I was just taught (comments) that any type of energy contributes to mass of the object. This must indeed include potential energy in gravitational field. But here, things cease to make sense, have a look:

  1. I have object at some distance $r$ from radial source of gravitational field.
  2. The potential energy is calculated like this: $E_p = m*a*r$ where $a$ is gravitational acceleration and $m$ is mass of your object.
  3. But that means, that the object is a bit heavier - because of the potential energy of itself - $m = m_0 + \frac{E_p}{c^2}$ ($m_0$ here is the mass without the potential energy)
  4. That would mean that the gravitational force is a bit stronger at higher distances.

Now, I do understand that the rules of physics are not recursive and the mass and force will be finite. But what is the correct approach to this situation? What is the correct equation for potential energy?


4 Answers 4


First let's start with Newtonian mechanics, no relativity. The equation $E_p=mgr$ is only valid when $g$ is approximately constant, as it is near the earth's surface. If $g$ is varying like $1/r^2$, then we get $E_p=-GMm/r$. This energy is not interpreted as energy that belongs to the mass $m$ or to the mass $M$. It's interpreted as energy that is stored in the gravitational field that surrounds both bodies, which equals the vector sum of their individual fields at any given point.

In relativity, you can't calculate gravitational effects just by adding $E/c^2$ to the mass; the source of gravitational fields in relativity is the stress-energy tensor, not the scalar mass-energy. Relativistically, an equation like $E_p=-GMm/r$ is only an approximation. To the extent that this approximation is valid, this energy will contribute to one component of the stress-energy tensor, and a distant observer will detect it through a reduction in the $(M,m)$ system's gravitational field, relative to what it would have been if $M$ and $m$ had been well separated and not interacting. Because $E_p$ isn't localized in $m$, there is no change in $m$'s gravitational or inertial mass.

Note that if you lift a rock, there is no change, even theoretically, in the distant field of the earth, since energy is conserved. All you've done is convert some chemical energy into gravitational and heat energy.

Although your argument is wrong in all the details, you have figured out an important idea about general relativity, which is that the theory is nonlinear.

  • $\begingroup$ In the above question the mass was held at a distance r from the gravitational field and no change in inertial mass was expected BUT if mass m is made to circle around with some velocity v then mass should change. (Not only the mass of m but also of the source of gravitational field because as the mass m will be forced to move around the gravitational field , the source of gravitational field will also gain some rotational energy.)According to GR what happens if a mass m is made to circle around a gravitational field with velocity v? $\endgroup$ Mar 27, 2016 at 9:25
  • $\begingroup$ It is a sorry state of affairs when the inability to even contemplate a change in the speed of light combined with a knowledge that the actual mass can't change leaves you with no option than to devise such a complicated scheme to explain a very simple phenomenon. I am a big fan of Einstein and of SR. I am also a big fan of the Equivalence Principle of GR. I would be happy if the rest of GR disappeared down a black hole. $\endgroup$
    – Alan Gee
    Feb 16, 2019 at 11:12
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    $\begingroup$ @AlanGee when one learns the mathematics of GR it is quite common to come to the judgement that it is one of the most elegant parts of the whole of physics, and a tremendous and beautiful cultural achievement as well. But of course even this would not suffice for it to stand, were it not for the wealth of detailed experimental confirmation. $\endgroup$ Jan 24, 2021 at 18:42
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    $\begingroup$ @AlanGee the "pointless mantra" to which you refer is the strong principle of equivalence, which at a stroke settles an enormous number of questions about physics in the presence of gravitation $\endgroup$ Jan 27, 2021 at 8:35
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    $\begingroup$ @Alan Gee in referring to frequency/wavelength it seems you may be describing redshift/blueshift rather than "traversal" speed (c)? Asking what gravity is has more to do with "what causes that shift," rather than whether or not it occurs, which we all agree it does. Furthermore if such change is all gravity were, as you say, that wouldn't account for its effect on time; clock speeds/etc, which are also concrete observables in daily use along with the red/blue shifts you mentioned. Nor would it help us model gravity in quantum terms with space/time probabilities. For those; curvature. $\endgroup$
    – jazamm
    Feb 21 at 23:29

I guess that it is more correct to say that "gravitational binding energy" contributes to a "decreased mass".

At least far away from a spherically symmetric mass distribution such as a planet, a star or a black hole you can assume that you are affected as if the mass of the spherical mass distribution is $E/c^2$. This mass concept is sometimes named "ADM mass" and the numerical value of this mass is the same as the mass $M$ in the Schwarzschild solution of general relativity.

Given that you have a homogenous "ball" consisting of a certain number of atoms of a certain distribution, the smaller and more compact the ball is, the less mass it will contain in the ADM sense, because a smaller "ball" is more tightly bound gravitationally, even though it contains the same distribution of atoms.

The higher the temperature of the "ball" the more energy and thus mass it will contain and other types of energy, such as chemical binding energy etc will also contribute. I guess that in a practical scenario the biggest and totally dominating difference in total mass of the body and the sum of the invariant mass of its individual constituents is due to "gravitational binding energy".

Let us say that you have a spherically symmetric planet of mass $M$ (mass in the ADM sense) and surface radius $R$ and you drop a small mass $m$, $m<<M$, onto it from being at rest infinitely far away. After the small mass has crasched into the big mass and the heat generated in the crash has radiated away the total mass will not be $M+m$ but $M+m\sqrt{1-\frac{2GM}{Rc^2}}$.

  1. If we assume that $m<<M$ and that there is no motion the energy can be written as $E=Mc^2+m\sqrt{1-\frac{2GM}{rc^2}}c^2$, where r is the radial distance between the center of the two masses. In weak fields we can write this as: $E\approx Mc^2+ mc^2-mGM/r$

3,4. Let us say that you have a set of two masses, A and B, that are apart from each other by a little bit. Now if you introduce a third body C far away, the gravitational force felt by C will in general be a little bit larger if A and B are a little bit more far apart than if they are closer together, given that A and B are at rest with respect to each other in both cases. (That is of course if you do not use energy stored in the A+B system to get them further apart)


Yes. All energy contributes to the overall mass. PBS Space Time has a really clear video on this subject: The Real Meaning of E=mc² In the video, Gabe points out that Einstein originally wrote the equation as m = E/c², and describes how all energy has the property of mass, including potential energy.

Edit to add: it's important to note (as is mentioned in the video) the potential energy of a particle within a system does not change the mass of the particle itself. Instead, it contributes to the mass of the system within which the particle has potential energy. Also note that a particle considered in isolation has no potential energy.

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    $\begingroup$ The video talks about mass of a composed system whose parts have potential energy, not mass of one part of such a system. In the first case there is change of mass, in the second there is not. $\endgroup$ Sep 27, 2016 at 18:42
  • $\begingroup$ Absolutely -- I wasn't thinking of the individual mass of the particle, but the mass that it contributes to whatever system it is in. The concept of potential energy has no meaning for an individual particle considered in isolation. I'll edit the answer to be more clear. $\endgroup$ Sep 29, 2016 at 16:13

The total energy of the test mass in the Schwarzschild gravitational field depends on the time-like component of the metric tensor $g_{00}$. This is how the gravitational redshift is derived (see, for example, Hartle J B, Gravity). In this case for the test particle (at rest) with the rest mass $m$ it is possible to write $$E=mc^2 \sqrt{g_{00}}$$ as the energy of the test particle residing in the gravity (as measured by an observer located in the center of the gravity). This would lead to $$E=mc^2 \sqrt{1-\frac{2GM}{rc^2}}\approx mc^2-\frac{GMm}{r} $$ But such energy measurement is highly dependant on the location of the observer. The expression, for example, would be different if the observer is located outside faraway from the gravity.


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