Zeroth law of black hole thermodynamics

[Antarres]

I was looking at the proof of zeroth law of thermodynamics from the original paper by Bardeen, Carter, Hawking, which can be found here.

Now, we have the Killing vector which is the generator of the horizon, we call it $l^\mu$, and auxiliary null vector field $n^\mu$, which we define to be normalized as $n^\mu l_\mu = 1$(note that this is the opposite convention of the paper, but it will just change the overall sign of the equations that follow, I used this one since I'm working in opposite signature).

Then we can define surface gravity by the equation
$$l^\mu_{;\nu}l^\nu = \kappa l^\mu$$
In the paper, they use the Newmann-Penrose convention of adopting a tetrad basis of 4 null vectors(the spacelike vectors at the horizon being combined into two complex null vectors), but I choose to just have two regular spacelike vectors at the horizon. We want to see how surface gravity changes, that is, we want to prove that it's constant. So we have:
$$\kappa = l_{\mu;\nu}n^\mu l^\nu$$
Call one of the spacelike vectors $p^\mu$. Then:
$$p^\rho \kappa_{;\rho} = (l_{\mu;\nu}n^\mu l^\nu)_{;\rho}p^\rho = l_{\mu;\nu\rho}n^\mu l^\nu p^\rho + l_{\mu;\nu}n^{\mu}_{\hphantom{\mu};\rho}l^\nu p^\rho + l_{\mu;\nu}n^\mu l^\nu_{\hphantom{\nu};\rho}p^\rho$$

Now for the second one we use the defining relation, and the Leibniz rule to swap out the covariant derivative:
$$l_{\mu;\nu}n^{\mu}_{\hphantom{\mu};\rho}l^\nu p^\rho = \kappa l_\mu n^{\mu}_{\hphantom{\mu};\rho}p^\rho = - \kappa l_{\mu;\rho}n^\mu p^\rho$$

However, then if we use the covariant derivative of the defining relation, we find:
$$l_{\mu;\nu\rho} l^\nu + l_{\mu;\nu}l^\nu_{\hphantom{\nu};\rho} = \kappa l_{\mu;\rho}$$

If we substitute this relation into the equation, we see that first and third term sum up, and all terms cancel out. This is all good, but in the paper, they cancelled second and third term, and then used Einstein equations and the dominant energy condition to get rid of the first term. So I must've made a mistake somewhere. I didn't use the complex vector notation, but I figured proving the relation along the direction of one real spacelike vector, and then the other(which should essentially be the same proof), the relation would be proven to hold.

 

[martinbn]

I was looking at the proof of zeroth law of thermodynamics from the original paper by Bardeen, Carter, Hawking, which can be found here.

Now, we have the Killing vector which is the generator of the horizon, we call it $l^\mu$, and auxiliary null vector field $n^\mu$, which we define to be normalized as $n^\mu l_\mu = 1$(note that this is the opposite convention of the paper, but it will just change the overall sign of the equations that follow, I used this one since I'm working in opposite signature).

Then we can define surface gravity by the equation
$$l^\mu_{;\nu}l^\nu = \kappa l^\mu$$
In the paper, they use the Newmann-Penrose convention of adopting a tetrad basis of 4 null vectors(the spacelike vectors at the horizon being combined into two complex null vectors), but I choose to just have two regular spacelike vectors at the horizon. We want to see how surface gravity changes, that is, we want to prove that it's constant. So we have:
$$\kappa = l_{\mu;\nu}n^\mu l^\nu$$
Call one of the spacelike vectors $p^\mu$. Then:
$$p^\rho \kappa_{;\rho} = (l_{\mu;\nu}n^\mu l^\nu)_{;\rho}p^\rho = l_{\mu;\nu\rho}n^\mu l^\nu p^\rho + l_{\mu;\nu}n^{\mu}_{\hphantom{\mu};\rho}l^\nu p^\rho + l_{\mu;\nu}n^\mu l^\nu_{\hphantom{\nu};\rho}p^\rho$$

Now for the second one we use the defining relation, and the Leibniz rule to swap out the covariant derivative:
$$l_{\mu;\nu}n^{\mu}_{\hphantom{\mu};\rho}l^\nu p^\rho = \kappa l_\mu n^{\mu}_{\hphantom{\mu};\rho}p^\rho = - \kappa l_{\mu;\rho}n^\mu p^\rho$$

However, then if we use the covariant derivative of the defining relation, we find:
$$l_{\mu;\nu\rho} l^\nu + l_{\mu;\nu}l^\nu_{\hphantom{\nu};\rho} = \kappa l_{\mu;\rho}$$

Why isn't there a derivative of kappa on the right handside?

If we substitute this relation into the equation, we see that first and third term sum up, and all terms cancel out. This is all good, but in the paper, they cancelled second and third term, and then used Einstein equations and the dominant energy condition to get rid of the first term. So I must've made a mistake somewhere. I didn't use the complex vector notation, but I figured proving the relation along the direction of one real spacelike vector, and then the other(which should essentially be the same proof), the relation would be proven to hold.

 

[Antarres]

Yes, I was just about to post that I saw that. I guess using the covariant derivative of the defining relation then makes it circular. I'm gonna look to correct it.

@Edit:I think I'm going to solve this soon, so I won't post anything until then, but if there is an obvious hint, feel free to post.

 

[martinbn]

@Edit:I think I'm going to solve this soon, so I won't post anything until then, but if there is an obvious hint, feel free to post.

Solve what? It is in the paper!

 

[Antarres]

@PeterDonis Aren't vectors tangent to the horizon(apart from the generators) spacelike? Those would be the vectors constructed from the angular part of the metric in usual models, in this case, the tetrad vectors that are orthonormal with respect to the induced metric.

 

[martinbn]

@PeterDonis Aren't vectors tangent to the horizon(apart from the generators) spacelike?

Yes. Why?

 

[Antarres]

I was replying to post by @PeterDonis, who said that the vectors with respect to which I should transport surface gravity are not spacelike, but the post isn't appearing now, maybe it was deleted.

And yes, it is in the paper, I'm just rederiving it to understand the proof well, that's what I meant by solving.

 

[PeterDonis]

@PeterDonis Aren't vectors tangent to the horizon(apart from the generators) spacelike?

I deleted the post you are implicitly responding to here.

 

[PeterDonis]

I was replying to post by @PeterDonis,

Which I have deleted as it was in error.

 

[Antarres]

I was busy these days, so it took me a while to write up the re-derivation of the algebra from the paper. I've used the following steps:
Let $p^\mu$, $q^\mu$ be the spacelike vectors on the horizon. Then the decomposition of the metric reads:
$$g_{\mu\nu} = l_\mu n_\nu + n_\mu l_\nu - p_\mu p_\nu - q_\mu q_\nu$$
In order to resolve the third term in the equation for parallel transport of surface gravity, I use the following two properties:

1. By Killing equation, we have $l_{\mu;\nu}p^\mu p^\nu = l_{\mu;\nu} q^\mu q^\nu = 0$.
2. Since the generator must satisfy Frobenius condition $l_{[\mu}l_{\nu;\rho]} = 0$, contracting it with $n^\mu p^\nu q^\rho$, we have $l_{\mu;\nu} p^\mu q^\nu = 0$.

Defining auxilliary scalars $F = l_{\mu;\nu}p^\mu n^\nu$ and $G = l_{\mu;\nu} q^\mu n^\nu$, I find the following decomposition into tetrad components:
$$l_{\mu;\nu}n^\mu = \kappa n_\nu + F p_\nu + G q_\nu$$
Then the third term becomes, using properties 1 and 2:
$$l_{\mu;\nu}n^\mu l^\nu_{\hphantom{\nu};\rho} p^\rho = \kappa l_{\nu;\rho}n^\nu p^\rho$$
And so the second and third term cancel:
$$p^\mu \kappa_{;\mu} = l_{\mu;\nu\rho}n^\mu l^\nu p^\rho = R_{\mu\nu\rho\sigma} n^\mu l^\nu p^\rho l^\sigma$$
The last equality follows from $l^\mu$ being a Killing vector.
Using the relation for transverse derivative on the horizon of the identity 2: $(l_{\mu;\nu} p^\mu q^\nu)_{;\rho}q^\rho = 0$, this relation can be further simplified using manipulations like the ones above. I won't write all this down since it is just a bit cumbersome, but we find, just like in the paper:
$$p^\mu \kappa_{;\mu} = - R_{\mu\nu}l^\mu p^\rho$$
Then by Einstein equations we get $p^\mu \kappa_{;\mu} = - 8\pi T_{\mu\nu}l^\mu p^\rho$.

Now to complete the proof the dominant energy condition should be invoked, asserting that $T_{\mu\nu}l^\nu$ is non-spacelike. At this point it was used that generators of the horizon have expansion scalar and shear equal to zero to conclude that $T_{\mu\nu} l^\mu l^\nu = 0$. This fact was used in the paper in the derivation that I just described, but I avoided it and used a tetrad expansion which seems correct to me. This is an intuitive fact, but proving it rigorously(looking at some other paper by Hawking) required some geometric construction that I wanted to avoid when reconstructing the proof(it's not unclear but it seemed to require much more writing).
I just wanted to employ the algebra I used before and to assert that if $T_{\mu\nu}l^\mu$ is non-spacelike, then it must must have an expansion on the horizon of the form:
$$T_{\mu\nu}l^\mu = Al_\nu + Bn_\nu$$
Then the conclusion that $\kappa$ is constant follows just by orthogonality of the tetrad basis. Is this approach good?