When you are solving the Einstein field equations (EFE), you basically have to input a stress–energy tensor and solve for the metric.

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

For a vacuum solution we have:

$$ T_{\mu \nu} = 0 $$ This yields: $$ R_{\mu \nu} = 0 $$

This means that the local curvature of an inertial frame of reference is zero.

But, setting the stress–energy tensor to zero, could be given in multiple situations: In flat spacetime, around a non-rotating black hole, around a planet, etc.

When I read about this equations in divulgation books, they portrait the Einstein field equations as:

$$ \text {Space-Time Curvature} = \text{Energy} $$

But with this interpretation, by setting $T_{\mu \nu} = 0$, you are saying that the energy is zero, hence no curvature, but you are able to get more solutions than the Minkowski metric (which is the only solution with truly no energy and with no curvature).

Are this books interpretations wrong or is there something I'm not getting from the true meaning of the equation? How would you distinguish, while solving the EFE, from a truly flat spacetime, from a locally flat spacetime?

  • 1
    $\begingroup$ I searched to see if divulgative was a word I didn't know, but it turns out it's just not a word. What did you mean? Pop science books? $\endgroup$
    – Jojo
    Dec 10, 2022 at 9:34
  • $\begingroup$ Yeah, sorry, I just translate it directly from my language and didn't search if I was right. $\endgroup$ Dec 10, 2022 at 12:11
  • $\begingroup$ It isn't ?!? How do you say it in english than? $\endgroup$ Dec 10, 2022 at 12:20
  • $\begingroup$ What did it mean? There is the English verb "divulge" - e.g. "To make public or known; to communicate to the public; to tell (information, especially a secret) so that it may become generally known." $\endgroup$ Dec 11, 2022 at 16:20
  • $\begingroup$ There is the Italian adjective divulgativo (feminine plural divulgative - the same as in revision 1-3) - "1. informative (that divulges information). 2. popular" $\endgroup$ Dec 11, 2022 at 16:25

5 Answers 5


It is common see Einstein's equations described as "curvature equals energy", but that is not a complete description. The full description is complicated, which is exactly why we express it in the language mathematics.

Remember that $R_{\mu\nu} = 0$ is a differential equation. The Ricci tensor, $R_{\mu\nu}$, is a complicated expression of derivatives of the spacetime metric. $R_{\mu\nu} = 0$ is a set of 16, second-order, nonlinear, partial differential equations.

The Ricci tensor being zero does not mean the curvature is zero. Consider this differential equation: $$\frac{d^2 f}{dx^2} + f = 0.$$ One valid solution is $f(x) = 0$ for all $x$. However, there are plenty of other solutions, like $f(x) = \sin(x)$. Which solution you get will depend on other factors, like the boundary conditions.

Both the Minkowski and Schwarzschild metrics are asymptotically flat, but they differ substantially at $r=0$. The spacetime singularity in the Schwarzschild metric is in some sense an extra boundary condition to apply.

There are other ways to express the curvature beyond the Ricci tensor, for instance the Ricci scalar, $R$, and the Kretschmann curvature scalar, $K$. The Minkowski metric has $R_{\mu\nu} = 0$, $R=0$, and $K=0$. However the Schwarzschild metric has $R_{\mu\nu} = 0$, $R=0$, but $K = \frac{48M^2}{r^6}$ (in geometric units where $G=c=1$). The most complete way to characterize the curvature of a spacetime is the Riemann curvature tensor $R_{\alpha\beta\mu\nu}$, which is a rank 4 tensor.

My feeling is that "curvature equals energy" is a zeroth-order description of the Einstein equations. It expresses one of the most important ideas and acts as a starting point for people who don't have the mathematical background.

A higher order description I use at the start of GR course is "second derivatives of the metric equal energy density". This is in direct analogy to the field equation of Newtonian gravity: $$\nabla^2 \phi = 4\pi G \rho.$$ This is a second-order, partial differential equation to find the gravitational potential $\phi$ from the mass density $\rho$.


Your first equation (the EFE) equates the Stress-Energy tensor with the Einstein tensor.

"Equating energy with curvature" is just a superficial description of that tensor equation; it is a simplification intended only for audiences unfamiliar with differential geometry (or tensor calculus).

In particular, "energy" is not the only component of the Stress-Energy Tensor. It also has other components, such as pressure and momentum, that are distinct from energy. Similarly, "the curvature of spacetime" is not a phrase with a precisely defined meaning; the Einstein tensor is just one among several different tensors (such as the Riemann tensor for example) that are functions of the local geometry.

When solving the EFE, you are always only solving for the local neighbourhood. The EFE cannot really tell you what the global structure of the spacetime manifold is. (For illustration, the EFE would not be able to distinguish a flat sheet from a tube or a möbius strip, since all have the same local geometry.) Instead, researchers make arguments about possible analytic extensions, applying additional constraints to favour one extension over another.


But with this interpretation, by setting Tμν=0, you are saying that the energy is zero, hence no curvature. But you are able to get more solutions than the Minkowski metric (which is the only solution with truly no energy and with no curvature). Are this books interpretations wrong or is there something I'm not getting from the true meaning of the equation?

The stress-energy tensor $T_{\mu\nu}$ in Einstein field equations describes the effect of matter on spacetime geometry [1]. It has not the same status as Ricci curvature tensor $R_{\mu\nu}$ that represents the gravity theory. Therefore, their equality does not mean their equivalence but merely specify which of matter properties to what extent change the spacetime curvature.

Regarding stress energy tensor I would like to quote [2], ... "Its components are related to the matter in the spacetime by

\begin{equation} T_{\mu}^{\nu} = \left( \begin{array}{c | c} \rho & S^{\nu} \\ \hline S_{\mu} & \pi_{i}^{j} \end{array} \right), \label{eq:Tab-compts} \end{equation} where $\rho$ is the energy density, $S_\mu$ is the energy-flux, and $ \pi_{i}^{j}$ is the stress ($i,j=$1,2,3). Typically, $S_{\mu}$ is considered a generalization of the Poynting vector and $ \pi_{i}^{j}$ is considered a generalization of the notion of pressure."

[1] https://arxiv.org/abs/1803.09872v1, Dennis Lehmkuhl, "How Einstein saw the role of the energy-momentum tensor in GR", page 5.

[2] https://arxiv.org/abs/2110.01121, Thomas Berry, Thesis, "Mimicking Black Holes in General Relativity", page 46.


To provide a more plain answer to your question:

No, energy and the curvature of spacetime are two different things. Energy is a physical quantity that is associated with the motion or arrangement of matter and radiation, while the curvature of spacetime is a property of space and time that is determined by the distribution of matter and energy within it. While the two concepts are related in some ways, they are not the same thing. For example, the curvature of spacetime can be affected by the presence of energy, but energy itself is not equivalent to the curvature of spacetime.


The curvature of spacetime is proportional to energy. Spacetime curves around that energy. If you imagine a sphere with energy at its centre then the curvature is the same at every point on the surface of the sphere.

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