Werner Israel proved in 1967 that a static, non-spinning black hole must have the Schwarzschild metric outside its event horizon, assuming some sanity conditions about the event horizon. That is, the event horizon has to be a perfect sphere.
Brandon Carter, David C. Robinson, and Stephen Hawking proved further results which say that a spinning black hole must have the must have the Kerr-Newman metric outside it.
Why is the Werner Israel 1967 result true?
Suppose that we have a black hole which satisfies the Scwarzschild metric. Let us drop a test mass m into it.
The black hole bends the field lines of the gravity field of the test mass m in such a way that when the test mass is relatively close to the horizon, its field seem to originate from a point which is very close to the center of the black hole.
Richard Hanni and Remo Ruffini calculated this for electric field lines in 1973.
This suggests that even if we try to make the gravity field of a black hole deformed by dropping a large number of test masses to one side of the black hole, the event horizon will stay perfectly spherical.
It is probably an "optical" phenomenon: the field of the black hole distorts the field of the test mass in such a way that the location of the test mass is hidden.
An electric charge close to the event horizon
___
/ \ ● charge Q
\____/
horizon
Suppose that we have an electric charge Q very close to the horizon of a black hole.
The charge is a source of the electric field E and there are no other sources or sinks of the field. The lines of force must extend to infinity.
Let ε₀ be vacuum permittivity. We assume that the electric field E becomes such that it minimizes the total field energy
W = ∫ 1/2 ε₀ E² * U dV,
entire
space
where we integrate over the entire space, and U is the gravitational potential at the volume element dV. Close to the horizon, U is very close to zero.
How should we draw the lines of force, so that the field energy would be the smallest possible?
Can we let lines of force go very close to the horizon? Let us calculate the field energy for such a configuration.
The Schwarzschild metric is (Wikipedia):
where r is the Schwarzschild radial coordinate and r_s is the Schwarzschild radius of the spherically symmetric mass M.
Let us denote
r = r_s + r'
where r' is a small positive distance.
r' dr
| |||
| ||| field lines
| |||
horizon layer
What is the energy of a configuration where we draw all the field lines from the charge Q very close along the horizon?
We assume that c = 1 and r_s = 1. The metric of time at a distance r' from the horizon is
sqrt(1 - 1 / (1 + r'))
= sqrt(r')
= U.
Let us pack the bunch of field lines into a "layer" whose thickness in the Schwarzschild radial coordinate is dr << r', as in the diagram.
The radial metric tells us the proper thickness of that dr. It is
dr / sqrt(r').
We pack, say, Q field lines into the layer. The field strength
E ~ sqrt(r') / dr,
and the (locally measured) energy density
E² ~ r' / dr².
The locally measured total energy of the field is ~ proper thickness * E², or
~ dr / sqrt(r') * r' / dr²
= sqrt(r') / dr.
Since U ~ sqrt(r'), we have that the total energy, as measured by an observer far away, is
~ r' / dr.
Let us choose
dr = b r',
where b is a small constant > 0. For example, b could be 0.01. The energy associated to putting Q lines of force to that layer is
constant * 100.
We can now divide the range r' > 0 into an infinite number of layers which converge to r' = 0.
horizon
| | | | |
| | | | |
... layer layer layer
Suppose that we want to have Q lines of force going along the horizon, close to the horizon. We divide Q lines into N layers. The field energy in one layer is
~ E²
~ Q² / N².
The total field energy in N layers is
~ Q² / N.
By choosing N large we can make the field energy as small as we like.
If we have a charge very close to the horizon, then we can lead the lines of force around the horizon, and the associated field energy of the field close to the horizon is negligible, as measured by a distant observer.
In order to minimize the field energy in the entire space, we need to draw the lines of force from the horizon to infinity. How can we minimize the energy? Intuitively, we have to minimize the field energy associated with the lines of force which leave the horizon.
lines of force
|
___ +
____ / \ ____ lines of force
\___/
metal sphere
|
If we have a charge Q on a metal sphere, the minimum energy electric field is the spherically symmetric uniform field. The field lines are normal to the surface of the sphere. This is the solution for a black hole, too.
Our proof is not rigorous, though. Since the charge Q is not exactly at the horizon, drawing the field lines downward almost to touch the horizon does involve some field energy. Anyway, we have now an intuitive explanation for the fact that a black hole hides the electric charge distribution in it, so that the field looks as if the charge would be uniformly distributed over the horizon.
Conclusion
We found a simple heuristic argument which explains the no-hair theorem for an electric charge in a black hole.
Let us next try to extend this to the gravity field. Is the gravity field of a non-rotating black hole spherically symmetric, even though the masses would not be? This is the claim of the Werner Israel 1967 theorem.
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