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center o bass
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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?
"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?