# Graviton and principle of equivalence

Is it possible that in a theory of gravitons, i.e., a quantum field theory of gravitation, general relativity's principle of the equivalence of gravitational mass and inertial mass, no longer holds?

• You'll need to clarify what you are asking. The graviton is the gauge boson associated with a quantum field theory description of gravity (though we don't know if such a description makes physical sense). You could for example ask if the equivalence principle is fully respected by all the QFT theories (there are several) of gravity. As it stands, just asking if gravitons break the equivalence principle is meaningless. – John Rennie Jul 10 '15 at 5:07
• Let's say maybe, but in that case it would be only a temporary effect (whatever it might be at that level of abstraction) and in its classical limit they should be the same. – gox Jul 24 '15 at 21:26

## 2 Answers

On the contrary Deser and others [1, and refs therein] have argued that trying to construct a theory of a graviton, that is, a massless spin-2 field in a flat background, consistent with special relativity, then "[c]onsistency [leads] us to universal coupling, which implies the equivalence principle" . The argument is summarized by MTW [2, Box 17.2.5, see also Box 18.2, and refs therein]. More concretely, the conclusion is that the field equations for the graviton field must be the Einstein field equations with the source being the stress-energy tensor.

References

1. Deser, S. (1970). Self-interaction and gauge invariance. General Relativity and gravitation, 1(1), 9-18.
2. Misner, C. W., Thorne, K. S., & Wheeler, J. A. (1973). Gravitation. Macmillan.
• The trouble with that is that "the energy of the gravitational field shall act gravitatively in the same way as any other kind of energy". See The Foundation of the General Theory of Relativity. And that energy has a mass-equivalence. The field is not massless. – John Duffield Jul 10 '15 at 12:27
• @JohnDuffield: A massless field in the quantum field theoretic sense is one without a mass term in the Lagrangian (sometimes protected from acquiring a mass term by gauge invariance). The classical mass-energy equivalence has nothing to do with it. – ACuriousMind Jul 10 '15 at 22:48
• @ACuriousMind : thanks for the info. But are you saying that we then have a massless field which has mass? – John Duffield Jul 11 '15 at 6:15
• @JohnDuffield The field has energy but has no mass. – my2cts Jun 22 at 12:50
• @my2cts : the above is nearly 4 years old! – John Duffield Jun 23 at 15:15

Yes and no. It depends which of the equivalence principles you have in mind. There are two formulations.

So called Einstein equivalence principle reads:

The outcome of any local non-gravitational experiment in a freely falling laboratory is independent of the velocity of the laboratory and its location in spacetime.

This is nothing, but a compact reformulation (by Haugan and Lammerzahl) of the original one presented by Einstein in:

A. Einstein, Annalen Phys. 35, 898 (1911) [Annalen Phys. 14, 425 (2005)]. doi:10.1002/andp.200590033

For instance, considering a quantum theory of the gravitational field, you don't necessary have to worry about the equivalence principle if you are interested only in the semiclassical limit of your theory. This may happen when you are going to find a curved background on which you will later rigorously quantize a matter field. Personally I disagree with this interpretation, but a lot of physicists follow it. A good example here is the model called Loop Quantum Cosmology.

So called strong equivalence principle reads (again, I'm citing Haugan and Lammerzahl):

The outcome of any local experiment (gravitational or not) in a freely falling laboratory is independent of the velocity of the laboratory and its location in spacetime.

In this case you have to consider the equivalence principle in the case of any classical or quantum fields.

Two more comments. Even if you don't assume the equivalence principle for gravitational field, you have to consider general covariance for any field, also for the gravitational one. Finally, in the case of the Einstein-Hilbert action and related equations of motion, classically the equivalence principle is satisfied.

Writing the answer, I've cited this article: Mark P. Haugan and C. Lammerzahl, Lect. Notes Phys., 562:195–212, 2001. [arXiv:gr-qc/0103067]