Is it possible to draw some kind of picture to illustrate why the Einstein tensor has zero divergence? I would guess not, because the only curved manifolds we can fully visualize are 2D surfaces with positive-definite metric, and the Einstein tensor is identically zero there anyway, right? But I'd like to know if anyone has a way to see this, at least partially. Like, maybe from 1+1 GR if that makes it easier. Or even just a way to describe it that makes it more intuitive. I mean, it is certainly one of the most important mathematical identities in modern physics, but I don't really have any intuition for it. Thanks in advance.
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$\begingroup$ Is it cheating to instead intuit why the stress–energy tensor is divergenceless? $\endgroup$– J.G.Sep 16, 2021 at 14:52
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1$\begingroup$ @J.G. Yes! Because the profound thing, as I understand it, is that GR removes the need for energy/momentum conservation to stem from the laws of physics. The only necessity is to identify matter/energy with the quantity that is already conserved mathematically. Granted, that identification is itself a law of physics, but to me it seems much simpler than the emergence of conservation from Newton's laws, even via Noether's theorem. That's why I want to understand it geometrically. $\endgroup$– Adam HerbstSep 16, 2021 at 15:54
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$\begingroup$ Thanks for that response. I've posted an answer I hope is of use. If it would be of even greater use if it went through the proof to which it links, let me know to make such an edit. But the main goal is to make sense of what's happening. $\endgroup$– J.G.Sep 16, 2021 at 16:08
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A folded away proof of $\nabla_aG^{ab}=0$ appears here. In particular, it follows from Riemann tensor identities, not the EFE. Since you asked for a "kind of picture", such a proof may miss the point of your question. But an instructive analogy is to Maxwell's equations. If we compare $G^{ab}=\kappa T^{ab}$ (where the cosmological constant is included on the RHS as a matter term to make this true in general) with $\nabla_aF^{ab}=\mu_0j^b$, both source equations are on-shell results, but the sourceless equations $\nabla_aG^{ab}=0,\,\nabla_{[a}F_{bc]}=0$ are off-shell results. Put another way, they're not physics, they're tautologies.
