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Hawking Mass in Schwarzschild Spacetime

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  1. Aug 23, 2013 #1
    1. The problem statement, all variables and given/known data

    Metric signature: [itex] - + + + [/itex]

    Schwarzschild metric:

    [itex]
    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}
    [/itex]

    Second fundamental form:

    [itex]
    h_{ij}=g_{kl}\Gamma^{k}_{ij}n^{l}
    [/itex]

    where:

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

    Mean curvature:

    [itex]
    H=g^{ij}h_{ij}
    [/itex]

    Hawking mass:

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

    2. Relevant equations

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

    2) How to find the normal vector [itex]n^{l}[/itex] (as shown above)?

    3. The attempt at a solution

    I have tried to find the Hawking mass but it's not equal to [itex]M[/itex]. Maybe it's because I used the wrong normal vector?
     
  2. jcsd
  3. Aug 23, 2013 #2

    mfb

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    Staff: Mentor

    To find an error, it would be useful if you show what you did and what you got as result.
     
  4. Aug 23, 2013 #3
    Here is what I did:

    Second fundamental form:

    [itex]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}[/itex]
    [itex]h_{11}=g_{11}\Gamma^{1}_{11}n^{1}[/itex]
    [itex]h_{11}=-\frac{M}{(r-2M)^{2}}[/itex]

    [itex]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}[/itex]
    [itex]h_{12}=g_{11}\Gamma^{1}_{12}n^{1}[/itex]
    [itex]h_{12}=0[/itex]

    [itex]h_{21}=h_{12}=0[/itex]

    [itex]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}[/itex]
    [itex]h_{22}=g_{11}\Gamma^{1}_{22}n^{1}[/itex]
    [itex]h_{22}=-r[/itex]

    Mean curvature:

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

    [itex]H^2=(\frac{(M-r)}{r(r-2M)})^2[/itex]
    [itex]H^2=\frac{M^{2}-2Mr+r^2}{(r^2)(r^{2}-4Mr+4M^{2}}[/itex]

    Area:

    [itex]S_{r}=\int^{2\pi}_{0}\int^{\pi}_{0} r^{2} sin\theta d\theta d\phi[/itex]
    [itex]S_{r}=4 \pi r^{2}[/itex]

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

    Hawking mass:

    [itex]m_{H}(S_{r})=\sqrt{\frac{Area S_{r}}{16\pi}}(1-\frac{M^{2}-2Mr+r^2}{4 (4M^{2}-4Mr+r^2)})[/itex]
    [itex]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)})[/itex]
    [itex]m_{H}(S_{r})≠M[/itex] ???Why???
     
  5. Aug 23, 2013 #4

    TSny

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    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.
     
  6. Aug 23, 2013 #5
    Well, I just did some calculation with those changes, but then [itex]m_{H}(S_{r})=0[/itex]:

    Christoffel Symbols:

    [itex]\Gamma^{0}_{01}=\Gamma^{0}_{10}=-\frac{M}{2Mr-r^{2}}[/itex]

    [itex]\Gamma^{1}_{00}=\frac{M(-2M+r)}{r^{3}}[/itex]

    [itex]\Gamma^{1}_{11}=\frac{M}{2Mr-r^{2}}[/itex]

    [itex]\Gamma^{1}_{22}=2M-r[/itex]

    [itex]\Gamma^{1}_{33}=(2M-r)(sin \theta)^2[/itex]

    [itex]\Gamma^{2}_{12}=\Gamma^{2}_{21}=\Gamma^{3}_{13}=\Gamma^{3}_{31}=\frac{1}{r}[/itex]

    [itex]\Gamma^{2}_{33}=-sin \theta cos \theta[/itex]

    [itex]\Gamma^{3}_{23}=\Gamma^{3}_{32}=cot \theta[/itex]

    Normalized Normal Vector:

    [itex]\hat{n}^l=\frac{n^l}{|n^l|}=(0,1,0,0)=n^l[/itex]

    Second fundamental form:

    [itex]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}[/itex]
    [itex]h_{22}=g_{11}\Gamma^{1}_{22}n^{1}[/itex]
    [itex]h_{22}=-r[/itex]

    [itex]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}[/itex]
    [itex]h_{23}=g_{11}\Gamma^{1}_{23}n^{1}[/itex]
    [itex]h_{23}=0[/itex]

    [itex]h_{32}=h_{23}=0[/itex]

    [itex]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}[/itex]
    [itex]h_{33}=g_{11}\Gamma^{1}_{33}n^{1}[/itex]
    [itex]h_{33}=-r[/itex]

    Mean curvature:

    [itex]H=g^{22}h_{22}+g^{23}h_{23}+g^{32}h_{32}+g^{33}h_{33}[/itex]
    [itex]H=-\frac{2}{r}[/itex]

    [itex]H^2=(-\frac{2}{r})^2[/itex]
    [itex]H^2=\frac{4}{r^2}[/itex]

    Area:

    [itex]S_{r}=\int^{2\pi}_{0}\int^{\pi}_{0} r^{2} sin\theta d\theta d\phi[/itex]
    [itex]S_{r}=4 \pi r^{2}[/itex]

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

    Hawking mass:

    [itex]m_{H}(S_{r})=\sqrt{\frac{Area S_{r}}{16\pi}}(1-1)[/itex]
    [itex]m_{H}(S_{r})=0[/itex]
     
    Last edited: Aug 24, 2013
  7. Aug 24, 2013 #6

    TSny

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    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##.]
     
  8. Aug 24, 2013 #7
    Oops sorry, for [itex]h_{33}[/itex] I made a typo and thank you I got the result now XD
     
  9. Aug 24, 2013 #8
    Now with the same method I try to prove that the Hawking Mass in Reissner-Nordstrom spacetime is equal to [itex](M-\frac{q^2}{2r})[/itex], but the result I got doesn't agree with it.

    Christoffel Symbols:

    [itex]\Gamma^{0}_{01}=\Gamma^{0}_{10}=-\frac{q^{2}+Mr}{r(r^{2}-2Mr+q^{2})}[/itex]

    [itex]\Gamma^{1}_{00}=\frac{(Mr-q^2)(r^2-2Mr+q^2)}{r^5}[/itex]

    [itex]\Gamma^{1}_{11}=\frac{q^2-Mr}{r(r^2-2Mr+q^2 )}[/itex]

    [itex]\Gamma^{1}_{22}=-\frac{(r^2-2Mr+q^2 )}{r}[/itex]

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

    [itex]\Gamma^{2}_{12}=\Gamma^{2}_{21}=\Gamma^{3}_{13}=\Gamma^{3}_{31}=\frac{1}{r}[/itex]

    [itex]\Gamma^{2}_{33}=-sin \theta cos \theta[/itex]

    [itex]\Gamma^{3}_{23}=\Gamma^{3}_{32}=cot \theta[/itex]

    Normalized Normal Vector:

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

    Second fundamental form:

    [itex]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}[/itex]
    [itex]h_{22}=g_{11}\Gamma^{1}_{22}n^{1}[/itex]
    [itex]h_{22}=-\frac{(r^2-2Mr+q^2)}{r}[/itex]

    [itex]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}[/itex]
    [itex]h_{23}=g_{11}\Gamma^{1}_{23}n^{1}[/itex]
    [itex]h_{23}=0[/itex]

    [itex]h_{32}=h_{23}=0[/itex]

    [itex]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}[/itex]
    [itex]h_{33}=g_{11}\Gamma^{1}_{33}n^{1}[/itex]
    [itex]h_{33}=-\frac{(r^2-2Mr+q^2)sin^2⁡θ}{r}[/itex]

    Mean curvature:

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

    [itex]H^2=(-\frac{2(r^2-2Mr+q^2)}{r^3})^2[/itex]
    [itex]H^2=\frac{4(r^2-2Mr+q^2)(r^2-2Mr+q^2)}{r^6}[/itex]

    Area:

    [itex]S_{r}=\int^{2\pi}_{0}\int^{\pi}_{0} r^{2} sin\theta d\theta d\phi[/itex]
    [itex]S_{r}=4 \pi r^{2}[/itex]

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

    Hawking mass:

    [itex]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})[/itex]
    [itex]m_{H}(S_{r})=M-\frac{q^4}{2r^3}[/itex]
     
  10. Aug 24, 2013 #9

    TSny

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    Check the normalization factor. ##g_{\mu \nu}n^\mu n^\nu## gives the square of the norm.
     
  11. Aug 24, 2013 #10
    I've checked it:

    [itex]|n^l|=g_{μ\nu}n^{μ}n^{\nu}[/itex]
    [itex]|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}[/itex]
    [itex]|n^l|=0+g_{11}n^{1}n^{1}+0+0[/itex]
    [itex]|n^l|=g_{11}n^{1}n^{1}[/itex]
    [itex]|n^l|=\frac{r^2}{r^2-2Mr+q^2}(1)(1)[/itex]
    [itex]|n^l|=\frac{r^2}{r^2-2Mr+q^2}[/itex]
     
  12. Aug 24, 2013 #11

    TSny

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    You calculated ##|n|^2##, not ##|n|##.
     
  13. Aug 24, 2013 #12
    Ahh thank you so much, now I get it!
     
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