# 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
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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.

@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: 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.

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@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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