The gravitational binding energy or self-energy of a system is the minimum energy which must be added to it in order for the system to cease being in a gravitationally bound state. Equivalently, the gravitational self-energy of a body is defined as the work done by an external agent in assembling the body from the infinitesimal elements that are initially at an infinite distance apart. A gravitationally bound system has a lower (i.e., more negative) gravitational potential energy than the sum of the energies of its parts when these are completely separated.

Is this type of energy a source for gravity? In order words, does a star have a smaller gravitational field due to its negative gravitational self-energy?


2 Answers 2


Is gravitational binding energy or gravitational self-energy a source of gravity?

Yes. If gravitational theory satisfies the strong equivalence principle (SEP) then the gravitational binding energy would contribute to the gravitational field of a system in exactly the same way as other types of energy. General relativity does satisfy SEP, however there are alternative theories of gravity that violate SEP and for them gravitational binding energy would contribute differently. For such alternative theories there could be observable consequences such as (hypothetical) Nordtvedt effect.

Observations of pulsar PSR J0337+1715 in a triple star system provide currently the most accurate test of SEP with results fully consistent with GR predictions.

  • $\begingroup$ While your answer is correct, the connection to the strong equivalence principle is not obvious. $\endgroup$
    – Yukterez
    Aug 31, 2023 at 23:10
  • $\begingroup$ If this was true, how would you incorporate gravitational binding energy into $T_{\mu\nu}$? E.g. for one body in a central gravity field. $\endgroup$ Aug 31, 2023 at 23:13
  • $\begingroup$ Cf. en.wikiversity.org/wiki/… -- The metric in GTR depends only on the matter and the electromagnetic field in the considered reference frame. $\endgroup$ Aug 31, 2023 at 23:22
  • $\begingroup$ @JánLalinský how would you incorporate gravitational binding energy into $T_{μν} $? That is what the pseudotensor of gravitational field does. It allows us to recast Einstein equation in a form that has $T^{\mu\nu}+t_\text{LL} {} ^{\mu\nu}$ as a source in the rhs with linear differential operator in the lhs. The inherent ambiguity in pseudotensor definition is the same ambiguity in separating energy into different ( e.g. gravitational and nongravitational) parts $\endgroup$
    – A.V.S.
    Sep 1, 2023 at 6:22
  • $\begingroup$ @Yukterez If gravitational binding energy contributes differently to gravitational field of a system then experiments involving systems containing different fractions of gravitational energy in them would demonstrate violations of universality of free fall, this violating SEP. $\endgroup$
    – A.V.S.
    Sep 1, 2023 at 6:30

To work out the gravitational mass of a star you would have to include the rest mass of its components, their internal energy and their gravitational potential energy. The sum of the latter two terms is negative for a bound star, so the gravitational mass of the star is lowered.

e.g. For the Sun, the virial theorem tells us that twice the internal energy plus the gravitational potential energy is zero. Thus the binding energy is about half the gravitational potential energy, roughly $$ E_B \simeq -\frac{3GM_\odot^2}{10R_\odot}$$ and the gravitational mass is reduced approximately by a fraction $$\frac{|E_b|}{M_\odot c^2} \simeq \frac{3GM_\odot}{10R_\odot c^2} = 3\times 10^{-6}\ , $$ but the effect is far bigger white dwarfs and neutron stars.

  • $\begingroup$ What is internal energy? $\endgroup$
    – Manuel
    Sep 1, 2023 at 23:57
  • 1
    $\begingroup$ @Manuel e.g. the kinetic energy of particles in the gas. $\endgroup$
    – ProfRob
    Sep 2, 2023 at 4:54
  • $\begingroup$ Why is internal energy negative? $\endgroup$
    – Manuel
    Mar 24 at 0:12
  • $\begingroup$ @Manuel it isn't. $\endgroup$
    – ProfRob
    Mar 24 at 7:37

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