# Zero-Point Energy and Gravitational Influence

• I
frankinstien
TL;DR Summary
Given that Zero-point energy is everywhere wouldn't it have a gravitational influence?
I ran into another article demonstrating the Casimir effect and it hit me that zero-point energy is real mass and therefore would have a gravitational influence on our universe. Is there something wrong with this idea, am I missing something?

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2022 Award
Summary: Given that Zero-point energy is everywhere wouldn't it have a gravitational influence?

I ran into another article demonstrating the Casimir effect and it hit me that zero-point energy is real mass and therefore would have a gravitational influence on our universe. Is there something wrong with this idea, am I missing something?
Energy is not mass but it IS part of the stress energy tensor which is what determines gravity in GR: 2022 Award
It is tempting to try to identify the energy of the vacuum with the cosmological constant, in fact. They should behave similarly. But as I understand it nobody really knows how to do the maths and our best plausible estimates suggest that vacuum energy density should be around ##10^{120}## times higher than the observed upper bound on the cosmological constant.

So, in summary, maybe. But if so there's something we don't understand.

• frankinstien
Mentor
Moderator's note: Thread moved to the Beyond the Standard Model forum.

Mentor
I ran into another article demonstrating the Casimir effect

zero-point energy is real mass and therefore would have a gravitational influence on our universe
As @Ibix notes, our best current understanding is that our universe has a positive cosmological constant, because the expansion of our universe is accelerating, and that is what the gravitational influence of a positive cosmological constant looks like.

An intuitively natural way of accounting for the positive cosmological constant is that it is an energy density associated with vacuum, which is similar conceptually to what you are calling "zero point energy"; however, attempts to calculate the value of this energy density based on our best current understanding of quantum field theory either give an infinite answer or give an answer that is some 120 orders of magnitude larger than the actual value we observed. So while there is probably something here, it is not something we currently understand very well.

• bhobba and frankinstien
Staff Emeritus
I ran into another article demonstrating the Casimir effect and it hit me that zero-point energy is real mass and therefore would have a gravitational influence on our universe. Is there something wrong with this idea, am I missing something?
Yes, and this is a problem creating a theory of quantum gravity.

I ran into another article demonstrating the Casimir effect

Which has nothing to do with zero-point energy.

Gold Member
It is tempting to try to identify the energy of the vacuum with the cosmological constant, in fact. They should behave similarly. But as I understand it nobody really knows how to do the maths and our best plausible estimates suggest that vacuum energy density should be around ##10^{120}## times higher than the observed upper bound on the cosmological constant.

So, in summary, maybe. But if so there's something we don't understand.
There's always something we don't understand, the universe is a mysterious monster to be observed...

More stuff to learn, an endless quest...

• frankinstien
Gold Member
I am being a little bit philosophical here... Gold Member
I ran into another article demonstrating the Casimir effect and it hit me that zero-point energy is real
Casimir effect does not imply that zero-point energy is real.
https://arxiv.org/abs/1702.03291

• • PeroK and bhobba
Mentor
QFT is an area I would say I only have intermediate knowledge of to the level of Student Friendly Introduction to QFT. Highly reccomended BTW as an approachable introduction after you have done QM:

http://www.quantumfieldtheory.info/

I am endeavouring to work up to Wienberg, but something always seems to come up.

Anyway, my understanding for several years now is normal ordering solves the issue ie there is no ZPE:

As explained above, you would want normal ordering, so Wicks's Theorem can be applied. No Wicks Theorem - no Feynman Diagrams - but those with a better understanding than I can comment on that - the books I have read always use normal ordering and Feynman Diagrams.

Thanks
Bill

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Anyway, my understanding for several years now is normal ordering solves the issue:
It's not that simple. See the paper in #9, the paragraph after Eq. (53).

• bhobba
Gold Member
As explained above, you would want normal ordering, so Wicks's Theorem can be applied. No Wicks Theorem - no Feynman Diagrams - but those with a better understanding than I can comment on that - the books I have read always use normal ordering and Feynman Diagrams.
There is also a version of Wick theorem without normal ordering. That creates additional Feynman diagrams, which however just change the phase of the S-matrix amplitude, without physical consequences.

• PeroK and bhobba
Gold Member
It is tempting to try to identify the energy of the vacuum with the cosmological constant, in fact. They should behave similarly. But as I understand it nobody really knows how to do the maths and our best plausible estimates suggest that vacuum energy density should be around ##10^{120}## times higher than the observed upper bound on the cosmological constant.

So, in summary, maybe. But if so there's something we don't understand.
In a recent paper I argue that the quantum vacuum energy does not contribute to cosmological constant if the diffeomorphism invariance of general relativity is emergent, rather than fundamental. https://arxiv.org/abs/2301.04448

• • ohwilleke, PeroK and bhobba
Gold Member
There is also a version of Wick theorem without normal ordering. That creates additional Feynman diagrams, which however just change the phase of the S-matrix amplitude, without physical consequences.
Now I noticed that the above was slightly wrong. There is no version of Wick theorem without normal ordering. There is a version without time ordering, but that's irrelevant here. What I really meant is that the interaction Hamiltonian can be taken either with or without normal ordering. Taking it without normal ordering creates additional Feynman diagrams, the bubble diagrams, which just change the phase of the S-matrix amplitude, without physical consequences.

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