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For a killing vector $k^a$, one can get the surface gravity of a Schwarzschild black hole by calculating $k^b \nabla_b k^a$ (which is supposed to equal $\kappa k^a$), setting $r=2m$, and then comparing the LHS and RHS to find an expression for $\kappa$.

Here, $k^a$ is the time translation killing vector, and in $(t,r,\theta,\phi)$ coordinates, we have $k^a = [1, 0, 0, 0]$.

If one works out $k^b \nabla_b k^a$, you will get $\displaystyle \left[0 ,\frac{m(r-2m)}{r^3},0,0 \right]$. Plugging in $r=2m$ into this however makes the result the zero vector, and it impossible to extract $\kappa = 1/4m$ from this. Regardless of it being the zero vector, the components of $k^b\nabla_b k^a$ and $k^a$ don't line up at all, so even if there wasn't a problem plugging in $r=2m$ it is still impossible to get an expression for $\kappa$ from this.

Does anyone know what I am doing wrong?

EDIT: While I know there is a formula for $\kappa$ involving square roots of covariant derivatives, I would like to derive $\kappa$ using this method.

EDIT2: To work out $k^b\nabla_bk^a$, one needs to work out the Christoffel symbols. We have $$\nabla_bk^a = \partial_b k^a+\Gamma^a_{bc}k^c = \Gamma^a_{b1},$$where we have made use of the fact that the derivatives of $k^a$ are all zero and that when you sum over the $c$ index only the first component of $k^c$ contributes.

One can calculate the Christoffel symbols easily by using the regular formulas, but the only nonzero ones that matter to us are $$\Gamma^{2}_{11} = \frac{m(r-2m)}{r^3}, \text{ and } \Gamma^{1}_{21} = \frac{m}{r(r-2m)}.$$

So, summing over $k^b\nabla_bk^a$, we get $$k^1\nabla_1 k^a + k^2\nabla_2 k^a + k^3\nabla_3 k^a + k^4\nabla_4 k^a.$$

Using the above information along with the fact that the only nonzero component of $k^a$ is $k^1$, the only nonzero term in this expression is $k^1\nabla_1 k^2$. Working this out, we get $$k^1\nabla_1 k^2 = k^1 \Gamma^{2}_{11} = \frac{m(r-2m)}{r^3},$$ and hence, $$ k^b \nabla_b k^a = \left[ 0, \frac{m(r-2m)}{r^3}, 0, 0 \right],$$ which is not proportional to $k^a$.

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  • $\begingroup$ Can you show how you worked out $k^b \nabla_b k^a$? $\endgroup$
    – knzhou
    Commented May 1, 2018 at 14:28
  • $\begingroup$ Sure, I'll edit it into the main body. It was verified using Maple as well. $\endgroup$
    – Sorey
    Commented May 1, 2018 at 14:29
  • $\begingroup$ Added in the details of the calculations. $\endgroup$
    – Sorey
    Commented May 1, 2018 at 14:41
  • $\begingroup$ Well, it looks perfectly right to me. The equation you're using simplifies to $0 = \kappa 0$, which is perfectly correct. $\endgroup$
    – knzhou
    Commented May 1, 2018 at 14:53
  • $\begingroup$ I don’t think so. If I look at the first component of the last equation above, it’s 0. But $\kappa k^1$ is not zero. $\endgroup$
    – Sorey
    Commented May 1, 2018 at 14:58

2 Answers 2

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As indicated, the surface gravity, $\kappa$, is calculated from the equation $$ k^b \nabla_b k^a = \kappa k^a $$ when it is $r=2M$. However, you have two ways to solve this equation: either you directly solve as you did, which gives you $0=0$, or you can solve it by lowering the index, $a$, by the metric and solve it as $$ \tag{1} k^b \nabla_b k_a = \kappa k_a $$ where you have $$k_a = (-1+\frac{2M}{r},0,0,0)$$ for $k^a=(1,0,0,0)$ because of the time component of the Schwarzchild metric.

Therefore, the derivatives of $k_a$ would not vanish and Eq.(1) would be a first order differential equation in r where you can calculate $\kappa$ directly to be $1/4M$.

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    $\begingroup$ So, at $r=2m$, you get that $k_a = [0, 0, 0,0 ]$, and $k^b\nabla_b k_a = [0,1/4m,0,0]$. How do these work out? $\endgroup$
    – Sorey
    Commented May 1, 2018 at 16:24
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    $\begingroup$ As you notice the time component of the metric would also be $0$ at $r=2M$. This divides the Schwarzchild solution into interior and exterior parts that are completely disconnected. This is called a coordinate singularity which occured because of the bad choice of coordinates. You can work in another coordinate system that has a regular surface at $r=2M$. For instance, Eddington–Finkelstein coordinates (or categorically tortoise coordinates), or Lemaitre coordinates etc. You can follow a formal and didactic discussion in the book "Spacetime and Geometry" by Sean Carroll. $\endgroup$ Commented May 1, 2018 at 17:18
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    $\begingroup$ I am familiar with Eddington-Finkelstein coordinates, and I am sure that the solution would work out in those coordinates. Are you saying that the solution I am looking for will not be able to be properly achieved in Schwarzschild $(t,r,\theta,\phi)$ coordinates? $\endgroup$
    – Sorey
    Commented May 1, 2018 at 17:21
  • $\begingroup$ No I am not saying that. You can indeed obtain the solution, $\kappa=1/4M$, as I stated in the answer. Just, one shouldn't compare the vectors and their duals living in the surface since the metric in Schwarzschild coordinates has a (fake) singularity there. $\endgroup$ Commented May 1, 2018 at 17:37
  • $\begingroup$ @OktayDoğangün So comparing $V^a$ is different than $V_a$ because $g_{ab}$ has a singularity and since the metric seeps through covariant derivative we have something like delta function and gauss law inconsistency for $\frac{\hat{r}}{r^2}$ potential. $\endgroup$
    – aitfel
    Commented Jan 4, 2020 at 18:06
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The following answer can be viewed as a more detailed explanation of the comments given by others. The main goal is if one performs the calculations in the original Schwarzschild coordinate system, as Sorey did, how to understand the apparent contradiction.

I read through the question and found myself quite confused. I mostly repeated Sorey's calculations and found them indeed correct. However, I cannot see, as some others pointed out, why Sorey's results are equivalent to $0=0$. To be explicit, the l.h.s. vanishes on the horizon according to the calculations, but the r.h.s does not suppose to vanish, as neither $\kappa$ nor $k^a$ vanishes on the horizon. As during the course of the calculations, one does not encounter any singularity, I simply did not understand why the calculations MUST be performed in a different coordinate system, such as Eddington-Finkelstein or Kruskal-Szekeres coordinates, for the results to be meaningful.

Then I sought help from a friend, who kindly pointed me to Padmanabahan's textbook Gravitation: Foundations and Frontiers. The following discussions are largely based on the discussions given in the textbook.

The main point is that due to the nature that Killing vector is a null vector on horizon, it turns to another direction if expressed in the original Schwarzschild coordinates. This can be explicitly shown by making use of the Kruskal-Szekeres coordinates as follows.

The Kruskal-Szekeres coordinates $X^a\equiv (U,V)$ read $$U=-e^{-u/4M},\ \ V=e^{v/4M}$$ with $$u=t-r_*, \ \ v=t+r_*$$ where $r_*$ is the tortoise coordinate $$r_*=r+2M\ln \left(\frac{r-2M}{2M}\right) .$$ We note that the definition involves $\ln(\cdots)$ in such a way that on the horizon $UV=0$, it is negative outside the horizon and positive inside.

Now it is straightforward to show that on the future horizon where $U=0$, the 1-form regarding the Killing vector reads $$\xi|_\mathcal{H}=\left(\frac{\partial}{\partial t}\right)_\mathcal{H}=\kappa V\frac{\partial }{\partial V}=\frac{\partial}{\partial v}$$ where, for simplicity, we have denoted $\kappa=1/4M$ which, is to be recognized as the surface gravity later.

In Sorey's (correct) resultant expression, one encounters an expression which, instead, possesses a nonvanishing $r$ component. So let us also evaluate the following quantity $$m=\frac{\partial}{\partial r}=\frac{\partial X^a}{\partial r}\frac{\partial}{\partial X^a}=\frac{\partial X^a}{\partial r^*}\frac1f\frac{\partial}{\partial X^a}=\frac{\kappa}{f}\left(U\frac{\partial}{\partial U}+V\frac{\partial}{\partial V}\right)$$ On horizon, one finds the relation $$fm|_\mathcal{H}=\kappa\left(U\frac{\partial}{\partial U}+V\frac{\partial}{\partial V}\right)_\mathcal{H}=\xi|_\mathcal{H} .$$

This relation shows that although $m$ diverges, $fm$ is a well-defined quantity. The nonvanishing components on the r.h.s. and l.h.s. of the above relation are different in the original Schwarzschild coordinates as a result of the properties of the future horizon. It is precisely what we need to enclose the loop!

To be specific, Sorey's calculation gives $$\xi^a\nabla_a\xi^b=\xi^c\xi^a\Gamma^b_{ca}=\Gamma^b_{00}=\frac12 ff'\delta^b_r=\kappa fm^b|_\mathcal{H}=\kappa\xi^b|_{\mathcal{H}} $$ where we have made use of $\kappa=\frac{1}{4M}=\frac12 f'(r_H)$ and the above relation.

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