Can we figure out a similar trick with which to over-charge a black hole which is extremal with respect to an electric charge?
The Reissner-Nordström metric of an electrically charged black hole
Here M is the mass-energy of the black hole, Q is its electric charge, G is the gravitational constant, c is the speed of light, and ε₀ is vacuum permittivity.
The event horizons are where the metric for dr diverges:
For an extremal black hole, the formula within the square root is zero. We see that the radius of the event horizon is exactly a half of the corresponding uncharged black hole with the same mass-energy M.
The mass-energy of the static electric field
Demetrios Christodoulou and Remo Ruffini (1971) state in the abstract that 50% of the energy of an extremal charged black hole can be extracted.
_____
/ \
| ● black hole Q
\______/
shell -Q
Let the charge of the extremal black hole be Q. We put a zero mass charged spherical shell with the opposite charge -Q around it and lower the shell slowly down. In this way we can extract the energy of the electric field of Q.
What is the energy of the electric field, from the viewpoint of a faraway observer?
The time component g₀₀ of the metric tensor is:
At the coordinate radius r, the time runs slower, causing a "redshift of energy" by the factor
sqrt(g₀₀),
but that is compensated by the stretching of the radial metric of r, which makes a volume element larger by the factor:
1 / sqrt(g₀₀).
The energy of the electric field, as seen by a faraway observer, is as if the metric on the time t and the radial distance r would be Minkowski.
Let us calculate the energy of the static electric field for an extremal black hole. An extremal black hole has 2 r_Q = r_s, which implies
Q = sqrt(G / k) * M.
We denote Coulomb's constant by
k = 1 / (4 π ε₀).
The electric field strength E at a distance r is
E = k Q / r².
The energy density D is
D = 1/2 ε₀ E²
= 1 / (8 π k) * E²
= k / (8 π) * Q² * 1 / r⁴.
The electric field energy W from r to infinity is
W = k / (8 π) * Q²
∞
* ∫ 1 / r⁴ * 4 π r² dr
r
= k / 2 * Q² * 1 / r.
For an extremal black hole, we have
Q² = G / k * M²,
and the event horizon is at r = G M / c².
W = k / 2 * G / k * M² * c² / (G M)
= 1/2 M c².
We conclude that the static electric field contains exactly 1/2 of the whole mass-energy of the extremal black hole.
Let us then calculate the energy of the electric field per a short distance dr:
dW / dr = k / 2 * Q² * 1 / r²
= k / 2 * G / k * M² * 1 / r²
= 1/2 G M² / r².
At the extremal black hole event horizon r = G M / c², we have
dW / dr = 1/2 c⁴ / G,
or in terms of the mass of the electric field,
dm / dr = 1/2 c² / G.
For an uncharged black hole, the Schwarzschild radius is
r = 2 G M / c²,
or
M / r = 1/2 c² / G.
The ratio M / r tells us how much mass we have to add if we want increase the Schwarzschild radius by a unit length.
We see that the ratio dm / dr at the horizon of an extremal black hole is almost enough to increase the Schwarzschild radius.
We get the following diagram for the extremal black hole whose total mass is M:
| M / 2 | M / 2 ...
r = 0 r = G M / c²
irreducible mass mass of electric field
The horizon r = GM / c² is at the location where it would be if we would remove the charge and the associated electric field energy, leaving just 1/2 M of the mass in the system.
Can we use over-charging to shrink the horizon of an uncharged Schwarzschild black hole?
Short answer: no.
Let us start from a neutral Schwarzschild black hole whose Schwarzschild radius is R. We assume that we have a shell of matter whose entire mass-energy is in its static electric field. Its "intrinsic" mass is zero.
We charge the black hole by pushing in these charged shells. The black hole under the newly built electric field does not change.
Let us assume that we are able to over-charge the black hole. We assume that we have used concentric shells whose charges are Q and -Q, to add the new electric field only to the radii
R ... R + dR,
where dR is small. The configuration is like this:
Q -Q
| | |
R R + dR
event
horizon
But now the black hole and the static electric field between R and R + dR contain too much mass-energy, as seen by a distant observer. The event horizon has advanced to R + dR or beyond.
If we would have the electric field also from R + dR to infinity, that does not change the fact that the event horizon has risen up. An event horizon in a spherically symmetric system is not affected if we add mass outside the sphere, at least as long as the horizon does not rise outside the spherically symmetric system.
Our Minkowski & newtonian viewpoint on overcharging
Let us analyze the problem using our own model of gravity. We ask if we can destroy a Schwarzschild black hole by pushing electric charge in.
The answer is obviously no. The volume inside and very close to the forming event horizon is essentially "frozen". Clocks tick extremely slow there, and the local speed of light is very slow, as seen by a faraway observer.
Pushing electric charge to the surface does not produce any force inside the black hole and cannot change anything. One cannot use that method to shrink the event horizon.
Conclusions
Stephen Hawking and other researchers claimed in the early 1970s that it is impossible to shrink the event horizon of a black hole. They, certainly, looked at over-charging a black hole, too.
Our previous blog post suggests that one may be able to disassemble a black hole by over-spinning it. A "centrifugal force" may be the way to reverse the direction of the force close to the horizon and make objects, which were falling in, to fly out. If this is the case, then it is a big step forward in showing that black holes do not contain a one-way membrane, and behave like any normal system in everyday physics.
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