My understanding of the mathematics of General Relativity is quite limited. So I am not expecting any mathematically rigorous answers, but what I do understand is that energy curves spacetime, which in turn dictates the motion of energy and mass. I've seen a lot of videos which explain why einstein thought of the universe as 4D space but non of them seem to explain how he related energy, pressure and stress to the curvature of spacetime

  • I understand why einstein proposed that energy curves spacetime, but how did he know how much mass curves how much spacetime? I could randomly say that the curvature is linear with radius for a spherical mass which is not true. Where did einstein get his constraint from? that relates energy and curvature?

Is this the conservation of some quantity?


3 Answers 3


Ultimately the justification for Einstein's equation was experimental.

The assumption that the equivalence principle holds strongly suggests that gravity has to be described by a metric theory, that is the theory relates the curvature of spacetime to some property of the matter and energy present. The first such theory was Nordström's theory of gravity proposed in 1913, in which the field equation is simply:

$$ R = 24\pi T $$

where $R$ is the Ricci scalar and $T$ is the trace of the stress-energy tensor. This is a perfectly good theory of gravity with all the features we expect. It respects the equivalence principle and is derivable from an action principle. But it was unable to account for the perihelion shift of Mercury and predicted that gravitational lensing did not occur, while Einstein's equation:

$$ R_{\mu\nu} - \frac{1}{2} R\,g_{\mu\nu} = 8 \pi T_{\mu\nu} $$

does correctly describe Mercury's orbit and gravitational lensing.

Einstein's equation is not the unique theory relating curvature and mass/energy but it is the simplest one that works. That's why Einstein chose it.


There is a gravitational constant $\mathbf{G}$ in Einstein field equations:

$$ R_{\mu\nu} - \frac{1}{2} R g_{\mu\nu} + \Lambda g_{\mu\nu} = \frac{8 \pi \mathbf{G}}{c^4} T_{\mu\nu} $$

For non-relativistic setting the predictions must match existing observations, i.e. Newton's gravitation law. This is why Newton's gravitational constant is there.

  • $\begingroup$ Was that the only constraint he worked with? But didin't he make modifications to his theory by messing around with the cosmological constant? $\endgroup$
    – Chandrahas
    Sep 28, 2017 at 18:48
  • $\begingroup$ Reproducing results of Newton's gravity law in low energy regime is one of the major constraints for any theory of Gravity. Of course, it played the most important role. Cosmological constant has very little effect in this regime due to being significantly smaller than $\mathbf{G}$. $\endgroup$
    – Darkseid
    Sep 28, 2017 at 18:56

So as somebody above pointed out, Einstein's field equations are; $$ G^{\mu \nu} = \frac{8 \pi G}{c^{4}} T^{\mu \nu} $$

Where $G^{\mu\nu}$ is a mathematical object that describes the curvature of the $4D$ spacetime. The object $T^{\mu\nu}$ is called the stress-energy tensor, and describes the energy, pressure, matter content of the spacetime. So essentially we have: $$ \left(\mathrm{Curvature} \right) \ = \ \frac{8 \pi G}{c^{4}} \times \left(\mathrm{Energy\ and\ Stuff}\right) $$

The story I have heard is the following:

  • We know that energy and momentum and so on must all be conserved: this condition can be very compactly written as $\nabla_{\nu} T^{\mu\nu} = 0$ (in simple terms it's basically saying that the derivative of $T$ is zero, and so it's overall energy/momentum content stays constant).

  • The mathematics involved with the development of this $G^{\mu\nu}$ object requires that $\nabla_{\nu} G^{\mu\nu} = 0$

I suppose because of his thoughts on the equivalence principle (??) Einstein suspected that energy and curvature are related to each other. Since $\nabla_{\nu} T^{\mu\nu} = 0 = \nabla_{\nu} G^{\mu\nu}$, he said that $T^{\mu\nu} \propto G^{\mu\nu}$ and came up with the above equation (after putting in some constants to make the units work).

Interesting side note: $G^{\mu\nu}$ is the simplest curvature object which obeys $\nabla_{\nu}G^{\mu\nu}=0$, but there are other more complicated ones that could have been used.

  • $\begingroup$ Lots of things are "unchanging", I don't see why that would lead to any implication they are related - or indeed unrelated. This is a gold standard answer for clarity in my view. $\endgroup$
    – JMLCarter
    Sep 28, 2017 at 22:52
  • $\begingroup$ But then how did he mess around with the cosmological constant and modify his equations? $\endgroup$
    – Chandrahas
    Sep 29, 2017 at 2:00
  • $\begingroup$ @Chandrahas there is $\Lambda$ in my answer below, which is cosmological constant. It enters the equation in pretty much the same way as the curvature itself. Could you make your question more specific, please? $\endgroup$
    – Darkseid
    Sep 29, 2017 at 2:35

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