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I always understood that gauge invariance of general relativity comes from the fact that the physical observables and states are the same regardless of the coordinates we choose to express them in. I may use Cartesian coordinates, I may use spherical coordinates, I may use whichever coordinate system I choose and the physical results will be unchanged. They all represent the same spacetime geometry.

It was thus a shock to me when I read in arXiv: 1312.6871 [gr-qc] that

This seems odd from the perturbative point of view, since in first order perturbation theory the only gauge invariant quantity is the perturbed Weyl tensor.

Well, this seems a odd remark to make. It seems to me that the perturbed metric, for example, would be just as gauge/coordinate-invariant as the Weyl tensor.

So what is the difference between coordinate invariance and gauge invariance in general relativity?

Qmechanic
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Níck Aguiar Alves
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1 Answers1

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In the context of perturbation theory in general relativity, gauge transformations are something distinct from coordinate transformations. To make the distinction clear lets introduce the notion of gauge dependence in a coordinate independent way.

Suppose we have two manifolds $M_1$ and $M_2$ and two tensor fields $T_1$ and $T_2$ (of the same rank and type) living on $M_1$ and $M_2$ respectively. Now, we want to know if $T_1$ and $T_2$ are similar. We cannot just directly compare them because they live in different mathematical spaces. The first thing we will need is a diffeomorphims $\phi: M_1\to M_2$, that associates the points of $M_1$ to the points of $M_2$. This map then induces a map between the tensor bundles on $M_1$ and $M_2$ allowing us to construct the pulled-back $\phi_{*}T_2$ as a tensor field living on $M_2$. Consequently, we can now study $\delta{T}= \phi_{*}T_2-T_1$ to say things about how similar the two are (or aren't).

Now the construction of $\delta{T}$ depends on the choice of $\phi$, and we could have chosen a different diffeomorphism $\phi'$. In general, $\phi$ and $\phi'$ will differ by and automorphism $\psi: M_1\to M_1$ such that $\phi' = \phi\circ\psi$. The value of $\delta{T}$ will consequently differ by $\psi_{*}T_1-T_1$. This is the gauge freedom in determining the difference $\delta{T}$.

For infinitesimal automorphisms $\psi$ is generated by a vector field $\xi$ and the freedom in $\delta{T}$ is given simply by the Lie derivative $\mathcal{L}_\xi T_1$.

In perturbation theory, you compare a perturbed spacetime (plus the tensor on it) $(M,g)$ to some background spacetime $(M^0,g^0)$. The perturbed metric,e.g. , is given by $h = \phi_{*}g - g^0$, and is ambiguous up to gauge transformations $\mathcal{L}_\xi g_0$.

Now to get back to the statement in the paper mentioned in the OP. In that paper the authors consider perturbations around Minkowski space $(\mathbb{R}^4,\eta)$. Minkowksi space has a Weyl tensor $C_0$ that is identically zero. Consequently, $\mathcal{L}_\xi C_0 =0$ and the Weyl tensor $C$ of the perturbed space there is not ambiguous under (infinitesimal) gauge transformations. This in contrast to $h$ which is only determined up to gauge transformations $(\mathcal{L}_\xi \eta)_{\mu\nu} = \partial_{(\mu}\xi_{\nu)}$.

TimRias
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