Binding energy and invisible mass

The gravitational binding energy of a sphere is: $U=\frac{3GM^2}{5r}$, the mass defect is $\Delta E=\Delta m c^2$. Putting: $M=\frac{4}{3}\rho\pi r^3$, we get: $$U=\frac{16}{15}G\rho^2\pi^2 r^5$$. Now if we put: $U=\Delta E$, we get for $r$: $$r=\frac{1}{2}\frac{\sqrt5c}{\sqrt{G\rho\pi}}$$ that means: $$r\approx\frac{2.3\times10^{13} \, \mathrm{kg}^{1/2\, }\mathrm{m}^{-1/2}}{\sqrt\rho}$$ Assuming $\rho=1$, we can conclude that if we have a spherical mass of radius $r$ given by the last equation and density $1$ this mass 'desappears'. So our universe could be full of big invisible masses because the binding energy is equal to the mass defect. Is it this conclusion correct?

• I'm having a lot of trouble following you to your second equation. If you're substituting in Delta_E, where did the Delta_m go? For one, there's no mass in your U expression to cancel with to begin with, and two, the Delta_m and M here are very very different things. Relating r and rho with nothing else doesn't make any sense. These are r and rho for a ball of mass in what circumstance? I own a tennis ball and a bowling ball, and the (r,rho) combination for either can be whatever I want it to be. – Alan Rominger Jun 23 '14 at 14:54

No, it is not correct because of relativity.

Because you use $E=mc^2$, clearly, you are using relativistic effects so you need to take the full theory of relativity into account. While you might think that you have only used the special theory of relativity, you are also considering gravitational effects – the gravitational binding energy – and the general theory of relativity is the only (and right) viable theory of gravity among those that are compatible with the special theory of relativity.

In general relativity, it is still true that the (absolute value of the) gravitational binding energy never exceeds the original total latent energy $E=mc^2$ of the system. Up to the purely numerical factors that are corrected by general relativity, your relationship between $r$ and $\rho$ is exactly the relationship we find for black holes. (In 3+1 dimensions, $R=2GM/c^2$ for the Schwarzschild. When combined with $M=C\rho R^3$, we get $R=2CGR^3\rho/c^2$, exactly your relationship, for a dimensionless numerical value of $C$.)

The only special value of $r$ for a given $\rho$ is the black hole radius. The total mass/energy of a black hole is a subtle thing. In general relativity, the total mass/energy cannot really be written as a volume integral of a nicely behaving quantity. But it may be defined asymptotically, using a probe of the gravitational field at infinity (ADM mass). And the ADM mass of the black hole may be seen to be positive.

Negative-mass black holes are prohibited because they have naked singularities etc. They also violate various energy conditions and shouldn't arise in a consistent theory because the theory would be in conflict with causality (causes preceded their effects). The marginal case is the $M=0$ black hole, and one may check that this massless black hole inevitably has $R=0$, too.

Classically, using your argument, you might indeed think that the gravitational binding energy converted to mass may ultimately "beat" the mass you started with. But general relativity slows down how much mass you may "subtract" in this way when the gravitational binding gets really strong, and at most, you may converge closer to a black hole which guarantees that the original mass you started with is never cancelled (and surely never exceeded).

Next to the obsious problem with the relativity (as Luboš Motl said), I think there is another problem as well.

The gravitational binding energy is calculated to the infinite distance as reference.

Thus your calculation doesn't mean, that the mass "disappeared" - your calculation means only, that disintegrating this sphere to infinitesimal particles, and bringing these particles infite distance from eachother, costed so many energy, as the mass-equivalent of this sphere.