# Computation of surface gravity with Kruskal coordinates

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?

• @G.Smith I edited the question. Commented Feb 24, 2021 at 22:10
• @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 Commented Feb 24, 2021 at 22:25

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,

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

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,

$$$$\frac{1}{4M}(U,V)=\kappa (-U,V),$$$$

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

$$$$\kappa^{\pm}= \pm \frac{1}{4M}.$$$$