In David Tong's lectures on general relativity the interpretation of the $M$ which appears in Schwarschild metric:

$$ds^2=-\left(1-\frac{2GM}{r}\right)dt^2+\left(1-\frac{2GM}{r}\right)^{-1}dr^2+r^2\sin^2(\theta)d\theta ^2+r^2d\phi^2$$

Is the Komar charge (or Komer mass) with respect to the Killing vector $K_{\mu}=(g_{tt},0,0,0)=-\left(1-\frac{2GM}{r}\right)(1,0,0,0)$ and thus can be viewed as the Black hole's mass according to section 6.1.

According to section 4.3.3, Komar charge can be written in terms of differential geometry notation: $$Q_{\text{Komar}}=-\frac{1}{8\pi G}\int_{\partial \Sigma}\star F$$ $F$ is a 2-form written in terms of the killing vector $K^{\mu}$ in the following way: $$F=dK=\left( \nabla_{\mu} K_{\nu}-\nabla_{\nu} K_{\mu}\right)dx^{\mu} \land dx^{\nu}$$

It is known that the Schwartzschild metric has another Killing vector given by: $$\tilde{K}_\mu=(0,0,0,g_{\phi\phi})=r^2(0,0,0,1)$$ In this case, $F$ has the following form: $$\tilde{F}=\partial_{r}g_{\phi\phi}dr\land d\phi = 2rdr\land d\phi $$ The Hodge dual should be: $$\star F = 2r\star(dr\land d\phi)=-2r\sqrt{g_{tt}g_{\theta\theta}}dt\land d\theta= - 2r^2\sin\theta\sqrt{\frac{2GM}{r}-1}dt\land d\theta$$ The Komar charge is:

$$Q_{\text{Komar}}=\frac{r^2}{4\pi G}\sqrt{\frac{2GM}{r}-1}\int^{t}_{0}\int^{\pi}_{0}dt'd\theta\sin\theta=\frac{t \cdot r^2}{2\pi G}\sqrt{\frac{2GM}{r}-1}$$ But this object looks peculiar, and in particular not look like an invariant object to translations in $r$ and in $t$ for example.

Is there a right way to do this analysis for the second Killing vector of the Schwarzschild metric?


1 Answer 1

  • Your spherical part of the metric is wrong, it should be $r^2\,d\theta^2+r^2\sin^2\theta\,d\phi^2$.
  • As a result, your formula for the azimuthal Killing covector is wrong. The azimuthal Killing vector field is $\frac{\partial}{\partial \phi}$, so the associated covector field is $K=r^2\sin^2\theta\,d\phi$.
  • We then have $F:=dK=2r\sin^2\theta\,dr\wedge d\phi+2r^2\sin\theta\cos\theta\,d\theta\wedge d\phi$.
  • The Hodge dual of this is $\star F=-2(r-2M)\sin\theta\,dt\wedge d\theta+2\cos\theta\,dt\wedge dr$, which you can easily check is closed, unlike your computed $\star F$ (and we must get something closed since Schwarzschild is Ricci-flat). Another thing I should mention is that your formula for computing Hodge duals seems wrong, I feel like you’re also missing some determinants in the denominator for the $r,\phi$ parts (i.e if we happened to consider the $2$-form $H=2r\,dr\wedge d\phi$, then the correct Hodge dual is$\star H=-\frac{2(r-2M)}{\sin\theta}\,dt\wedge d\theta$).
  • I’m not sure what you’re doing with the integrals either (especially the $t$ integral, since you should be integrating along spacelike hypersurfaces). For the $\star F$ we see above, notice that because of the $dt$ term present in both summands, if we pull it back to a constant $t$ hypersurface $\Sigma_{t_0}$, then $\star F$ vanishes, so all integrals also vanish. This should be reasonable because $\frac{\partial}{\partial \phi}$ is the generator of azimuthal rotations, and our calculation shows the associated Komar charge vanishes, i.e we interpret this as vanishing angular momentum (which makes sense since Schwarzschild is the $a=0$ special case of the rotating Kerr spacetimes).

A Remark on Computing Hodge Duals.

Like I said above, I’m not sure which formula you’re using to compute Hodge duals, but it’s wrong. Here is the general calculational rule:


Let $(V,g)$ be an $n$-dimensional oriented pseudo inner-product space with $\{\omega_1,\dots,\omega_n\}\subset V^*$ being a positively-oriented, $g$-orthonormal coframe (so the volume element is $\omega_1\wedge\cdots\wedge \omega_n$). Fix a set of distinct indices $I=(i_1,\dots, i_k)$ from $1$ to $n$, and let $I’$ be the complementary set of indices $\{1,\dots, n\}\setminus I$. Then, denoting $\omega_I=\omega_{i_1}\wedge\cdots\wedge \omega_{i_k}$, we have \begin{align} \star(\omega_I)&=(-1)^{\sharp(I)}\cdot\epsilon_{I,I’}\,\omega_{I’}, \end{align} where $\sharp(I)$ is the number of minus signs in the numbers $g^{i_1i_1},\dots, g^{i_ki_k}$, and $\epsilon_{I,I’}$ is the sign of the permutation. Also, no summation convention in the formula above.

In words, you simply take an orthonormal coframe, and then to compute the Hodge star, you simply take all those which have not yet appeared, and then modify by a suitable sign $\pm 1$. So, for Schwarzschild, we have the metric \begin{align} g=-(\omega_t)^2+(\omega_r)^2+(\omega_{\theta})^2+(\omega_{\phi})^2, \end{align} where \begin{align} \begin{cases} \omega_t&=\sqrt{1-\frac{2M}{r}}\,dt\\ \omega_r&=\dfrac{1}{\sqrt{1-\frac{2M}{r}}}\,dr\\ \omega_{\theta}&=r\,d\theta\\ \omega_{\phi}&=r\sin\theta\,d\phi. \end{cases} \end{align} and we declare $\{\omega_t,\omega_r,\omega_{\theta},\omega_{\phi}\}$ to be positively oriented. So, with this, we have that our $2$-form $F$ from above is \begin{align} F&=2r\sin^2\theta\,dr\wedge d\phi+2r^2\sin\theta\cos\theta\,d\theta\wedge d\phi\\ &=2\sqrt{1-\frac{2M}{r}}\sin\theta\,\omega_r\wedge \omega_{\phi}+2\cos\theta\,\omega_{\theta}\wedge\omega_{\phi}. \end{align} The Hodge dual of the first term is for instance obtained by replacing $\omega_r\wedge\omega_{\phi}$ with $\omega_t\wedge\omega_{\theta}$, and then the correct sign is $(-1)^0\cdot\epsilon_{r\phi,tr}$ since there are no minus signs in the metric for the $r,\phi$ terms. Likewise for the other term. Putting it all together yields \begin{align} \star F&= 2\sqrt{1-\frac{2M}{r}}\sin\theta\cdot (-1)^0\cdot \epsilon_{r\phi, t\theta}\cdot\omega_t\wedge \omega_{\theta}+2\cos\theta\, (-1)^0\cdot\epsilon_{\theta\phi,tr}\cdot \omega_{t}\wedge\omega_{t}\\ &= -2(r-2M)\sin\theta\,dt\wedge d\theta+2\cos\theta\,dt\wedge dr, \end{align} as claimed above.

  • $\begingroup$ Thank you for your great answer! How do I calculate the number of minus signs according to $♯(I)$ ? For example $ω_r ∧ ω_t=(−1)^{♯(I)}ϵ_{rt,θϕ}ω_θ∧ω_ϕ$ We know that $ωr=(1−\frac{2M}{r})^{−1}$ and $ω_t=(1−\frac{2M}{r})$ So there are two minutes? $\endgroup$ Jan 14 at 8:40
  • $\begingroup$ And in general for two-dimensional systems: $ds^2=-h(r,t)dt^2+g(r,t)dr^2=-(\omega_t)^2+(\omega_r)^2$ does $\star(\omega_r \wedge \omega_t)=1$? $\endgroup$ Jan 14 at 8:40
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    $\begingroup$ @DanielVainshtein For the first comment, note that $g=\color{red}{-}(\omega_t)^2+(\omega_r)^2+(\omega_{\theta})^2+(\omega_{\phi})^2$, so $\omega_t$ is the only one which has a minus sign as a coefficient in the metric (i.e its inner product with itself is $-1$ rather than $+1$ (and I mean the induced inner product in the dual space)). So, if $I=\{t,r\}$, then $\sharp(I)=1$, and in fact $\sharp(I)=1$ if and only if $t\in I$, and is $0$ otherwise (since $\omega_r,\omega_{\theta},\omega_{\phi}$ come with plus signs). $\endgroup$
    – peek-a-boo
    Jan 14 at 14:19
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    $\begingroup$ there is another minus sign coming from the sign of the permutation: $\epsilon_{rt,\theta\phi}=-\epsilon_{tr\theta\phi}=(-1)\cdot 1=-1$, because I have declared $\epsilon_{t,r,\theta,\phi}=1$. $\endgroup$
    – peek-a-boo
    Jan 14 at 14:21
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    $\begingroup$ For the second comment, let’s say we declare $\{\omega_t,\omega_r\}$ to be positively oriented, so $\epsilon_{tr}=+1$. Then $\star (\omega_r\wedge\omega_t)=(-1)^1\cdot \epsilon_{rt}=(-1)\cdot (-\epsilon_{tr})=+1$. But again, the question of whether it is plus or minus $1$ depends on the orientation you decide (through the $\epsilon$ symbol). $\endgroup$
    – peek-a-boo
    Jan 14 at 14:27

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