# Komar mass simplificiation

1. Jan 24, 2006

### pervect

Staff Emeritus
I've been trying to express the Komar mass formula in component notation for a general static metric.

I'm finding that the expression for

$$\nabla_c \xi_d$$

is reasonably simple, where $\xi^{\mu}$ is a timelike Killing vector, but the formula calls for

$$\epsilon_{abcd} \nabla^c \xi^d$$

and this is very messy.

(We have to integrate the above two-form over some surface to get the mass and multiply by an appropriate constant).

Is it kosher to re-write the formula for the Komar mass as

$$-\frac{1}{8 \pi} \int_S \epsilon^{abcd} \nabla_c \xi_d$$

and to do so, would I be expressing the surface to be integrated by one-forms rather than vectors?

Last edited: Jan 24, 2006
2. Jan 28, 2006

### George Jones

Staff Emeritus
I only know a little about integration with differential forms, so my comments may be well wide of the mark and/or not very useful.

Are you trying to do the integral, or are just trying to get a component expression under the integral sign?

I have a vague idea what

$$-\frac{1}{8 \pi} \int_S \epsilon_{abcd} \nabla^c \xi^d$$

means, but I don't know what

$$-\frac{1}{8 \pi} \int_S \epsilon^{abcd} \nabla_c \xi_d$$

means.

Could you leave things in the form

$$-\frac{1}{8 \pi} \int_S \epsilon_{abcd} g^{ec} g^{fd} \nabla_{e} \xi_{f}?$$

Regards,
George

Last edited: Jan 28, 2006
3. Jan 28, 2006

### pervect

Staff Emeritus

What I'm trying to do is to explain the Komar intergal conceptually, and relate it to Gauss's law.

The ideal explanation would be detailed enough to allow someone to compute the correct answer to the intergal (at least in a simple enough case), without knowing any more than advanced calculus (i.e. they wouldn't need to know about covariant derivatives or Killing vectors).

We can get rid of Killing vectors by insisting on a static space-time, i.e. we have metric coefficients that are not functions of time.

To actually carry out the intergal, for example in the Schwarzschild metric, I work by rote. We find that $\nabla^a \xi^b$ is equal to

$m/r^2 dr \wedge dt$

Multiplying it by the Levi-Civita tensor essentally takes the dual, so we now have

$m/r^2 \sqrt{g} d\theta \wedge d\phi$

Substituting sqrt(g) = r^2 |sin(theta)| gives us the intergal

$$\int_0^{2 \pi} \int_0^{2 \pi} (m/r^2) r^2 |sin(\theta)| d\theta d\phi$$

which gives the right answer for the mass, modulo the sign issue.

Last edited: Jan 28, 2006