>> Up to this point in these lectures,
I've only discussed black holes
and gravitational collapse using the
idealization of spherical symmetry.
In this lecture, I want to talk about
black holes in my general terms.
Let's begin by describing the geometry
of a rotating black hole in vacuum.
[ Writing sounds ]
This is the Kerr black hole
or the Kerr solution.
[writing sounds] We can write the
metric and coordinates that are referred
to as Boyer-Lindquist coordinates.
[ Writing sounds ]
And the coordinates are t,r,theta and phi.
In these coordinates, the metric is minus 1
minus 2mr over rho squared, and I'll define rho
in a minute, times dt squared minus 4a,
I'll define a in a moment times
r sine squared theta divided
by rho squared d phi dt plus
rho squared over delta,
I'll define delta in a moment times dr squared
plus rho squared d theta squared plus r squared
plus a squared plus 1 over rho squared
times 2mr a squared sine squared theta,
all of that times sine squared
theta d phi squared.
Now, rho squared, which appears in several
places in this metric, in this line element,
is defined as r squared plus a
squared cosine squared theta.
And delta, which appears here, is defined
as r squared minus 2mr plus a squared.
Now, these other parameters you see are
m, which is the mass of the black hole,
and this parameter a, which
is defined as j over m
where j is the angular momentum
of the black hole.
[ Writing sounds ]
[ Silence ]
One thing we can check right away is the special
case of which the angular momentum vanishes.
All right, this should be a
non-rotating black hole in that case.
You'll notice that if j is 0,
then this parameter a is 0,
the angular momentum per unit mass is 0 and that
means rho squared is just equal to r squared
and delta is equal to r squared
minus 2m times r. So,
this coefficient of dt squared is
just minus, 1 minus 2m divided by r,
that's the same coefficient of dt squared
that we have the Schwarzschild
metric and Schwarzschild coordinates.
This coefficient goes to 0 because a is 0.
This coefficient is r squared divided by
r squared minus 2mr, so that's 1 divided
by 1 minus 2m over r, the
same as the grr component
of the Schwarzschild black hole
and Schwarzschild coordinates.
This is r squared d theta squared and
finally here, we have these terms vanish,
so this is r squared sine squared d phi squared.
So, you can see that when j is equal to 0,
the Kerr black hole is just
a Schwarzschild black hole.
[writing sounds] In particular, the Kerr metric
and Boyer-Lindquist coordinates just
becomes the Schwarzschild metric
and Schwarzschild coordinates.
So, how do we know that these
parameters, m and j, represent the mass
and angular momentum the black hole?
[ Writing sounds ]
So, how do we identify m and j
as mass and angular momentum?
Well, we can identify m as the mass the same
way we did for the Schwarzschild black hole,
by releasing a satellite near infinity
in the asymptotically flat region
and measuring its orbital period,
then applying Kepler's third law.
But identifying the angular
momentum is a bit more difficult.
What we can do is examine the metric far
outside a weakly gravitating body and show
that it has the same form as the Kerr
metric in the r goes to infinity limit.
[writing sounds] So, we examine the
metric outside a weakly gravitating mass
or weakly gravitating body and
we show that it has the same form
as the Kerr geometry [writing sounds] in
the asymptotic region as r goes to infinity.
When we do this, when we relate the
Kerr geometry and the asymptotic region
to the metric outside a weakly gravitating
body, we find that the Kerr parameter m is given
by the integral of t00 integrated over
all space, so this is the 00 component
of the stress energy momentum tensor
for the weakly gravitating body and it's
in a nearly Newtonian frame, so and
we're in the rest frame of the body,
so this is the energy density of the body.
So, the integral of the energy density is
the total energy of a total mass of the body.
And in a similar way, we can identify the Kerr
parameter j [writing sounds] as the integral
of x times t 0y minus y times t 0x, and here
x and y refer to the Cartesian coordinates
in the nearly Newtonian frame of this weakly
gravitating body integrated overall space.
So, this in, in this frame, t 0y is just
the momentum density in the y direction,
this is the momentum density in the x direction,
so this is just the total
angular momentum of the body.
And that's how we can identify m and j as the
mass and angular momentum by their effects
on the geometry in the asymptotic region.
Now let me point out some of the
key features of the Kerr geometry.
You'll notice that in Boyer-Lindquist
coordinates,
the metric components are
independent of the coordinate t,
and they're independent of the coordinate phi.
So, that means we have two Killing
vector fields, d by dt and d by d phi.
[ Writing sounds ]
So, one of the Killing vector fields is
d by dt and this vector field is time
like outside the black hole, so this corresponds
to a time translation symmetry and shows us
that the black hole is stationary but in fact
is not static as the Schwarzschild black hole is
because this vector field is
not hypersurface orthogonal.
The other Killing vector field
we identified was d by d phi,
so this tells us that the Kerr black hole
has an axisymmetry, in other words a symmetry
under rotation is about what we
sometimes refer to as the z axis.
But it's not spherically symmetric
like the Schwarzschild black hole.
[ Writing sounds ]
So, these two Killing vector fields, which we
easily identified from the Kerr metric written
in Boyer-Lindquist coordinates, these are
in fact the only two Killing
vector fields of the Kerr geometry.
One of the physical consequences
of a black hole's rotation is
that it drags the inertial frames
in the direction of rotation.
[ Writing sounds ]
So, recall that an inertial frame is a
reference frame of a freely falling observer.
So, let's consider an observer who
falls freely from r equal infinity.
[ Writing sounds ]
From r equals infinity with no angular motion
initially towards the rotating black hole.
[ Writing sounds ]
We can solve for the motion of this
observer by solving the geodesic equations.
[writing sounds] So, let x mu dot
be the 4 velocity of the observer,
that's t dot r dot theta dot phi dot
or dot denotes differentiation
with respect to proper time.
So, this is the 4 velocity and let's restrict
the motion to the theta equal pi over 2 plane.
[writing sounds] So, then in
particular theta dot would be 0
and now because we have two Killing vector
fields, namely d by dt and d by d phi,
we have two constants of the motion.
[ Writing sounds ]
Namely, what we'll call little e, which
is the inner product of d by dt and x dot
and the 4 velocity and little l, which is the
inner product of d by d phi and the 4 velocity.
So, these are first integrals
of the geodesic equations
and we can evaluate these constants anywhere
along the geodesic, since they're constant.
So, in particular, if we look initially at
the initial time the observer is at rest
at r equal infinity, so r dot and phi dot are
0 and t dot is just 1 because at infinity,
the Kerr geometry's asymptotically flat
and the Boyer-Lindquist coordinate's
just reduced to Minkowski coordinates.
So, in particular, the Boyer-Lindquist
time coordinate t,
coincides with the proper time
for an observer who's at rest.
So, t dot is 1 initially.
So, let's use this initial data to
evaluate e and l. First for e, we have,
it's the inner product of d by dt, that's
the vector with components 1, 0, 0, 0.
And the inner product uses the metric, which
at r equal infinity is just a Minkowski metric,
minus 1, 1, 1, 1, down the diagonals.
In an x dot, initially is just 1, 0, 0, 0.
So, this is minus 1.
E is minus 1.
And now for l, we have vector d by d phi, which
has components 0, 0, 0, 1 times the metric
at infinity times the initial 4 velocity and
that's just equal to 0, so little l is 0.
So, the geodesic equations for our observer
who's falling towards the black hole have the
first integrals of these equations
given by the inner product of d by dt
with the 4 velocity is equal to minus 1.
The inner product of d by d phi with
the 4 velocity is 0 and of course,
there's one other first integral
of the equations of motion,
that is the normalization condition that the 4
velocity should have be normalized to minus 1.
Now, these three equations can
be used to solve for t dot r dot
and phi dot, remember theta dot is 0.
The results are t dot equals r cubed plus
a squared r plus 2a squared m divided
by r times r squared minus 2mr plus a squared.
R dot is equal to minus the square
root of 2m over r times the square root
of 1 plus a squared over r squared.
And phi dot, remember theta dot
is 0, this is equal to 2am divided
by r times r squared minus 2mr plus a squared.
[silence] Now, these equations are pretty
straightforward to solve numerically,
but let's see what we can learn by just
examining the form of these equations.
And remember, these equations describe
an observer who's falling freely
into the black hole starting near r equal
infinity, and near r equal infinity,
r dot is close is close to 0, but it's negative,
so the r value of the observer decreases
and it continues to decrease
without any limit until,
in these coordinates what happens is the
integration breaks down when this factor,
r squared minus 2m over r plus a squared, which
appears in the denominator here as well as here,
when that factor goes to 0, and that value of
r that makes this 0 is the black hole horizon.
So, what happens is the observer hits the
horizon, the observer passes through the horizon
but in these coordinates, the integration
breaks down, so these are not good coordinates
to describe the geodesic infall
all the way through the horizon,
but up until the observer hits the
horizon, these equations are perfectly fine.
Now, the other thing you'll notice about
these equations is the phi dot equation,
when r is near infinity, phi dot is close to 0
but as r decreases, phi dot begins to increase,
so the observer develops some angular motion
in the direction given by the sine of a,
that's the sine of angular momentum.
So, this is the effect that we refer
to as the dragging of inertial frames.
[ Writing sounds ]
So, let me draw a picture of the black hole,
this is a top view looking down the z axis,
so this is the horizon of the black
hole, it's rotating in this direction.
Now, this is a phi equal constant
coordinate line and our observer,
it begins out near infinity falls in
towards the black hole, but is swept along
by the rotation of the black hole.
The observer hits the horizon, but the
trajectory is swept away from this radial line
by the rotation of the black hole,
and that's referred to as the dragging
of the inertial frames, the inertial frame
defined by this freely falling observer.
Let's take another look at the Kerr
metric and Boyer-Lindquist coordinates.
You'll notice that some of these metric
components vanish or go to infinity
when rho squared is 0 or when delta is 0.
So, we need to examine the properties of the
geometry when rho squared is 0 and delta is 0.
[ Writing sounds ]
In particular, we want to know if these
correspond to true curvature singularities
or simply coordinate singularities.
And one way to do that is to
examine curvature and variance.
So, for example, we can look at
the square of the Riemann Tensor.
So, r mu nu sigma rho r mu nu sigma rho.
If we compute this for the Kerr geometry,
what we find is expression that looks
like a complicated function of r and theta
in the numerator [writing sounds]
divided by rho to the 6th power.
So, this shows that when rho goes to 0,
there's a true curvature singularity.
On the other hand, when delta equals 0,
there's no problem with the curvature,
and we can examine other curvature
and variance and find the same result,
namely that delta equals 0 is
merely a coordinate singularity.
[ Writing sounds ]
Whereas, rho squared equals 0
is a true curvature singularity.
[ Writing sounds ]
So, let's examine this curvature singularity.
[ Writing sounds ]
So, curvature singularity occurs when rho
squared is equal to 0 and if we recall,
rho is defined as r squared plus
a squared cosine squared theta.
So, rho squared equals 0 requires both
r equals 0 and theta equals pi over 2.
Now, to understand the nature
of this singularity,
we need to examine the r equals 0
subspace within a t equal constant surface.
So, let's examine r equals 0 in a t
equal constant surface of the spacetime.
So, setting r equals 0 and t equal constant,
we have dr and dt are both 0 and also
from their definitions, rho squared is a squared
times cosine squared theta and delta is equal
to just a squared and then the
line element, or the metric,
becomes a squared times cosine squared theta,
d theta squared plus a squared
sine squared theta d phi squared.
So, this is the geometry of r equals
0 within a t equal constant surface.
And we can write this in a little bit more
familiar looking form by changing variables.
Let's let, let's define a radial coordinate,
and I don't want to use the letter r,
let's say r tilde, let's let r tilde
equal sine theta then dr tilde is equal
to cosine theta d theta and this metric becomes
ds squared equals a squared times dr tilde
squared plus, I'll factor out the a squared,
plus r tilde squared times d phi squared.
So, this is just the metric for a flat
two-dimensional plane in polar coordinates.
And you'll notice that as theta ranges from 0 to
pi, r tilde ranges from 0 up to 1 and then back
down to 0 again, so the maximum value of
r tilde is 1, so that means this is a disc
of radius a [writing sounds] and in fact,
these coordinates cover the disc twice,
once as theta ranges from 0
up to pi over 2 and then again
as theta ranges from pi over to 2 to pi.
But the rim of the disc or the edge of the
disc [writing sounds] at theta equal pi over 2,
is a circle or a ring of radius
a and this is the singularity.
Singularity is the theta equal pi over
2, the edge of this r equal 0 disc,
so this is referred to as the ring singularity.
[ Writing sounds ]
The singularity of a Kerr rotating black
hole is called the ring singularity.
[silence] Now, we noted before that the
Kerr geometry has a coordinate singularity
at delta equal 0.
[ Writing sounds ]
And just like in Schwarzschild,
the coordinate singularity,
it's not a curvature singularity,
but it still has significance.
It's the horizon of the black hole, so delta was
defined as r squared minus 2mr plus a squared,
so if we solve this equation, we find
there are two roots, they are m plus
or minus the square root of
m squared minus a squared,
so they are actually two black hole horizons,
an inner horizon and an outer horizon
and I've labeled those two
horizons as r plus and r minus.
And you'll notice in the limit as a goes to 0 as
the angular momentum of the black hole vanishes,
the r plus horizon is 2m, that's
the Schwarzschild horizon.
The r minus horizon is just 0, so it doesn't
exist for the Schwarzschild black hole,
or said another way, that second
inner horizon just coincides
with the singularity for Schwarzschild.
Now, from the perspective of an outsider
observer, it's the outer horizon that matters,
so that's what we'll refer
to as the event horizon.
[writing sounds] Is this r plus horizon,
which is given by m plus the square
root of m squared minus a squared.
Now, this horizon plays the same
role as the Schwarzschild horizon.
If any particle observer crosses this
horizon to smaller values of little r,
then they can never return and an outside
observer can never receive any signals
or information from the interior,
inside this even horizon.
Now, one thing you'll notice about this
expression for the event horizon is
that it requires m squared
to be greater than a squared.
So, it exists only if a squared is less
than m squared, so if a squared is greater
than m squared, then we have no
horizon, but we still have a singularity.
[ Writing sounds ]
So, in this case, the singularity is not
hidden behind an event horizon and it could,
in principle, be seen by
observers out near infinity.
This is referred to as a naked singularity.
[ Writing sounds ]
I'll have more to say about naked
singularities in a few minutes.
For now, let me give you a quick summary.
The Kerr geometry [writing sounds] describes a
stationary [writing sounds] rotating black hole
[writing sounds] in vacuum and it's
parameterized by two parameters, the mass, m,
and the angular momentum [writing
sounds], j. Now, if instead of vacuum,
if we had electromagnetic fields,
then we could write down what's known
as the Kerr Newman geometry [writing sounds],
which describes a stationary [writing
sounds] rotating black hole that's also
electrically charged.
[writing sounds] So, this black hole is
parameterized by mass, m, angular momentum, j,
and electric charge, q. So, this is a
three-parameter family of black hole solutions,
that is, solutions of the Einstein equations
with the stress energy momentum tensor
given by the electromagnetic field.
Now, the Kerr Newman geometry that describes
a charged black hole is pretty interesting
theoretically, but from a physical
or astrophysical standpoint,
it's probably not very relevant.
After all, matter is pretty much neutral
on average, electrically neutral,
so we would expect any black hole
that formed by gravitational collapse
to have charge equal to 0 or very close to 0.
So, what do we know about black holes in general
apart from the particular example of Kerr
and the special case of Schwarzschild?
[ Writing sounds ]
Well, one thing that we can
do is write down solutions
that correspond to small perturbations of Kerr.
[writing sounds] So, we can construct solutions.
[ Writing sounds ]
That are small perturbations.
[ Writing sounds ]
Of the Kerr solution.
And what always happens if we
run this forward in time is
that the perturbations radiate
away as gravitational waves.
The waves either radiate to infinity or radiate
down across the horizon into
the black hole interior.
So, the perturbed Kerr solution always settles
down to a simple unperturbed Kerr black hole.
But are there other types of black holes,
black holes that aren't just
simply perturbations of Kerr?
[ Writing sounds ]
Well, what we know and what's been proven is
that a horizon always contains a singularity.
[ Writing sounds ]
Now, by singularity here, I don't
necessarily mean a curvature singularity,
what is meant by singularity is that
timelike geodesics can't be extended
for an infinite proper time
once they cross the horizon.
[writing sounds] So, timelike geodesics
end in finite proper time [writing sounds].
So, of course our generic examples
of the Kerr black hole, Kerr Newman
and Schwarzschild all have timelike
geodesics ending at a curvature singularity.
Moreover, we don't know of
any cases of a horizon
that contains a non-curvature singularity.
So, it appears likely that all horizons
contain a curvature singularity.
[ Writing sounds ]
Now, we can turn this around
and ask the question,
is a curvature singularity
always hidden behind a horizon?
[ Writing sounds ]
A curvature singularity that's not hidden
behind a horizon but can be seen by observers
out to infinity is referred
to as a naked singularity.
[ Writing sounds ]
Now, we already found from our
study of the Kerr solution,
that if the angular momentum per unit mass
squared is greater than the mass squared,
the solution corresponds to a naked singularity.
[writing sounds] So, we have at least
this one example of the solution
of the Einstein equations that corresponds to
a naked singularity, but the question remains
as to whether or not such a
solution is relevant physically
if naked singularity actually occurs in nature.
So, we have evidence from both
theoretical and numerical studies
that a naked singularity won't form
from natural physical processes.
So, for example, we can start with a Kerr
black hole with a squared less than m squared
and try to spin it up by throwing in
matter with a lot of angular momentum
and invariably what happens is it doesn't work.
You can increase a, the angular momentum,
but you also increase the mass in such a way
that a squared always remains
less than m squared.
So, it appears that you can't spin up
a Kerr black hole [writing sounds],
that would be a black hole with a squared less
than m squared, to produce a naked singularity.
[ Writing sounds ]
Now, in 1969, Roger Penrose formulated what's
known as the Cosmic Censorship Conjecture.
[ Writing sounds ]
Which says that naked singularities
can't exist in nature.
[ Writing sounds ]
With the possible exceptions of
the Big Bang and the Big Crunch,
so this conjecture would require any
singularity to be hidden behind a horizon.
Now, these questions of naked
singularities and cosmic censorship continue
to be an area of active research.
At this point, I think it's fairly clear that
the Cosmic Censorship Conjecture is false
in the sense that Einstein equations do allow
for the possibility of naked singularities,
but that still leaves open the question
of whether naked singularities play
a significant role astrophysically.
In other words, we still don't know if
there are any naked singularities floating
around in the universe and we don't know
if naked singularities can be formed
by any natural physical processes.
So, now let me turn to the related question of
gravitational collapse and ask the question,
what do we know about gravitational
collapse in general?
[ Writing sounds ]
So, in particular, without making
the assumption of spherical symmetry.
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Well, we don't have a bunch of
theorems to address this question,
but we do have expectations that are supported
by theoretical calculations
and numerical simulations.
[ Writing sounds ]
So, the expectation is that an isolated
star [writing sounds] when it collapses,
[writing sounds], it will quickly
settle down to form a Kerr black hole.
[writing sounds] This is referred to as
the No-hair Conjecture [writing sounds],
or the No-hair theorem, and
it's really quite remarkable.
Remember, a Kerr black hole is
characterized by just two parameters,
the mass and the angular momentum.
So, this theorem is telling us that no matter
how complicated the star might be initially,
after it collapses, it's characterized
by just two parameters, m and j. So,
here's our star initially, and it contains lots
of detailed structure, different types of matter
and different layers and various processes are
occurring, nuclear fusion is causing distortions
and waves and flares on this
star, but when it collapses,
all of this detailed structure,
all of this hair, disappears.
It's either radiated out to infinity
or swallowed up into the black hole.
So, the only properties of the star that
remain are its mass and its angular momentum.
[ Silence ]
