• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,064 questions , 2,215 unanswered
5,347 answers , 22,728 comments
1,470 users with positive rep
818 active unimported users
More ...

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

+ 0 like - 0 dislike

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 ]
Most voted comments show all comments

@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.


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.

Most recent comments show all comments

@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.

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights