Gravitational Binding Energy in GR

[StateOfTheEqn]

What is the gravitational binding energy in GR in the spherically symmetric case?

I calculate $E=mc^2(1-\frac{1}{\sqrt{1-\frac{r_s}{R}}})$

where $m$ is the mass of the body, $r_s$ is the Schwarzschild radius, and $R$ is the area radius as in the Birkhoff theorem.

 

[WannabeNewton]

For a spherically symmetric static star it is $E = M_p - M$ where $M_p$ is the total proper mass of the star and $M$ is the total mass of the star.

 

[pervect]

What is the gravitational binding energy in GR in the spherically symmetric case?

I calculate $E=mc^2(1-\frac{1}{\sqrt{1-\frac{r_s}{R}}})$

where $m$ is the mass of the body, $r_s$ is the Schwarzschild radius, and $R$ is the area radius as in the Birkhoff theorem.

We need a bit more information, were you assuming a constant density interior?

 

[StateOfTheEqn]

We need a bit more information, were you assuming a constant density interior?

We can assume the mass of a star is concentrated at a single point and a body of mass $m$ is a distance $R$ from the point. The binding energy is the energy required to move the body of mass $m$ from $R$ to infinity.

Essentially I am asking what is the GR counterpart to the Newtonian formula $-\frac{GMm}{r}$.

In my calculations I derived $E=mc^2(1-\frac{1}{k})$ where $k=\sqrt{1-\frac{r_s}{R}}$ and $r_s$ is the Schwarzschild radius.

 

[pervect]

MTW's gravitation has a formula for binding energy on pg 604, but their formula won't converge for the case which you describe. Certainly if you plug 0 into the Newtonian formula you won't get a finite binding energy.

Note: MTW's expression uses geometric units, where G=1.

I believe a taylor series expansion of MTW's formula will show that it gives an answer which is always greater than the Newtonian one, so I don't see how you can get any value other than infinity when r=0.

MTW's formula:
$$-\int_0^r \rho \left[ \left( 1-2m/r \right)^{-\frac{1}{2}} -1 \right] \,4 \pi r^2 dr$$

$\rho$ is the density, which in general a function of r that depends on the equation of state of whatever is composing the mass. m is not the total mass, but the mass inside radius r, so it's also a function of r. $\rho$ is the proper density, i.e. the density in a locally Minkowskii frame.

[add]
$m(r) = \int 4 \pi r^2 \rho dr$ see pg 603. This equation for m(r) is only valid in Schwarzschild coordinates.

Taylor expansion for m/r < 1
$$-\int_0^r \rho \left[ \frac{1}{2} \frac{2m}{r} + \frac{3}{8} \left( \frac{2m}{r} \right) ^2 + \frac{5}{16} \left( \frac{2m}{r} \right)^3 + ... \right] 4 \pi r^2 dr$$

If you take only the first term of the taylor series expansion, you get the Newtonian result.