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.
We will use Boyer-Lindquist coordinates along with the following definitions:
The specific charge e which will be the charge in units of distance be related to the standard charge q by:
What I will call the specific angular momentum "a", is sometimes used synonymously with rotation parameter. Rotation parameter is a term I will reserve 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:
also retain Eqn. 10.1.5
128 Chapter 11 Charged and Rotating Black Holes and Their Thermodynamics
and we define a by
The angular velocity w defines the angular velocity df/dt of locally nonrotating observers which are observers with a zero angular momentum parameter Lz as defined by eqn(11.1.43).
The Kerr-Newman or charged and rotating black hole invariant interval for general relativity is:
This can be equivalently expressed by
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:
And the contravariant metric tensor is given by:
Where we chose the index order as
x0 = ct
x1 = r
x2 = q
x3 = f
In the case that:
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:
The second type of horizon, called the inner event horizon is given by a distance :
These are simply the locations where D = 0.
For general relativity another mathematical surface of interest is called the static limit. It is a surface described by:
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
df/dt = w ± (Rc/v)
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 charged rotating black hole to include a nonzero cosmological constant simply by modifying the definition of a above to
a= 1 - r0/r + e2/r2 - (1/3)lr2
One can then change to coordinates appropriate for an observer at any finite r location.
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
mgcosmological = mc2(1/3)lr.
a= 1 - r0/r + e2/r2, The Kerr-Newman solution is the exact
Ricci-scalar = 0 solution to Einstein's field equations for the stress-energy of the electromagnetic field outside a collapsed spinning charge.
In the case that e = 0, the Kerr-Newman solution with
a= 1 - r0/r, becomes an exact vacuum solution for the region outside collapsed spinning matter called, "the Kerr black hole" solution and can be expressed as
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:
It is currently thought that no black hole will form under normal gravitational collapse without obeying the inequality
[Though fermions result 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
For the Reissner-Nordstrom charged but nonrotating J=0 black hole the complete exact equations of geodesic motion for a neutral test mass in the spacetime are
where g is a constant of the motion equal to the g of special relativity at r=¥ and is a constant of the motion that can be thought of as the test particle's angular momentum divided by its mass.
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 rTotc2 = gmnTmn 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 bm 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 an 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
Of these, the only nonzero G lmn for the case q = 0,p is G 001.
So Eqn 11.1.26 reduces to
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
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
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
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
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
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 uf = 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
t1orbit = [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 = ect, and F = ef.
The first gives us the following conserved quantity for geodesic motion called the energy parameter Econs for which we define g by, g = Econs/mc2.
gc = gmnTmUn
gc = g00T0U0 + g0fT0Uf
gc = g00(dct/dt) + g0f(df/dt)
This results in
The second killing four-vector results in another conserved quantity for geodesic motion called the angular momentum parameter, lz = Lz/m
- lz = gmnFmUn
- lz = gffFfUf + gf0FfU0
lz = - gf0(dct/dt) - gff(df/dt)
This results in
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 = (df/dt)|lz=0. In this texts notation, it is given by
where w is given by equation 11.1.8
These observers are simply those with Lz = 0 as defined above.
The angular velocity of the hole is defined as,
WH = ac/(r+2 + a2)
Finally lets solve for geodesic equatorial motion.
Using equations 11.1.42 and 11.1.43, it is easy to work out the time travel equation. This is
Likewise they can be used to work out
Notice here that when the black hole has spin, that is a is not zero, that things with zero angular momentum have nonzero angular velocity. This is frame dragging.
Since we are looking at equatorial motion, q=p/2 so w, v2, and these reduce to
and the line element reduces to
Dividing 11.1.52 by dt2 and inserting 11.1.50 and 11.1.51 and solving for (dr/dt)2 yields
So in terms of the mass and charge the equations of equatorial geodesic motion for a neutral test mass can be written
The polar equations of geodesic motion from earlier in this section can likewise be written
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?
Find the force felt by an observer hovering on the polar axis axis. Compare all regions.
Draw a picture of the horizons and the static limit according to a remote observer's perspective.
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.
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.
Two of the four metric terms for this spacetime were correct in a published paper by
Nese Ozdemir, Exact metric around a wiggly cosmic string 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.
(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.
kbT » Dpc
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
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
138 Chapter 11 Charged and Rotating Black Holes and Their Thermodynamics
The power radiated from the hole is
P = sT 4A
Where A is the surface area of the event horizon given by
A = 4pr02 = 4p(2GM/c2)2
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
The solution is
M0 = initial gravitational mass of the hole.
s= the Stefan-Boltzmann constant
kb = 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
We will define the letter s for this section by
Consider how it varies
Inserting the temperature expression and writing it in terms of the energy of the hole results in
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
WH = ac/(r+2 + a2)
The work term from Penrose energy extraction is WHdJ. The first law of thermodynamics equation for Kerr black holes then becomes
dE = Tds + WHdJ
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)
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 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.
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.
What is the entropy of a one solar mass Schwarzschild hole, and how long will it take to radiate away?
What is the expression for A for a black hole of zero charge and maximal spin?
11.3 Classically Radiative Black holes and Modeling Gravitational Collapse
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
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
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
But, no choices for the functions yield radiative vacuum solutions. At best we could reduce it to describing purely electromagnetic radiation as
akin to how equation 11.3.2 reduces to the Vaidya solution.
The exact stress energy tensor corresponding to equation 11.3.4 is
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.
These wave solutions each have cases that when expressed in x,y,z coordinates take a common family equation form. The metrics for those cases may be written as
where u is a distance coordinate along which the radiation flows and is the three components of a unit vector along that direction.
For example, in x,y,z the Vaidya solution with f=-Rc2/2r, u=r, can be written
Oppenheimer-Snyder Collapse was the first model for the collapse of a ball of chargeless pressureless matter using known exact solutions to Einstein's field equations for the exterior and interior of the matter. They merely met boundary conditions for the surface on the metric to connect the Schwarzschild solution for the exterior to the closed Friedmann solution for the interior to describe the collapse of the surface with the time dependant part of the Friedmann solution chosen to model pressureless matter. The Interior line element is given by
And the result for the Scharzschild radial coordinate radius of the ball of matter as a function of the time coordinate t" is
Schwarzschild time at the surface becomes singular as the surface approaches Rs=2GM/c2 and as such is not actually a good choice of coordinate for a remote observer trying to observe the collapse as the surface nears there, but the relation of the time coordinate t" to Schwarzschild time is
And the resulting Kerr-Schchild coordinate speed (v=dr/dtks) for the collapsing surface is then
This is a more meaningful end result using time appropriate for a remote observer which does not become singular at the horizon, a time reversal of the tw in the expression for Vaidya's solution and in problem 11.3.1, than one would arrive at using Schwarzschild time. See problem 11.3.2
Since a line element modeling a closed universe was used to model the matter undergoing the collapse the interior line element of course looks like the line element for a universe leading some to speculate that black holes are baby universes and that the formation of a black hole in another universe could have been the cause of our big bang.
The exact solution to Einstein's field equations for spherically symmetric electromagnetic radiation from a point source whose stress-energy is perturbed by the emission of a coupled spherically symmetric gravity wave from the point source is
This reduces to the Vaidya solution for a=1 and the gravity wave a can perturb the stress-energy all the way to a vacuum solution when it is defined so that the mass is proportional to its square,
What is interesting about this is that if you take the case that globally M=ma2, what you get is a frame transformation of the Schwarzschild solution. This is consitent with Birkoff's theorem, but in light of this expresion the question arises "where" is the mass.
Here is how the solution transforms to Schwarzschild's for M=ma2:
The line element can be written as
V = ct-r
a. Show that the time coordinate transformation given by
transforms equation 11.3.1 into the Schwarzschild solution for
R(r-ctw) = r0 = constant.
b. Find k for an observer using tw who is at some finite radial coordinate "a" when the times are synched at 0.
a. Show that when a remote observer is given, r = a » ¥ , ct and ctw standards of time for problem 11.3.1 come to run at the same rate, i.e. Limr® ¥ (dt/dtw) = 1, and note that the line element becomes
ds2 » dctw2 - dr2 - r2dW 2 there and thus ctw is just as valid a standard for a remote observer's time.
b. Seperate variables for equation 11.4.11 to find an integral expression for tks in terms of Rs. In terms of tks which is as appropriate a time for a remote observer as any for this problem, about how long does it take for the surface to collapse from Rs(0) to 2GM/c2 ? (tks = -tw)
c. Find the coordinate speed of outward moving light for the Kerr-Schild metric
ds2 = (1-2GM/rc2)dctks2 - (4GM/rc2)dctksdr -(1+2GM/rc2)dr2 - r2dW2 and discuss how long this implies it will take for the information about the matter crossing the horizon to actually reach the remote Kerr-Schild observer.
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