# Why is gravity sensitive to absolute energy?

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1. Aug 19, 2015

### "Don't panic!"

As far as I understand it gravity is sensitive to absolute energies, as seen directly through Einstein's equation $$G_{\mu\nu}=8\pi GT_{\mu\nu}$$ Thus the local geometry of spacetime is directly affected by the local energy density (and not just differences in energy).
So whenever gravity is taken into consideration we must include all energy contributions.

What confuses me though is, if what I've said above is correct why is it that in Minkowski (flat) spacetime only energy differences matter? In particular, why is it possible to neglect the vacuum energy in QFT and instead just measure energy differences relative to it?

2. Aug 19, 2015

### George Jones

Staff Emeritus
Absolute mass, not relative mass, is a source for Newtonian gravity. Relativity allows transformations between mass and energy, so, in relativistic gravity, mass, energy, flow of mass and energy, etc., are all sources of gravity,

Minkowski spacetime is a good example. Minkowski spacetime is devoid of matter and energy, and there is no gravity. Strictly speaking, fields in Minkowski spacetime should be bootstrapped into sources of gravity, in which case spacetime is no longer Minkowski spacetime.

3. Aug 19, 2015

### "Don't panic!"

So is what I put in the first part of my original post essentially correct?

How does one deal with this situation in QFT which is done on a Minkowski background? Why are only energy differences relevant in this case when special relativity (and hence mass-energy equivalence) still applies?

4. Aug 19, 2015

### Staff: Mentor

Strictly speaking, it isn't; it's just that, for many purposes, we can get away with it. In most situations where QFT is used, because we are dealing with situations in which the effects of gravity are negligible. For example, if I use QFT to analyze a particle physics experiment at CERN, I don't have to worry about gravitational interactions between the particles, or even between the particles and the Earth; they're way, way too small to matter. (So is the energy of the vacuum, aka the cosmological constant, in this situation--but see below.)

However, in cosmology, it turns out that we can't ignore the energy of the vacuum; it has gravitational effects, which show up as a nonzero cosmological constant. (At least, that's the simplest way of accounting for a nonzero cosmological constant.) The problem then becomes, why is the cosmological constant so small? If we do a "naive" calculation of how much energy there should be in the vacuum, we get either infinity, or (if we are a little more sophisticated) a number about 120 orders of magnitude larger than the actual cosmological constant we observe. So another answer to your question as you ask it is, we can neglect vacuum energy in most situations because it's so small--but we don't understand why it's so small.

5. Aug 19, 2015

### "Don't panic!"

In the canonical formulation of QFT though, when redefining the vacuum such that we neglect the vacuum energy there is no mention of it being small, it is infinitely large, but it always seems to be argued that only energy differences are physically observable (if one neglects gravity) and so we can simply measure energies relative to an arbitrary reference point (enabling us to redefine the vacuum such that it has zero energy).

Also, as George Jones wrote, isn't gravity sensitive to absolute energies before we even consider the vacuum energy, and this is why we cannot neglect the vacuum energy when gravity is "switched on" as all forms of matter and energy should gravitate and so a priori there is no reason why the vacuum energy shouldn't?

6. Aug 19, 2015

### Staff: Mentor

Which ignores gravity, as you noted. But you're asking about how all this works with gravity, so obviously you can't use a formulation of QFT that ignores gravity.

Yes. The vacuum energy does gravitate: that's what the cosmological constant is.

7. Aug 19, 2015

### "Don't panic!"

Sorry, this part was really a separate question. If we ignore gravity, why is it that only energy differences are physically observable?

Sorry again, I didn't word this part very clearly either. What I meant is, forget that vacuum energy exists for the moment, then it is still true that gravity is sensitive to absolute energies (as implied in Einstein's equations and the mass-energy equivalence), right?
Then, given that this is the case, if we now take into account that the vacuum does indeed have a non-zero energy associated with it we must necessarily include its contribution to the energy-momentum tensor, manifesting as a cosmological constant.

8. Aug 19, 2015

### Staff: Mentor

Are you asking only about QFT, or about special relativity in general?

In QFT, the answer is, because you're neglecting gravity, which is the only way in which absolute energies (as opposed to differences in energies) manifest themselves.

In SR more generally, the premise of the question is wrong. We can certainly measure "absolute energies" in SR; as you say, mass-energy equivalence applies, so any time we measure the mass of something, we are measuring an absolute energy. We can get away with neglecting gravity in such cases, even though we are measuring absolute energies, because gravity is so weak.

Yes.

Yes.

9. Aug 19, 2015

### "Don't panic!"

Sorry, yes the question was in relation to QFT. What is the exact argument in the theoretical formulation for why only energy differences are physical and hence we can "subtract off" the vacuum energy contribution in the Hamiltonian? (Why is gravity sensitive to absolute energies and other forces aren't)?

10. Aug 19, 2015

### Staff: Mentor

That's a QFT question and should be asked in a separate thread in the Quantum Physics forum. You'll get better answers there.

Because gravity is connected to the geometry of spacetime and other forces aren't. Or, to put it another way, "absolute energies" act as a source in the Einstein Field Equation, so, strictly speaking, as soon as we put anything into spacetime that has energy, spacetime can no longer be flat Minkowski spacetime (as George Jones pointed out), and gravity will be present. But, because gravity is so weak, the effects on spacetime of ordinary energies and masses can almost always be ignored, and spacetime can be treated as flat as an approximation, even though it really isn't. In this approximation, there is no gravity at all, and the other forces can be described using QFT in flat spacetime, in which only energy differences are physically observable.

(Note that this approximation also requires the energy of the vacuum--the cosmological constant--to be small enough to be ignored in almost all cases; otherwise its gravitational effects would always be present and spacetime could not be approximated as flat. As I said before, we don't understand why the cosmological constant is so small, but measurements show that it is, so empirically the approximation of flat spacetime is a good one for the purposes we use it for.)

Last edited: Aug 19, 2015
11. Aug 19, 2015

### "Don't panic!"

Ok, thanks I'll do that then.

Thanks for your help on the matter, I think it's cleared things up for me.