'm getting off topic and further into areas I'm unlikely to really grasp...

Anyhow, to the original question - I'm assuming then that regardless of the configuration of the plates and their velocity, Lorentz invariance holds (unless it doesn't hold more generally for some reason)?

6. Aug 25, 2016

### Demystifier

The vacuum is defined as the state of lowest energy. Energy can be defined as an eigenvalue of Hamiltonian. But which Hamiltonian? There are many different Hamiltonians used in physics, and hence there are many different "vacuums".

Some Hamiltonians are meant to be fundamental, describing "all" physics, or at least a large part of physics. Other Hamiltonians are merely effective Hamiltonians, describing only a small subset of all physical phenomena. Consequently, some vacuums are supposed to be fundamental, while other vacuums are merely effective vacuums.

The Casimir vacuum is one such effective vacuum. It is the lowest energy state for the system with Casimir plates at a fixed distance y. Clearly, such a vacuum is not fundamental, because in a fundamental vacuum there are no Casimir plates at all. The Casimir vacuum is the vacuum for the effective Hamiltonian, not for the fundamental Hamiltonian.

In what sense is this Hamiltonian not fundamental? In several senses. In this Hamiltonian
1. Only EM fields are quantized, while charged matter is treated as classical. (Technically, dielectric function $\epsilon({\bf x},\omega)$ is not a quantum operator.)
2. Only EM fields are dynamical, while charged matter is treated as a fixed background. (Technically, dielectric function $\epsilon({\bf x},\omega)$ is a fixed function.)
3. Even the distance y between the plates is treated as a fixed parameter. (Technically, it is a consequence of 2.)

The fact 3. is particularly important for the physical interpretation of Casimir effect. As long as y is fixed, there is no Casimir effect at all. The Casimir effect is the force in the y-direction, and the existence of such a force requires y to be a dynamical variable, not a fixed parameter. To make y dynamical, one must add a new kinetic term to the Hamiltonian. But with this new Hamiltonian, the Casimir energy is no longer the lowest possible energy. Instead, the energy can be further lowered by decreasing y. So to describe Casimir effect by a Hamiltonian, Casimir energy cannot be a vacuum energy for that Hamiltonian.

The preceding paragraph can also be rephrased as follows. Parameter y cannot simultaneously be fixed and non-fixed. If it is fixed then Casimir energy can be interpreted as effective-vacuum energy, but in that case there is no Casimir effect. If it is not fixed then there is Casimir effect, but in that case Casimir energy cannot be interpreted as effective-vacuum energy. So Casimir effect cannot be consistently interpreted as being due to effective-vacuum energy.

Finally, a note on Lorentz invariance: The fundamental Hamiltonian is supposed to be Lorentz invariant (at least in the sense that the corresponding Lagrangian is Lorentz invariant). But effective Hamiltonian need not be Lorentz invariant. Consequently, effective vacuum also does not need to be Lorentz invariant.

http://arxiv.org/abs/hep-th/0503158
http://arxiv.org/abs/1605.04143

Last edited: Aug 25, 2016
7. Aug 25, 2016

### Demystifier

At the fundamental microscopic level of description, Casimir force is produced by van der Waals forces between microscopic charged constituents of metal plates. But for practical calculations such a description turns out to be too complicated (except in a diluted approximation), so in practice one achieves better agreement with measurements when effective theories are used in which only EM fields are dynamical and quantized. When one interprets successful effective theory as "true" theory, then one may have a reason to say that Casimir force is "due to vacuum fluctuations". But strictly speaking, it is not.

Last edited: Aug 25, 2016
8. Aug 25, 2016

### Staff: Mentor

You need to look into something called normal ordering:
https://en.wikipedia.org/wiki/Normal_order

Vacuum energy by definition is a big fat zero.

Thanks
Bill

9. Aug 25, 2016

### vanhees71

The VEV of the Higgs fiels is non-zero :-).

10. Aug 25, 2016

### vanhees71

Well, I guess many physicists just have read about the Casimir effect from introductory chapters of some QED textbooks that unfortunately treat it in this way. In fact it's a limit where the em. coupling constant is sent to $\infty$.

https://arxiv.org/abs/hep-th/0503158

11. Aug 25, 2016

### Mentz114

Thanks for the reference to that excellent paper. My own suspicion and mistrust of the 'standard' calculation is vindicated.

12. Aug 25, 2016

### Staff: Mentor

It's the typical QFT stuff.

To get to grips with the formalism all sorts of intuitive 'views' such as virtual particles are actual particles are promulgated.

Here, when we try to correct misconceptions, its easy to be dismissive its value. Developing intuition is in fact VERY important.

After all its why Feynman displaced Schwinger - he was in possession of much more powerful methods that solved problems in a few hours it took others years.

Thanks
Bill

13. Aug 26, 2016

### vanhees71

Indeed, the diagrammatic technique is of utmost importance as a calculational tool. However, one should be careful with building too classical intuitions on them. Indeed they are just a very clever notation for evaluating the perturbative Dyson series for S-matrix elements and many other things, also going beyond perturbation theory. If any intuition is at place, it's usually more save to think in terms of fields than particles. The fundamental building block of relativistic QT are fields, and that's reflected by the meaning of the diagrammatical elements: external lines in Feynman diagrams symbolizing S-matrix elements stand for the asymtptotic free wave functions of the initial an final state of the considered scattering event, internal lines stand for propagators, and vertices for interactions in the Lagrangian/Hamiltonian of the field theory. Then a lot of confusion can be avoided and there is indeed a quite intuitive picture developing in ones mind when becoming familiar with the math that is well hidden in the elegance of Feynman's ingenious notation, which is indeed at the basis of all modern applications of QFT, including the systematic solution of the renormalization problem by Bogoliubov, Parasiuk, Hepp, and Zimmermann in the 60ies, which is entirely based on the diagram formalism, including the sometimes cumbersome treatment of the appropriate counter terms for subdivergences and the solution of the problem of overlapping divergences, all culminating in Zimmermann's forest formula, which is entirely formulated in the language of diagrams and their topology. The applications today reach much farther: Feynman diagrams are not only used in relativistic vacuum QFT to evaluate S-matrix elements from the Standard Model and beyond but also in the entire range of many-body theory, reaching from the non-relativistic realm condensed-matter physics (including funny exotic materials like graphene or ultracold Bose and Fermi gases showing all kinds of fascinating QFT objects like anyons, fractional charges/magnetic moments, magnetic monopoles, Weyl fermions,...) to the relativistic thermal QFT heavily used in heavy-ion physics and stellar astrophysics (including the structure of neutron stars, neutron-star mergers, neutron-star collisions...) to learn about the state of strongly interacting matter with its yet not too well understood phase diagram of QCD.

14. Feb 14, 2017

### Demystifier

Now a more elaborated version of this is available:
https://arxiv.org/abs/1702.03291

15. Feb 14, 2017

### A. Neumaier

The plates already break Lorentz invariance. The Casimir effect is caused by the electromagnetic field emanating from the plates.

16. Feb 15, 2017

### Demystifier

With the important caveat that the average electromagnetic field is zero
$$\langle\psi |{\bf E}|\psi\rangle = \langle\psi |{\bf B}|\psi\rangle =0$$

17. Feb 16, 2017

### vanhees71

That's why I'd put it as: "The Casimir effect is caused by the fluctuations of the electromagnetic field due to the presence of electric charge within the plates." In our context it's important to emphasize that there are no vacuum fluctuations but fluctuations of quantum fields (in this case the electron and photon fields in "minimal QED").

18. Feb 16, 2017

### Demystifier

Sure, but as I stress in the paper above, it depends on what exactly one means by "vacuum". In particular, the relevant state can be called vacuum because it is annihilated by certain lowering operators. However those are not photon lowering operators. Those are operators for polaritons - quasi-particles which can be thought of as mixtures of photons and quanta of polarization field.

19. Feb 16, 2017

### vanhees71

What's a "polarization field". Are you referring to the problem to define asymptotic free states in models containing massless excitations a la Kulish&Faddeev?

20. Feb 16, 2017

### Demystifier

It's electric dipole moment per unit volume, ${\bf P}({\bf x},t)$, appearing in electrodynamics of continuous media (see e.g. Jackson) in the equation
$${\bf D}({\bf x},t)={\bf E}({\bf x},t)+{\bf P}({\bf x},t)$$