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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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Níckolas Alves
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  • They likely mean "gauge covariant". This is the same reason why it is OK to talk about the field strength in non-Abelian gauge theories even though it is not gauge invariant. – Prahar Mar 06 '24 at 12:18
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    A tensor is not a coordinate-dependent object. The thing with indices is the tensor component (its components given a choice of coordinates). – Ben H Mar 06 '24 at 13:24
  • @BenH That is an excellent point. Maybe it is the bit I'm missing – Níckolas Alves Mar 06 '24 at 14:12
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    Yes it is something that the physics description of General Relativity misses (or at least does not keep at the front of mind). Choosing a coordinates means to choose a basis of vectors at every point, like $\left{ \frac{\partial}{\partial x^\mu} \right}$. Then, the metric tensor, which takes two vectors and gives back a scalar, $g: V \times V \rightarrow \mathbb{R}$, will have components, $g_{\mu \nu} := g\left(\frac{\partial}{\partial x^\mu}, \frac{\partial}{\partial x^\nu}\right)$. Check out Frederic Schuller's Multilinear Algebra lecture. – Ben H Mar 06 '24 at 14:18
  • @BenH Ironically, that is something I'm aware of and often emphasize when discussing GR with younger people, but I overlooked it this time lol. – Níckolas Alves Mar 06 '24 at 14:19
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    Ha, yes I can empathize with that! – Ben H Mar 06 '24 at 14:20
  • See https://physics.stackexchange.com/q/346793/50583, https://physics.stackexchange.com/a/706483/50583, https://physics.stackexchange.com/a/759904/50583 for various parts of the puzzle around symmetries, coordinates and GR – ACuriousMind Mar 06 '24 at 16:17
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    Re. the quote in the question. The perturbed metric may be gauge invariant but it is not “in first order perturbation theory” since $g=\eta+h$, the perturbed metric is the sum of zeroth and first order terms and those terms are not gauge invariant individually. – A.V.S. Mar 07 '24 at 03:54

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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