# Would a spin-2 particle necessarily have to be a graviton?

I'm reading often that a possible reason to explain why the Nobel committee is coping out from making the physics Nobel related to the higgs could be among other things the fact that the spin of the new particle has not yet been definitively determined, it could still be 0 or 2.

This makes me wonder if the spin would (very very surprisingly!) finally be discovered to be 2, this then necessarily would mean that the particle has to be a graviton? Or could there hypothetically be other spin-2 particles? If not, why not and if there indeed exist other possibilities what would they be?

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I don't think so, since the observed particle is massive and the graviton is massless (unless there is some mechanism ala Higgs which gives it mass). If it was spin 2, it would be much more likely that its a bound states of some particles rather than the graviton. But I don't think you can construct any spin 2 bound states of standard model particles, such as quarks, with so low mass. – Heidar Oct 10 '12 at 11:18
I think there is a pretty good argument in Weinberg that no other fundamental particle can have spin-2 besides the graviton (I can't recall it atm), but composite particles can certainly have spin-2, which I think is what is one possible exciting result if Higgs is spin-2, but that is so far away... – kηives Oct 10 '12 at 14:09

There are theoretical arguments that a massless spin-2 particle has to be a graviton. The basic idea is that massless particles have to couple to conserved currents, and the only available one is the stress-energy tensor, which is the source for gravity. See this answer for more detail.

However, the particle discovered at LHC this year has a mass of 125 GeV, so none of these arguments apply. It would be a great surprise if this particle did not have spin 0. But it is theoretically possible. One can get massive spin 2 particles as bound states, or in theories with infinite towers of higher spin particles.

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Of course, there could be an as-yet undiscovered conserved current in, say, the dark matter sector. – Jerry Schirmer Oct 10 '12 at 14:48
Only if the dark matter sector violates the assumptions of the Coleman-Mandula theorem, which would be very weird indeed. – user1504 Oct 10 '12 at 14:59
@Dilaton: Yes, you can have towers of higher spin particles. Most of the known towers come out of string theory, where the spin & mass increase together along Regge trajectories. – user1504 Oct 10 '12 at 19:28
Other examples are the Vasiliev "higher spin gauge theories". – Mitchell Porter Oct 11 '12 at 6:27
To the anonymous editor reverted by Neuneck: The field strength F decomposes as a sum of $(1,0)$ and $(0,1)$ representations. This is not what most people mean by spin 2. – user1504 Dec 16 '13 at 0:55

It depends on your definition of "particle" . In the particle data group listings there exist a number of spin 2 resonances. These ultimately will be built up by quarks.

f_2(1270 MeV) page 9

a_2(1320) page 11

etc

there is a pi_2 (1670) page 16

So the bump called now "the Higgs" could turn out to be one more resonance. Not the graviton as it is envisaged in possible theories of everything, since gravity is long range and it should be massless, I believe.

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Dear Anna, would it be possible for such a compound spin-2 "particle" with a mass of 125 GeV to be produced by the same processes, decay into the same channels, and generelly behave in the same manner (similar to a SM higgs) is it is measured so far at the LHC ? Would spin 2 of the measured particle mean that we have to look for something else that breaks the EW symmetry or could ist still do the job? Sorry if I'm confused maybe I should just shut up and read the link you gave to me :-) – Dilaton Oct 10 '12 at 15:53
@Dilaton To identify it as a Higgs, they would have to nail the spin as zero, measure the branching ratios and find them consistent with the values predicted by the SM. If the spin turns out to be 2 we will have to look elsewhere for the Higgs. A spin 2 cannot do the job of the Higgs. Then the game is open. It will certainly still be very interesting to see how the phenomenologists deal with a spin 2 bump at 125GeV. Personally I think it will turn out to be the Higgs, but we have to wait for the statistics to say so unequivocally. – anna v Oct 10 '12 at 16:11
Thanks Anna, I rather think it is the higgs too ... – Dilaton Oct 10 '12 at 16:14

A massive spin 2 particle must have five modes: helicity $\pm 2$, $\pm 1$, 0. If a massless spin 2 particle has only helicity $\pm 2$ modes without other modes and has a dispersion $\omega = c k$, then such a massless spin 2 particle must be graviton (at least at linear order).

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... this reminds me slithly of the argument about why the photon has only two polarizations, is it an extension of this ...? – Dilaton Oct 11 '12 at 10:40
No. It is the reverse. If a particle has only helicity $\pm 1$ modes without other modes and has a dispersion $\omega=ck$, then it has to be massless photon. But a massless particle with $\omega=ck$ can have helicity $\pm 1$, 0 modes. The phonon in 3D crystal is such massless particle. – Xiao-Gang Wen Oct 11 '12 at 13:31
@Xiao-GangWen Is there any symmetry that connects those excitations? The reason why photons have two polarizations —rather than being two independent particles— is that photons only participate in electromagnetic interactions —which are invariant under parity transformations— and parity connects the +1 helicity with the -1 helicity. – Diego Mazón Mar 9 '13 at 6:37
@Xiao-GangWen Particles are classified according to the irred. repr. of the proper orthocronus Poincare group. The helicity is invariant under these transformations (it is a pseudo-scalar). The only reason why both helicities are considered the same particle —although strictly speaking are not— is that photons just interact through interactions that respect parity and both helicities are related by a parity transformation. If photons also interacted through weak interactions —which are parity invariant—, $\pm 1$ helicities would be considered different particles, – Diego Mazón Mar 14 '13 at 22:52
Sorry: where it says "weak interactions —which are parity invariant—", it should obviously say "weak interactions —which are not parity invariant—" – Diego Mazón Mar 15 '13 at 9:21