According to Newtonian gravity, when dealing with multiple sources of gravity, the net gravitational field equals the sum of the individual gravitational fields of each source.

Does the same hold for general relativity (GR)? Because gravity is described by a spacetime metric, and not by a field, is it accurate to say that given multiple sources of gravity, each source corresponding to some metric, the "net metric" at a given point equals the sum of the individual metrics of each source?


5 Answers 5


One of the most interesting, and complicated, features of general relativity is the fact it is a non-linear theory, i.e., adding solutions together won't yield a solution.

One example of that behavior is a Schwarzschild black hole, which means a black hole with no charge and no rotation (a spinning black hole is a bit more complicated, but it would work as well). The Schwarzschild solution is what is known as a vacuum solution: there is no matter in the spacetime. At any point in spacetime you look, there won't be matter. Still, there certainly is gravity.

One of the pictorial ways of interpreting this is by noticing that the gravitational field itself possesses "energy" (in quotation marks, because the notion of energy in general relativity is complicated, as mentioned in this blog post by Sean Carroll). By means of $E=mc^2$, having energy means, in some sense, having mass, and hence the gravitational field is a source of even more gravitational field. The gravitational energy "creates" more gravity, which leads to more gravitational energy, and then more gravity and... And this is essentially what we call non-linearity. The effects start piling up on each other and the description gets quite complicated. Far more complicated than what one has in Newtonian gravity or Maxwellian electromagnetism, both of which are linear theories.

Notice then that if you add two solutions together, you'll be increasing the amount of gravitational energy. That is a source of more gravity, and hence you'll need to account for this extra gravity, which was not present in the two original solutions.

Disclaimer: notice that this "energy creates more gravity" view is pictorial, and meant only to bring more intuition. There is no way of assigning an "adequate" notion of energy to the gravitational field (see the Sean Carroll blog post for some more detail). While this picture can be used as a way of getting intuition and interpretation, it does have limitations and should be taken with a grain of salt.

Black Holes are made of Vacuum

I noticed this bit of the answer caused some discomfort on the comments, so maybe I should add some more resources on it. I believe Kip Thorne is someone particularly famous who quite often mentions how black holes are made of warped spacetime, instead of compact matter. His comments appear in this site now and again. Here are some instances:

I should also have added before that all of my answer should be understood in the context of General Relativity, which means I'm neglecting all quantum effects. Within the framework of General Relativity, spacetime is a differentiable Lorentzian manifold, which means it must have a well-behaved metric at all points. This prevents the singularity of Schwarzschild spacetime from being a point in the manifold, since a curvature scalar blows up "there". Hence, in the description provided by General Relativity, there is not a single point in Schwarzschild spacetime where there is matter. All points are at vacuum.

"But near the singularity, quantum gravity effects should kick in and—" I agree. This description is not necessarily final, and most likely it will be modified by quantum effects. However, it is important to distinguish what happens in the actual Universe—in which the Schwarzschild solution doesn't even exist, since we had a Big Bang and we have a positive cosmological constant—and what is described by General Relativity. I discussed similar issues (the difference between theory and reality) in this post about Classical Electrodynamics. It is one thing to ask whether an actual black hole corresponds to complete vacuum, and another thing to ask whether a GR solution is a vacuum solution.

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    $\begingroup$ I disagree with this answer's use of the Schwarzschild metric as an example which doesn't have mass but does have gravity. If you regularize the Schwarzschild metric and find the corresponding stress energy tensor, it gives a regularized delta function centered on $r=0$. $\endgroup$
    – user196574
    Aug 4, 2022 at 18:12
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    $\begingroup$ @user196574 $r=0$ is a singularity, so it is not in the spacetime manifold (quoting Wikipedia: "A gravitational singularity, spacetime singularity or simply singularity is a condition in which gravity is so intense that spacetime itself breaks down catastrophically. As such, a singularity is by definition no longer part of the regular spacetime and cannot be determined by 'where' or 'when'."). In E&M, it is fair to consider a field that blows up in a point, but in GR the field is spacetime itself, so things get more complicated. $\endgroup$ Aug 4, 2022 at 19:08
  • $\begingroup$ I don't like calling the Schwarzschild solution a vacuum solution. I think it's best to say that GR is agnostic about the mass in this case. Certainly the theory breaks down at the singularity which isn't part of the manifold, so you can't really say that the mass is there in the mathematical model, but at the same time the theory is incomplete and a better description (string theory?) would probably resolve the singularity and would perhaps make speaking of mass and its location more sensible. My intuition is that surely the mass of say a collapsing star doesn't just disappear into a vacuum. $\endgroup$
    – J_P
    Aug 5, 2022 at 13:56
  • $\begingroup$ The gravitational field is not source of more gravitational field, but less gravitational field, since if you assign an energy to it that is negative (if you have a particle of rest mass m hovering stationary close above the horizon of a black hole with mass M its negative potential energy makes it contribute less than m to the total mass, so that is less tham M+m (if the particle of mass m falls in with the escape velocity kinetic and potential energy cancel out and it contributes exactly m to the mass of the black hole). $\endgroup$
    – Yukterez
    Aug 5, 2022 at 16:21
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    $\begingroup$ @Níckolas Alves - I don't think that "gravity generates more gravity" is a watertight formulation, I also read that in popular science articles but I think that is misleading, a body will not attract you more after it collapsed to a black hole, but that's what it sounds like. $\endgroup$
    – Yukterez
    Aug 6, 2022 at 7:50

Nope. GR is a non-linear theory.


While I am no expert on GR - here's another perspective.

While the forces in Newtonian gravity do follow super-position, the potential energy does not: the force experienced by a test mass in the presence of multiple sources is same as the sum of forces experienced when only individual sources were present but the potential energy in the presence of multiple sources isn't the sum of the potential energies of the individual cases - there is an additional contribution - the mutual potential energy of the multiple sources.

Note that this "energy of the sources"$^1$ is a constant contribution regardless of where the test mass is. Further, the field depends on the gradient of the potential so constant contributions don't matter and so fields still add up linearly.

Now, in GR, energy acts as a source of gravitational force$^2$. From this alone, it follows that when we try and super-position the solution of individual sources, a new gravitational source is activated - the mutual energy of the sources. This means the metric being calculated for the source as a whole will be different$^3$ than the "sum of the metrics"$^4$ of the individual sources.

In one-line, this lack of superposition is a synonym for non-linearity, as tersely put by @ConnorBehan.

$^1$ aka. binding energy

$^2$ it contributes to the stress-energy tensor, which is the source term in the Einstein field equations.

$^3$ What about the gradient logic of Newtonian gravity? Why doesn't that apply here? Well, in GR the quantity of interest, the metric, isn't a simple gradient or any linear function of the sources so it doesn't apply.

$^4$ I am not sure what that would mean but sum in some sense


GR is a non-linear theory, and that is enough to answer the OP's question. Regarding how a test particle possessing mass affects the gravitational field in GR, search the literature on "Back Reaction".


According to Newtonian gravity, it doesn't account for the strong and weak equivalence principle, where you cannot just aggregate all the sum of their gravitational fields and derive a solution, it doesn't make sense nor does it have any pragmatic solution. The same as Einstein's spacetime metric version of gravity. To be clear about understanding the significance of "gravitation" there is no one-standard universal law unless you incorporate the SEP, and WEP. The Newtonian version of gravity coupled with Herr Einstein's field equations can best be described, as the force of attraction between two masses divided by the square of their distance in spacetime curvature. The second universal axiom, the SEP and WEP are directly proportional to the product of the sum of their masses or matter fields, and inversely proportional to their distances. The third axiom, gravity is a nonlinearized scalar-tensor-vector matter field.


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