Chapter 11

Return to Modern Relativity

11.1 The Kerr-Newman Solution

We will here display the solutions for the invariant intervals for charged and rotating black holes for general relativity.

Let

The specific charge e be related to the charge q by:

(11.1.1)

Even though it shouldn't be, the specific angular momentum "a" is sometimes used synonymously with rotation parameter. Rotation parameter really should be reserved only for the quantity ac²/GM. The specific angular momentum "a" will be related to the angular momentum J and gravitational mass of the hole M by:

(11.1.2)

also retain Eqn. 10.1.5

127 

128 Chapter 11 Charged and Rotating Black Holes and Their Thermodynamics

and we define a by

(11.1.3)

We will use these in Boyer-Lindquist coordinates:

(11.1.4)

(11.1.5)

along with the following definitions

(11.1.6)

(11.1.7)

(11.1.8)

The angular velocity w defines the angular velocity df/dt of locally nonrotating observers which are observers with a zero angular momentum parameter L_z as defined by eqn(11.1.43).

The Kerr-Newman or charged and rotating black hole invariant interval for general relativity is:

(11.1.9a)

This can be equivalently expressed by

(11.1.9b)

where R is the gravitational red shift factor.

R is given by

Consider the frequency of light observed remotely f ' from emitted with a local proper frame frequency f₀ from a locally nonrotating source also constant with respect to r and q . The light will be gravitationally red shifted by

f ' = Rf₀

11.1 The Kerr-Newman Solution 129

The covariant metric tensor for the Kerr-Newman spacetime of general relativity is given by:

(11.1.10)

And the contravariant metric tensor is given by:

(11.1.11)

Where we chose the index order as

x⁰ = ct

x¹ = r

x² = q

x³ = f

In the case that:

(11.1.12)

there will be more than one horizon.

130 Chapter 11 Charged and Rotating Black Holes and Their Thermodynamics

The outer event horizon for a rotating charged black hole is not defined by the Schwarzschild radius r₀ . The outer event horizon distance r₊ is given by:

(11.1.13) 

The second type of horizon, called the inner event horizon is given by a distance :

(11.1.14)

For general relativity another mathematical surface of interest is called the static limit. It is a surface described by:

(11.1.15)

Remote observers can observe an object cross through the static limit, but they can't observe it cross the event horizon unless the event horizon is changing size. The region between the static limit and the event horizon is known as the ergosphere. Objects in this region must rotate with the Kerr black hole.

The static limit equation for general relativity can be derived as follows. Consider light instantaneously moving in the ± f direction. Eqn 11.1.9b becomes

0 = R²dct² - v ²(df - wdt)²

This yields

df/dt = w ± (Rc/v)

(11.1.16)

Everything is constrained to move at an angular velocity within these boundaries of the coordinate angular speed of light. The static limit occurs where the coordinate speed of light moving opposite the direction of rotation is at zero coordinate velocity. This results in

0 = w - (Rc/v)

Solving for r,

wv/c = R

w²v²/c² = R²

w²v²/c² = [(D - a²sin²q)/r²] + w²v²/c²

D - a²sin²q = 0

a² + r²a - a²sin²q = 0

r²a + a²(1 - sin²q) = 0

r²a + a²cos²q = 0

r² - r₀r + e² + a²cos²q = 0

(1/2)r² - (GM/c²)r + (1/2)(a²cos²q + e²)

The quadradic equation solving for r then yield Eqn 11.1.15

One can easily modify the Kerr-Newman spacetime for general relativity for a model approximation of a charged rotating black hole in a universe with a nonzero cosmological constant simply by modifying the definition of a Eqn.11.1.3. page 128. Modify a to

a = 1 - r₀/r + e²/r² - (1/3)lr² + r₀/R + (1/3)lR² - e²/R²

(11.1.17)

In the case of zero spin, this Kerr-Newman-DeSitter spacetime corresponds to a choice of coordinates for an observer hovering at r = R.

The mass term corresponds to an inverse distance squared attraction. The charge term corresponds to a inverse distance cubed repulsion. And the cosmological term corresponds to a spring-like force which if l is positive is repulsive given by

mg_cosmological = mc²(1/3)lr.

 (11.1.18)

The Kerr black hole solution (e = 0) for an uncharged but rotating hole for general relativity can be more simply expressed as

 (11.1.19)

11.1 The Kerr-Newman Solution 131

A Reissner-Nordstrom black hole is a charged but non-rotating hole J = 0.

The Reissner-Nordstrom solution for the invariant interval of general relativity is more simply expressed:

(11.1.20)

 It is currently thought that no black hole will form under normal gravitational collapse without obeying the inequality

(11.1.21)

[Though fermions have been modeled by a Compton Radius Vortex see- http://www.innerx.net/personal/tsmith/Sidharth.html , which is a Kerr-Newman space-time that results in a union of relativity and Bohmian Mechanics and for which in the case of fermions results in a² + e² > (GM/c²)²]

Next lets derive the geodesic motion of a neutral particle falling along the polar axis of a Kerr-Newman black hole with a static metric in general relativity. It will be seen that spin and charge each give rise to repulsion. It will also be seen that if the black hole has only mass and electric charge, but no spin there will exist a gravitational repulsion due to the charge that will throw the particle back out past both horizons in a finite proper time.

The equation for geodesic polar motion of a neutral particle in a Kerr-Newman black hole space-time for general relativity is

And the differential time travel equation relating proper time to remote frame time is

Where g is an integration constant usually known by the energy parameter.

132 Chapter 11 Charged and Rotating Black Holes and Their Thermodynamics

For a Kerr-Newman black hole for general relativity we have 11.1.3,

Eqn. 11.1.22 becomes

And for general relativity for a Reissner-Nordstrom black hole J = 0 Eqn. 11.1.24 becomes

Since it is currently thought that the charge, mass and spin are the only distinguishing characteristics of a static state black hole Eqn. 11.1.3 can be taken as the definition of a . However, one might consider the introduction of massive fields (massive in that r_Totc² = g_mnT^mn ¹ 0) where a is not restricted to these. There are problems related to the stress energy tensors of such spacetimes for a Universe with a single hole, but in considering the possibility of a hole with other such fields, Eqn 1a and 1b would still obtain. However, for a expressed as a Laurent series truncated for b_m such that m > 2 results in a point physical singularity in Eqn 1a at the origin instead of a ring singularity. Also, for m > 2, the sign of the last term in the series will determine whether the neutral particle in free fall will encounter a final gravitational attraction or repulsion as it approaches r = 0. If a Taylor series is added to a and the series is truncated at any a_n n > 0, equation 1b vanishes at infinity. Since we typically put our remote observers there, such a generalization should demand a be normalized to 1 at infinity. Therefore, in any addition of a Taylor series to the definition of a must either not be truncated or we must not be physically able to put remote observers there.

The invariant interval in this metric is Eqn. 11.1.9

11.1 The Kerr-Newman Solution 133

Since we consider here only polar motion, dq = 0, q = 0 or p , q (r) = 0 and v(r) = 0, . Eqn. 11.1.9 thus reduces to

ds = dct , so this can be rewritten as

From here all that is needed in order to find the equation of motion is to find dt/dt . The geodesic equation is used to find this. We use the geodesic equation in the form Eqn. 5.2.8.

x⁰ = ct, so the one of these that has the information we want to replace dt/dt in Eqn 11.1.27 is

Written out in terms of our coordinates and recalling that dq = 0 for polar motion, this is

(11.1.29)

Of these, the only nonzero G ^l_mn for the case q = 0,p is G ⁰₀₁.

So Eqn 11.1.26 reduces to

(11.1.30)

134 Chapter 11 Charged and Rotating Black Holes and Their Thermodynamics

The equation for the affine connection for general relativity is

Referring to the covariant metric tensor we see that

g₀₁ = g₃₁ = g₁₀ = 0. Also for q = 0 or p, g₃₀ = 0 at any r and so g₃₀,₁ = 0 so Eqn 11.1.31 reduces to

Inserting this into Eqn 11.1.30 results in

Simplifying:

Now we notice that for q = 0, p , g⁰⁰ = r²/D = 1/g₀₀ so Eqn 11.1.34 becomes

Separation of variables, integration replacing g⁰⁰ with r²/D and simplification results in

The g is a constant from the integration known as the energy parameter.

 11.1 The Kerr-Newman Solution 135

For q = 0,p this can also be written as Eqn.11.1.23

Inserting Eqn 11.1.23 into Eqn 11.1.27 results in

For q = 0,p Eqn 11.1.37 can also be written

Differentiating Eqn 11.1.38 with respect to t and use of the chain rule and simplification results in

Placing in and r₀ = 2GM/c² for the Kerr-Newman Black hole of general relativity and doing the differentiation with respect to r results in

(11.1.24)

136 Chapter 11 Charged and Rotating Black Holes and Their Thermodynamics

For a Reissner-Nordstrom black hole J = 0 so a = 0 so this becomes

Lets consider a state of extreme charge where and e » GM/c². Both of the event horizons coincide at

r_± = GM/c²

(11.1.39)

and equation 4 becomes

The inverse distance cubed repulsion dominates over the inverse distance squared attraction for small r. It begins to dominate where d²r/dt² = 0. This occurs at

r = GM/c².

This distance is also where the event horizons occur. Since the repulsion doesn't become dominant until the particle reaches the location of the event horizons, the particle must fall past both outer and inner horizons. According to the equation of motion the particle is then thrown back out.

To recap, we have seen that the presence of charge in a black hole leads to a repulsion in the gravitational field that dominates over the attraction due to mass at small distance from the center. And we have derived the geodesic motion for a neutral particle falling along the polar axis. At a small enough distance the spin's effects on gravitation dominate over both attraction due to mass and repulsion due to charge. And we have seen that in the absence of spin, the particle's geodesic motion leads back out of the hole in a finite proper time.

Next consider circular motion about a Schwarzschild black hole of general relativity. In Newtonian orbital mechanics, Kepler's third law relates the period of an orbit to the semi-major axis (equal to the mean distance from the focus) by

T ² = (4p²/GM)a³

(11.1.41a)

Consider a circular orbit of a test particle in a Schwarzschild spacetime for general relativity. If r is the radial coordinate and T is the coordinate time according to a remote observer, it is interesting that Kepler's equation

T ² = (4p²/GM)r³

(11.1.41b)

Holds valid at all distances from the hole. However, orbits below the photon sphere radius would require an orbital speed greater than the remote coordinate speed of light which means that there are no circular orbits beneath r = 3GM/c². There are a few different ways to go about obtaining Kepler's equation from the general relativistic case. One way is to insert the affine connections (equations 10.4.13) into equation 10.4.12, use u^f = 2p/T, and simplify. Here is another way which makes use of the principle of maximal proper directly. For equatorial motion

ds² = (1 - r₀/r)dct² - r²df ²

Using 2p/T = df/dt, and ds = dt , this becomes

dt = [1 - r₀/r - r²(2p/T)²/c²]^1/2dt

Integration over a single full orbit results in

t_1orbit = [1 - r₀/r - r²(2p/T)²/c²]^1/2T

To maximize the proper time for a fixed orbital time one merely needs to take the derivative of [1 - r₀/r - r²(2p/T)²/c²]^1/2 with respect to r and set it equal to zero. Using r₀ = 2GM/c² and doing this results in

(2GM/r²c² - 2r(2p/T)²/c²)(1/2) / [1 - r₀/r - r²(2p/T)²/c²]^1/2 = 0

Simplify and it finally becomes the above mentioned form of Kepler's third law eqn 11.1.37b.

T ² = (4p²/GM)r³

Next lets derive a couple of constants of geodesic motion for a Kerr-Newman spacetime.

The metric is independent of both t and f yielding two killing four-vectors right off the bat

T = e_ct, and F = e_f.

The first gives us the following conserved quantity for geodesic motion called the energy parameter E_cons for which we define g by, g = E_cons/mc².

gc = g_mnT^mUⁿ

gc = g₀₀T⁰U⁰ + g_0fT⁰U^f

gc = g₀₀(dct/dt) + g_0f(df/dt)

This results in

(11.1.42)

The second killing four-vector results in another conserved quantity for geodesic motion called the angular momentum parameter, l_z = L_z/m

- l_z = g_mnF^mUⁿ

- l_z = g_ffF^fU^f + g_f0F^fU⁰

l_z = - g_f0(dct/dt) - g_ff(df/dt)

This results in

(11.1.43)

Within the static limit one must rotate with the hole and so there are no static observers beneath this limit. General Relativity by Robert M. Wald defines the closest state to a nonrotating observer which he calls a "locally nonrotating observer" by those with a coordinate angular velocity of W. In this texts notation, it is given by

W = w

(11.1.44)

where w is given by equation 11.1.8

These observers are simply those with L_z = 0 as defined above.

The angular velocity of the hole is defined as,

W_H = ac/(r₊² + a²)

(11.1.45)

Exercises

Problem 11.1.1

Multiply Eqn.11.1.44 by m, the mass of the test particle. Compare this to the Newtonian conservation of energy equation. Which term represents kinetic energy? Which represents total Newtonian conserved energy? Which term corresponds to potential energy?

Problem 11.1.2

Find the force felt by an observer hovering on the polar axis axis. Compare all regions.

Problem 11.1.3

Draw a picture of the horizons and the static limit according to a remote observer's perspective.

Problem 11.1.4

Use equations 11.1.42 and then 11.1.43 to solve for dt/dt and df/dct and use the result in 11.1.9 to write the a general equation for geodesic motion, then reduce it to equatorial motion for q = p/2.

Problem 11.1.5

This vacuum solution was discovered by me, David Waite, Oct 26,2003

a. Show that the Einstein tensor for the following spacetime is zero

ds² = [dct²/(1+x/X)⁴] - [dx²/(1+x/X)⁸]- [(1+x/X)²dy²] - [dz²(1+x/X)⁴]

to verify that therefor it is a vacuum field solution.

b. Solve for the Riemann tensor.

c. Note that the metric is independent of three coordinates and use that fact to obtain three independent Killing vectors.

d. Use the results of part c and the equation for ds² to completely solve for geodesic motion.

Problem 11.1.6

Two of the four metric terms for this spacetime were correct in a published paper by Nese Ozdemir:

http://www.arxiv.org/abs/gr-qc/0010043

Gen.Rel.Grav. 33 (2001) 603-606

The other two would not be correct given an incorrect condition in the paper relating two constants. Corrections yielding this vacuum field solution were worked out by me, David Waite.

.

Where "a" and "b" are related by

- 4b² - 4b(2a + m + T) + (m - T)² = 0

Note how this reduces to the metric of special relativity at r=R.

The paper claims this is a cosmic string solution. In actuality it is a domain wall. Take

f = y/R, r/R = 1 + x/R and you'll see its actually a plane. I present my correct exact charged cosmic string solution in section 1 of chapter 7, and the exact solution for an infinite line of negligible mass and charge radiating electromagnetic radiation radially in section 3 of this chapter.

a. Calculate the Riemann tensor and use Einstein's field equations to calculate from this the Einstein tensor and then from this get the stress-energy tensor.

b. Note that the metric is independent of three coordinates and use that fact to obtain three independent Killing vectors.

c. Use the results of part b and the equation for ds² to completely solve for geodesic motion.

Problem 11.1.7

(This vacuum field solution was discovered by me, David Waite on Oct 29, 2003)

Show that the more general vacuum field solution in the form

ds² = [(1+x/X)^a]dct² - [(1+x/X)^f]dx² - [(1+x/X)^g]dy² - [(1+x/X)^b]dz²

has the following conditions.

b² + bg -2b - fb - fg + g² - 2g = 0

a² + ag -2a - fa - fg + g² - 2g = 0

ab + ag + bg = 0

- 2b - 2a + b² + a² + ab - fa - fb = 0

This solution contains within it each of the above two problem's spacetimes.

11.2 Hawking Radiation 137

  The vacuum of space is not actually empty but is in continuous production and annihilation of particle/anti-particle pairs. A photon is its own anti-particle and there is always a possibility that two photons with opposite four-momentum will form. At this point it is useful to recall the relation between energy and frequency. One photon will have a positive energy and due to conservation of four-momentum the other photon would have negative energy/frequency. Normally the negative energy/frequency photon can only exist for a small time given by a quantum mechanics energy-time uncertainty principle within which these will annihilate with each other shortly after creation. This negative frequency would be considered positive and thus the energy would be observed as positive if we were to do a time reversal. Recall that for an observer in free fall, there is a time reversal at the horizon of a black hole with respect to external observers. Sometimes these photons will form in the region of the event horizon of a black hole. In which case the positive energy photon might escape in the same instance that the negative energy/frequency photon becomes trapped by the hole. Because of the time reversal, in the internal region it is the remotely negative energy/frequency photon that has positive energy and positive energy and frequency according to local free fall observers. Thus the remotely negative energy/frequency photon has a chance of becoming absorbed and behaving real so that it does not immediately annihilate due to the energy-time uncertainty principle. The absorption of the negative energy/frequency photons have the effect of decreasing the gravitational mass of the hole, and the positive energy photons have the effect of giving the black hole an apparent temperature.

A simple way to approximate this temperature is to recognize that the thermal energy is proportional to the momentum of the photons.

k_bT » Dpc

(11.2.1)

Use the Heisenburg uncertainty principle, DpDr = h/2p . Then recognize that Dr will be on the order of the event horizon radius. Dr » 2GM/c². Putting these together results in

(11.2.2)

Hawking showed that the exact solution for this is that a Schwarzschild black hole of general relativity acts as a black body radiator with a temperature of

(11.2.3)

138 Chapter 11 Charged and Rotating Black Holes and Their Thermodynamics

The power radiated from the hole is

P = sT ⁴A

(11.2.4)

Where A is the surface area of the event horizon given by

A = 4pr₀² = 4p(2GM/c²)²

(11.2.5)

Most scattering actually occurs far enough outside the hole so that the exact physical optics is needed to accurately describe the wave propagation where geometric optics approximation fails. As a result the use of Stephan's law here is really only an order of magnitude approximation. But using it we go on and use this power radiated as the rate of change in the energy contained by the hole. P = d(Mc²)/dt This leads to a separable differential equation for the mass of the hole as a function of time

(11.2.6)

(11.2.7)

The solution is

(11.2.8)

M₀ = initial gravitational mass of the hole.

(11.2.9)

s = the Stefan-Boltzmann constant

k_b = the Boltzmann constant

11.2 Hawking Radiation 139

From these we notice that the temperature of the hole diverges as it radiates away its mass to the Hawking radiation. Because of this divergent temperature, some speculate that the black hole may end in a final burst of energy at some critical mass just prior to the evaporation time

(11.2.10)

We will define the letter s for this section by

(11.2.11)

Consider how it varies

(11.2.12)

(11.2.13)

Inserting the temperature expression and writing it in terms of the energy of the hole results in

(11.2.14)

140 Chapter 11 Charged and Rotating Black Holes and Their Thermodynamics

But it turns out that this is the first law of thermodynamics in the absence of work terms as long as s is the entropy of the hole. Therefor black holes have an entropy related to the surface area of the event horizon given by Eqn.11.2.11

According to Penrose energy extraction, energy can be extracted from a rotating hole by reducing its spin. Likewise it can be done by reducing its charge. Any variation in these must be included as the missing work terms in the first law of thermodynamics equation. If we consider a Kerr black hole without charge, then we can include variation in spin in the following way. The angular velocity of the hole can be defined as eqn 11.1.41

W_H = ac/(r₊² + a²)

The work term from Penrose energy extraction is W_HdJ. The first law of thermodynamics equation for Kerr black holes then becomes

dE = Tds + W_HdJ

(11.2.15)

Here the entropy is also related to the area in the same way, but for the Kerr hole, the area is given by

A = 4p(r₊² + a²)

(11.2.16)

Note that this is not the area of the outer horizon.

Another difference is that for a Kerr-Newman or charged and rotating hole of general relativity, the temperature is related to the mass charge and angular momentum. It is given by

(11.2.17)

11.2 Hawking Radiation 141

There is another paradigm for the origin of Hawking radiation that has equivalent results. Energy in frame dependent. Instead of thinking of the negative frequency trapped virtual particles as having negative energy, one may consider the time reversal on in-falling matter in accordance with the sign reversal of Eqn. 11.1.20 under the event horizon. Choosing to look at it from the appropriate frame, the energy of the external photon as well as the trapped photon both have positive energy. The separation of one from the other can be thought of as an effect of the gravitational fields tidal gradient and the work done against the external photon as it moves off to infinite separation is done by the gravitational field of the hole which has the end result of a decrease in the gravitational mass of the hole.

Problem 11.2.1

Show for e = 0 that in the limit as "a" tends to GM/c² the temperature tends to zero. Then show that for a = 0, in the limit as e tends to GM/c² the temperature tends to infinity.

Problem 11.2.2

What is the entropy of a one solar mass Schwarzschild hole, and how long will it take to radiate away?

Problem 11.2.3

What is the expression for A for a black hole of zero charge and maximal spin?

_________________________________________________________________________

11.3 Classically Radiative Black holes

The previous sections modeling of a process for radiating energy from black holes was a quantum mechanical process theoretically proposed by Stephen Hawking. What is less well known is that classically radiative black hole solutions exist. The first exact solution to Einstein's field equations discovered for a radiative black hole is the Vaidya solution

(11.3.1)

 This solution has a zero Ricci-scalar corresponding to massless radiation it emits as it radiates its mass. Generalizing to include charge I have found that

(11.3.2)

also exactly has a zero Ricci-scalar even for arbitrary angular dependence of the functions it contains.

I point to my finding concerning this solution for matter corresponding to a zero Ricci-scalar here because it surprised me that once I was able to find a radiative metric with a zero Ricci-scalar with arbitrary angular dependence on the functions it contained, I found one can not set the behavior of the functions to yield a radiative vacuum solution. I was expecting it to yeild a gravity wave solution for some choice of angular dependence of the functions. It was obvious that the Vaidya solution will not yield gravity waves independent of its electromagnetic radiation because of its spherical symmetry. The Birkoff theorem states that a spherically symmetric vacuum solution is the Schwarzschild solution in some choice of coordinates. Due to Vaidya's spherical symmetry, the case that makes it vacuum also reduces it to a frame transformation of the Schwarzschild solution. However, I was expecting that should I find such a solution that allowed massless radiation corresponding to a zero Ricci-scalar, but allowed arbitrary angular dependence and kept the Ricci-scalar zero, it would yeild vacuum cases that would correspond to gravity waves. It did not.

Note the expression for the exact solution for pp waves equation 9.1.34

ds² = [1 + h(x,y,z-ct)]dct² - 2hdctdz - [1 - h(x,y,z-ct)]dz² - dy² - dx²

and the similarity it bears to the above radiative black hole solutions.

In a similar effort I have likewise found that the radiation field of an infinite radiative line of negligible mass and charge has a zero Ricci-scalar corresponding to massless radiation described by the line element

ds²=[1-f(z,theta,r-ct)+(g(z,theta,r-ct)/r)]dct²+2[f(z,theta,r-ct)-(g(z,theta,r-ct)/r)]dctdr-dz²-[1+f(z,theta,r-ct)-(g(z,theta,r-ct)/r)]dr²-r²dq²

(11.3.3)

But, no choices for the functions yield radiative vacuum solutions. At best we could reduce it to describing purely electromagnetic radiation as

ds²=[1-f(r-ct)]dct²+2[f(r-ct)]dctdr-dz²-[1+f(r-ct)]dr²-r²dq²

(11.3.4)

akin to how the metric I found regarding the radiative black hole reduces to the Vaidya solution.

The exact stress energy tensor corresponding to equation 11.3.4 is

(11.3.5)

where u = r-ct.

As a result of these findings, I conjecture that gravity waves do not exist in nature independent of coupled electrimagnetic radiation. I suspect our gravity wave detectors have been getting null results because the matter containing them shields them from electromagnetic radiation and thereby shield them from any coupled gravity waves.

Problem 11.3.1

Show that the time coordinate transformation given by

transforms equation 11.3.1 into the Schwarzschild solution for

R(r-ct_w) = r₀ = constant.

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