In Newtonian Physics there is an equation that for the Gravitational Flux of an object known as Gauss's Law For Gravity. Gauss's Law for Gravity describes the number of Gravitational Field Lines coming from an object with mass.

Is there something similar to Gauss's Law For Gravity in General Relativity? Is there an equation similar to Gauss's law for Gravity in General Relativity?

  • $\begingroup$ Possible duplicate of Gauss law for gravitational field $\endgroup$
    – ShankRam
    Dec 9, 2015 at 12:46
  • 3
    $\begingroup$ @ShankRam: Not a duplicate, because the alleged duplicate is about Newtonian gravity, not general relativity. $\endgroup$
    – ACuriousMind
    Dec 10, 2015 at 0:07
  • 1
    $\begingroup$ See also the notion of gravitoelectromagnetism, cf. e.g. Wikipedia. $\endgroup$
    – Qmechanic
    May 31, 2016 at 7:49

4 Answers 4


There is no perfect analogue because gravitational flux is not a (common) feature of GR.

Recall that Gauss' integral law $$\oint_{\partial V} \mathbf{g}\cdot\mathrm{d}\mathbf{A}=-4\pi G\underbrace{\int_V\rho\,\mathrm{d}V}_M$$ is just the differential law $$\operatorname{div}\mathbf{g}=-4\pi G\rho$$ integrated + the divergence theorem. So the two are equivalent.

Now, there is a pretty good analogue for the differential law in GR, namely the Einstein equations. It is well known that the time-time component of the Einstein equations reduces to the Poisson equation $\Delta\phi=4\pi G\rho$ for "nearly Newtonian" spacetimes. (The Poisson equation is equivalent to Gauss' law because $\mathbf{g}=-\operatorname{grad}\phi$.)

But it is not so simple to integrate the Einstein equations and get an integral law.

For asymptotically flat spacetimes, we have the ADM momentum equation $$P^\mu=\int_\Sigma\sqrt{-g}\star [g^{\mu\sigma}(T_{\sigma\nu}+t_{\text{LL}\sigma\nu})\theta^\nu]$$ (Here $\Sigma$ is a spacelike hypersurface, $\star$ is the Hodge dual, $T$ is the energy-momentum tensor, $t_{\text{LL}}$ is the Landau-Lifsitz pseudotensor and $\theta^\mu$ is a dual tetrad.) The time component of this equation gives the "mass" of the spacetime, like the RHS of Gauss' law above. For for details, cf. e.g. N. Straumann, General Relativity (2013).


Birkhoff's_theorem is sometimes called the relativistic analogue to Gauss's Law, because it says that only the interior energy density effects the local gravitation. ( see Lindesay: Foundations of Quantum Gravity, Grupen: Astroparticle Physics)

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    $\begingroup$ Note that it is not a perfect analogue, because Birkoff's theorem requires vacuum and spherical symmetry to work. $\endgroup$ Jul 12, 2017 at 17:16

There is an analogue to Gauss' law for gravity in general relativity, which applies to stationary space-times, which admit a one-parameter group of isometries with timelike Killing vector whose orbits are diffeomorphic to $\mathbb R$ and span the manifold. In such a case, the metric is,

$$ds^2 = -\psi^2(dt + a_i dx^i)^2 + g_{ij}dx^i dx^j.$$

One then has that the flux at infinity of $\mathrm{grad} \, \psi$, which we denote $\Phi$, is given by the integral,

$$\Phi = \lim_{\rho\to\infty}\int_M \psi \rho^t_t \, \sqrt{|g|} \, d^4x$$

assuming a source $\rho$ and furthermore that the space-time is asymptotically Euclidean. This can be derived by looking at the $(t,t)$ component of the Einstein field equations.

  • $\begingroup$ You're describing a measure of mass for a stationary, asymptotically flat spacetime. Isn't this unique and equal to the ADM mass? $\endgroup$
    – user4552
    Jul 12, 2017 at 19:08
  • $\begingroup$ This special case of a stationary spacetime is what sent me here, and it took me a while to figure out that 1. the exact counterpart of Gauss' Law is Komar's theorem: the Komar mass is $ -{1\over 8 \pi G} \oint_{\partial V} \nabla^\mu \xi^\nu d_{\mu\nu}S $ with $\xi$ the time-like Killing vector, regardless of the exact shape of the 2-D surface $\partial V$. It is only required to be space-like and to lie entirely in the surrounding vacuum i. e. to enclose all of the source mass; $\endgroup$ Sep 14, 2021 at 14:52
  • $\begingroup$ and 2. the ADM equation is also available as a (2-D) surface integral, even if the spacetime is not stationary; only in the limit $\partial V \to \infty $, though. In a stationary spacetime, it coincides with the Komar mass. See e. g. [Eric Gourgoulhon, Lect. on Rotating Stars][1]. [1]: arxiv.org/pdf/1003.5015.pdf $\endgroup$ Sep 14, 2021 at 14:54

Yes, and relativistic gravity neatly explains Gauss's law.

The simple way to think of gravity resulting from spacetime curvature is: a 3 dimensional volume of space containing mass M expands over time at a volume acceleration of 4πGM. That's essentially a relativistic re-expression of Gauss's Law where instead of describing a gravitational field it describes the deformation of space itself.

For example: for earth, 4πGM is 5M cubic kilometers/sec/sec, so the apparent acceleration of gravity at earth's surface is that rate divided by earth's surface area of 510M square kilometers. The quotient is 9.8 m/s/s, so objects near earth's surface appear to accelerate downward at that rate, 1g. (Accelerometers show the earth's surface actually accelerates upward, as if its volume was expanding at that rate, while "falling" objects nearby have no forces on them and aren't really accelerating - confirming Einstein's 'elevator' epiphany that gravity is a fictitious force.)

Thus relativistic spacetime curvature uncovers the true meaning of Gauss's Law of Gravity and explains Newton's Inverse Square Law. You can find details here.

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  • $\begingroup$ Okay, is it General Relativity related? $\endgroup$
    – user36790
    May 31, 2016 at 3:21

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