In Newtonian physics, we come across different forms of energy, such as kinetic energy, potential energy etc. But in general relativity, we find only the total energy that is obtained from the energy-momentum tensor. Are the various forms of energy not valid in general relativity?


4 Answers 4


The $T^{\mu\nu}$ on the right side of the Einstein field equations includes all non-gravitational forms of energy, including mass-energy, kinetic energy, electrostatic potential energy, the energy of static magnetic fields, of electromagnetic waves, of gluons, of the Higgs field, of a possible inflaton field, etc.

The curvature of spacetime doesn’t depend on what kind of energy density is causing the curving, but only on the amount. (And actually the curvature is determined by all the components of the energy-momentum-stress tensor, not just the $T^{00}$ component that is the energy density.)

  • $\begingroup$ Yes, exactly! However, I want to know whether analogs to kinetic and potential energy exists in GR, i.e., whether it is possible to decouple the various forms of energy. $\endgroup$
    – AnnaD
    Dec 28, 2020 at 10:43
  • $\begingroup$ I’m not sure what “decouple” means. If you are asking whether you can distinguish, say, kinetic energy from potential energy by observing only their gravitational effects, the answer is no. $\endgroup$
    – G. Smith
    Dec 28, 2020 at 18:02

General relativity is a classical field theory, and so is classical E&M. In these classical field theories, the notion of potential energy isn't particularly useful. For example, when you accelerate an electron from the cathode to the anode of a vacuum tube, what's really happening is that the total energy of the electric field is decreasing, and you're trading this for the increased kinetic energy of the electron.

As an example of how this applies to the stress-energy tensor, you can take a look at the stress-energy tensor for an electromagnetic field: https://en.wikipedia.org/wiki/Electromagnetic_stress%E2%80%93energy_tensor

Potential energy is a more appropriate concept when you're dealing with Newtonian instantaneous action at a distance. The only reason you can get away with this notion as an approximation, in a case like the cathode ray tube, is that the motion of the electron is slow compared to the speed of light, so you can pretend that changes in the fields aren't propagating at some finite speed.

As pointed out in G. Smith's answer, we don't include the gravitational field's energy in the stress-energy tensor, but based on your comments this doesn't seem to be what you were asking about.


Are the concepts of kinetic energy, potential energy etc not valid in general relativity?

Somewhat valid.

Kinetic energy becomes relativistic kinetic energy in the form : $$ E_{\text{k}}=mc^{2} \left( \gamma - 1 \right) $$

Gravitational potential energy is replaced by metric tensor :

$$ g_{\mu \nu } = {\begin{pmatrix}g_{00}~g_{01}~g_{02}~g_{03}\\g_{10}~g_{11}~g_{12}~g_{13}\\g_{20}~g_{21}~g_{22}~g_{23}\\g_{30}~g_{31}~g_{32}~g_{33}\\\end{pmatrix}} $$ Which is related to Newtonian gravitational potential $\Phi$ for a unit mass in the following way :

$$ g_{00} = -\left(1+2 \Phi\right) $$

It's important to notice that this tensor relation is only valid for weak gravitational fields, in strong gravity- like the one near black hole - this approximation doesn't make sense anymore, otherwise we would not need relativity, isn't it ?

Other forms of "classical energy" are also superseded by relativistic counterparts.

  1. In Newtonian physics, when you are talking about kinetic energy and potential energy of a system, say $L_p=\dot x^2/2 -V(x)$, you refer to a kinetic energy $T=\dot x^2/2$ and a potential energy $V(x)$ of a particle. Even if the potential is generated by some kind of fields $L=L_p+L_F$, the only thing you need to care is the particle part $L_p$ (the equation of particle), in order to define the kinetic energy and potential energy.
  2. For general relativity, let us first define a kinetic energy for a special case, i.e. a free falling particle, whose Lagrangian is $2\mathcal{L}=g_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\mu}{d\tau}$. From my side, the kinetic energy is obvious, it is $\mathcal{L}$. If you can not see what I see, you can write the geodesic equation, then you will find the first integral of motion, in other words, $\mathcal{L}$ is a constant along the geodesic, which can be regarded as "kinetic energy" of this particle. Meanwhile it is consistent with Newtonian limit up to a constant.
  3. Now let us turn to a more general case, where there is a "potential" generated by a kind of field, say EM. First, the "kinetic energy" would not change; Second, the only thing new is Lorentz force, which can be cast in the form of potential (this does not bother me, we can simply cast in a formal integral). If you write down a geodesics equation, the demonstration in this item is not hard to be understood.
  4. However the definitions "difference of Hamiltonian with Lagrange as 2 times potential energy, and the difference of Hamiltonian with potential energy as kinetic energy" are more challenging. For free falling particle, $\mathcal{L}=\mathcal{H}$, thus there is only kinetic energy; for the case in item 3, this definition is also consistent with above one.
  5. For definitions of the kinetic and potential energies of classical fields, the situation is similar.
  6. In Einstein's general relativity, gravity is a background geometry, rather than a force/potential, thus there is no need to separate the particle with its background. Therefore the kinetic and potential energies in item 4 refer to the particles with non-gravitational force.

Edit: 7. How to separate the "Gravitational potential" in the free falling particle case in the sense of Newtonian physics?

I think a lot of people did this problem in his beginning course of GR. All we need is two approximations: 1, slow moving condition and 2, weak field condition. Based on these two conditions, the potential energy can be formally extracted from the equation of motion, which is relevant to the 0-0 component of perturbed metric.

Edit: 8. In fact, if you get rid of energy-momentum tensor, you even can define "kinetic" and "potential" energies for gravity via its Hamiltonian constraint.


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