# Proper mass vs Schwarzschild Mass

1. Dec 25, 2014

### Matterwave

Hello guys,

I am reading through Wald chapter 6 section 2 on the interior Schwarzschild solution. In it he states that matching the interior solution to the exterior (Schwarzschild) solution gives a Schwarzschild mass of $$M=4\pi\int_0^R \rho(r)r^2 d$$ This would presumably be the same mass $M$ that appears in the Schwarzschild metric. However, Wald then notes that the proper mass is actually $$M_P=4\pi\int_0^R \rho(r)r^2\left[1-\frac{2m(r)}{r}\right]^{-1/2} dr$$ Where $m(r)=4\pi\int_0^r \rho(r')r'^2 dr'$.

Wald then goes to say that $M_P\gt M$ and that the difference $M_P-M$ can be interpreted as a "gravitational binding energy".

This interpretation seems screwy to me since I would have expected the "gravitational binding energy" to source gravitation as well, and I would have expected the Schwarzschild mass to account for this. Indeed, one usually talks about the contribution of "potential energy" to the "rest mass" of an object. In Astrophysical contexts, we talk about how, for example, a supernova releases 99% of the gravitational binding energy of the core into the shock wave and that this energy accounts for ~10% of the rest mass of the core. In light of this, why does the "Schwarzschild mass" which dictates gravitation not include the "gravitational binding energy" of the mass?

I would have expected the situation to be exactly the opposite. That the "proper mass" would only account for the "mass due to the integrated local mass density", while the Schwarzschild mass would include the gravitational binding energy.

2. Dec 26, 2014

### WannabeNewton

The gravitational binding energy $E_B$ should be negative so $M_p > M$ would imply exactly what you have stated above i.e. that $M$ takes into account $E_B$ whereas $M_p$ doesn't. It seems to me this is indeed what Wald is saying but in a convoluted way by defining $E_B$ to be the negative of what one usually calls the gravitational binding energy so that if one removes from $M_p$ the mass equivalent of $E_B$ then one gets $M$.

3. Dec 26, 2014

### Staff: Mentor

Yes, it is.

It does, in the sense that, as WannabeNewton pointed out, gravitational binding energy is negative, so an object that is gravitationally bound has less gravitational mass (i.e., is a smaller source of gravity) than the same amount of matter would be if it were not gravitationally bound, or not bound as tightly. See further comments below.

Think about what this means. Before the supernova happens, we have a star with Schwarzschild mass $M$, which includes some amount of (negative) binding energy. That means that, if we took all the individual particles in the star and added up their rest masses (i.e., the rest masses they would each have as isolated objects), we would get an answer larger than $M$. In fact, this is what Wald's "proper mass" is doing--it's adding up the rest masses of each small piece of the star, counting each piece as if it were an isolated system (i.e., not gravitationally bound). Physically, this means that in order for the star to form in the first place from its constituents, an amount of energy equal to $M_B$, the gravitational binding energy, had to be released.

Then the supernova happens, and we end up with a new object (presumably a neutron star) with a Schwarzschild mass $M'$, which will be less than $M$. The difference between $M$ and $M'$ is just the change in gravitational binding energy, since the new object (neutron star) is more tightly bound gravitationally than the original star was. This difference in energy gets released as radiation, shock wave energy, etc.

4. Dec 26, 2014

### Matterwave

I am trying to wrap my head around this. I understand that a binding energy is negative, but it still seems to me that this binding energy should make the mass of the whole more than the mass of the constituents. For example, the mass of three bound individual quarks is much much less than the mass of the proton.

So it seems if I went in and added up all the individual constituents, and got $M_P$, I should get a number that is less than if I include the binding energy in my mass, $M$. But this is not so.

If I take the mass of quarks vs mass of protons analogy, would not the sum of the individual quark masses be analogous to $M_P$ while the mass of the proton be analogous to $M$? The binding energy (due to strong force) would then be $M_P-M$ which simply turns out to be a negative number in this case.

5. Dec 26, 2014

### pervect

Staff Emeritus
If you can get a hold of MTW, you might try reading their treatment of the same topic.

The proper mass is equal to the proper volume element (as seen by static observers - note that this is NOT 4 pi r^2 dr) multiplied by the density. If you can stomach imagining that the object is made of incompressible material , you'll conclude (and read in MTW) that the proper volume and proper mass of each piece doesn't change during an assembly/dissasembly process wherein the pieces are removed to infinity.

However, if you imagining assembling the object from pieces at infinity, you'll see that the assembly process generates energy, while the dissassebly process requires energy. For instance you might imagine the pieces are blocks being lowered by cranes, which generate energy via a motor-generator when the blocks are lowered, but require energy to lift the blocks up to infinity,.

This energy generated during the assembly has the magnitude of the binding energy. The sign of the binding energy is negative, the asembled mass is less than the unassembled mass.

The assembled mass (which is less than the disassembled mass at infinity) is equal to the Schwarzschild mass parameter. It's also equal (I beleive) to the Bondi and/or ADM masses.

If you have doubts about the consistency of the idea of the object being incompressible during assembly/dissasmebly, the thought experiment becomes much more complex :(. Having the doubts is probably sensible, but I'm not sure how then to provide a simple explanation of what's going on.

6. Dec 26, 2014

### Staff: Mentor

But the binding energy in this case is due to the strong force. The strong force has some significantly different properties from the other fundamental interactions. One key property is that we can never observe free quarks, which means the "masses" of the individual quarks that add up to much less than the mass of a proton are not numbers we can ever measure directly, we can only infer them. So the thought experiment I described (and pervect described), where you start with the individual constituents and assemble them into the bound object, can't even be modeled, since the constituents--the quarks--can never be free, even in theory. They only occur in bound states.

If you want a better analogy, think of a hydrogen-1 atom, a bound state of a proton and electron. The total energy of this state is less than the mass of the proton plus the mass of the electron; forming the atom from a free proton and a free electron releases energy, and ionizing the atom, i.e., going from the bound state to the free state, costs energy.

7. Dec 27, 2014

### Matterwave

Ok, this post and pervect's cleared things up for me. Thanks guys! :D