Einstein went to a lot of trouble linking the stress-energy tensor to the Ricci tensor (and curvature scalar).

Fast forward to using General Relativity for the Schwarzschild solution. - Those beautiful field equations are reduced to zero on both sides i.e. Ricci goes to zero. - Symmetry arguments are used to create some limits for the metric. ... And we simply plug in the answer! (weak field approximation)

Is that deeply unsatisfying to anybody else? I had thought of General Relativity as: Energy (T tensor) generates the curvature (R tensor) that makes the geodesics change and, heyho, we see gravity. But the Schwarzschild solution has gravity with Ricci at zero. How can that be? And what was the point of Einstein perfecting his field equations?

  • $\begingroup$ The point was that not all solutions are vacuum solutions, let alone vacuum solutions with very high symmetry. And for vacuum solutions the equations place strong constraints on the curvature (namely that $R_{ij} = 0$). $\endgroup$
    – user107153
    Commented Mar 7, 2017 at 11:08
  • $\begingroup$ Can you imagine a real world version of the Schwarzchild solution, with no angular momentum? The point of this approach, as far as I know, to to teach the basics of a toy model, and what terms in the metric can be associated with particular physical ideas, such as the mass of the star. It also was the first solution of the application of the Field Equations, so i t has historical significance. Things, as you know, have progressed from 100 years ago, to the recent LIGO experiments. $\endgroup$
    – user146020
    Commented Mar 7, 2017 at 11:50
  • $\begingroup$ Fair enough Counto10. I realise the model is a model. But with Ricci at zero, how does one skew G00 without some effect from T00 at a distance? I know I have this WRONG. I really do. I just want to understand how Einstein's equations account for force at a distance (without just plugging in the answer as happens with Schwarz..) $\endgroup$
    – Napasai
    Commented Mar 7, 2017 at 22:30
  • $\begingroup$ An extra comment is that the Ricci tensor is not zero everywhere: it is infinite at the point mass. $\endgroup$ Commented Mar 11, 2017 at 22:39
  • $\begingroup$ Wiskundelief.. Absolutely. That makes sense. I am struggling with the mechanism that allows that infinite Ricci at one point to affect the curvature of spacetime at another point. The Einstein field equations seem very local in nature. $\endgroup$
    – Napasai
    Commented Mar 13, 2017 at 11:49

1 Answer 1


The Einstein field equations are,

$$R_{\mu\nu} -\frac12 g_{\mu\nu}R = 8\pi \, T_{\mu\nu}$$

for some matter described by a stress-energy tensor, $T_{\mu\nu}$. The Schwarzschild solution describing a non-rotating, neutral black hole corresponds to a Ricci-flat ($R_{\mu\nu} = 0$) solution of the Einstein field equations and can be derived with a spherically symmetric ansatz.

None of this means the Einstein field equations are redundant; remember that $R_{\mu\nu} = 0$ which is the Einstein field equation for a vacuum solution, imposes conditions on the ansatz for the Schwarzschild metric and is required for the derivation.

So what would make you think the general Einstein field equations are redundant? Say I give you some stress-energy $T_{\mu\nu}$; what are you going to use to find $g_{\mu\nu}$? In general, it will be the Einstein field equations alongside perturbation theory.

There are other advanced techniques to describe solutions, but these are based on for example Lie symmetries of the Einstein field equations, or generation techniques which rely on knowledge of the behaviour and characteristics of solutions to the Einstein field equations.

Regardless of what method you may use, they can all be somehow linked to something requiring the fact that $G_{\mu\nu} = 8\pi \, T_{\mu\nu}$ .

  • $\begingroup$ Thanks so much Jamal. I realise I am missing something. What I struggle with is likely due to my ignorance. I don't know how to use the symbol typing so forgive my clunky writing. $\endgroup$
    – Napasai
    Commented Mar 7, 2017 at 18:03
  • $\begingroup$ Thanks so much Jamal. I realise I am missing something. What I struggle with is due to my ignorance. I don't know how to use the symbol typing so forgive my clunky writing. My issue is that T00 is zero at any distance from a point mass. So R00 is zero. I worked through the weak field assumption and see that R00 is linked to the second derivative of g00 so that is zero too. That means the first derivative of g00 must be a constant (or zero). But surely the first derivative of g00 needs to vary with 1/r^2 for gravity? How do Einsteins equations account for force at a distance. $\endgroup$
    – Napasai
    Commented Mar 7, 2017 at 18:28

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