# Hawking Mass in Schwarzschild Spacetime

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1. Aug 23, 2013

### darida

1. The problem statement, all variables and given/known data

Metric signature: $- + + +$

Schwarzschild metric:

$dS^{2}=-(1-\frac{2M}{r})dt^{2}+(1-\frac{2M}{r})^{-1}dr^{2}+r^{2}d\theta^{2}+r^{2}(sin\theta)^{2}d\phi^{2}$

Second fundamental form:

$h_{ij}=g_{kl}\Gamma^{k}_{ij}n^{l}$

where:

$i=1,2$
$j=1,2$
$n^{l}=(0,1,0,0)=$normal vector

Mean curvature:

$H=g^{ij}h_{ij}$

Hawking mass:

$m_{H}(\Sigma)=\sqrt{\frac{Area \Sigma}{16\pi}}(1-\frac{1}{16\pi}\int_{\Sigma}{H^2}d\sigma)$

2. Relevant equations

1) Prove that in the Schwarzschild metric, the Hawking mass of any sphere $S_{r}$ about the central mass is equal to $M$.

2) How to find the normal vector $n^{l}$ (as shown above)?

3. The attempt at a solution

I have tried to find the Hawking mass but it's not equal to $M$. Maybe it's because I used the wrong normal vector?

2. Aug 23, 2013

### Staff: Mentor

To find an error, it would be useful if you show what you did and what you got as result.

3. Aug 23, 2013

### darida

Here is what I did:

Second fundamental form:

$h_{11}=g_{00}\Gamma^{0}_{11}n^{0}+g_{11}\Gamma^{1}_{11}n^{1}+g_{22} \Gamma^{2}_{11}n^{2}+g_{00}\Gamma^{3}_{11}n^{3}$
$h_{11}=g_{11}\Gamma^{1}_{11}n^{1}$
$h_{11}=-\frac{M}{(r-2M)^{2}}$

$h_{12}=g_{00}\Gamma^{0}_{12}n^{0}+g_{12}\Gamma^{1}_{12}n^{1}+g_{22} \Gamma^{2}_{12}n^{2}+g_{00}\Gamma^{3}_{12}n^{3}$
$h_{12}=g_{11}\Gamma^{1}_{12}n^{1}$
$h_{12}=0$

$h_{21}=h_{12}=0$

$h_{22}=g_{00}\Gamma^{0}_{22}n^{0}+g_{12}\Gamma^{1}_{22}n^{1}+g_{22} \Gamma^{2}_{22}n^{2}+g_{00}\Gamma^{3}_{22}n^{3}$
$h_{22}=g_{11}\Gamma^{1}_{22}n^{1}$
$h_{22}=-r$

Mean curvature:

$H=g^{11}h_{11}+g^{12}h_{12}+g^{21}h_{21}+g^{22}h_{22}$
$H=\frac{(M-r)}{r(r-2M)}$

$H^2=(\frac{(M-r)}{r(r-2M)})^2$
$H^2=\frac{M^{2}-2Mr+r^2}{(r^2)(r^{2}-4Mr+4M^{2}}$

Area:

$S_{r}=\int^{2\pi}_{0}\int^{\pi}_{0} r^{2} sin\theta d\theta d\phi$
$S_{r}=4 \pi r^{2}$

$\frac{1}{6 \pi}\int_{S} H^{2} d \sigma$
$=\frac{1}{6 \pi}\int^{2\pi}_{0}\int^{\pi}_{0} H^{2} r^{2} sin\theta d\theta d\phi$
$=\frac{M^{2}-2Mr+r^2}{4 (4M^{2}-4Mr+r^2)}$

Hawking mass:

$m_{H}(S_{r})=\sqrt{\frac{Area S_{r}}{16\pi}}(1-\frac{M^{2}-2Mr+r^2}{4 (4M^{2}-4Mr+r^2)})$
$m_{H}(S_{r})=\sqrt{\frac{4 \pi r^{2}}{16\pi}}(1-\frac{M^{2}-2Mr+r^2}{4 (4M^{2}-4Mr+r^2)})$
$m_{H}(S_{r})≠M$ ???Why???

4. Aug 23, 2013

### TSny

I am not familiar with Hawking mass. But I will just make a couple of observations.

You are considering a 2D spherical surface surrounding the mass M. The coordinates intrinsic to this surface are $\theta$ and $\phi$. So I would think that in defining the second fundamental form for this surface that the indices $i$ and $j$ would take on values of 2 or 3 (instead of 1 or 2).

Also, I think you need to normalize the normal vector $n^l$.

If I make these changes, I find that your expression for the Hawking mass reduces to M.

5. Aug 23, 2013

### darida

Well, I just did some calculation with those changes, but then $m_{H}(S_{r})=0$:

Christoffel Symbols:

$\Gamma^{0}_{01}=\Gamma^{0}_{10}=-\frac{M}{2Mr-r^{2}}$

$\Gamma^{1}_{00}=\frac{M(-2M+r)}{r^{3}}$

$\Gamma^{1}_{11}=\frac{M}{2Mr-r^{2}}$

$\Gamma^{1}_{22}=2M-r$

$\Gamma^{1}_{33}=(2M-r)(sin \theta)^2$

$\Gamma^{2}_{12}=\Gamma^{2}_{21}=\Gamma^{3}_{13}=\Gamma^{3}_{31}=\frac{1}{r}$

$\Gamma^{2}_{33}=-sin \theta cos \theta$

$\Gamma^{3}_{23}=\Gamma^{3}_{32}=cot \theta$

Normalized Normal Vector:

$\hat{n}^l=\frac{n^l}{|n^l|}=(0,1,0,0)=n^l$

Second fundamental form:

$h_{22}=g_{00}\Gamma^{0}_{22}n^{0}+g_{11}\Gamma^{1}_{22}n^{1}+g_{22} \Gamma^{2}_{22}n^{2}+g_{33}\Gamma^{3}_{22}n^{3}$
$h_{22}=g_{11}\Gamma^{1}_{22}n^{1}$
$h_{22}=-r$

$h_{23}=g_{00}\Gamma^{0}_{23}n^{0}+g_{11}\Gamma^{1}_{23}n^{1}+g_{22} \Gamma^{2}_{23}n^{2}+g_{33}\Gamma^{3}_{23}n^{3}$
$h_{23}=g_{11}\Gamma^{1}_{23}n^{1}$
$h_{23}=0$

$h_{32}=h_{23}=0$

$h_{33}=g_{00}\Gamma^{0}_{33}n^{0}+g_{11}\Gamma^{1}_{33}n^{1}+g_{22} \Gamma^{2}_{33}n^{2}+g_{33}\Gamma^{3}_{33}n^{3}$
$h_{33}=g_{11}\Gamma^{1}_{33}n^{1}$
$h_{33}=-r$

Mean curvature:

$H=g^{22}h_{22}+g^{23}h_{23}+g^{32}h_{32}+g^{33}h_{33}$
$H=-\frac{2}{r}$

$H^2=(-\frac{2}{r})^2$
$H^2=\frac{4}{r^2}$

Area:

$S_{r}=\int^{2\pi}_{0}\int^{\pi}_{0} r^{2} sin\theta d\theta d\phi$
$S_{r}=4 \pi r^{2}$

$\frac{1}{16 \pi}\int_{S} H^{2} d \sigma$
$=\frac{1}{16 \pi}\int^{2\pi}_{0}\int^{\pi}_{0} H^{2} r^{2} sin\theta d\theta d\phi$
$=H^{2}\frac{1}{16 \pi}\int^{2\pi}_{0}\int^{\pi}_{0} r^{2} sin\theta d\theta d\phi$
$=H^{2}\frac{1}{16 \pi}4 \pi r^{2}$
$=\frac{4}{r^2}\frac{1}{16 \pi}4 \pi r^{2}$
$=1$

Hawking mass:

$m_{H}(S_{r})=\sqrt{\frac{Area S_{r}}{16\pi}}(1-1)$
$m_{H}(S_{r})=0$

Last edited: Aug 24, 2013
6. Aug 24, 2013

### TSny

You need to use the metric to normalize $n^l$. $|n|^2 = g_{\mu\nu}n^\mu n^\nu$.

When finding $h_{33}$, shouldn't there be a factor of $sin^2\theta$ that comes from $\Gamma^1_{33}$? [Edit: The $sin^2\theta$ factor will later get canceled out by $g^{33}$ when calculating $H$.]

7. Aug 24, 2013

### darida

Oops sorry, for $h_{33}$ I made a typo and thank you I got the result now XD

8. Aug 24, 2013

### darida

Now with the same method I try to prove that the Hawking Mass in Reissner-Nordstrom spacetime is equal to $(M-\frac{q^2}{2r})$, but the result I got doesn't agree with it.

Christoffel Symbols:

$\Gamma^{0}_{01}=\Gamma^{0}_{10}=-\frac{q^{2}+Mr}{r(r^{2}-2Mr+q^{2})}$

$\Gamma^{1}_{00}=\frac{(Mr-q^2)(r^2-2Mr+q^2)}{r^5}$

$\Gamma^{1}_{11}=\frac{q^2-Mr}{r(r^2-2Mr+q^2 )}$

$\Gamma^{1}_{22}=-\frac{(r^2-2Mr+q^2 )}{r}$

$\Gamma^{1}_{33}=-\frac{(r^2-2Mr+q^2)sin^2⁡θ}{r}$

$\Gamma^{2}_{12}=\Gamma^{2}_{21}=\Gamma^{3}_{13}=\Gamma^{3}_{31}=\frac{1}{r}$

$\Gamma^{2}_{33}=-sin \theta cos \theta$

$\Gamma^{3}_{23}=\Gamma^{3}_{32}=cot \theta$

Normalized Normal Vector:

$\hat{n}^l=\frac{n^l}{|n^l|}=\frac{(0,1,0,0)}{\frac{r^2}{r^2-2Mr+q^2}}$
$\hat{n}^l=(0, \frac{r^2-2Mr+q^2}{r^2},0,0)$

Second fundamental form:

$h_{22}=g_{00}\Gamma^{0}_{22}n^{0}+g_{11}\Gamma^{1}_{22}n^{1}+g_{22} \Gamma^{2}_{22}n^{2}+g_{33}\Gamma^{3}_{22}n^{3}$
$h_{22}=g_{11}\Gamma^{1}_{22}n^{1}$
$h_{22}=-\frac{(r^2-2Mr+q^2)}{r}$

$h_{23}=g_{00}\Gamma^{0}_{23}n^{0}+g_{11}\Gamma^{1}_{23}n^{1}+g_{22} \Gamma^{2}_{23}n^{2}+g_{33}\Gamma^{3}_{23}n^{3}$
$h_{23}=g_{11}\Gamma^{1}_{23}n^{1}$
$h_{23}=0$

$h_{32}=h_{23}=0$

$h_{33}=g_{00}\Gamma^{0}_{33}n^{0}+g_{11}\Gamma^{1}_{33}n^{1}+g_{22} \Gamma^{2}_{33}n^{2}+g_{33}\Gamma^{3}_{33}n^{3}$
$h_{33}=g_{11}\Gamma^{1}_{33}n^{1}$
$h_{33}=-\frac{(r^2-2Mr+q^2)sin^2⁡θ}{r}$

Mean curvature:

$H=g^{22}h_{22}+g^{23}h_{23}+g^{32}h_{32}+g^{33}h_{33}$
$H=-\frac{2(r^2-2Mr+q^2)}{r^3}$

$H^2=(-\frac{2(r^2-2Mr+q^2)}{r^3})^2$
$H^2=\frac{4(r^2-2Mr+q^2)(r^2-2Mr+q^2)}{r^6}$

Area:

$S_{r}=\int^{2\pi}_{0}\int^{\pi}_{0} r^{2} sin\theta d\theta d\phi$
$S_{r}=4 \pi r^{2}$

$\frac{1}{16 \pi}\int_{S} H^{2} d \sigma$
$=\frac{1}{16 \pi}\int^{2\pi}_{0}\int^{\pi}_{0} H^{2} r^{2} sin\theta d\theta d\phi$
$=H^{2}\frac{1}{16 \pi}\int^{2\pi}_{0}\int^{\pi}_{0} r^{2} sin\theta d\theta d\phi$
$=H^{2}\frac{1}{16 \pi}4 \pi r^{2}$
$=\frac{(r^2-2Mr+q^2)(r^2-2Mr+q^2)}{r^4}$

Hawking mass:

$m_{H}(S_{r})=\sqrt{\frac{Area S_{r}}{16\pi}}(1-\frac{(r^2-2Mr+q^2)(r^2-2Mr+q^2)}{r^4})$
$m_{H}(S_{r})=M-\frac{q^4}{2r^3}$

9. Aug 24, 2013

### TSny

Check the normalization factor. $g_{\mu \nu}n^\mu n^\nu$ gives the square of the norm.

10. Aug 24, 2013

### darida

I've checked it:

$|n^l|=g_{μ\nu}n^{μ}n^{\nu}$
$|n^l|=g_{00}n^{0}n^{0}+g_{11}n^{1}n^{1}+g_{22}n^{2}n^{2}+g_{33}n^{3}n^{3}$
$|n^l|=0+g_{11}n^{1}n^{1}+0+0$
$|n^l|=g_{11}n^{1}n^{1}$
$|n^l|=\frac{r^2}{r^2-2Mr+q^2}(1)(1)$
$|n^l|=\frac{r^2}{r^2-2Mr+q^2}$

11. Aug 24, 2013

### TSny

You calculated $|n|^2$, not $|n|$.

12. Aug 24, 2013

### darida

Ahh thank you so much, now I get it!