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Why is a massless spin-2 automatically a graviton

  1. Mar 19, 2014 #1
    The Wikipedia page on the graviton
    http://en.wikipedia.org/wiki/Graviton
    contains the following sentence:
    "Additionally, it can be shown that any massless spin-2 field would give rise to a force indistinguishable from gravitation, because a massless spin-2 field must couple to (interact with) the stress–energy tensor in the same way that the gravitational field does"
    Can anybody explain why this should be so (I don't have access to the source cited in the article)

    According to Misner-Thorne-Wheeler, §18.1, to get GR from a QFT, you have to assume that a massless spin-2 exists (that part I understand) and that it couples to the stress-Energy-Tensor as a field. Why is this second assumption not really necessary, i.e. why must any massless spin-2 particle "automatically" couple to the stress-energy-tensor?
     
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  3. Mar 19, 2014 #2

    atyy

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    I've never understood this properly, but apparently it is an argument made by Weinberg.

    http://arxiv.org/abs/1007.0435 (p8 & p33) discusses it as "Weinberg’s low energy Theorem".

    There is also the classical argument about the self-interaction of gravity http://arxiv.org/abs/gr-qc/0411023, but I don't think this gives the interaction with other particles.
     
  4. Mar 19, 2014 #3

    Bill_K

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    Weinberg gives an argument in his QFT book which addresses this question, but with one added assumption: the spin-2 particle leads to a long-range interaction.

    The traditional route is to start with a field and see what particle states you can build. Weinberg goes in the opposite direction. He starts with particle states, which in this case are massless with definite helicity, and attempts to build fields from their creation and annihilation operators..

    Starting with helicity ±1, he has no trouble building an antisymmetric field tensor Fμν, but finds it impossible to build a (genuine) four-vector. But when it comes to interactions, Fμν falls off as 1/r2, i.e. is not long range. To get a 1/r interaction it's necessary to introduce a quantity Aμ whose transformation includes a gauge, and this gauge symmetry implies that the source Jμ must be conserved.

    For helicity ±2, the field tensor is Rμνστ which falls off like 1/r3, and is again not long range. To get a 1/r interaction one must introduce a quantity hμν and a corresponding gauge symmetry. The latter implies that the source Tμν must be conserved, and the only conserved symmetric tensor known to exist is the stress energy tensor.
     
  5. Mar 19, 2014 #4
    Thanks a lot.
    So if I understand this correctly, it is possible to have a massless particle of spin 2, but this particle cannot act as a particle that transmits a force (1/r potential). Is there any idea what the physics of such a particle would/could look like?
     
  6. Mar 19, 2014 #5

    Bill_K

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    I don't think the existence of such a particle is implied.
     
  7. Mar 19, 2014 #6
    Sorry, probably I did express this badly. If such a particle were to exist (as a thought experiment), how would it interact with other particles (since the standard way of transmitting a long-range interaction at low ebergies would not work)? Would it act like a short-range force even though massless?
     
  8. Mar 19, 2014 #7

    dextercioby

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    The Lagrangian of the massless spin-2 field was shown by Pauli and Fierz way back in 1939 that coincides with the linearized limit of the H-E Lagrangian computed by Einstein some 20 years before (an argument you can find in the book on gravity by Feynman). It was proven at least by 10 people that the only consistent self-coupling of the massless spin-2 field is the perturbative series of the H-E Lagrangian wrt the coupling constant.

    This question should belong to the FAQ.
     
  9. Mar 19, 2014 #8

    Bill_K

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    If you think that this question was resolved definitively back in 1939, perhaps it does belong in the FAQ.
     
  10. Mar 20, 2014 #9
    @dextercioby
    Thanks. IIUC, this means that a massless spin-2 field has to couple to its own stress-energy tensor.
    From what Bill_K writes, this does not necessarily imply a coupling to the SET of all fields; for this, we need the additional assumption that a long-range interaction is mediated.
    So my question still stands, I think (but I may have misunderstood something): Is it possible that there is a massless spin-2 particle that does not couple to other fields like a graviton would? And if so, what would the physics of such a particle be? Would it transfer a short-range force between the particles it couples with?
     
  11. Mar 21, 2014 #10
    There are two massless spin-1 particles. They are very different from each other, and neither of them works on everything. These are photon and gluon.

    So what rules out existence of massless spin-2 particles other than graviton?
     
  12. Mar 21, 2014 #11

    atyy

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    I don't understand this very well either, but http://arxiv.org/abs/hep-th/0007220 addresses the question.

    Edit: They also address coupling to matter in section 9.
     
    Last edited: Mar 22, 2014
  13. Mar 21, 2014 #12

    dextercioby

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    Atty's linked article by the Brussel group just above nails my quoted statement on the head.
     
  14. Mar 22, 2014 #13
    @snorkack
    If I understand the Weinberg argument correctly, for a spin-1 particle you don't get a unique coupling from Lorentz invariance, but you do get the condition that the sum over all coupling constants (which is nothing but the sum over all charges) is constant. In other words, if a massless spin-1 particle couples to some particles, the corresponding charge is a conserved quantity. There is, however, nothing to rule out different couplings for different particles and thus different conserved quantities.
    Both electric charge and quark color charge are conserved quantities.
     
  15. Mar 22, 2014 #14

    Bill_K

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    It should be nailed. As late as 1964, Feynman tackled the problem of writing the perturbative series for gravity and failed at the two-loop level. Not summing them, mind you, just writing down the terms. It wasn't until the development of Faddeev-Popov ghosts in the late 60's that the problem was solved.
     
  16. Mar 22, 2014 #15

    samalkhaiat

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    The representation theory of the Poincare’ group contains massless spin-2 particle. By imposing certain constrain on the algebra we can write the field equation for such particle. Now, leave Poincare’ group and go to Einstein’ field equation. Write the Reimannian metric as a small deviation from the Minkowski metric. Substitute that in Einstein equation, with a bit of algebra you obtain equation that look exactly like the field equation you obtained for the spin-2 particle. Is that just a coincidence? May be, some people said. People (Oppenheimer, Pauli and others) knew that back in the 1930’s. The unshakable link between the two came in the 1960’s when Weinberg showed that GR is the only theory one can have for massless spin-2 particles that is consistent with Poincare’ invariance and the principle of general covariance.

    Sam
     
  17. Mar 22, 2014 #16

    samalkhaiat

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    In 1964, I believe (for personal reason*) that Weinberg produced his best contribution in theoretical physics. In his classic paper, Weinberg proved the following three statements:
    1) Maxwell’s theory is the most general Poincare’ and gauge invariant theory of massless spin-1 particle.
    This can be easily proved by writing down the most general Poincare invariant amplitude for emitting a single photon in the so-called soft limit. Then, by demanding gauge invariance, you get charge conservation.

    2) Einstein’s GR is the most general Poincare’ and generally covariant theory for massless spin-2 particle.
    The proof is similar to photon case. You write the most general amplitude for emitting a soft graviton. Then you see what happen when you demand general covariance, i.e., demand that the amplitude is “gauge” invariant. If you do that, you find that the equivalence principle pops up. That is, all particles couple to the massless spin-2 particle with equal strength.
    From this, he also concluded that the coupling strength most vanish when the spin of the massless particle is greater than 2. To some extent, this explains why we don’t see this kind of particle.

    3) Yang-Mill’s theory is the most general Poincare’ and gauge invariant theory for massless, self interacting spin-1 particles.
    In this case, if you do the same exercise, you find that the coupling strengths of the interaction satisfy the Lie algebra of a compact group.

    (*) I believe every physicist should study this paper. I don’t know about the others, but I describe that work as the magic that made a theorist out of me.

    Sam
     
  18. Mar 22, 2014 #17

    atyy

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    Does Weinberg's proof include the cosmological solutions of GR, or GR with a cosmological constant?
     
  19. Mar 22, 2014 #18

    samalkhaiat

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    No, and he does not need to include it in considering the amplitude for soft graviton. The proof based on writing down the biggest contributions to the amplitude.
     
  20. Mar 22, 2014 #19

    dextercioby

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  21. Mar 22, 2014 #20

    atyy

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    Does that mean that if one says GR is implied by massless spin 2, that form of GR does not contain cosmological solutions or a cosmological constant?
     
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