>> This lecture is a continuation of
the previous lectures on black holes.
And the focus for this lecture
will be on gravitational collapse.
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The most simple black hole is given
by the Schwarzschild solution --
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-- which describes a static
spherically symmetric black hole.
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Using the curve shield coordinates
that I introduced in the last lecture,
we can write the metric as the ds squared equals
minus 1 minus 2m over r dt squared plus 4m
over r dtdr plus 1 plus 2m over r dr
squared plus r squared d Omega squared
where d Omega squared is shorthand notation
for d theta squared plus sine
squared theta d5 squared.
Now in these coordinates, these curve
shield coordinates, r is aerial radius --
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-- r equal 2m is the horizon of the black hole,
and r equals zero is the black hole singularity.
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We can plot the horizon and the
singularity on a tr diagram.
So let me bring this plot
in from the last lecture.
This is a plot of the tr plane.
So this is, the horizontal
axis is the aerial radius,
and the vertical axis here is the t coordinate.
The heavy black line is the horizon at 2m --
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-- and the singularity is it at r equals zero.
So I'm going to put a little
squiggly line to remind ourselves
that the curvature's infinite
along r equals zero.
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Now, keep in mind, each point in this diagram is
a two-dimensional sphere with aerial radius r,
and the curves here show the
directions for knoll rays [phonetics],
the knoll geodesics in the rt plane.
And you'll notice the inside the black hole
in this region where r is less than 2m,
less than the horizon size, that all
of these knoll rays are directed
in towards the singularity.
And since nothing can travel faster than
the speed of light, any observer or particle
that passes, say this event right here,
must stay within future light cone,
which is tipped towards the singularity.
So this observer or our particle is destined to
hit the singularity somewhere in this region.
Now in earlier lectures, we
use the Schwarzschild metric
to describe the geometry outside a static
spherical body of mass m. For example,
our sun is approximately static
and spherically symmetric.
So this metric describes the
geometry in spacetime outside the sun.
On a diagram like this, the surface
of the sun would be, say right here.
So this is, the radius of the sun, which
is approximately 700,000 kilometers.
So this metric applies to the region outside
the surface of the sun, and inside the surface
of the sun, the metric is something
else that depends on the details
of the matter that the sun is made from.
So for the sun, we need to throw away this
part of the diagram to the left of the,
of this heavy black line and
replace it with something else.
Now, you might wonder how I know to draw the
surface of the sun out here, outside the horizon
and not inside the horizon, and that's
because if we compute 2m, where m is the mass
of the sun, we obtain approximately
1.5 kilometers;
much less than the actual radius of the sun.
So the radius of the sun would have to be less
than 1.5 kilometers in order for the surface
of the sun to be inside the horizon.
But what would happen if
the sun began to collapse?
So the surface of the sun, instead of
following a world line of constant aerial radius
like this, the surface of
the sun follows a world line
like this with a decreasing aerial radius.
Well, the first thing that you
might wonder is, can we still assume
that the geometry outside the surface of the
sun is described by the Schwarzschild geometry?
For example, you can imagine this collapsing
body produces some sort of perturbation
on the gravitational field that
propagates outward into the exterior region
and alters the exterior geometry
so it's no longer exactly given
by the Schwarzschild metric.
But that actually doesn't happen.
When a body collapses like this, the
exterior region remains Schwarzschild,
and what happens is the body
simply uncovers more and more
of that Schwarzschild geometry as it collapses.
In fact, the body can collapse
clear through the horizon,
and at that point, the body
becomes a black hole.
So how do we know that the geometry outside
a spherical collapsing body is given
by the Schwarzschild geometry?
Well, the answer is contained in
what's called Brinkhoff's theorem.
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Brinkhoff's theorem tells us that the most
general spherically symmetric vacuum solution
of Einstein equations is
the Schwarzschild solution.
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A couple of lectures ago, I
derived the Schwarzschild solution
by assuming the spacetime was vacuum
spherically symmetric, and static.
And what Brinkhoff's theorem shows is
that you really didn't even need to assume
that the spacetime was static,
all we needed to assume was
that it was spherically symmetric, and vacuum.
So when a body collapses, as long
as it's spherically symmetric,
then the only possible exterior solution where,
the region where there's vacuum
is the Schwarzschild solution.
And as the body collapses, it uncovers more
and more of the Schwarzschild geometry.
Let's take a few minutes to
recall the lifecycle of the star.
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So a typical star is made mostly of hydrogen --
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-- and the star will burn hydrogen, it will fuse
together four hydrogen atoms to create a helium
of nucleus, and in the process release energy.
And this energy supplies the thermal pressure
that holds the star up and prevents it
from collapsing under its own weight,
under its own gravitational pull.
Now, when the hydrogen is gone --
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-- the star can begin to fuse the helium
nuclei together to form heavier elements.
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And this process can continue
all the way up to iron.
You'll recall iron has the largest
binding energy per nucleon of any element.
So it's not possible to fuse together
iron atoms and have a release of energy.
So as the star continues this fusion process
in its outer layers, the core of iron builds
up on the inside and the
iron core continues to grow
until it doesn't have enough thermal
support to keep it from collapsing.
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Now, the details of what happens depend
most crucially on the mass of the star.
So if the star is small, so a mass
less than about the mass of the sun,
then the end product is what's
called a white dwarf.
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Now a white dwarf is supported
from further gravitational collapse
by electron degeneracy pressure.
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Electron degeneracy pressure is a
result of the Pauli Exclusion Principle,
which tells us that electrons
can't occupy the same state.
This effectively limits the physical
space occupied by the material in the star
and provides the pressure to support
the star from further collapse.
The typical mass of the white dwarf star
is about 1 times the mass of the sun
and typical radius of the
surface is about 1,000 kilometers.
So this is still well below 2m.
So 2m for a white dwarf would
be about 1.5 kilometers.
This would be the horizon of a
black hole with the same mass.
Now, if the star has a mass
greater than the mass of the sun,
then electron degeneracy pressure isn't strong
enough to prevent it from further collapse.
So we have a medium star, medium size star --
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So we'll take the medium size star to
have a mass around 1 or 2 solar masses.
The end result is a neutron star --
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Now in a neutron star, the electrons
and protons have been forced together
to inverse beta decay to form neutrons.
So a neutron star is basically a big
bowl of neutrons, and it's supported
from further collapse by
neutron degeneracy pressure.
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So like electron degeneracy pressure, this
comes from the Pauli Exclusion Principle
and effectively limits the physical
space occupied by the neutrons.
Now a typical size for neutron star --
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-- would be about 10 kilometers.
And this is larger than the value
of 2m, which would be about 2
or 3 kilometers, a few kilometers.
Now if the mass of the star is greater
than about two times the mass of the sun,
neutron degeneracy pressure isn't strong enough
to prevent the star from further collapse.
So for a large star --
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-- which we'll take to be
roughly 2 solar masses or more.
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Then the end result, we believe,
is a black hole.
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Or said a little more precisely, when the
mass is greater than about two times the mass
of the sun, there's no known physical mechanism
that can prevent the star from
collapsing to a black hole.
And of course there are lots of stars
in the universe with masses greater
than two times the mass of
the sun; in some cases,
much greater than twice the mass of the sun.
And those stars eventually
run out of nuclear fuel.
So this process of collapse to forming
a black hole is really quite common.
So what does the collapse look
like on a curve shield diagram?
So here's the aerial radius r, here's
the curve shield time coordinate t,
and this is the Schwarzschild geometry.
Every point in this diagram
is a two-dimensional sphere.
Now let me draw the surface of the star.
So let's begin here.
The star begins to collapse.
This is the singularity, squiggle line here.
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And of course 2m is the event horizon.
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Now the interior of the star
is this region in here in the,
we're discarding the Schwarzschild geometry in
here because the geometry is something else,
depends on the details of the matter.
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So even though we don't know what the geometry
is in here, we do know what the geometry is
out here, it's the Schwarzschild geometry.
So in particular, the surface,
once it crosses the horizon,
it has no choice but to run
into the singularity.
In fact, if you like, you can think of a
particle that's riding along the surface,
that particle must always move
within the forward light cone.
And the forward light cones here
are tipped towards the singularity.
So that particle riding along the surface
has no choice but to hit the singularity,
and it can never escape to the exterior region.
So you see from this diagram
that once the surface
of the star crosses its own event
horizon, then nothing can prevent it
from full collapse to singularity.
What I'd like to do next is spend a
little bit of time talking about a model
for the interior geometry for the
solution in the interior of the star.
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Now, the Einstein equations are much
too complicated to solve analytically
for any realistic model of the star's interior.
So this model I'm about to show you is a
highly idealized model, will not only assume
that the star is spherically symmetric,
but also assume that the material
is a simple pressureless dust.
So we'll assume that the stress energy momentum
tensor is the mass density Rho times u mu u nu,
where u is the four velocity of the dust.
So we can imagine that earlier times,
prior to this t equals zero surface,
the star had some pressure that
supported it from collapse.
Then at t equals zero, the pressure support
was lost and the star begins to collapse.
Then we model the material in the star
beyond t equals zero as a simple dust,
and it's this region of the
spacetime that I want to describe.
So I'll first write down the
interior solution and we'll verify
that it satisfies the matter equations
in motion in the Einstein equations.
And then we'll worry about how that interior
solution is joined together along the surface
of the star with the exterior
solution, which is Schwarzschild.
So this will be the solution
for a collapsing ball of dust.
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So we'll assume the ball
is spherically symmetric --
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-- and dust means the stress
energy momentum tensor is given
by the mass density times
the four velocity u mu u nu.
I'll draw a picture of the
collapsing ball as a wedge.
Every point in this wedge
is a 2 dimensional sphere.
This is the same wedge as
we showed in this diagram.
And I'm going to use coordinates
to describe the geometry in here,
which I'm going to call chi and eta
So we can think of chi as a radial coordinate.
So it monotonically increases as the aerial
radius increases, starting with chi equals zero
out to the surface of the star, and eta,
we can think of it as a time coordinate.
It starts at zero at the moment
that star begins to collapse,
and it has some maximum value
before the singularity forms.
Here's the singularity.
And the maximum value of eta is actually
pi, and the maximum value of chi,
the value of chi that corresponds to the
surface of the star, is sine inverse or arc sine
of the square root of 2m
over some constant r naught.
And now let me draw a few
of the coordinate lines.
So these lines are lines of constant chi.
And the surface of the star itself is a
surface of constant chi with the value given
by sine inverse of square root 2m over r naught.
And the surfaces of constant
eta, this is 8 equals zero,
and those surfaces look something like this.
So the coordinates are --
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eta, chi, theta, and phi.
Remember each point in this
diagram is a 2 dimensional sphere
with coordinates theta and phi.
So the metric for this collapsing
ball of dust --
-- is ds squared equals r naught cubed divided
by 2m times cosine to the fourth power of eta
over 2 times the quantity minus d eta squared
plus d chi squared plus sine squared chi d
Omega squared.
And the density of the dust is equal to 6m
divided by 8 pi r naught cubed times cosine
to the minus 6 power of eta over 2.
And the four velocity of the dust
is given by the zero component
of the eta component is square root
of 2m over r naught cubed times cosine
to the minus 2 power of eta over 2.
And the other components are zero.
So you can see from this
form of the four velocity
that the dust particles travel along these
chi equals constant coordinate lines.
So the dust just travels along these
lines, so these are the dust world lines,
and they all converge here
at the tip of the wedge.
And you can see here when eta is equal
to pi, the density goes to infinity.
So this would be the cosine of pi over 2 is
zero, zero to the minus 6 power's infinite.
Now we still need to verify
that this metric density
and four velocity satisfy
the equations of motion.
So first let's look at the dust equations.
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If we remember for any perfect fluid, the
equations of motion are just the equations
of local energy momentum conservation.
So for the case of dust,
t mu nu is Rho u mu u nu.
We can write this out in detail as partial mu
of Rho u mu u nu plus gamma mu mu sigma Rho
u sigma u nu plus gamma nu mu sigma Rho u mu
u sigma.
And now, if we remember that, only the eta
component of u is nonzero, this becomes partial
with respect to eta of a Rho u eta
u nu, so the free index is nu here,
plus gamma mu mu eta Rho u eta u nu
plus gamma nu eta eta Rho u eta u eta.
Now, if we examine these Christoffel
symbols, we have some over mu.
So this includes things like gamma eta eta,
and it turns out also gamma chi chi eta,
it turns out that all of these
are equal to one another.
So gamma theta theta eta
equals gamma phi phi eta.
These are all equal to minus
tangent of 8 of eta over 2.
And for these Christoffel symbols, we already
know that gamma eta eta eta is minus tangent 8
over 2, but all the others vanish.
So chi eta eta gamma theta eta eta gamma
phi eta eta, these are all equal to zero.
Now let's put all this together and you'll
notice that when the index nu is equal to chi --
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-- then this term is zero because the chi
component of u is zero, here the chi component
of u is zero, so this term is zero.
And also this term is zero
because gamma chi eta eta is zero.
And similarly, when the index
nu is equal to theta or phi,
we just get zero equals zero from this equation.
So the only nontrivial case is when nu is equal
to eta and we obtain zero equals partial eta
of Rho u eta squared plus gamma mu mu eta
is the sum of all these Christoffel symbols,
so that's minus 4 tangent of eta
over 2 times Rho u eta squared.
And this term with nu equal to eta,
gamma eta eta eta is minus tangent of eta
over 2 times these factors, Rho u eta squared.
This of course simplifies to partial eta
of Rho u eta squared minus 5 times tangent
of eta over 2 times Rho u eta squared.
And now if we remind ourselves
to form that Rho and u eta,
take they're written down right here --
We find that Rho u eta squared is equal
to 3/2m squared over pi times r naught
to the sixth power times cosine to
the minus tenth power of eta over 2.
Now we can insert this into our equation, which
becomes zero equals partial eta of 3m squared
over 2 pi r naught to the sixth power times
cosine to the minus tenth power of eta
over 2 minus 5 times tangent of eta over
2 times 3m squared over 2 pi r naught
to the sixth times cosine to the
minus tenth power of eta over 2.
Of course, these factors 3m
squared divided by 2 pi r naught
to the sixth cancel in this equation.
And we're just left with the
derivative of cosine minus 10.
So that would be minus 10 times cosine
to the minus 11th power times the
derivative of cosine, which is minus sine.
And then we have another factor of 1/2
coming from differentiation eta over 2.
And this term is minus 5 times,
let's write tangent sine over cosine.
So this is sine eta over 2 times cosine
to the minus 11th power of eta over 2.
And now you can see that these 2 terms cancel.
We have 2 minus sines, gives
an overall plus sine,
and 10 divided by 2 is 5, here we have minus 5.
And both of these terms have
factors of sine eta over 2,
and cosine to the minus 11th
power of eta over 2.
So we have now verified that the matter
equations a motion hold for this solution.
And now we need to verify that
the Einstein equation holds.
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That's g mu nu, Einstein tensor
equals 8 pi times t mu nu.
Now, when we compute the Einstein
tensor for the metric that I gave,
you find the only nonzero components --
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-- are when mu and nu are both equal to eta.
And what we you obtain then for ga to
eta is 12m squared divided by r naught
to the sixth power times cosine
to the tenth power of eta over 2.
For the stress energy momentum tensor,
that's Rho times u mu u nu,
the only nonzero components --
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-- are when mu and nu are both equal to eta.
And then this is equal to Rho times u eta
squared, which you might recall is equal
to 3m squared divided by 2 pi r
naught to the sixth power times cosine
to the minus tenth power of eta over 2.
So you can see if we multiply this by 8
pi, and the pis cancel, 8 times 3/2 is 12
and that's precisely what we have here,
12m squared over r naught to the sixth
and then cosine to the minus tenth power.
So Einstein equations are satisfied.
So we've now verified that this metric density
and four velocity is a solution of the matter
and Einstein equations of motion.
And this solution describes
a collapsing ball of dust.
And by the way, this is referred
to as the Oppenheimer-Snyder --
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-- dust solution.
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Of course, this is just the interior solution.
Outside this ball of dust with the star,
we have the Schwarzschild geometry.
And we need to know that the interior solution
and the exterior solution can be glued
together in a continuous fashion.
So in particular, we need to know that the
geometry along the surface of the star,
we need to know that that's the same
geometry as viewed from the interior
as it is as viewed from the exterior.
So this is the surface of the star.
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And from the interior it's given by the
coordinate value chi equal to sine inverse
of square root of 2m over r naught.
So the 3 dimensional geometry of this
surface is obtained by inserting a chi equal
to this value into the line element here.
So if we do that, then of course d
chi squared vanishes and sine squared
of chi is just the square of
square root of 2m over r naught.
So that's 2m over r not, which
partially cancels these factors.
And what we're left with then
for the surface geometry --
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-- this is as viewed from
the interior of the star --
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-- is just ds squared equals r naught cubed
over 2m times cosine to the fourth power of eta
over 2 times minus d eta squared plus r naught
squared times cosine to the fourth power
of eta over 2 times d Omega squared.
Now the exterior geometry --
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-- that's just Schwarzschild, and remember
we've been using curve shield coordinates.
So the line element is minus 1 minus 2m over r
dt squared plus 4m over r dtdr plus 1 plus 2m
over r dr squared plus r
squared d Omega squared.
Now the surface of the star
is viewed from the outside,
from the exterior, is some
curve in the tr plane.
So it's this curve, and we can parameterize
the curve using the coordinate values eta.
So each point along this curve has
a value of eta that it inherits
from the coordinates on the inside.
So we'll describe the surface
of this star from the outside
as some parameterized curve, t of eta, r of eta.
Then the surface geometry
as seen from the outside --
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-- this is obtained by inserting
t as a function of eta
and r as a function of eta
into this line element.
What we obtaine then is ds squared equals minus
1 minus 2m over r, now that's r as a function
of eta, times -- and dt I'm
going to write t dr, times d eta,
and I'll put the d eta outside
of all of these terms.
So this is t dot squared plus 4m divided by r
of eta times t dot r dot plus 1 plus 2m over r
of eta times r dot squared,
all of that time d eta squared.
So that's all of these terms.
And then of course we have r
squared, that's r as a function
of eta squared times d Omega squared.
So now we have the surface
geometry is viewed from the interior
and the surface geometry is viewed from the
exterior, and of course these need to match.
So these 2 line elements
should be equal to one another.
So we first equate the coefficients
of d Omega squared
and that tells us r is a function of eta.
And then we equate the coefficients
of d eta squared.
So that gives us the following 2 results.
We have r of eta equals capital R naught
times cosine squared of eta over 2.
And the other equation that comes
from equating the coefficients
of d eta squared is minus 1 minus 2m
over r times t dot squared plus 4m
over r t dot r dot plus 1 plus 2m over r times
r dot squared equals minus r naught cubed
over 2m times cosine to the
fourth power of eta over 2.
Now we can insert this result for
r of eta everywhere we see r here,
and also differentiate this with respect to
eta and insert that for r dot here and here.
Then solving this equation for t
dot gives t dot equals m times sine
of eta minus r naught times the square root
of r naught over 2m minus 1 times cosine
to the fourth power of eta over
2, all of that divided by 2m
over r naught minus cosine
squared of eta over 2.
And now we can integrate this equation
to obtain t as a function of eta.
And the result is 1/2 square root of r naught
over 2m minus 1 times the quantity r naught
plus 4m times eta plus r naught times sine
of eta plus 4m times the natural log of cosine
of eta over 2 plus sine of eta over 2 divided
by square root of r naught over 2m minus 1.
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So these calculations are
straightforward in principle,
but they're actually a bit
trickier than you might suspect.
It took me the better part of a day to obtain
this answer in this nice relatively simple form.
And by the way, you'll notice I've chosen the
integration constant here so that t is equal
to zero when eta is equal to zero.
So these two equations define
the surface of the star as viewed
from the exterior Schwarzschild geometry.
And by construction, the geometry of the
surface matches precisely the geometry
of the surfaces viewed from the inside,
from the interior dust solution.
So the interior and exterior
geometries match at the surface.
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And that means that we can connect the
interior and exterior continuously.
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But that does not necessarily mean
that the geometries connect smoothly.
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Let me give you a simple example
to illustrate the issue here.
We can take a cylinder.
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So this is a cylinder.
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And we can put a top on it, a disk.
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So here we have a disc, a flat
disc, and we have a cylinder,
and they can be connected continuously.
So they connect along the rim here, this circle.
So the geometry of the circle
is the same as viewed either
from the disc or from the cylinder.
So the connection is continuous,
but in this case it's not smooth.
So a continuous connection requires
the geometries to match at the surface
and the geometries are intrinsic geometries.
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So we examine the intrinsic geometry of
the three dimensional surface from interior
and exterior and made sure that
those intrinsic geometries match.
To determine whether the matching is smooth,
we have to examine the extrinsic geometry --
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-- or the extrinsic curvature.
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I discussed briefly in some of the earlier
lectures in this series the distinction
between intrinsic and extrinsic
geometries or curvature.
The intrinsic geometry always refers to
the metric, it's what the metric defines.
Extrinsic curvature always refers to
the way that a surface is embedded
in a higher dimensional space or spacetime.
So the idea is that we have a surface --
-- and this might be a one dimensional surface
in a two dimensional space like I've drawn here,
or it might be a three dimensional surface
in a four dimensional space or spacetime.
So in order to quantify the way
this is bent or shaped with respect
to the embedding spacetime is
we examine the normal factors.
So these are unit normals.
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And to compare them, we parallel transport.
For example, we will take this normal vector
and parallel transport it so that it sits
at the same point as this vector.
So here's the parallel transported version
of this vector, and then we subtract.
And the result is a vector that
tells us how bent this surface is
within the embedding space or spacetime.
So let's call these normal
vectors N, components N mu.
And let's also introduce a set
of basis vectors in the surface.
So here's a vector e with components e mu.
So the way the extrinsic curvature is defined --
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-- is by parallel transporting the normal,
and that's done with the covariate derivative
of N along the direction of some basis
vector e. So this is what we'll define
as the extrinsic curvature and is
the vector for each basis vector.
So in general, if we have several basis vectors,
say if the surface is three dimensional,
then we'll have three basis vectors,
we'll label them with a subscript i.
And the extrinsic curvature will
be a set of vectors, k mu sub i.
And these extrinsic curvature vectors
tell us how the surface is curved
or bent within the embedding space for
each one of the basis vector directions.
Now let's verify that the interior
dust geometry connects smoothly
to the exterior Schwarzschild geometry.
So let me draw this, the wedge, once again --
-- here's the singularity.
This is the surface of the star.
And we need to construct, at
each point along the surface,
we need to construct a unit normal --
-- in mu, and we need to construct a
set of basis vectors in the surface.
So here's one of those basis
vectors, let's call it e zero.
And there are two other basis vectors
that are perpendicular to this diagram.
Remember, each point in this
diagram is a two dimensional sphere.
So we'll call these other
basis vectors e2 and e3.
So the set of basis vectors is e zero, e2,
and e3, and of course the normal vector is n.
So we will use these vectors to construct
the extrinsic curvature of the surface,
but the extrinsic curvature will depend on
whether this surface is viewed as a subspace
of the interior geometry inside the dust,
or as a subspace of the exterior
geometry, the Schwarzschild geometry.
So we'll have to compute the extrinsic curvature
for both cases and then show that they match,
and then we'll know that the matching of the
two geometries, inside and outside, is smooth.
So let's begin by computing the
extrinsic curvature of the surface
as a subspace of the interior geometry.
So the surface as subspace --
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-- of the interior.
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Let's start by constructing
this unit vector e zero.
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This is easy.
This is a unit vector along
the surface of the dust
in the plane perpendicular to the two spheres.
So this is just the four velocity
of the dust at the surface.
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If you'll recall, that's equal to square
root of 2m over r naught cubed times cosine
to the minus 2 power of eta over 2.
That's the eta component and
the other components are zero.
And now we can construct the unit vector n, it's
a vector that is perpendicular to this vector,
it's in the plane orthogonal to the two spheres.
So it only has components in the eta
and chi slots and it's normalized.
So those conditions uniquely defined this
normal vector as zero comma squared of 2m
over r naught cubed times cosine minus
2 of eta over 2 comma zero comma zero.
And now the other two basis vectors are e2 is
zero zero 1 over r naught cosine squared eta
over 2 comma zero, and e3 is zero zero zero
comma 1 divided by r naught times cosine squared
of eta over 2 times sine of theta.
So you can verify that all of
these vectors are normalized
and they're all orthogonal to one another.
And now we compute the extrinsic curvature
of course using the interior metric.
So I'll remind you the metric inside the ball
of dust is r naught cubed over 2m times cosine
to the minus 4 to the plus 4 power eta
over 2 times minus d eta squared
plus d chi squared plus sine squared
of chi d Omega squared.
Now the extrinsic curvature --
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-- is defined as k mu sub i
equals e nu i dell nu N mu.
And this is now straightforward calculation
using this metric and these basis vectors.
And the result is the following, the zero
extrinsic curvature vector is just zero.
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The second one, the extrinsic curvature
vector associated with e2 is zero zero 1
over r naught squared times square root
of 1 minus 2m over r naught times cosine
to the minus fourth power
of eta over 2 comma zero.
And k3 is zero zero zero 1 over r
naught squared times squared 1 minus 2m
over r naught times cosine to
the minus fourth power of eta
over 2 times sine inverse,
that's 1 over sine of theta.
So this is the extrinsic curvature of the
surface as a subspace of the interior geometry.
And now we turn to computing
the extrinsic curvature
of the surface as a subspace of the exterior.
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Once again, we need to construct
the basis vectors --
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-- e zero, e2, e3, that span the
surface and also the normal vector --
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-- N mu. So we'll start with e zero.
Remember this is just a four
velocity of the dust at the surface.
Now the four velocity is written as a derivative
operator as 2m over r naught cubed times cosine
to the minus 2 power of eta over 2 d by d eta.
And we can use this to find the
components of u in the coordinates used
in the exterior in the following way.
This is 2m over r naught cubed square root times
cosine minus 2 power of eta over 2 times t dot d
by dt plus r dot d by dr. And the dot here
refers to differentiation with respect to eta.
We wrote down t dot and r dot explicitly before,
so this factor times t dot is the t component
of u, this factor times r dot
is the r component of you.
In other words, those are the t and r components
of our vector, our basis vector e zero.
So when we put all that together, we
find that e zero has the first component,
the t component is 2m over r
naught cubed times m times sine
of eta minus r naught times the square root
of r naught over 2m minus 1 times cosine
to the fourth power of eta over 2.
All of that divided by cosine
squared of eta over 2 times 2m
over r naught minus cosine
squared of eta over 2.
So that's the t component.
The r component is minus square root of 2m
over r naught times the tangent of eta over 2
and the theta and phi components are both zero.
The next step is to construct a normal vector in
this vector lies in the plane that's orthogonal
to the two spheres, so it's data and
phi components of N are both zero.
This vector is orthogonal to
e zero and it's normalized
and that completely determines
the vector to be the following;
N mu is equal to 4m times the square root of
1 minus 2m over r zero minus the square root
of 2m times r zero times the sine of eta.
All of that divided by minus 4m plus 2 r
zero times cosine squared of eta over 2.
That's the t component.
The r component is square root
of 1 minus 2m over r naught,
and theta and phi components are both zero.
And now it's easy to construct
the remaining 2 basis vectors.
This is the basis vector that points
along the d by d theta direction.
It's zero comma zero comma 1 over r naught
cosine squared of eta over 2 comma zero,
and e3 is zero zero zero 1
divided by r naught cosine squared
of eta over 2 times sine of theta.
And now we compute the extrinsic curvature
of course using the exterior metric
which is minus 1 minus 2m over r dt
squared plus 4m over r dtdr plus 1 plus 2m
over r dr squared plus r
squared d Omega squared.
So the extrinsic curvature is --
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-- k mu sub i equals ei nu dell nu N mu.
Now the only thing that's tricky about this
calculation is here we have derivatives
with respect to the coordinates t and r, but the
components of the vector N are written in terms
of this parameter eta rather than t
and R. So when you differentiate N along
the e zero direction, you need to use --
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-- e zero nu partial nu N mu equals square
root of 2m over r naught cubed times cosine
to the minus 2 power of eta over 2 times
d by d eta acting on the components N mu.
This is just the vector e zero written
as a derivative operator that's
proportional to d by d eta.
So with this hint, I'll leave it
as an exercise for you to verify
that the extrinsic geometry using these
basis vectors and normal vectors turn
out to be exactly the same as they were for the
surface as a subspace of the interior geometry.
So you should obtain exactly the same results.
That shows that the extrinsic curvature on the
interior and exterior of the surface match,
which shows that the two geometries, interior
and exterior, match together smoothly.
