The Einstein field equations are given by (with assuming $\Lambda = 0$), $$ R_{ab} - \frac{1}{2} R g_{ab} = \kappa T_{ab}. $$ The principle of general covariance states that the form of these equations are invariant under diffeomorphisms. These equations can be linearized, by introducing $$ g_{ab} = \eta_{ab} + h_{ab} $$ and letting $h_{ab}$ be small. The coordinate change $x'^\mu = x^\mu + \xi^\mu(x^\mu)$ changes the perturbation metric $h$ as $$ h'_{ab} = h_{ab} + \mathcal{L}_\xi\eta_{ab} = h_{ab} + \partial_a \xi_b + \partial_b \xi_a. $$ In the linear theory, which can be readily checked, the Riemann tensor is invariant under the transformation mentioned above, which also means that the energy-momentum tensor $T_{ab}$ is invariant to first order in $h$.

So, now comes my question:

In the full non-linear theory, the gauge freedom is given by general changes of coordinate systems. My understanding goes as follows; the Ricci/Riemann tensors express curvature, which is an intrinsic property of the manifold, and as this curvature is invariant of coordinate changes (or diffeomorphisms?). The "curvature" property of these tensors have to be invariant. Still, they are tensors, and should therefore act as tensors under rotations, boosts, etc., so they can't be invariant either.

I have only seen the gauge transformation properties discussed in the linear theory, and I am having problems understanding what it "means" in this setting to posses an intrinsic property which should be invariant of coordinate system. All of $R_{abcd}$, $R_{ab}$ and $T_{ab}$ posesses some physcial meaningful value, and should therefore have some "invariant part". They should possess some equivalence classes that are different, and not possible to transform into one another by coordinate changes. What does this mean, and what is this "invariant part"? Also, is there a fundamental difference between a gauge transformation/diffeomorphism and a global transformation (such as a rotation/boost)?

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    $\begingroup$ Scalars like Kretschmann scalar have the invariant property you're looking for, since they have no indices. Also, Riemann tensor is not even invariant under first order gauge transformations when the background is curved (it's covariant). $\endgroup$
    – Avantgarde
    Commented Mar 28, 2019 at 11:15
  • $\begingroup$ I am trying to explain why a physical energy-momentum tensor can't be gauge dependant, as it describes a physical property. Specifically, when looking at gravitational waves in flat spacetime you can obtain a pseudo energy-momentum tensor that depends quadratically on h, and is a source for second order corrections in the metric. This tensor is not gauge invariant, and thus can't be a valid energy-momentum tensor, so you have to avrage. Sc. 9.2 in web.phys.ntnu.no/~mika/week10.pdf . Does this requirement of gauge independance come from the invariance of R_ab in the linear theory? $\endgroup$
    – Ehinda
    Commented Mar 28, 2019 at 13:10

1 Answer 1


Einstein's field equations (EFE) in nonlinear form are not a gauge theory. Attempts have been made to formulate it as a gauge theory, but this requires the use of the tetrad's formalism so that the EFE can be formulated in a coordinate-independent way. As tentative gauge group the local Poincare' -invariance is required. However, as far as I know the EFE cannot be formulated completely gauge -invariant under this group. Actually different attempts exist, but the formulation of the EFE as gauge theory is out of scope here. On the other hand the EFE are coordinate-covariant which can be expressed simply by the fact that the form of the EFEs remains the same under any (at least differentiable) coordinate transformations (this is what sometimes is called invariance under diffeomorphisms)


and each of its tensors $R_{ik}$, $g_{ik}$ and $T_{ik}$ transforms in the same way (I take $R_{ik}$ as an example):

$$R_\overline{ik} = \frac{\partial\overline{x^i}}{\partial x^l}\frac{\partial\overline{x^k}}{\partial x^m} R_{lm}$$

As all tensors transform in the same way, it is guaranteed that the EFE don't change their form. Another detail: whatever coordinate transformation is carried out, a tensor is only zero if it is zero in all coordinate systems. Actually, the coordinate change $x'^i = x^i+\xi^i(x)$ is a linearization of the more general coordinate transformations $\overline{x^i}=\overline{x^i}(x)$, so the gauge transformation of the tensor $h_{ik}$ in your post just expresses the more general coordinate covariance.

  • $\begingroup$ "EFE...are not a gauge theory." Wasn't that the whole point of the ADM formalism? The Poisson bracket with the constraints generate gauge transformations, as one would expect. Or do you mean they don't look like a Yang-Mills theory? $\endgroup$ Commented Mar 28, 2019 at 13:52
  • $\begingroup$ Hm.. ok. This is interesting. (BTW I think you transformation fractions are inverted). So EFE is only a gauge theory in the linearized form? But in the linear theory we use the gauge freedom in h to remove 8 out of the 10 d.o.f, and end up with two polarizations of GWs. Should this not also hold in the non-linear case, where we have 2 degrees of radiative freedom, where the rest is removed by a gauge equivalence? $\endgroup$
    – Ehinda
    Commented Mar 28, 2019 at 14:00
  • $\begingroup$ Actually, Deser and Isham showed in the '70s it is possible to formulate GR using Tetrads as a gauge theory "completely"... $\endgroup$ Commented Mar 28, 2019 at 14:10
  • $\begingroup$ In my post I in particular refer to the gauge properties of the EFEs under gauge group of Poincare'. As it is well-known it also contains translations apart from the "Lorentz-rotations". I think, the EFEs cannot be made gauge-invariant under translations. $\endgroup$ Commented Mar 28, 2019 at 14:21
  • $\begingroup$ Please look into this post physics.stackexchange.com/questions/437787/… . In the answer of A.V.S it is said the theory cannot be made fully gauge-invariant. $\endgroup$ Commented Mar 28, 2019 at 14:27

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