# What is the energy integration constant from time symmetry in general relativity?

If the metric components $$g_{\mu\nu}$$ are independent of the time coordinate $$x^0=t$$, then it's common to say there is a constant of motion called "energy" begotten from considering the Euler-Lagrange equations of a free (test?) particle. That is, from the Lagrangian $$L = \frac{1}{2}m g_{\mu\nu}d\dot{x}^\mu d\dot{x}^\nu$$, the equation for $$t$$ is $$\frac{d}{d\tau} \frac{dL}{d\dot{t}} = \frac{dL}{dt} = 0$$ as long as $$g_{\mu\nu}$$ is independent of $$t$$. This implies that $$dL/d\dot{t} =$$ const $$\equiv E$$.

I wanted to ask specifically what energy this is? Is this a purely kinetic energy or some sort of total energy? I think what's hard for me to wrap my head around is that usually we only consider test masses, which are practically massless, so it acts like light in that it's energy without mass but we also want to use it to find time-like (not null-light) geodesics.

Also, if the metric components are independent of an "$$\phi$$" coordinate like spherical coordinate, there is a conserved "angular momentum" $$L$$. If $$E$$ is total energy, is this $$L$$ a portion of that energy $$E$$? If so, then does there always exist a maximum of the ratio $$E/L$$? My apologies if this is in the literature.

A priori, the interpretation of this energy is not obvious. It is the covariant t-component of the momentum 4-vector, but by itself it does not say much, other than that in flat Minkowski space it will reduce to the familiar notion of kinetic energy.

For an asymptotically flat black hole, this interpretation will also still be true for a test particle starting at infinity. However, this particle will accelerate as it approaches the black hole, presumably increasing its kinetic energy, yet since it is following a geodesic $$E$$ remains constant, suggesting that $$E$$ should be interpreted as the total (kinetic+potential) energy of the particle, and this is also what it reduces to in the Newtonian limit.

For bound orbits around a black hole, we do not have the luxury of an asymptotic region to help us with the interpretation of $$E$$, and it is not immediately obvious that the previous interpretations. In this case, one needs to examine the perturbation to the gravitational field caused by the test particle. This allows us to calculate the the ADM energy of the system as a whole. Doing so tells us (after a not so trivial calculation) that the total energy of the system is $$Mc^2 + E$$ (where $$M$$ is the mass of the black hole). This confirms the interpretation of $$E$$ as the total energy contributed to the system by the particle, including its kinetic energy and binding energy. (The orbit being bound further means that we must have that $$E< mc^2$$.)

In a similar way, $$L$$ is associated to the ADM angular momentum (along the symmetry axis) contributed by the particle. For any fixed value of $$L$$ there is a minimum value of $$E$$ (which is realized for circular orbits). Similarly for fixed values of $$E there is a maximum value of $$L$$ (again realized for circular orbits). For $$E>mc^2$$, $$L$$ can take any value.

• Thanks for your great answer! Could you elaborate just a bit why E<mc^2 in a bound orbit? It's not entirely obvious to me. – UrsaCalli79 Dec 2 '20 at 15:12
• If $E< mc^2$ the ADM mass is smaller than the sum of the component masses. This means that you have to add energy to the system to separate the masses. I.e. the masses are bound together. – mmeent Dec 2 '20 at 15:27

That is nothing but the temporal component of the four-momentum, i.e., the general notion of energy in GR in the presence of a timelike Killing symmetry ($$\frac{\partial}{\partial t}$$ is a Killing vector in your hypotheses). It works also for lightlike geodesics obviously. It is not only kinetic as it includes the gravitational field when the metric is not flat.

For instance, that is the notion of energy used in Penrose's extraction procedure to extract energy from a rotating black hole.

You can obtain the result directly from the Killing equation and the geodesic equation.

The result generalizes to the case where there is a generic Killing vector tangent to one coordinate (for instance in the presence of spherical or translational symmetry). In case of rotational symmetry you indeed get the angular momentum around the symmetry axis.