# Does the density of virtual particles decrease when space expands due to dark energy? [closed]

I understand that vacuum (i.e. QED vacuum) is theorized to contain spontaneously appearing and annihilating pairs of virtual particles. Suppose the average density of those particles at a given moment in time and a particular location in space is Dinit.

Suppose we wait an amount of time such that dark energy expanded the space at that location and the density of virtual particles is measured to be Dexp.

Will the density at which the virtual particles appear at that location be the same or different? In other words, will Dinit = Dexp or Dinit > Dexp?

Edit: Maybe my understanding is incomplete, this is what I would like to know:

Suppose at some time tinit, the universe has a volume Vinit, and some experiment is performed that estimates that, on average, in some window of time, 2 electron-positron pairs appear and annihilate. The "pair density" Dinit is computed to be 2/Vinit.

Then time tlong passes, dark energy expands space, and the universe now has volume Vnew. The same measurement is performed, estimating the average number of virtual particle pairs appearing and disappearing in the new expanded universe.

Question, would A or B be expected to be observed:

A: New avg number of pairs conserved at ~2, thus pair density Dexp=2/Vnew, and thus Dexp < Dinit?

OR

B: New avg pairs >2, thus pair density Dexp= >2/Vnew, and thus pair density conserved and Dexp ~= Dinit?

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• Dark energy doesn't change local physics. Having said that, I don't think virtual particle density is even a physical quantity. – CuriousOne Sep 15 '15 at 4:29
• @CuriousOne. Isn't that what renormalization does, lower the density? – Jitter Sep 15 '15 at 4:47
• @Jitter No. Renormalization is connecting the mathematical parameters in your equations to the correct experimentally determined parameters as opposed to corresponding the wrong things. Its like if you forgot about expenses, then included them and realized the thing you were calling sales was actually profit and sales are larger when you pay attention to expenses rather than just looking at your bank balance head upwards. – Timaeus Sep 15 '15 at 5:32
• @Jitter You have mathematical parameters that you call m and q and such. And experimentally you measure mass and charge. But then you realize the mathematical parameters in your formulas are not equal to what you measured in the lab. There's a nontrivial relationship between the m and the mass and between the q and the charge. So you have the bare quantity and the measured quantity and you realize the bare ones have to be whatever is required to make the experimental predictions match the lab results. The point is that a finite number of experimental results can nail down the bare quantities. – Timaeus Sep 15 '15 at 5:58
• – Justas Sep 15 '15 at 22:08

Grossly, the "positing" of virtual particles is connected with the Heisenberg Uncertainty principle, HUP .

The expansion of space and the incoming dark energy will be changing the limits within which the virtual particles can manifest. The delta(x) will be changing but as virtual particles are just mathematical formulae within the bounds of the HUP, a density cannot be defined, imo.

Edit after question edit:

The virtual particles are not countable in the sense of your drawings. Their existence in the vacuum can only be seen by the changes in measurable interactions. The calculations need this vacuum virtual particles in order to accurately describe some data. Please see my answer to the question are-virtual-exchange-particles-real-or-just-mathematical

• Is the best definition greater than zero but less than infinity? I thought the universe would tear apart at a certain point? – Jitter Sep 15 '15 at 6:50
• I do not think you can define density for mathematical functions, and virtual particles are mathematical functions. As you see above, the delta(e) depends on the delta(t) so even if you wanted you could not use the energy intervals to give the envelopes of the infinity of virtual particles under calculation within the bounds. – anna v Sep 15 '15 at 18:07
• Now that a way has been found to directly observe virtual particles will you need to change your answer a bit? – Jitter Oct 5 '15 at 9:17
• @Jitter what are you talking about? – anna v Oct 5 '15 at 16:40
• @Jitter zero point energy is also seen in the Casimir effect. It does not contradict what I am saying: virtual particles describing/fitting zero point energy are a mathematical construct. – anna v Oct 6 '15 at 15:37

I understand that vacuum (i.e. QED vacuum) is theorized to contain spontaneously appearing and annihilating pairs of virtual particles.

I'm afraid your understanding is wrong. You can read in an awful lot of places that virtual particles are short-lived real particles that pop in and out of existence. But it just isn't true. Virtual particles are field quanta. It's like you divide a field up into abstract chunks and say each is a virtual particle. See anna's answer to do virtual particles actually physically exist? The answer is no. They don't physically exist. "Virtual particles exist only in the mathematics of the model used to describe the measurements of real particles". Also see Matt Strassler's article: "The best way to approach this concept, I believe, is to forget you ever saw the word “particle” in the term. A virtual particle is not a particle at all".

Suppose the average density of those particles at a given moment in time and a particular location in space is Dinit.

Vacuum fluctuations are real. They're the electromagnetic equivalent of the little ripples on the surface of the sea. I'll continue on the assumption that you're talking about vacuum fluctuations.

Suppose we wait an amount of time such that dark energy expanded the space at that location and the density of virtual particles is measured to be Dexp Will the density at which the virtual particles appear at that location be the same or different? In other words, will Dinit = Dexp or Dinit > Dexp?

I'm confident that the energy density will reduce. So the rate and/or wavelength of the vacuum fluctuation will reduce. I can't prove that that will happen, but conservation of energy is one of the most important tenets of physics. There are no perpetual motion machines.

Question, would A or B be expected to be observed?

Neither I'm afraid. We just don't observe electron-positron pairs popping in and out of existence.