Special relativity and diffeomorphism invariance

In studying general relativity (GR) we learn that the Einstein-Hilbert (EH) action $$S_{EH}=\int_{M}\mathrm{d}v_{g}R$$ (where $$\mathrm{d}v_{g}=\mathrm{d}^{4}x\sqrt{-g}$$, with $$g$$ the metric tensor) is diffeomorphism invariant, i.e. under a diffeomorphsim $$\phi^{\ast}:M\to M$$ (where $$(M,g)$$ is the spacetime manifold): $$\phi^{\ast}S_{EH}=S_{EH} \iff \mathcal{L}_{\xi}S_{EH}=0$$, where $$\xi$$ is the vector field generating the diffeomorphism. As far as I understand it, one can show this is true as follows: $$\delta_{\xi}S_{EH}=\epsilon\mathcal{L}_{\xi}S_{EH}=\epsilon\int_{M}\big[\mathcal{L}_{\xi}(\mathrm{d}v_{g})R+\mathrm{d}v_{g}\mathcal{L}_{\xi}(R)\big]=\epsilon\int_{M}\big[\mathrm{d}v_{g}\nabla_{\mu}\xi^{\mu}R+\mathrm{d}v_{g}\xi^{\mu}\nabla_{\mu}R\big]\\ =\epsilon\int_{M}\mathrm{d}v_{g}\nabla_{\mu}(\xi^{\mu}R)=0\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\;$$ assuming that $$\xi$$ has compact support, such that it vanishes on the boundary $$\partial M$$.

This is all well and good, but then how does one show that the action of a theory in the context of special relativity (SR) is not diffeomorphism invariant? Take, for example, electromagnetism (EM), which has the following action $$S_{EM}=-\int\mathrm{d}^{4}x\frac{1}{4}F^{\mu\nu}F_{\mu\nu}$$ where we have noted that $$\sqrt{-\eta}=1$$. Isn't it still the case that $$\mathcal{L}_{\xi}(\mathrm{d}v_{\eta})=\mathrm{d}v_{\eta}\partial_{\mu}\xi^{\mu}$$ and $$\mathcal{L}_{\xi}(F^{\mu\nu}F_{\mu\nu})=\xi^{\mu}\partial_{\mu}(F^{\mu\nu}F_{\mu\nu})$$? In which case it would follow that $$\phi^{\ast}S_{EM}=S_{EM}$$, which would contradict the statement that SR is not diffeomorphism invariant? My understanding of diffeomorphism invariance is that solutions to the Einstein Field equations (EFEs) $$G=\text{Ric}(g)-\frac{1}{2}gR(g)=0$$ are mapped into solutions, i.e. if $$g$$ is a solution to the EFEs, related to $$g'$$ via a diffeomorphism $$g=\phi^{\ast}g'$$, then $$G(g)=G(\phi^{\ast}g')=(\phi^{\ast}G)(g')=0.$$ However, SR only holds under isometries $$\phi^{\ast}\eta=\eta\iff\mathcal{L}_{\xi}\eta =0$$.

I have gotten myself very confused over the issue. Any help on the matter would be very much appreciated.

P.S. I get that so-called active (which we are discussing here) and passive diffeomorphisms are essentially the same in coordinate form (since any infinitesimal diffeomorphism can be expressed in coordinate form as an infinitesimal coordinate transformation), however, the two are very different (at least in a physical sense). There must be a way to understand this without resorting to the coordinate argument?!

• Posting this as a comment because I'm not sure it addresses the issue. The EH action is understood to be a functional of the metric field $g_{ab}$, so we can vary the action with respect to $g_{ab}$. But in the EM action in SR, the metric is fixed: it's just a collection of coefficients in the integrand, not a collection of dynamic variables. Of course, we can generalize the EM action so that the metric is variable, and then the EM action is diffeomorphism-invariant (if we do it right). So maybe this isn't a problem with your calculations as much as clarifying what can be varied. – Chiral Anomaly Jan 26 at 19:02
• @DanYand Ah yes, you're on to something there. I wasn't really thinking about the fact that the background is fixed in SR, i.e. the metric is not dynamical, and therefore we can't vary it: this will generically change the metric such that it is no-longer Minkowski, and thus SR cannot be diffeomorphism invariant by construction. Would this be correct? – Will Jan 26 at 19:05
• Yeah, that's the way I'm thinking about it. – Chiral Anomaly Jan 26 at 19:06
• @DanYand Cool. Also, is my calculation showing that GR is diffeomorphism invariant correct? Have I understood what is meant by diffeomorphism invariance correctly in the last of the display equations in my OP? – Will Jan 26 at 19:08
• It would take me some time to think through the subtleties of the compact notation, but the gist of it (and the words) seems to be correct. Given one solution of $G=0$, any other metric obtained from that one by a diffeomorphism (active or passive, doesn't matter) will be another solution. That's what physicists usually mean by diffeomorphism invariance. Of course, if other fields are involved (such as the EM field in the Einstein-Maxwell coupled system), then we need to apply the diffeomorphism to all of the fields in order to still have a solution. – Chiral Anomaly Jan 26 at 19:13

The "diffeomorphism invariance" of general relativity is a figure of speech that more properly refers to the invariance of the theory under $$\mathrm{GL}(1,n-1)$$ gauge transformations that act on all fields like the pushforward under a diffeomorphism but that do not act on the coordinates at all. That is, the gauge transformations of GR are not diffeomorphisms but transformations on the fields that just look like the transformations induced by a diffeomorphism.
• .... What confuses me further is the usual mismatch between mathematician's and physicist's terminology. In particular, in physics an isometry is a transformation for which $\phi^{\ast}g=g$, but it is said that GR is invariant under diffeomorphisms that correspond to isometries in (as far as I understand) the mathematician's sense, i.e. if the metrics $g_{1}$ and $g_{2}$ are related by $g_{2}=\phi^{\ast}g_{1}$, then they are isometric, that is they describe the same physical geometry. Is this correct in any way? – Will Jan 27 at 15:27
• So would it be correct to say that two metrics are isometric (and thus "physically equivalent") if there is a pullback such that $g_{2}=\phi^{\ast}g_{1}$? Also, how does one show that they both satisfy the same equations of motion, i.e. Einstein's equations? Does one simply note that if $g_{1}$ satisfies the EFEs, i.e. $G(g_{1})=0$, then taking the pullback of this, we have that $\phi^{\ast}G(g_{1})=G(\phi^{\ast}g_{1})=G(g_{2})=0$, and so $g_{2}$ also satisfies the EFEs? – Will Jan 27 at 15:43