Gravitational effect of virtual particles

1. Jul 31, 2014

trendal

If virtual particles are constantly popping in and out of existence all around us, what gravitational effect does this have? Even if they are here for the briefest of moments they should be effected by gravity and have their own gravitational effect on other matter...shouldn't they?

2. Jul 31, 2014

Chronos

At the particle level, as in atomic and subatomic, the gravitational force is so incredibly weak compared to the EM and nuclear forces, it can be safely ignored. Only at the macroscopic level does the force of gravity begin to flex its muscles.

3. Jul 31, 2014

Staff: Mentor

I believe the energy of virtual particles are accounted for as vacuum energy, so they shouldn't have any gravitational effects themselves. But the nature of virtual particles is extremely complicated. They aren't "real" particles and so they don't behave they way you would think they would.

4. Jul 31, 2014

Staff: Mentor

Well, vacuum energy does have a "gravitational" effect, in the sense that it works like dark energy, which causes the expansion of the universe to accelerate. That makes it a sort of "gravitational repulsion" instead of attraction, but I would still call it "gravitational" in the sense that it has to do with spacetime itself.

5. Aug 1, 2014

timmdeeg

Hmm, but how would you argue if someone insists that the positive vacuum energy acts gravitational attractive?

6. Aug 1, 2014

Staff: Mentor

I would point out that the math and actual observations say otherwise. Positive vacuum energy appears in the Einstein Field Equation as a positive cosmological constant, which, if you work out the math, produces repulsive "gravity", not attractive gravity. And this math matches actual observations (dark energy causing the expansion of the universe to accelerate).

In other words, you can't do physics by simple intuitive reasoning like "positive energy produces attractive gravity". You have to actually look at the math and the observations. Sometimes the math and the observations can be summarized in a simple intuitive rule; but sometimes they can't--the actual rule is more complicated, and the simple intuitive rule can lead you astray.

7. Aug 1, 2014

timmdeeg

Thanks. I understand that the observations require a cosmological constant, at least something that exerts negative pressure. But I wonder - besides its empirical evidence - which theory (quantum field?) predicts said negative pressure?
Please allow me another question regarding the quantum vacuum in this context. As virtual particles and also photons due to annihilation of virtual particle / antiparticle pairs are not real, why then is the result 'real' energy density (regardless of its pressure)?

8. Aug 1, 2014

Staff: Mentor

The Einstein Field Equation with cosmological constant predicts it: you just move the cosmological constant term to the RHS and interpret it as a stress-energy tensor associated with the vacuum. Then it's easy to show that, for a positive cosmological constant (i.e., positive when it's on the LHS of the equation), this SET has positive energy density and negative pressure.

Viewing vacuum energy density as resulting from virtual particles is actually backwards. Vacuum energy density, as a prediction of quantum field theory, does not depend on virtual particles; virtual particles are an interpretation that is put on the underlying math to give some sort of intuitive reason why there is vacuum energy density. If that intuition doesn't work for you, that just means that virtual particles aren't necessarily a good interpretation. It doesn't change the underlying prediction of QFT.

9. Aug 1, 2014

timmdeeg

So I was totally mislead thinking of a theory dealing with that independently of General Relativity, thanks again.

Yeah, that's exactly what I was looking for.

10. Aug 1, 2014

George Jones

Staff Emeritus
I want to expand (pun intended) on some of the things that Peter has written.

As the universe expands, the energy density of the quantum vacuum remains constant, since there is just more of the same type of vacuum with the same energy density (just as two bars of gold has the same density as one bar of gold). Below, the first law of thermodynamics is used to show that this means a positive energy density vacuum has negative pressure.

Einstein's equation applied to FLRW universes gives (note the negative sign)

$$\frac{d^2 a}{dt^2} = -\frac{4}{3} \pi a \left(3P + \rho\right),$$
where $a$ is the scale factor of the universe. In Einstein's theory of gravity, pressure $P$ is a source of gravity. Positive pressure causes gravitational attraction that slows down the rate of expansion of the universe, while negative pressure causes gravitational repulsion that speeds up the rate of expansion of the universe.

Putting everything together, a positive energy quantum vacuum has negative pressure, and this negative pressure causes gravitational repulsion which in turn causes the expansion of the universe to speed up (accelerate).

(It is important not to confuse the gravitational properties of pressure, with the mechanical force/area properties of pressure. Think of a gas with positive pressure caused by molecules whizzing around. The molecules have masses that cause gravitational attraction, and also kinetic energy that (roughly through the equivalence of energy and mass) causes gravitational attraction.)

First Law for Constant Density Expansion

When a material of pressure $P$ undergoes a volume change $\Delta V$, a work (energy) equal to $P \Delta V$ is required. This affects the (internal) energy $E$ of the material, so the energy of the material changes by an amount $\Delta E$. The first law of thermodynamics (conservation of energy) gives

$$0 = \Delta E + P \Delta V.$$
Consider a "material" that keeps its energy density $\rho =E/V$ constant as it expands. Because $E = \rho V$ and energy density $\rho$ is constant, the change in energy $\Delta E$ is related to the change in volume $\Delta V$ by

$$ΔE = \rho \Delta V$$
Combining this with the conservation of energy equation gives

$$0 = \Delta E + P \Delta V = \rho \Delta V + P \Delta V = \left(\rho + P\right) \Delta V,$$
which is only true if $P = -\rho$. If a material that maintains constant (positive) density expands, it must have negative pressure!

Last edited: Aug 6, 2014