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In terms of General relativity we have as a matter of principle that anything that has inertial mass contributes to gravity. All forms of potential energy have inertial mass, it follows that the potential gravitational energy of a star must contribute to the gravitational field of that star. Depending on the amount of matter that is present, (the remains of) a dying star may contract to a white dwarf, a neutron star, or a black hole.

Let me compare contraction to a white dwarf to contraction to a neutron star.

As a thought experiment, let's say a known amount of gas is distributed in a region of galactic space, and something triggers that amount of gas to contract to a star. From a thought experiment point of view, how much of that star's gravity must be attributed to arising from its internal gravitational potential energy?

Here's why I ask:
In the case of ending up as a neutron star there is significantly more contraction, as gravity overcomes the electron degeneracy pressure. It seems to follow that in the case of contraction to a neutron star more potential gravitational energy is released.

Let's say that in this thought experiment the number of atoms (and types of atoms) of the star is known, and the flow of atoms/particles being ejected from the star is known. So theoretically you can give an expression for the amount of potential energy per unit of rest mass.

The critical case is the Chandrasekhar limit, of course. How much gravitational potential energy should be attributed to a star that is just at the Chandrasekhar limit?

In the real world you never know the amount of atoms in a star, of course. All you can assess is the strength of the gravitational field of that star (the amount of spacetime curvature.)


[Later edit]

As pointed out by Trimok, I had failed to appreciate that there is no explicit expression for potential energy in the equations, explained by Stan Liou. I have rewritten the title of the question accordingly.

So the question only touches interpretation of the theory. It would seem a star that is just above the Chandrasekhar limit will collapse deeper than an star just just below it. In the interpretation, does that make a difference as to the amount of potential gravitational energy that is to be attributed to the system?

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  • $\begingroup$ In tge Einstein field equations, the right hand side stress-energy tensor $T_{\mu \nu}$ does not contain a potential gravitational energy contribution. This non-gravitational $T_{\mu \nu}$ is the source of the gravitational field $g_{\mu \nu}$ (the metrics). Gravitational effects like gravitational potential energy, are interpreted from the expression of the metrics (ant its derivatives). $T_{\mu \nu}$ is not a conserved tensor see Wiki $\endgroup$ – Trimok Oct 19 '13 at 9:13
  • $\begingroup$ ....Which is conserved is the sum of the non-gravitational tensor $T_{\mu\nu}$ and the gravitational stress-energy pseudo-tensor. $\endgroup$ – Trimok Oct 19 '13 at 9:19
  • $\begingroup$ Why don't you just do a rough Newtonian calculation? It will give you an answer that's on the right order of magnitude. @Trimok: The question is perfectly sensible if you interpret it as being a question about, say, the Komar mass. $\endgroup$ – Ben Crowell Oct 19 '13 at 15:10

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