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Aug 30, 2013 at 16:17 history edited Abhimanyu Pallavi Sudhir
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Aug 17, 2013 at 18:50 comment added user4552 I think you're confused on multiple points, and I don't think we're going to straighten this out in comments. If you like, you could start a new question such as "How do we define the total energy of a black hole?" or "How do we define the total energy of an isolated object in GR?"
Aug 17, 2013 at 18:40 comment added Trimok @BenCrowell : Let $E = M$ be the non-gravitational energy of a spherical object of radius $R$, of mass $M$. The negative gravitational energy is of order $- \frac{GE^2}{R}$. The total energy is $E_{TOT} = E - \frac{GE^2}{R}$. This total energy must be positive. The limit case is $E_{TOT}=0$, this is the black hole case, this corresponds to a radius $R_S$ of order $R_S = GE =GM$
Aug 17, 2013 at 18:18 comment added user4552 I agree this is a naive definition, but I like it. What definition? You still haven't explained what you mean by gravitational versus nongravitational energy. Because, in a classical point of view, the total energy of a isolated physical object cannot be negative. By "classical" do you mean nonrelativistic? Either way, this statement is false.
Aug 17, 2013 at 16:45 comment added Trimok @BenCrowell : I agree this is a naive definition, but I like it. The black hole is a limit case. Why? Because, in a classical point of view, the total energy of a isolated physical object cannot be negative. The black hole is a limit case, because the classical total energy is zero, negative gravitationnal energy is compensating positive non-gravitationnal energy.
Aug 17, 2013 at 16:39 comment added user4552 Sorry, I don't understand what you mean by that. What definition of energy are you using? For example, the Komar energy is simply equal to the mass of the black hole (the $m$ appearing in the Schwarzschild metric).
Aug 17, 2013 at 16:30 comment added Trimok @BenCrowell : Well, from my point of view, the total energy of a black hole is zero, that is : the positive non-gravitationnal energy is compensated by the negative gravitationnal energy.
Aug 17, 2013 at 16:19 comment added user4552 What do you mean by "non-gravitationnal energy?" I don't see more than one type of energy in a Schwarzschild spacetime.
Aug 17, 2013 at 9:08 answer added Luboš Motl timeline score: 4
Aug 17, 2013 at 8:51 history asked Trimok CC BY-SA 3.0