3
$\begingroup$

Kip Thorne, in this public lecture, says that a black hole is kept together by the energy of the warping of space.

Quote around 8:00-8:30 :

"in this case, the energy [which keeps a black hole together] is contained in the warping itself. So the black hole is held together by nonlinear self interaction. The energy tied up in the warping, produces the warping."

Does this mean the curvature contains energy? Does this mean "gravitational energy" can itself cause (more) curvature?

I am confused because I have understood that in General Relativity gravity (space time curvature) does not contain energy--with kind of exception of gravitational waves, but I guess that is not localized energy (not my main question here anyway). The energy momentum tensor accounts for every other kind of energy except "gravitational energy". Or in more technical terms: there is no tensor for Gravitational energy.

I best explanation I found was so called ADM mass If a black hole is just warped spacetime, then where is the electric charge?. I do not understand much about ADM, but apparently there is a way to assign "energy" to curvature.

$\endgroup$
1
  • $\begingroup$ 'Held together'? It is even collapsing by its self energy, if there is such a thing in GRT. $\endgroup$
    – my2cts
    Feb 24 at 8:07

1 Answer 1

1
$\begingroup$

Does this mean the curvature contains energy? Does this mean "gravitational energy" can itself cause (more) curvature?

In short, that is a way of saying it. However, it should be taken with a grain of salt: no one is saying that you can find a stress-energy tensor for the gravitational field itself. Think of it as an intuitive view.

Pick a Schwarzschild black hole, for example. It is a vacuum solution of the Einstein Field Equations, so there is no matter anywhere in spacetime to bend spacetime. Yet, it is bent. Furthermore, if we compute the mass of the spacetime (for example, the ADM mass), we'll find a non-vanishing value. A convenient way of understanding this is that it is as if there was gravitational energy sourcing this bending. Notice I'm not saying there is a gravitational stress-energy tensor nor giving a coordinate-invariant, local definition of gravitational energy. Rather, I'm saying this is a way of interpreting in an intuitive manner what is going on.

Similarly, one can do this things to interpret, for example, redshifts. In a black hole, it is common to understand the redshift of a photon as it "climbs" out of the potential well: the photon loses energy to the gravitational field and is redshifted as a result, and we state is as if the gravitational field was gaining energy even though it does not have a stress-energy tensor.

Another example is with an expanding Universe. In Cosmology, energy is not generally conserved. However, there is a way of interpreting the equations of General Relativity and keep energy constant at cosmological scales: add on gravitational energy. This is comfortable because it allows us to interpret things in terms of constant total energy, but uncomfortable because gravity does not have a stress-energy tensor, and so on. Sean Carroll mentions it in one of his blog posts.

As a summary, Thorne is nowhere stating that there is a local, coordinate-independent notion of gravitational energy. The point is that it is sometime convenient to understand the nonlinear effects of General Relativity as being due to some sort of gravitational energy. To be fair, one can derive the Einstein–Hilbert action in this way (see e.g., Zee's Einstein Gravity in a Nutshell, Chap. IX.5 if I recall correctly).

$\endgroup$
1
  • $\begingroup$ Good answer, but this is misleading: “it is common to understand the redshift of a photon as it "climbs" out of the potential well” - It is a Newtonian interpretation. There is no potential energy or potential well in GR, so ascending photons don’t redshift or lose energy. A fundamental principle of energy conservation is measuring energy before and after in the same reference frame, because energy is frame dependent. For example, in the Schwarzschild coordinates, a photon is emitted already “redshifted” and doesn’t lose energy in flight. Comparing energy in different frames is meaningless. $\endgroup$
    – safesphere
    Feb 26 at 20:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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