# Why is GR ghost-free?

I wonder how one can show that general relativity is ghost-free? By ghost I mean the negative norm state that breaks the unitarity. I think it is a well-known "fact" but I just couldn't find any derivation or proof in any reference. Thanks!

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Could clarify a little. General relativity is not a quantum theory. – MBN Apr 4 '11 at 17:31

Heng, the answer is a part of the "linearized general relativity" that is used when one derives the polarizations of the gravitational waves.

The result of the derivation is that only positively-normed "purely transverse" polarizations exist. In $D=4$ spacetime dimensions, there are two physical positive-normed polarizations which may be represented as the $\pm 2$ helicity eigenstates.

The metric tensor may be chosen to be $$g_{\mu\nu} = \epsilon_{\mu \nu} \exp(i k \cdot x).$$ At the linearized level, Einstein's equations act as the wave equation or massless equation for the components, implying that $k^2=0$ i.e. the momentum vector has to be light-like. However, the constraint part of the equations - the combination of Einstein's equations that contains no second time derivatives - says that $$k_\mu \epsilon^{\mu\nu} = 0$$ while the general covariance makes another identification for the polarization tensor, $$\epsilon_{\mu\nu} \equiv \epsilon_{\mu\nu} + k_\mu \delta x_\nu + k_\nu \delta x_\mu.$$ He, I suggestively used the symbol $\delta x^\mu$ for the infinitesimal coordinate variations - the parameters of gauge symmetry of general relativity. Those conditions and identifications allow us to choose the most general $\epsilon_{\mu\nu}$ to be in the purely transverse form, $$\epsilon_{ij}, \quad i,j=1,2,\dots D-2.$$ On this space, there is an obvious positively definite metric for the states. Because a residual gauge symmetry may also be used to eliminate the trace (sum of $(D-2)$ diagonal components) of the tensor $\epsilon_{ij}$, this tensor has $$\frac{(D-2)(D-1)}{2} - 1$$ components which is equal to two components in $D=4$.

However, the last step - the elimination of the trace (corresponding to the overall volume form) - may be a bit subtle. In the Einstein-Hilbert action, this component of the metric actually does correspond to a bosonic field with the wrong sign of the kinetic term, a ghost.

This mode still ends up being unphysical. But if all spatial coordinates were compactified on a torus, one could actually see the wrong-sign scalar. That's e.g. where the Lorentzian structure in the paper

http://arxiv.org/abs/hep-th/9811194

comes from.

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As an addition, note that you can get ghost modes with a different choice of gauge--just like you can get ghost modes if you choose a non-covariant gauge in Yang Mills theory. – Jerry Schirmer Apr 4 '11 at 18:21
Hi Lubos Thank you very much for your detailed answer. I can understand the truly dynamical variables of GR are the transverse-traceless (TT) part of the metric. But I want to know why the linearized Hilbert-Einstein action is classically stable, that is, why the Hamiltonian of it is positive definite? I guess this is a equivalent way to say that the theory is ghost-free. – Michael Shaw Apr 5 '11 at 14:35