Total energy/mass in stationary GR

1. Oct 25, 2007

fnesti

Hello,
does anybody know a proof of the equivalence (or differences) between the "total mass" integral \int T_munu \xi^mu dV^nu and the "Komar mass" integral \int R_munu \xi^mu dV^nu ?

(here \xi is the timelike killing vector, dV is a 3-volume, and spacetime is assumed asymptotically flat, in addition to stationary)

Recall that both these integrals are constant of motion, and for Schwarzschild they are both equal to "m"....

The only reference to this proof I know, is exercise n.4 in cap.11 of Wald, but that assumes weak field... Then a paper by ashtekar (JMath Phys 1979) also proves the equality of the ADM and Komar integrals.. but still the total mass is not mentioned.

Does anybody have some insight or a good reference to this point?

2. Oct 25, 2007

pervect

Staff Emeritus
First off, I've used the latex features on this board to fix up the equations. You can "quote" this post to see the details of how I've done this.

I'm scratching my head, because the Schwarzschild solution is a vacuum solution, so that $T_{\mu\nu}=0$ everywhere. Of course, so is $R_{\mu\nu}.$ The problem is the singularity, so the volume integral is ill-defined, and one needs to use the surface integral form (Wald 11.2.9) instead.

There's also a factor of 1/4 pi needed in the Komar mass integral, assuming you use units where G=1, (Wald 11.2.10).

I suppose the way around this is to consider a Schwarzschild interior solution where the volume integral is defined - I haven't done this.

Sorry, no. I've never heard of the "total mass" integral before, where did you run across it in the first place?

There is a derivation of how the Komar mass is equal to
$$\int \sqrt{|\xi^a \xi_a|} \left(\ T_{00} + T_{11} + T_{22} + T_{33} \right) dV$$ on the Wikipedia which may be of some interest, though it's basically just something I wrote up. Note that this is equal to the integral of (rho+3P) when the pressure is isotropic - the length of a time-like Killing vector in a Minkowski space-time is unity.

Last edited: Oct 25, 2007
3. Oct 25, 2007

Stingray

The two notions of mass are not generally equal. Even barring the topological issues associated with Schwarzchild (where, as pervect mentions, your first mass is ill-defined), use Einstein's equation in the Komar expression. You get (see Wald's 11.2.10)
$$M_{Komar} = 2 \int_{\Sigma} (T_{ab} - \frac{1}{2} T g_{ab}) n^a \xi^b dV \neq \int_{\Sigma} T_{ab} n^a \xi^b dV$$

4. Oct 25, 2007

Stingray

Given any Killing vector $\xi^a$, you can define an object
$$C = \int_{\Sigma} T_{ab} \xi^a dS^b$$

If the hypersurface completely cuts through the support of the stress-energy tensor and $\nabla_a T^{ab}=0$, Stoke's theorem immediately implies that $C$ is independent of $\Sigma$. It's therefore a constant of motion. A simplified verison of this is usually used when calculating geodesics in simple spacetimes. You write down energies and angular momenta (of test particles) in this form.

5. Oct 26, 2007

fnesti

Yes thanks, that was my first post, now I try :)

Yes, this was indeed the question: what difference between the two?

The exercise n.4 of wald, $$\int T_{ij}d^3x=0$$ is quite confusing, since you are led to think that the integral of pressure is zero... but this is true only in minkowsky, where the conservation has ordinary derivatives: $$\partial_\mu T_{\mu\nu}=0$$. On the other hand in a real background the situation is different, see here below.

Yes you are right, I was confused and thought that for Schwarzschild $$T_{00}=m\delta^3(x)$$, so that you could just evaluate the total mass integral. This is not true, and one can make such integral only for the smooth "interior solution". (And btw I think that you can not easily find $$T_{\mu\nu}$$ in the limit of zero star size, since you hit the Chandrasekar limit... is this true?)

I also see now that $$\sqrt{|\xi^a \xi_a|}$$ is unity in both minkowsky and Schwarzschild (!), while it is not unity in the interior solution of a star (Wald ch.6). So the two integrals are actually
$$M_{Total}=\int\rho\sqrt{|\xi^a \xi_a|}d^3x\qquad M_{Komar}=\int(\rho+3p)\sqrt{|\xi^a \xi_a|}d^3x\,.$$
As we know, the Komar mass can be calculated easily as a surface integral at infinity (because the integrand is a total divergence) and the result is the $$m$$ parameter of the exterior schwarzschild solution.
The total mass instead can not be computed in this way and one has to do the three-dimensional integral of $$\rho+3p$$. (to be done...!)

The difference however has probably a nice interpretation in newtonian lmit, since in equilibrium $$p'=\rho \phi'$$, where $$\phi$$ is the newtonian potential, so $$p=\int\rho \phi'(r)dr$$ that is some sort of work done to build the star shell by shell, $$dE/dv$$. Therefore:
$$M_{Komar}-M_{Total}=\int 3p \sqrt{|\xi^a \xi_a|}\to_{newtonian} \int_0^R \frac{dE}{dv} r^2 dr\,.$$
This seems the total binding energy when one builds the star...

Therefore it seems that $$M_{Total}$$ represents the total mass of the body, while $$M_{Komar}$$ counts this mass plus the gravitational energy.

Is this reasonable?

Last edited: Oct 26, 2007
6. Oct 26, 2007

pervect

Staff Emeritus
Huh? the components of $\xi^a[/tex] are (1,0,0,0) for any metric where the [itex]g_{\mu\nu}$ are not functions of time (which includes Schw. and Minkowski), but it doesn't follow that $\xi^a \xi_a$ is unity. In fact, for a diagonal metric, $\xi_a =(g_{00},0,0,0)$ and $$\sqrt{|\xi^a \xi_a|} = \sqrt{|g_{00}|}$$ as mentioned in the webpage.

Furthermore T is in general is $T^a{}_a$, so some important steps in the derivation don't work unless you have a flat metric.

It's true that if we divide up the static star into little pieces, we can make the pieces so small that there aren't any significant deviations from local flatness within each piece. But we can't, unfortunately, come up with a consistent overall coordinate system that makes space-time Minkowskian for each piece of the star simultaneously, though we can come up with a coordinate system that makes any particular piece Minkowskian.

Last edited: Oct 26, 2007
7. Oct 29, 2007

fnesti

Yes, excuse me, I had included a 3-D volume factor.. $\sqrt{|g_{ij}|}$ and for Schwarzschild this is $\propto 1/\sqrt{g_{00}}$ (because -g_00=1/g_rr).
Therefore this cancels the radial dependence in the norm (but this happens out of the star only). Ok?

In other words, the surface integral 11.2.9 has an implicit three dimensional volume factor... ?
I will try recovering that with forms, to avoid confusion.

Wald's (11.2.10) should anyway work for any metric (asymptotically flat and stationary)..