Why is Killing vector field normal to Killing horizon?

[JunhoPhysics]

In p.244 of Carroll's "Spacetime and Geometry," the Killing horizon $\Sigma$ of a Killing vector $\chi$ is defined by a null hypersurface on which $\chi$ is null. Then it says this $\chi$ is in fact normal to $\Sigma$ since a null surface cannot have two linearly independent null tangent vectors.

But I can't follow the last statement. First, I have no idea what is wrong with having two linearly independent null tangent vectors in a null surface and, second, I have no idea how it is related to the fact that $\chi$ is normal to $\Sigma$.

One example related to this issue is the case of a Kerr BH. For a Kerr BH, we can think of a Killing vector $$\chi=\partial_t+\Omega_H\partial_\phi,$$ and it is straightforward to show $\chi^\mu\chi_\mu=0$ on $$\Sigma:~r=r_H=M+\sqrt{M^2-a^2},$$ so $\Sigma$ is a Killing horizon of $\chi$. But this $\chi$ doesn't look like normal to $\Sigma$, since the tangent vectors on $\Sigma$, such as $\partial_t,~\partial_\theta,~\partial_\phi$ which are obviously orthogonal to the normal vector of $\Sigma$, $$n_\mu=\nabla_\mu f(r)=(0,1,0,0)~~(f(r)=r-r_H)$$, are not orthogonal to $\chi$. The same issue can be rephrased as "$\chi_\mu$ doesn't look parallel to $n_\mu$."

I would really appreciate if you help me out with this trouble...! Even if you are not sure about the general argument in Carroll's, answer for the Kerr BH example would be just enough!

 

[PeterDonis]

I can't follow the last statement. First, I have no idea what is wrong with having two linearly independent null tangent vectors in a null surface and, second, I have no idea how it is related to the fact that $\chi$ is normal to $\Sigma$.

First, it might be helpful to recognize a counterintuitive property of null surfaces: any null vector that is normal to a null surface is also tangent to it!

Why is this the case? Consider the definition of "normal": a vector $\chi$ is normal to a surface $\Sigma$ if, given any tangent vector $\sigma$ in $\Sigma$, $\chi$ is orthogonal to it. Now consider the definition of "orthogonal": two vectors $\chi$ and $\sigma$ are orthogonal if $\chi \cdot \sigma = 0$.

Now consider what must be the case if $\chi$ and $\sigma$ are both null. Then we have $\chi \cdot \chi = 0$ and $\sigma \cdot \sigma = 0$. Plus, from the above, we have $\chi \cdot \sigma = 0$. I'll leave it to you to work out how all of these taken together imply that $\chi = \kappa \sigma$, where $\kappa$ is some constant--in other words, $\chi$ itself is a null tangent vector to the null surface. (If the null surface is a Killing horizon for the Killing vector $\chi$, the constant $\kappa$ is called the "surface gravity" of the horizon.)

Armed with this, you should now be able to see why there can't be two linearly independent null tangent vectors to a null surface. For that to be the case, we would have to have two vectors $\sigma$ and $\lambda$, both null, both tangent to the same null surface, but linearly independent, i.e., there would not be any constant $c$ such that $\lambda = c \sigma$.

Now, consider the inner product $\sigma \cdot \lambda$. Is it zero, or nonzero? If it's zero, then we just showed above that we must have $\lambda = c \sigma$ (just substitute $\lambda$ for $\chi$ and $c$ for $\kappa$ in the above). But if it's nonzero, you should be able to show that $\sigma$ and $\lambda$ can't possibly both be tangent to the same null surface. The easiest way I know of to show this is to work in a local inertial frame centered on a point on the surface, and pick the spatial axes so that one of the two vectors, say $\sigma$, has components (1, 1, 0, 0). Then consider the requirements on the components of $\lambda$, and show that it's impossible for $\lambda$ to satisfy all of them: null vector, linearly independent of $\sigma$, nonzero dot product with $\sigma$, but tangent to the same null surface.

One example related to this issue is the case of a Kerr BH.

In addition to the above, I would note that you have improperly identified the tangent and normal vectors to the surface $\Sigma$ for this case. The equation for that surface, i.e.., the surface where the Killing vector field $\chi$ vanishes, is not just a function of $r$; it's a function of both $r$ and $\theta$.

 

[JunhoPhysics]

Thanks! Now I got the answer for some part of my questions but not for all of them, so please let me summarize them.
1. Now I see why there are no two linearly indept null tangent vectors on a null hypersurface. (If $\sigma_1$ and $\sigma_2$ both are normal to a normal vector of the null hypersurface $n$, according to the 3rd paragraph of your answer, $n=k_1\sigma_1=k_2\sigma_2$ for some constants $k_1,~k_2$. So they can't be linearly indept.)
2. But I still can't see why it implies a Killing vector $\chi$ on its Killing horizon $\Sigma$ must be normal to it. Let's say, since $\Sigma$ is a null hypersurface by definition, there must be a null normal vector $n$ to $\Sigma$. And our claim is $\chi=k n$ for some constant k. But what I can tell now is just the following: $n$ itself is a null tangent vector on $\Sigma$, so $\chi$ can't be another null tangent vector on $\Sigma$ which is linearly indept with $n$ by 1. Then there are two options: case 1) $\chi$ is a null tangent vector on $\Sigma$ which is linearly dept with $n$, which is just our claim; case 2) $\chi$ is not a tangent vector on $\Sigma$ at all. My question is, how can we treat the case 2)...?
3. About your comment on the example of Kerr BH, I think you are thinking of a stationary limit surface, but here I am talking about the event horizon, which is just a function of $r$ only. $\Omega_H$ there is designed to make $\chi$ to vanish on the event horizon, not on the stationary limit surface.

 

[PeterDonis]

how can we treat the case 2)...?

By recognizing that it's impossible. Perhaps I didn't state the counterintuitive property of null vectors and surfaces strongly enough: for a null vector $\chi$ and a null surface $\Sigma$, the two statements "$\chi$ is normal to $\Sigma$" and "$\chi$ is tangent to $\Sigma$" are equivalent. Either one implies the other.

About your comment on the example of Kerr BH, I think you are thinking of a stationary limit surface, but here I am talking about the event horizon, which is just a function of $r$ only.

Ah, you're right, I was confusing the two.

 

[JunhoPhysics]

I am really sorry, but I still don't get it. I do understand that "$\chi$ is normal to $\Sigma~\Leftrightarrow~\chi$ is tangent to $\Sigma$" when $\Sigma$ is a null hypersurface. But my question is, the only information that we have on $\chi$ is that it is a Killing vector field and null on $\Sigma$. From this information, can we argue $\chi$ is normal, or equivalently tangent, to $\Sigma$??

Maybe I can make the necessary claim here: "Any null vector on a null surface must be tangent(or equivalently normal) to the null surface." If I can prove this, I think that will be enough...!

 

[PeterDonis]

Maybe I can make the necessary claim here: "Any null vector on a null surface must be tangent(or equivalently normal) to the null surface."

This claim is too strong; a null vector "on" a null surface (i.e., at a point on the surface) does not have to be tangent (= normal) to the surface. For example, consider the null surface $t = x$ in Minkowski spacetime, and the null vector with components $(1, -1, 0, 0)$ at any point on this surface.

The key point about a Killing vector on its Killing horizon is that the Killing horizon is a surface on which the norm of the Killing vector field is constant. You should be able to convince yourself that a Killing vector at a point must be tangent to the surface through that point on which the norm of the Killing vector field is constant. (Note that there can be only one such surface at any point.)

 

[JunhoPhysics]

Oh, so "Killing" was important! I mean, for any Killing vector field $\chi$ and its Killing horizon $\Sigma:\chi^\nu\chi_\nu=0$, the normal vector of $\Sigma$ can be written as $n_\mu=\nabla_\mu(\chi^\nu\chi_\nu)$. And then $\chi$ is orthogonal to $n$ since $\chi^\mu n_\mu=2\chi^\mu\chi^\nu\nabla_\mu\chi_\nu=0$, which is clear from symmetrization on $\mu\leftrightarrow\nu$ and the fact that $\chi$ is Killing. This means $\chi$ is tangent(so normal) to $\Sigma$. Thanks for your help!

Then the issue with a Kerr BH is quite interesting... Now I understand that in general $\chi$ must be normal(=tangent) to $\Sigma$. But $\chi=\partial_t+\Omega_H\partial_\phi$ doesn't look like normal to $\Sigma:r=r_H$... If you figure out what is going on here, please let me know!

 

[PeterDonis]

Then the issue with a Kerr BH is quite interesting

There are a couple of points here:

First, $\Sigma$ is a null surface, which is also a surface of constant $r$. What does that imply about the direction of the basis vector $\partial_r$ on that surface?

Second, on $\Sigma$, the vectors $\partial_t$ and $\partial_\phi$ are both spacelike. But the vector $\chi$, a linear combination of the two, is null. Can you show in detail how that works out?

 

[JunhoPhysics]

1. One thing I can tell about the direction of $\partial_r$ is that it is normal to $\Sigma:r-r_H=0$ by definition. Then since $\Sigma$ is a null surface, it would also be tangent to $\Sigma$. I am not sure what you mean by "the direction of the basis vector on that surface."

2. From the Kerr metric, $\chi=\partial_t+\frac{a}{2Mr_H}\partial_\phi~(r_H=M+\sqrt{M^2-a^2}~\to~r_H^2-2Mr_H+a^2=0)$ satisfies
$$
\begin{align}
\chi^2|_{r=r_H}&=g_{tt}|_{r=r_H}+2g_{t\phi}|_{r=r_H}\frac{a}{2Mr_H}+g_{\phi\phi}|_{r=r_H}\frac{a^2}{4M^2r_H^2}\nonumber\\
&=-\left(1-\frac{2Mr_H}{r_H^2+a^2\cos^2\theta}\right)-\frac{4Mar_H\sin^2\theta}{r_H^2+a^2\cos^2\theta}\frac{a}{2Mr_H}+\frac{\sin^2\theta}{r_H^2+a^2\cos^2\theta}((r_H^2+a^2)^2-a^2(r_H^2-2Mr+a^2)\sin^2\theta)\frac{a^2}{4M^2r_H^2}\nonumber\\
&=\frac{a^2\sin^2\theta}{r_H^2+a^2\cos^2\theta}-\frac{2a^2\sin^2\theta}{r_H^2+a^2\cos^2\theta}+\frac{\sin^2\theta}{r_H^2+a^2\cos^2\theta}(2Mr_H)^2\frac{a^2}{4M^2r_H^2}\nonumber\\
&=0\nonumber
\end{align}
$$

 

[PeterDonis]

One thing I can tell about the direction of $\partial_r$ is that it is normal to $\Sigma:r-r_H=0$ by definition. Then since $\Sigma$ is a null surface, it would also be tangent to $\Sigma$.

Exactly.

I am not sure what you mean by "the direction of the basis vector on that surface."

That it is tangent to the surface, as you have shown. And you have also shown that $\chi$ is null on that surface, which is a Killing horizon for $\chi$, so therefore $\chi$ is also tangent to the surface (and normal to it). So that means that $\chi$ and $\partial_r$ are parallel. And since they are both null, that means they are normal to each other (because of the counterintuitive property of null vectors that we already discussed). So $\chi$ is in fact normal to $\partial_r$, just as we desired to show.

 

[JunhoPhysics]

Yes, I absolutely agree with your general argument (except a minor thing, that is $n\neq\partial_r$). But my question was, when we try to check this general argument with a specific example,
1) On a Kerr BH's event horizon, $\chi$ and $n~(n_\mu=(0,1,0,0))$ both are null, normal(=tangent).
2) And I want to check if $\chi$ is truly a normal vector with an example. So I considered orthogonal vectors to $n$ such as $\partial_t,~\partial_\theta,~\partial_\phi$ and try to check if they are also orthogonal to $\chi$. But they are not orthogonal to $\chi$.
3) Here, I think that $\partial_t,~\partial_\theta,~\partial_\phi$ are tangent vectors on the event horizon since they are orthogonal to $n$. (If they are tangent vectors then they must be orthogonal to $\chi$.) But it looks like this statement can't be true for this situation to make sense, i.e. they are NOT tangent vectors on the event horizon EVEN THOUGH they are orthogonal to $n$... I thought that any vector orthogonal to the normal vector of a hypersurface is a tangent vector of the same hypersurface but now I guess it is not true in a null hypersurface...??
4) The same issue can be written as, $n$ doesn't look like parallel to $\chi$ for a Kerr BH. Note that $n_\mu=(0,1,0,0)$ but $\chi_\mu=g_{\mu\nu}\chi^\nu=(g_{tt}+g_{t\phi}\Omega_H,0,0,g_{\phi t}+g_{\phi\phi}\Omega_H)$.

 

[PeterDonis]

I want to check if $\chi$ is truly a normal vector with an example. So I considered orthogonal vectors to $n$ such as $\partial_t,~\partial_\theta,~\partial_\phi$ and try to check if they are also orthogonal to $\chi$. But they are not orthogonal to $\chi$.

You are right that $\partial_t$ is not orthogonal to $\chi$; but that's ok, because $\partial_t$ is not tangent to $\Sigma$. See below.

For the others, $\partial_\theta$ is obviously orthogonal to $\chi$, since there are no cross terms in the metric between $\theta$ and any other coordinates. And since $\chi = \partial_t + \Omega \partial \phi$, we have

$$
\chi \cdot \partial_\phi = \partial_t \cdot \partial_\phi + \Omega \partial_\phi \cdot \partial_\phi = g_{t \phi} + \Omega g_{\phi \phi}
$$

Now just substitute $\Omega = - g_{t \phi} / g_{\phi \phi}$ in the above.

I thought that any vector orthogonal to the normal vector of a hypersurface is a tangent vector of the same hypersurface but now I guess it is not true in a null hypersurface...??

That's correct, because of the counterintuitive property of null vectors and null surfaces. Heuristically, for a non-null 3-surface in 4-dimensional spacetime, a normal vector to the surface cannot be parallel to any of the tangent vectors of the surface. So the normal plus the 3 tangents uses up all four dimensions.

But for a null 3-surface, as we've seen, since the normal vector is null, it will be parallel to one of the three tangents. So the normal plus the three tangents still leaves one more dimension unused. That dimension in the Kerr case is $\partial_t$.

The correct condition for a vector $\chi$ to be normal to a surface $\Sigma$ is that $\chi$ must be orthogonal to all tangent vectors of $\Sigma$. To show this, it is sufficient to show that $\chi$ is orthogonal to any set of three linearly independent tangent vectors of $\Sigma$. And we have already shown that, because $\chi$ itself is tangent to $\Sigma$ and is orthogonal to itself, and we've also shown that it's orthogonal to $\partial_\theta$ and $\partial_\phi$, which are both tangent to $\Sigma$ and are linearly independent of each other and of $\chi$.

 

[JunhoPhysics]

Thanks a lot! And I just found that $\chi_\mu$ is in fact a zero vector (even though $\chi^\mu$ is not a zero vector) on the event horizon so it is obviously parallel to $n_\mu$. Finally everything perfectly makes sense. Again, I really appreciate for your help!

 

[PeterDonis]

I just found that $\chi_\mu$ is in fact a zero vector (even though $\chi^\mu$ is not a zero vector) on the event horizon

Huh? You showed in post #9 that $\chi^2 = 0$ on the horizon, i.e., that $\chi^\mu$ is a null vector. (I assume that's what you mean by "a zero vector".)

What is true on the horizon is that the $\phi$ component of $\chi_\mu$ vanishes, so $\chi_\mu$ has only a $t$ component. This is evident from the expression you wrote for $\chi_\mu$ in post #11.

 

[JunhoPhysics]

I thought $\chi_\mu=(g_{tt}+g_{t\phi}\Omega_H,0,0,g_{t\phi}+g_{\phi\phi}\Omega_H)=(0,0,0,0)$ is true. In post #9, I have shown $\chi^2=0=g_{tt}+2g_{t\phi}\Omega_H+g_{\phi\phi}\Omega^2$ on the horizon. But in the calculation I checked $g_{tt}+g_{t\phi}\Omega_H$ and $g_{t\phi}\Omega_H+g_{\phi\phi}\Omega_H^2$ vanish separately in fact. Is there any problem here...?

 

[JunhoPhysics]

?? I checked this just a few minutes ago. Also, if it does not vanish, can you explain what is wrong in the following?

$$\begin{align}
\chi^2=0&=g_{tt}+2g_{t\phi}\Omega_H+g_{\phi\phi}\Omega_H^2\nonumber\\
&=g_{tt}+g_{t\phi}\Omega_H+\Omega_H(g_{t\phi}+g_{\phi\phi}\Omega_H)\nonumber\\
&=\chi_\mu~\mbox{1st component}+\Omega_H(\chi_\mu~\mbox{4th component})\nonumber\\
&=\chi_\mu~\mbox{1st component}
\end{align}$$

 

[PeterDonis]

In post #9, I have shown $\chi^2=0=g_{tt}+2g_{t\phi}\Omega_H+g_{\phi\phi}\Omega^2$ on the horizon. But in the calculation I checked $g_{tt}+g_{t\phi}\Omega_H$ and $g_{t\phi}\Omega_H+g_{\phi\phi}\Omega_H^2$ vanish separately in fact.

Ah, I see. Yes, you're right, the $t$ component vanishes as well.

 

[JunhoPhysics]

Oh wait... This means $\chi$ is infact orthogonal to $\partial_t,\partial_r,\partial_\theta,\partial_\phi$ all of them. Haha... it looks like my original question about this Kerr BH was not a problem in the first place... But anyway I learned a lot from your help. Thanks!

 

[PeterDonis]

This means $\chi$ is infact orthogonal to $\partial_t,\partial_r,\partial_\theta,\partial_\phi$ all of them.

Hm, yes, that's right.

 

[PeterDonis]

it looks like my original question about this Kerr BH was not a problem in the first place

That's true, but I still think it's instructive to work through the details and see exactly how things are related. In particular, I think it's worth working through why the relationship of $\chi$ to $\partial_r$, $\partial_\theta$, and $\partial_\phi$ is relevant to the topic we have been discussing, while the relationship of $\chi$ to $\partial_t$ is not (for the reason I gave in post #12).