Hawking Mass in Schwarzschild Spacetime

[darida]

Homework Statement

Metric signature: - + + +

Schwarzschild metric:

<br /> 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}<br />

Second fundamental form:

<br /> h_{ij}=g_{kl}\Gamma^{k}_{ij}n^{l}<br />

where:

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

Mean curvature:

<br /> H=g^{ij}h_{ij}<br />

Hawking mass:

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

Homework 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)?

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?

 

[mfb]

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?

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

 

[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?

 

[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.

 

[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

 

[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 $H$.]

 

[darida]

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

 

[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}

 

[TSny]

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.


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)

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

 

[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}

 

[TSny]

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

 

[darida]

Ahh thank you so much, now I get it!