I often hear about how general relativity is very complicated because of all forms of energy are considered, including gravitation's own gravitational binding energy. I have two questions:

  1. In general relativity, objects following the motion of gravitation should simply be travelling by geodesics. In such 'free fall', why would there be any 'binding energy'?
  2. From Einstein's field equations, $$R_{\mu \nu} - {1 \over 2}g_{\mu \nu}\,R = {8 \pi G \over c^4} T_{\mu \nu},$$ isn't the curvature only coupled to the energy-momentum tensor? As far as I understand, potential energy is not included inside the energy-momentum tensor.
  • $\begingroup$ Are you asking about gravitational self energy? If so a quick Google will find you lots of stuff. Gravitational waves carry energy, so by extension gravitational fields do as well. $\endgroup$ Nov 26, 2012 at 17:39
  • $\begingroup$ Your first question seems unrelated to the others. It could be phrased better as "What's the potential energy experienced by a test particle following geodesic motion? Why would a geodesic experience 'binding energy'?" $\endgroup$ Nov 26, 2012 at 17:45
  • $\begingroup$ It would be helpful to have some quotes, to know what you are refering to. What exactly is meant by "potential energy" and "binding energy" in that context? $\endgroup$ Nov 23, 2022 at 15:19

2 Answers 2


1) I gather you mean gravitation potential energy of the test particle. Well, any such thing is only useful in so far as it is related to a constant of motion throughout the geodesic--in the case of gravitational potential, being part of the conserved mechanical energy, kinetic + potential. (Another example could be angular momentum.)

In GTR, these constants are given by a Killing vector field, which is an infinitesimal generator of an isometry: spacetime "looks the same" in the direction of a Killing vector. Most spacetimes do not have any, but by definition, a static spacetime has a timelike Killing vector field, and can always be put in the following form: $$ds^2 = -\lambda dt^2 + d\Sigma^2,$$ where $d\Sigma^2$ is the metric for any spacelike manifold and $\lambda$ is independent of $t$. The factor $\lambda^{1/2}$ is commonly called the gravitational redshift.

For example, for the Scwarzschild spacetime in the usual Schwarzschild coordinates, $\lambda = \left(1-\frac{2GM}{c^2R}\right)$, and the following is a constant of motion representing the specific (per-mass) energy of the freefalling particle: $$e = \left(1-\frac{2GM}{c^2r}\right)\frac{dt}{d\tau}.$$ This is the natural generalization of the total mechanical energy, including also rest-mass energy; for the Schwarzschild case, the spherical symmetry allows one to build an "effective potential" quite analogous to the Newtonian case, but that approach is less useful in general.

2) Gravitational energy cannot be explicitly included in the Einstein field equations because the equivalence principle--there is always a local inertial frame (the free-falling one) in which spacetime looks like the ordinary, flat, special-relativistic one. Hence if there was a frame-independent local notion of gravitational energy, i.e., a tensor, that tensor is zero in some local frame, and hence zero in every frame.

However, one can think of the non-linearity of the Einstein field equation as caused by gravitational energy itself interacting with spacetime. In this sense, gravitational energy is "implicitly" included. Another thing one can do is try to build another notion of gravitational energy that's not necessarily both local and frame-independent, e.g., Landau-Liftshitz pseudotensor and others.

  • $\begingroup$ +1. I ask for clarification: so you mean GENERALLY the concept of potential energy is invalid in GR? $\endgroup$
    – resgh
    Nov 27, 2012 at 10:53
  • $\begingroup$ Many (in some sense 'most') spacetimes don't have any symmetries (Killing vector fields), so you just plain can't do anything of the sort. If the spacetime is static, orbits get conserved energy, and Schwarzschild's effective potential actually requires spherical symmetry as well. There's a loosening of this requirement to stationary spacetimes (with only an asymptotically timelike Killing field), but then it needs more than one potential. In more complicated (non-stationary) cases, you're just out of luck. $\endgroup$
    – Stan Liou
    Nov 27, 2012 at 15:54
  • $\begingroup$ Many thanks. So this means applying GR practically, we cant really make anything of total conserved energy... $\endgroup$
    – resgh
    Nov 28, 2012 at 3:32

As it has been explained by Stan Liou, for a static space-time it is possible to define an "effective potential energy". I present here an explicit computation. Effectively, for any static space-time the metric can be reduced to the form:

$$\text{d}s^2 = -\lambda (x^1,x^2,x^3)\text{d}t\otimes\text{d}t + \text{d}\Sigma^2(x^1,x^2,x^3)$$

From fundamental energy-momentum relation, we have:

$$-\frac{E^2}{c^2\lambda} + p^2 = - m^2c^2$$

defining an "effective gravitational potential" $\phi_g$ such that:

$$1+\frac{2\phi_g}{c^2} = \lambda$$

We have:

$$E = \sqrt{1+(2\phi_g/c^2)} \sqrt{p^2c^2+m^2c^4} = mc^2 \sqrt{1+(2\phi_g/c^2)} \sqrt{1+\frac{p^2}{m^2c^2}}, \qquad (*)$$

we can rewrite equation $(*)$, using that $\sqrt{1+y} \approx 1+y/2$, as:

$$ E \approx \left(1+\frac{\phi_g}{c^2}\right)mc^2\left( 1+ \frac{1}{2}\frac{p^2}{m^2c^2}\right) \approx mc^2 + \frac{p^2}{2m} + m\phi_g$$

which is the classical expression for energy plus rest energy. So we can write for the relativistic case:

$$E \approx (1+\phi_g/c^2)\sqrt{p^2c^2+m^2c^4} $$

which is valid when $\phi_g << c^2$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.