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I have a puzzle I can not even understand. A graviton is generally understood in $D$ dimensions as a field with some independent components or degrees of freedom (DOF), from a traceless symmetric tensor minus constraints, we get:

  1. A massless graviton has $D(D-3)/2$ d.o.f. in $D$-dimensional spacetime.

  2. A massive graviton has $D(D-1)/2-1$ d.o.f. in $D$-dimensional spacetime.

Issue: In classical gravity, given by General Relativity, we have a metric (a symmetric tensor) and the Einstein Field Equations(EFE) provide its dynamics. The metric has 10 independent components, and EFE provide 10 equations. Bianchi identities reduce the number of independent components by 4. Hence, we have 6 independent components. However, for $D=4$, we get

  1. 2 independent components.

  2. 5 independent components.

Is the mismatch between "independent" components of gravitational degrees of freedom (graviton components) one of the reasons why General Relativity can not be understood as a quantum theory for the graviton?

Of course, a massive graviton is a different thing that GR but even a naive counting of graviton d.o.f. is not compatible with GR and it should, should't it? At least from the perturbative approach. Where did I make the mistake?

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Does this help? --… . I don't see any quantum mechanics in the question at all. It seems purely classical. – Ben Crowell Aug 16 '13 at 0:11
This question (v3) is also addressed in e.g this and this Phys.SE answers. – Qmechanic Aug 16 '13 at 8:22
@BenCrowell Well, I have certainly some confussion, that is why I asked. GR is a classical field theory for the metric (without torsion). Gravitational field is provided with the aid of a metric. Therefore, I am interested in the number of independent components of the "graviton" due to the Weinberg's formulae I wrote above. However counting independent d.o.f. does not match what I believed to. – riemannium Aug 16 '13 at 15:37

As far as counting d.o.f.s for GR, I believe it goes: Start with a symmetric tensor (10 d.o.f in 4-D). Throw out 4 because of the Bianchi identities (6 d.o.f left). Throw out another 4 because of invariance under space-time diffeomorphisms (in other words, GR is invariant under General Coordinate Transformations, so you have 4 unphysical d.o.f.s). Thus there are only two degrees of freedom left.

Regarding massive gravity, see: Theoretical Aspects of Massive Gravity by Kurt Hinterbichler [arxiv 1105.3735] which has a fairly readable introduction.

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Question: were not the Bianchi identities the 4 relationships due to Diff. invariance itself too? Anyway I think I understand this stuff better know... :). By the way, thanks for the reference. – riemannium Aug 20 '13 at 10:27

You can think of diff as bianchi id. The additional 4 dof is killed by the fact that 4 of the 4 of the EFE are constraints.

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As i can remember, for the case of classical electrodynamic the d.o.f counting start after the assumption of the Bianchi Identity, and at the end the desired result came all from the gauge freedom. In fact $\partial_{[\mu}F_{\nu\alpha]}=0$ only choose a suitable form for $F_{\mu\nu}$, for example $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu.$$ After this choise from the four components of $A_\mu$ (our starting point for d.o.f counting) we must take into account the gauge freedom in choosing the four-potential, i.e. $$ A_\mu\sim A_\mu+\partial_\mu\Lambda. $$ For example if we want $\partial_\mu A^\mu=0$, we must perform a gauge transformation with $\Lambda$ such that $$\color{red}{\Box\Lambda=-\partial_\mu A^\mu}.$$ At this point we are left with another possible gauge transformation (residual gauge freedom) such that $\partial_\mu A^\mu=0$ still holds, i.e. we must choose another $\Lambda$ with $$ \color{blue}{\Box\Lambda=0}. $$ It is the gauge freedom that fix the correct numbers of degree of freedom for the photon $$ 4-\color{red}{1}-\color{blue}{1}=2. $$ In the case of GR the path is the same: Einstein equation, differential Bianchi equation for $R_{\mu\nu\alpha\beta}$ and the gauge freedom (space-time diffeomorphisms). Again you can perform two of these transformation and you have as you say $10-4-4=2$ d.o.f.

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I used to count dofs in terms of spinor representation. There are two spinor representation of SO(3,1) (1/2,0) and (0,1/2) denoted by dot and no-dot indices. Vector (spin 1) representation in spinor indices is (1/2,0) X (0,1/2) = (1/2,1/2) which has 4 dofs. If the theory is massless, the theory is gauge invariant, there is one gauge dof, which can be eliminated by the gauge fixing condition. One dof is time-like one, which is not physical. So 4-1-1 =2 physical dofs remain. For symmetric tensor we have (1/2,1/2)X (1/2,1/2) = (1,0)+ (0,1)+ (1,1)+(0,0) The irreducible symmetric representation is just 9-dimensional. The general coordinate transformation eliminate 4 dofs (which essentially is 4 space-time coordinates). So the massive spin 2 has 9-4=5 dofs. The massless spin 2 theory will be gauge invariant, where we can eliminate 1 dof. Two dofs are time-like which do not propagate. So only 5-1-2 dofs remain. We can extend the argument to arbitrary dimension.

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