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

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


$$ \frac{\partial U}{\partial r} = - \frac{U}{4m}\frac{\partial u}{\partial r} = \frac{U}{4m}\frac{\partial r^*(r)}{\partial r}=\frac{U}{4m f(r)} = - e^{r/2m} \frac{r}{8m^2 V} $$ 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?

  • $\begingroup$ @G.Smith I edited the question. $\endgroup$
    – apt45
    Commented Feb 24, 2021 at 22:10
  • $\begingroup$ @G.Smith I don't think there is a typo in my equations. I think there is something wrong with my understanding of computing the surface gravity $\endgroup$
    – apt45
    Commented Feb 24, 2021 at 22:25

1 Answer 1


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



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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.