The Modern Geometric View of Black Holes

[center o bass]

Im trying to understand an argument from Hobson and Estathiou's book on GR where they argue that null-surfaces; i.e surfaces with a null-vector as a normal are in general horizons. Their argument goes as follows (page 316) in the book:

"Before discussing the particular case of a stationary axisymmetric spacetime,
let us briefly consider null 3-surfaces in general. Suppose that such a surface is
defined by the equation
$$f(x^\mu) = 0$$
The normal to the surface is directed along the 4-gradient $n_\mu = \partial_\mu f$
(remembering that f is a scalar quantity), and for a null surface we have
$$g^{\mu \nu} n_\mu n_\nu = 0$$
This last property means that the direction of the normal lies in the surface
itself; along the surface

$$df = n_\mu dx^\mu = 0$$

and this equation is satisfied when the
directions of the 4-vectors $dx^\mu$ and $n^\mu$ coincide. In this same direction, from the property $g^{\mu \nu} n_\mu n_\nu = 0$ we see that the element of length in the 3-surface is $ds = 0$. In other words, along this direction the 3-surface is tangent, at any given point, to the lightcone at that point. Thus, the lightcone at each point of a null 3-surface
(say, in the future direction) lies entirely on one side of the surface and is tangent
to the 3-surface at that point. This means that the (future-directed) worldline of a
particle or photon can cross a null 3-surface in only one direction, and hence the
latter forms an event horizon.
In a stationary axisymmetric spacetime the equation of the surface must take
the form

$$f(r, \theta) = 0$$
Moreover, the condition that the surface is null means that

$$\partial_\mu f \partial^\mu f = 0$$

which, for a metric of the form (13.4), reduces to

$$g^{\theta \theta} (\partial_{\theta} f)^2 + g^{r r}(\partial_r f)^2 = 0$$

This is therefore the general condition for a surface $f(r,\theta)= 0$ to be an event horizon.
We may, however, choose our coordinates r and $\theta$ in such a way that we can
write the equation of the surface as f(r)= 0, i.e. as a function of r alone. In this
case, the condition (13.8) reduces to

$$g^{rr} (\partial_r f)^2 = 0$$

from which we see that an event horizon occurs when $g^{rr} = 0$, or equivalently
$g_{rr} = \infty$. This is consistent with our analysis of the Schwarzschild metric, for
which $g = \infty$ at $r = 2M$."

I have a few questions to this argument:

A crucial step in their argument is the observation that $n^\mu$ and $dx^\mu$ coincide, but how can this be when $n_\mu dx^\mu = 0$? Have I missed something about four vectors when I think this is the condition for orthogonality?

Why does the condition $g^{\mu \nu} n_\mu n_\nu = 0$ mean that the direction of the normal lies in the surface?

And finally the author argues that one can choose coordinates such that the surface can be expressed as f(r) = 0 and the result is later applies to the Kerr metric. Why is it so that one can make this choice in the case of the Kerr metric?

 

[WannabeNewton]

Hi Bass! Let $\Sigma\subseteq M$ be the null surface given by $\zeta(x^a) = 0$ and $\nabla^{a}\zeta$ be the normal to $\Sigma$. Furthermore, let $x^{a}(\lambda)$ be a curve on $M$ such that for any point on the image of the curve, $\frac{\mathrm{d} x^{a}}{\mathrm{d} \lambda}\nabla_{a}\zeta = 0$ i.e. the tangent field to the curve is always orthogonal to the normal field. Note that then $\frac{\mathrm{d} \zeta(x^{a})}{\mathrm{d} \lambda} = 0$ so $\zeta(x^a) = 0$ all along the image of the curve which is of course only possible if the curve lies on $\Sigma$ implying that $\frac{\mathrm{d} x^{a}}{\mathrm{d} \lambda}$ is tangential to $\Sigma$ at every point on the image of the curve. We can then naturally broaden this to vector fields. Thus, since $\nabla_{a}\zeta \nabla^{a}\zeta = 0$ everywhere, $\nabla^{a}\zeta$ must be tangential to $\Sigma$.

As for your second question, what exactly is $dx^{\mu}$?

 

[Ilmrak]

[...] A crucial step in their argument is the observation that $n^\mu$ and $dx^\mu$ coincide, but how can this be when $n_\mu dx^\mu = 0$? Have I missed something about four vectors when I think this is the condition for orthogonality?

Why does the condition $g^{\mu \nu} n_\mu n_\nu = 0$ mean that the direction of the normal lies in the surface? [...]

Hello,
remember that g_{\mu\nu} is not strictly speaking a metric, it has signature (-+++) so there are vectors orthogonal to themselves: light-like vectors.

Ilm

 

[center o bass]

Hello,
remember that g_{\mu\nu} is not strictly speaking a metric, it has signature (-+++) so there are vectors orthogonal to themselves: light-like vectors.

Ilm

I'm agree, but I suppose that one means that when two vectors coincide that they are proportional to each other. Does the equations $n^\mu n_\mu =0$ and $n^\mu dx_\mu = 0$ imply that

$$ dx^\mu = \alpha n^\mu?$$

 

[center o bass]

Hi Bass! Let $\Sigma\subseteq M$ be the null surface given by $\zeta(x^a) = 0$ and $\nabla^{a}\zeta$ be the normal to $\Sigma$. Furthermore, let $x^{a}(\lambda)$ be a curve on $M$ such that for any point on the image of the curve, $\frac{\mathrm{d} x^{a}}{\mathrm{d} \lambda}\nabla_{a}\zeta = 0$ i.e. the tangent field to the curve is always orthogonal to the normal field. Note that then $\frac{\mathrm{d} \zeta(x^{a})}{\mathrm{d} \lambda} = 0$ so $\zeta(x^a) = 0$ all along the image of the curve which is of course only possible if the curve lies on $\Sigma$ implying that $\frac{\mathrm{d} x^{a}}{\mathrm{d} \lambda}$ is tangential to $\Sigma$ at every point on the image of the curve. We can then naturally broaden this to vector fields. Thus, since $\nabla_{a}\zeta \nabla^{a}\zeta = 0$ everywhere, $\nabla^{a}\zeta$ must be tangential to $\Sigma$.

As for your second question, what exactly is $dx^{\mu}$?

I guess you can think of it as $\frac{d x^{mu}}{d \lambda}$. Does your argument essentially say that since the normal vector is orthogonal to the surface and the normal vector is orthogonal to itself, it must be parallel with the surface?

 

[WannabeNewton]

Essentially yes, Bass, but keep in mind that just because a vector is orthogonal to a vector field at some arbitrary point and another vector is orthogonal to that same vector field at another arbitrary point doesn't mean the two vectors are parallel to each other. The key point here is that the normal field is orthogonal to the surface everywhere and that it is orthogonal to itself everywhere.

 

[WannabeNewton]

I honestly have no idea what $dx^{\mu}$ is (perhaps you could quote the relevant passage in the text, that might make it clearer). What is true, however, is that the integral curves of the normal field $\xi ^{a} = \alpha\nabla^{a}\zeta$ are in fact null geodesics. To see this, note that $\xi^{a}\nabla_{a}\xi^{b} = \beta\xi^{b} + \frac{1}{2}\nabla^{b}(\xi^{c}\xi_{c}) - \xi^{c}\xi_{c}\nabla^{b}\ln \alpha$ where $\beta = \xi^{a}\nabla_{a}\ln\alpha$. But we know that $\xi^{c} \xi_{c} = 0$ on $\Sigma$ and we can take $\nabla^{b}(\xi^{c}\xi_{c})$ to be in the direction of $\xi^{b}$ since $\xi^{c}\xi_{c} = 0 = \text{const.}$ on $\Sigma$ therefore $v^{a}\nabla_{a}(\xi^{c}\xi_{c}) = 0$ for any $v^{a}$ tangent to $\Sigma$ implying that $\nabla^{b}(\xi^{c}\xi_{c})$ is orthogonal to $\Sigma$ and since $\xi^{b}$ is null, we can just take $\nabla^{b}(\xi^{c}\xi_{c}) = \kappa \xi^{b}$.

This reduces the expression to $\xi^{a}\nabla_{a}\xi^{b} = \varsigma \xi^{b}$ for an appropriate function $\varsigma$. We know that we can always re-parametrize this to then take the form $\xi^{a}\nabla_{a}\xi^{b} = 0$. Hence, at every $p\in \Sigma$, there exists a unique null geodesic $x^{a}(\lambda)$ passing through $p$ such that $\xi^{a} = \frac{\mathrm{d} x^{a}}{\mathrm{d} \lambda}$ at $p$.

 

[center o bass]

As I said i guess you can think of $dx^\mu$ as $dx^\mu/d\lambda$. The author seem to be using it as an infinitesimal separation.

 

[WannabeNewton]

But what the author wrote makes no sense then. $df$ is the exterior derivative of the smooth function $f:M\rightarrow \mathbb{R}$. If $(U,\varphi)$ is a smooth chart on $M$ and $\{dx^{1},...,dx^{n}\}$ are the basis one-forms associated with the chart, we can write $df$ as $df = \nabla_{a}f dx^{a}$ but this is a sum over the basis one-forms, not the sum over components of some 4-vector!

"I guess you can think of..." is not really rigorous enough to justify what the author is writing. If possible, could you take a picture or quote the relevant passage?

 

[Ilmrak]

From the first post it seems that dx^\mu is the light-like element of a basis of the one-forms on the null surface. This would be consistent with the rest of the argument I think.

Ilm

 

[George Jones]

But what the author wrote makes no sense then. $df$ is the exterior derivative of the smooth function $f:M\rightarrow \mathbb{R}$. If $(U,\varphi)$ is a smooth chart on $M$ and $\{dx^{1},...,dx^{n}\}$ are the basis one-forms associated with the chart, we can write $df$ as $df = \nabla_{a}f dx^{a}$ but this is a sum over the basis one-forms, not the sum over components of some 4-vector!

"I guess you can think of..." is not really rigorous enough to justify what the author is writing. If possible, could you take a picture or quote the relevant passage?

Write $n_a = \nabla_{a}f$. Then $n^a$ is normal to the level surface $f\left(p\right) = 0$.

 

[WannabeNewton]

Write $n_a = \nabla_{a}f$. Then $n^a$ is normal to the level surface $f\left(p\right) = 0$.

That's fine but what exactly is $dx^{\mu}$? If the author meant to say $dx^{\mu}$ are the basis one-forms and not the components of a 4-vector then saying $df = 0$ amounts to saying $n_{0}dx^{0} + ... + n_{3}dx^{3} = 0$, where $\{dx^0,...,dx^3\}$ are the basis one-forms, but why would that follow from $n^{a}$ being normal to the null surface? If the author meant something else by $dx^{\mu}$ then I have no idea what, and why he/she writes $df$ as equal to that.

 

[dx]

along the surface

$$df = n_\mu dx^\mu = 0$$

and this equation is satisfied when the
directions of the 4-vectors $dx^\mu$ and $n^\mu$ coincide.

df as a 1-form is obviously not necessarily zero. The only way I can make sense of this is that the author means that if

ζ = ΔX^a∂_a

lies in the null surface, then

df(ζ) = 0

i.e.

ΔX^an_a = 0

Where n_a = ∂f/∂X^a

 

[dx]

So the authors are using dX^a to mean the components ΔX^a of the vector ζ which lies in the null surface, and not as basis 1-forms

 

[George Jones]

So the authors are using dX^a to mean the components ΔX^a of the vector ζ which lies in the null surface, and not as basis 1-forms

Exactly. Maybe I will write some more later, but, right now, I am off to lunch.

 

[WannabeNewton]

So the authors are using dX^a to mean the components ΔX^a of the vector ζ which lies in the null surface, and not as basis 1-forms

So it should really be written $df(\zeta) = 0$? Hence not $df = 0$, since the latter does not follow logically from the fact that $n$ is normal to the null surface. I can't say I'm particularly fond of the notation used by the author then.

 

[dx]

Me neither. I haven't read this particular book, but from the OP's quote it seems likely that the authors are speaking "old tensor."