The Weinberg-Witten theorem (disclaimer: I don't know this wikipedia entry) is usually mentioned as the reason why gravitons may not be composite particles. I do understand the proof of the theorem, but not the previous conclusion.

The theorem states that in an interacting and Poincare invariant quantum field theory, there are not massless, spin-2 particles unless there exists a gauge symmetry — which makes the energy-momentum tensor non-covariant (actually covariant up to a gauge transformation) under Lorentz transformations in the Fock space. So the immediate conclusion of the theorem is that the existence of a massless, spin-2 particle (like a graviton) requires linearized diffeomorphisms.

My question is: why do linearized diffeomorphisms imply that gravitons are elementary particles? Or, more in general, why the particle corresponding to a gauge field must be elementary (I know that a gauge symmetry must be exact, but why this implies that the corresponding particle must be elementary?).

  • $\begingroup$ Not quite sure I understand: aren't they just taking the definition of an elementary particle to be a quantum of a field (gauge or matter) appearing in the Lagrangian? To get particles/Fock spaces for gravity they have to linearize gravity to be able to work free of the interactions in the highly nonlinear EH Lagrangian. $\endgroup$
    – twistor59
    Jun 14, 2013 at 7:31
  • $\begingroup$ Thanks, @twistor59. I don't think so. All particles — regardless of whether they are elementary or not (think of pions or atoms, for example) — are as far as we know connected with fields appearing in a given Lagrangian, at least at the tested energies. On the other hand, I'm totally aware of the good properties that gauge interactions have, but does this imply that "gauge particles" are elementary? Could they be "made of" other particles, perhaps gauge too? $\endgroup$ Jun 14, 2013 at 18:26

1 Answer 1


I have followed this reference

The Weinberg-Witten theorem states that a theory containing a Poincaré covariant conserved tensor $T_{\mu\nu}$ forbids massless particles of spin $j > 1$ for which $P_\nu = \int T_{0\nu}dx$ is the conserved energy-momentum four-vector.

Consider a composite graviton made of $2$ particles of spin $1$.

Each of the spin-$1$ particles will be possibly have a non-vanishing charge current, in this case the Poincaré covariant conserved tensor $T_{\mu\nu}$ (this is authorized for a spin-$1$ particle)

But this means that the composite graviton, being the "sum" of these 2 spin-1 particles, will have also a non-vanishing Poincaré covariant conserved tensor $T_{\mu\nu}$

But this is forbidden by the Weinberg-Witten theorem, because the spin of the graviton is 2.

So the graviton cannot be a composite particle.

In the full General Relativity, the covariant stress-energy tensor $T_{\mu\nu}$ is not conserved, and the conserved stress-energy quantity $(T_{\mu\nu} + \tau_{\mu\nu})$, is not a full covariant tensor.

If we linearized the Einstein equation, so as to have a conserved stress-energy tensor, we have:

$$(G_{\mu\nu})_{linearized} = \chi [(T_{\mu\nu} + \tau_{\mu\nu})] $$

The gauge symmetries, for the linear graviton as :

$$h_{\mu\nu} \rightarrow h_{\mu\nu} + \partial_\mu \phi_\nu + \partial_\nu \phi_\mu$$

and could be interpreted as "linear diffeomorphisms".

But in fact, the $\tau_{\mu\nu}$ term is not invariant, by the gauge symmetry, so the full conserved stress-energy quantity $(T_{\mu\nu} + \tau_{\mu\nu})$ is not gauge-invariant, and so we escape from the Weinberg-Witten theorem.

  • $\begingroup$ Thanks @Trimok +1. I got it. But then it seems to me that there are extra assumptions: 1) The energy momentum tensor is the sum of the energy-momentum tensor of the constituents (is this true for collective d.o.f. such as phonons whose "constituents" are atoms?). 2) In your example, the theorem do not say anything about the energy-momentum tensor of the spin 1 particles. In the way I understand the theorem, the implication only works in one way. $\endgroup$ Jun 16, 2013 at 18:51
  • $\begingroup$ 1) If we admit, than, in the composite, there are no interactions between the constituents, the assumption seems correct. It there is an interaction, I think that it depends on the particular model, and probably, you are right, it would be necessary to refine the argument. 2) The WW theorem states does not forbid, for a spin 1, that there is a covariant conserved energy-momentum tensor, so, it is true, I have implicitely choosen the case where the spin-1 particle has effectively a covariant conserved energy-momentum tensor. Details of some specific composite model may demand precisions. $\endgroup$
    – Trimok
    Jun 17, 2013 at 8:56

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