# What is the energy required to create a gravitational field equivalent to that a mass $m$ shows?

If the mass of a neutron star in its collapse becomes a singularity, then the resting energy of this gravitational field must be $E = mc^2$ ($m$ = star mass).

Is this possibility wrong?

• It depends on what exactly you mean by m = star mass. There are several measures of this same mass that I can think of. Is it the sum total of all the masses of the particles in the star before collapse? If so, then no: there's extra energy from the pressure and heat in the neutron star (which came from the energy of its initial collapse into a star). Is it the mass you would deduce from measuring the gravitational field before the collapse? If so, and assuming nothing escaped from the collapse, then yes: those two masses are the same. You'll need to clarify. – Mike Aug 23 '18 at 18:50
• @Mike - I refer to your second option, and I ask you: Can only the resting energy of these two gravitational fields (before and after the collapse) be E = mc², considering that the singularity is only a pure gravity? – João Bosco Aug 23 '18 at 22:18
• I'm sorry, I don't really understand this last question. What do you mean by "only"? I'll add an answer that explains this more fully, but I might be missing part of what you're asking. – Mike Aug 23 '18 at 23:53

The rest energy of a star which collapsed to form a black hole is not represented by the gravitational field before and after the collapse. The gravitational field has no energy density and hence is not a source of gravity. It is caused by the mass in the singularity of the black hole.

• First - This I can't understand: "Gravitacional field is not a source of gravity" ? Can you get gravity without gravitacional field? Second - I think mass is a Gravitacional field effect (inertia) and no reason of nothing. – João Bosco Aug 23 '18 at 22:41
• Sources of gravity are components of the stress-energy tensor. You may convince yourself that the gravitational field is not contained. The reason is, as mentioned in my answer: the gravitational field has no energy density. It is the other way round, the gravitational field is caused by a source of gravity, the mass. The mass has energy density and therefor is a component of said tensor. -- The Schwarzschild spacetime as vacuum solution is described by the Weyl tensor, but that's another story. – timm Aug 24 '18 at 12:12

Let's consider the idealized case, where there's no radiation in any form. Furthermore, we usually even assume that this neutron star (and later black hole) is the only thing in the universe. Basically, this just means that we ignore the influence of any distant things so that we can get mathematically exact answers.

So you start off with a nice, quiet neutron star, and assume that we can ignore any radiation it's giving off. Just to make things extra simple (though this isn't strictly necessary), let's assume that it's not even spinning. If you look at the gravitational field outside of this star, Birkhoff's theorem tells us that it should look exactly like the Schwarzschild metric. And we define the mass of the neutron star to be the $M$ found in that metric.

Now there's a second way of defining the mass in this simple universe, called the Bondi mass. It turns out that, numerically, this mass is just the same as the $M$ we defined above. But the nice thing about Bondi mass is that there's a conservation law that says that the initial Bondi mass, $M$, is equal to the final Bondi mass plus whatever mass-energy that's radiated.

So if we assume that this neutron star collapses to a black hole without ejecting any matter, or radiating any type of mass-energy, then this conservation law tells us that the final Bondi mass is the same as the initial Bondi mass, $M$. And if the final object is a black hole, this means that the mass that goes into its Schwarzschild metric.

Now, as usual, the mass I'm talking about here is just the rest mass. And through Einstein's mass-energy equivalence, we have $E = M\, c^2$ — both before and after the collapse. Of course, all of this really relies on the assumption that absolutely nothing is radiated, which isn't realistic. Especially with rotation, there's likely to be some mass ejection, and even more likely to be some electromagnetic radiation, which would both reduce the final mass-energy of the black hole.

But one important point is that the total mass does not depend on what form that mass takes. Whether in the puffy matter hypergiant star, super-dense neutron star matter, a black hole singularity, or a truly enormous collection of squeaky clown noses, mass is mass all the same.

• Yes, I agree! You also said "...mass is mass all the same". This is clear to me. But about this I think too " gravity is gravity all the same" before or after the colapse. This allows me to conclude that the resting energy E = mc² of any material object, planet, star or black hole, belongs only to its gravitational field. This thought reduces matter to mere effect status, and raises the gravitational field to single cause status. My question is about this relationship.. – João Bosco Aug 24 '18 at 4:00