Computation of surface gravity with Kruskal coordinates

Question

The surface gravity of a Schwarzschild black hole (BH) is

$$ \kappa = \frac{1}{2}f'(r)\Big|_{r=2m} $$

where $f(r)=1-2m/r$ is the component $g^{rr}$ of the contravariant metric in Schwarzschild coordinates. I am trying to compute the same quantity in Kruskal coordinates but I get a different result. So, I am wondering whether is something fundamentally wrong with my way of computing the surface gravity. Let me explain better.

There are several definitions of surface gravity. The one that I use is

$$ \tag{1} -\nabla_a(\xi_b \xi^b) = 2\kappa\, \xi_a $$

where $\xi^\mu$ is the Killing vector which becomes null on the horizon. For the Schwarzschild BH, $\xi^\mu = dx^\mu/dt$.

To do the computation of $\kappa$, I adopt the Kruskal coordinates $(U,V)$ that are defined as

$$ U = \mp e^{-u/4m}\,,V = e^{v/4m} $$

where the minus (plus) sign holds for $r>2m$ ($r<2m$) and

$$ u = t-r^*(r)\,,\qquad v= t+r^*(r) \,,\qquad r^*(r) = \int \frac{dr}{f(r)} = r +2m \ln\Big|\frac{r}{2m}-1\Big|\,. $$

The region of the spacetime described by the Schwarzschild metric corresponds to the region $U<0,V>0$ and the future horizon corresponds to $U=0$. Indeed, the Kruskal coordinates satisfy

$$ \tag{2} UV = -e^{r/2m} \frac{r}{2m} f(r)\,. $$

Let's focus on the radial part of the Schwarzschild metric so that I can work with 2D vectors. The Killing vector in Schwarzschild coordinates is $\xi^\mu = (1,0)$, while in Kruskal coordinates $(V,U)$ it is

$$ \xi^A = (\xi^V, \xi^U) = \frac{1}{4m}(V,-U)\,,\qquad \xi_A = (\xi_V, \xi_U) = - \frac{4 m^2}{r}e^{-r/2m} (-U,V)\,. $$

The modulus of the Killing vector is $\xi_A \xi^A = -f(r)$, which indeed vanishes on the horizon.

I want to compute the surface gravity on the future horizon $U=0$, so the only non-vanishing component of $\xi_a$ entering in Eq.(1) corresponds to $a=U$.

Let's choose $a=U$ in Eq.(1) then,

$$ \tag{3} \partial_U (f(r)) = 2\kappa \times \left(-\frac{4m^2}{r}e^{-r/2m} V\right)\,. $$

The left side is just

$$ \text{L.H.S} =\frac{\partial r}{\partial U}f'(r) = \frac{1}{\frac{\partial U}{\partial r}}f'(r) $$

but

$$ \frac{\partial U}{\partial r} = - \frac{U}{4m}\frac{\partial u}{\partial r} = \frac{U}{4m}\frac{\partial r^*(r)}{\partial r}=\frac{U}{4m f(r)} = - e^{r/2m} \frac{r}{8m^2 V} $$ where the last equality is implied by Eq.(2) and shows that the derivative of $U$ wrt to $r$ is smooth at $U=0$.

Eq.(3) then becomes $$ -\frac{8 m^2}{r} f'(r) e^{-r/2m} V = \kappa \times \left(-\frac{8 m^2}{r}e^{-r/2m} V\right) $$

which, simplified and evaluated at $r=2m$ gives

$$ \kappa = f'(2m). $$

This differs by a factor of two with the correct answer. Why is there this discrepancy when performing the computation in Kruskal coordinates?

Answer 1

Using your definitions for the killing vector fields in Kruskal coordinates,

$\xi^a=(\xi^V,\xi^U)=\frac{1}{4M}(V,-U)$,

$\xi_a=(\xi_V,\xi_U)=-\frac{4M^2}{r^2}e^{-\frac{r}{2M}}(-U,V)$.

We can compute the surface gravity as you did,

$d(\xi^a\xi_a)|_{U=0} \equiv -\kappa \xi_b|_{U=0}$.

Computing the left hand side gives,

\begin{equation} d(\xi^a\xi_a)|_{U=0}= d(\frac{4M^2}{r^2}e^{-\frac{r}{2M}}UV) = (-\frac{M}{r^3}-\frac{1}{2r})e^{-\frac{r}{2M}}dr \,UV +\frac{M}{r^2}e^{-\frac{r}{2M}}d(UV). \end{equation}

Now evaluating at either surface $U=0$ or $V=0$ gives $\mathcal{H}^{+}$, $\kappa^{+}$, or $\mathcal{H}^{-}$, $\kappa^{-}$, respectively, both which set the term $UV=0$.

With $d(UV) = U\, d(V)+ V\, d(U)$, we obtain a vector equation,

\begin{equation} \frac{1}{4M}(U,V)=\kappa (-U,V), \end{equation}

which we can finally find the surface gravity at the two surface $U=0$ and $V=0$:

\begin{equation} \kappa^{\pm}= \pm \frac{1}{4M}. \end{equation}