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Analysis of the Mechanics of Black Holes

The culmination of Stephen Hawking's analysis in the early 1970's of the physics of black holes was in the letter he published in the journal Nature. It is short enough to be quoted in its entirety.

Letters to Nature

Nature 248, 30 - 31 (01 March 1974)

Black hole explosions?


Department of Applied Mathematics and Theoretical Physics and Institute of Astronomy University of Cambridge

QUANTUM gravitational effects are usually ignored in calculations of the formation and evolution of black holes. The justification for this is that the radius of curvature of space-time outside the event horizon is very large compared to the Planck length (Għ/c 3)1/2 ≈ 10−33 cm, the length scale on which quantum fluctuations of the metric are expected to be of order unity. This means that the energy density of particles created by the gravitational field is small compared to the space-time curvature. Even though quantum effects may be small locally, they may still, however, add up to produce a significant effect over the lifetime of the Universe ≈ 1017 s which is very long compared to the Planck time ≈ 10−43 s. The purpose of this letter is to show that this indeed may be the case: it seems that any black hole will create and emit particles such as neutrinos or photons at just the rate that one would expect if the black hole was a body with a temperature of (κ/2π) (ħ/2k) ≈ 10−6 (M/M)K where κ is the surface gravity of the black hole1. As a black hole emits this thermal radiation one would expect it to lose mass. This in turn would increase the surface gravity and so increase the rate of emission. The black hole would therefore have a finite life of the order of 1071 (M/M)−3 s. For a black hole of solar mass this is much longer than the age of the Universe. There might, however, be much smaller black holes which were formed by fluctuations in the early Universe2. Any such black hole of mass less than 1015 g would have evaporated by now. Near the end of its life the rate of emission would be very high and about 1030 erg would be released in the last 0.1 s. This is a fairly small explosion by astronomical standards but it is equivalent to about 1 million 1 Mton hydrogen bombs.

The symbol M represent the mass of the Sun and (M/M) would be the mass of the black hole relative to the mass of the Sun.


The work of Hawking and his associates presumes that a black hole is a singularity; in effect, a point particle. There are many considerable problems involved with point particle singularities, such as the following:

For the first item above consider a stationary electron. The electrical field intensity E at a distance r from the center of the electron is given by

E(r) = (1/(4πε0))e/r²

where ε0 is a constant called the permitivity of space and e is the charge of the electron. For an electron in empty space the energy density U is

U = ½ε0E² = (1/(32π²ε0))e²/r4

The energy in a spherical shell of radius r and thickness dr is equal to 4πr²(e²/(32π²r4) which reduces to e²/(8πε0r²). The integration of these terms from R to ∞ gives

ER = ∫RU(4πr²dr = e²/(8πε0)/R

As R→0, ER → ∞. Thus a charged point particle, if such existed, would have infinite energy. And likewise for the gravitational field of a point mass, althoug there is some debate on this matter. Since mass is attracted to mass, in contrast to charge which is repelled by like charge, the gravitational field of a point mass would have an energy of negative infinity.

Conventionally singularity is another term for a point particle with charge increasing without bound as distance goes to zero.

The theorems which Stephen Hawking and Roger Penrose prove are in the nature of showing that singular solutions are compatible with Einstein's field equations. In other words, Einstein's equations do not prohibit sinularities. But that does not mean singularities exist; they may be prohibited by other considerations such as their infinite energies.

Further consideration of the matter of singularity of black holes is at Singularity of Black Holes.

The Metric Tensor

The metric tensor is given by the elements gμν. They give the interval in spacetime. For example in ordinary space and time the interval is given by

ds² = dt² − (dx²+dy²+dz²)

This means that g0,0=+1 and g1,1=g2,2=g3,3=−1 and gμ,ν=0 for μ≠ν. This presumes that the dimensions of space and time are such that the speed of light is equal to unity.

The Schwarzschild solution for the space outside of a black hole gives the interval as

ds² = −Hdt² + H−1dr² + r²(dθ² + sin²(θ)dφ²)

where (r, θ, φ) are the spherical coordinates and

H = 1 − (2mγ/r)

In this material it is assumed that the natural system of units is used; i.e.,, the one such that c, the speed of light, is equal to unity and h-bar, Planck's constant divided by 2π is also equal to unity. Later the important formulas will be given in the conventional system of units.

The event horizon radius is the radius which make H=0 and thus the coefficient of dr², H−1, goes to infinity; i.e., RH=2mγ.


The analogue of thermodynamic temperature in the mechanics of black holes is the gravitational field intensity κ at the event horizon; i.e.,

κ = γm/RH² = ½/RH
but RH=2mγ so
κ = 1/(4mγ)


Initially it was believed that black holes had zero entropy but then some theorists recognized that such would imply a violation of the second law of thermodynamics because when a physical system of positive entropy disappears into a black hole the entropy of the Universe would decrease.

In thermodynamics energy E, temperature T and entropy S are related by the relation

dE = TdS

Let B denote the analogue of entropy in the mechanics of black holes. In a system in which c=1, E=mc² reduces to E=m and dE to dm.

The relation analogous to the thermodynamic dE=TdS is then

dm = κdB

When this relation is multiplied by m and κ is replaced by 1/(4mγ) the result is

mdm = (1/(4γ))dB

When this is integrated the result is

½m² = (1/(4γ))B
and thus
B = 2γm²

However m is equal to RH/(2γ) so

B = RH²/(2γ)

The area A of the event horizon is 4πRH² so the entropy of a black hole is proportional to the area of its event horizon. More specifically

B = A/(8πγ)

The scale of entropy is arbitrary so the entropy of a black hole could be said to be equal to the area of its event horizon. This is a surprise because in the thermodynamics of ordinary objects entropy depends upon their volumes.

If two black holes merge then their masses add, but this means

(m1+m2)² > m1² + m2²
and therefore
A1∪2 > A1 + A2
and furthermore
B1∪2 > B1 + B2

Thus the analoge of the second law of thermodynamics is satisfied.

In the conventional system of units

B = kBA/lP²

where kB is Boltzmann's constant and lP is the Planck length. The Planck length is gien by

lP = (γhc3)½

Gravitational intensity κ at the event horizon is given by the formula

κ = c4/(4γm)

The Hawking temperature TH of a black hole is given by

TH = κ/(2π) = hc3/(8πkBγm)

where kB is Boltzmann's constant,

Electrostatic Charge in Black Holes

There is a great amount of difference in the magnitude of the force between charged particles due to their electrical charges and their masses. Consider the case of two electrons:

FE/FG = [ηe²/r²]/[γm²/r²] = (η/γ)(e/m)²
≅ 1042

It does not take much in the way of charge to overwhelm the graitational attraction between masses and prevent the formation of a black hole. The limit on the charge Q of a black hole is given by:

|Q | < mγ½

Most of the mass of a black hole must come from electrically neutral entities such as neutrons, hydrogen atoms or plasma in which the nuclei and electrons of atoms are separated.

Angular Momentum

The angular momentum L of a system is defined as

L = Jω

where J is the moment of inertia of the system and ω is its angular rate of rotation in radians per unit time. The moment of inertia of a spherical ball is equal to (2/5)mr² where m is its mass and r is its radius. A point particle has a radius of zero and so no matter how fast is its rate of rotation its angular momentum is zero.

(To be continued.)

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