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  Could the cosmological constant be due to vacuum fluctuations in a box, i.e., in a finite universe?

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Assumption: If the universe were a finite box whose boundary is the cosmological horizon, then there would be a zero-point energy inside that box.

Consequence 1: This zero-point energy would be given by the size of the box. The calculated energy value is very similar to the measured cosmological constant. 

Consequence 2: The zero-point energy would have been larger when the universe was smaller. The cosmological constant would not be a constant, but decay in time. 

Question: Could that be the case?

asked Mar 3, 2019 in Theoretical Physics by Benny [ no revision ]

Maybe, who knows? But note that actually "the box" has a lot of suff in it; in other words, the box state is not the ground state, but highly excited one. Finally, we cannot apply QM to the whole Universe, but only to a small part of it.

One issue would be the sign of the casimir force. As the casimir force turns out to be attractive, it can not be used to model a positive cosmological constant.

@VladimirKalitvianski I see no obstruction to applying QM to the universe as a whole, what should be the problems with this?

@Dilaton: Casimir force is a particular case of interaction of two neutral material bodies. Generally is it kind of Van des Waals force that can be a complicated function of the distance and orientation of two neutral bodies. The "borders of the Universe" are not such interacting  bodies.

The Universe itself is not that microscopic object to safely apply QM. On the contrary, it is a macroscopic object, or a macroscopic process, as a matter of fact. It is not "repeatable" and it is unknown to us, to tell the truth. Those who dare apply QM, QFT, string theory, etc. to the whole Universe pay Ein Taler.

@VladimirKalitvianski  I am afraid, your interpretation of the Casimir force is quite far from being complete. This force can arrise in any quantum system with boubdaries or interfaces (i.e. surfaces of volumes where quantum fields have different properties) Solid bodies are just one example of such boundaries. 

@IgnatRU: As I said, Casimir force is nothning but a van der Walls force. For its existence, there is no need of boubdaries, interfaces, etc. It is only necessary for the "distance" $R$ to be more or less well defined quantity.

@VladimirKalitvianski I must say you seems contradicting yourself. To have a well defined distance, as you put it, you do need boundaries or interfaces. Otherwise, between what do you measure the distance?  "Neutral material bodies" you mentioned earlier, are well defined precisely because each of them have a very well defined boundary, a boundary which separates two regions of space where electromagnetic field have very different effective behaviour. 

Moreover, the relation between Casimir and van der Waals forces is quite opposite. It is the van der Waals force that is the short distance, non-retarded electromagnetic version of the much more general Casimir effect. 

@IgnatRU: Then look into the "Quantum Electrodynamics" by LL (Berestetsky, Lifshitz, Pitaevsky), where they calculate the long-distance interactions of atoms. You will see the retarded interaction contribution. The inequality that defines well the distance $R$ is $R\gg a_0$ for example. It is clear that an ensemble of atoms interact in the same way: there are charges and electromagnetic field in the play. Some things can be simplified and reduced to effective "boundaries" with specific properties, but it is still a QED calculation.

@VladimirKalitvianski It is a QED calculation indeed, WHEN it is about electromagnetic interactions. But just searching arXiv for "casimir" will bring up a plethora of effects which are NOT about electromagnetic interactions, and thus not about wan der Waals force.

@IgnatRU: You are trying to change the subject, in vain. Electromagnetic or not, there must be something that we replace with effective boundary conditions to simplify our calculations.

@VladimirKalitvianski Any quantum filed confined to a box, or otherwise being subject to boundary (or matching) conditions will show the Casimir effect, i.e. the dependence of its vacuum energy on the geometry and/or dimensions of the boundary or distance between them. It is by no means limited to QED and/or "real" atoms, electrons, etc.

Your statement that it does not apply to possible boundaries of the universe is false.

@IgnatRU:

Any quantum filed confined to a box, or otherwise being subject to boundary (or matching) conditions...

Because any QFT is about a small part of the Universe. QFT is about quasi-particles of a complicated interacting system, and the boundaries are created with the remaining matter.

@VladimirKalitvianski  I am sorry, where did you get that QFT is only about a small part of the universe? 

@IgnatRU: I am sorry, but this is another subject. It is about Physics and its relationships with Mathematics. There was a prominent physicist, Julian Schwinger, who mastered math well and who knew QFT as one of its creators. He wrote a work called "Particles, Sources, and Fields" from a physical point of view. In this work he clearly oulined the domain of applicability of QFT. Maybe this answers your question.

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