[SQUEAKING]
[RUSTLING]
[CLICKING]
SCOTT HUGHES: So
we're switching gears.
We're done with the large scale
structure of the universe.
Now we're done with cosmology.
And I'm going to move
into the topic that
will dominate the last
several lectures of 8.962.
Let's look at how to
construct the spacetime
of a compact body.
In particular, we're
going to start by focusing
on a compact spherical body.
What I mean by compact-- so
spherical, I hope, is clear--
but what I mean by compact
is that the body occupies
a finite spatial region.
And it has a surface with
an exterior that is vacuum.
So what I'm going
to do is design
some kind of a spacetime.
I'm imagining that
there is a source that
fills some compact region.
I'm going to make it
strictly symmetric.
And the exterior of the source
will be t mu nu equals 0.
So if you think about
this, this is telling you
that if it's
spherically symmetric,
your metric can only
depend on radius.
And so the most
general form, at least
for a static spherically
symmetric spacetime--
we'll talk a little
bit about a few things
related to non-static
situations a little bit later.
So the most general static
form of a spherically symmetric
compact body tells me
that I can write my metric
as some function of r, which
for reasons of convenience
I will write that is e
to the 2 phi, t squared.
Some other function
of r, dr squared.
And then some function of
r times my angular sector.
You might wonder why
there's no cross-term.
In particular, why can't
I have a cross-term
between my dt and my dr?
I have a few lines
on this in my notes.
I won't go through
them in detail here,
but I'll just describe.
Basically, what you
find is that if there
is a cross-term between,
say, the t and the r pieces,
you can get rid of it by a
coordinate transformation.
This way of doing
things so that it
is diagonal amounts to choosing
a good time coordinate.
We can simplify
the angular sector
by choosing a good
radial coordinate.
The word "good" is a
little bit loaded here.
I'm going to
actually in a moment
describe a slightly different
choice of radial coordinate,
which is also good, but
for different purposes.
The one I'm going
to use is one where
this function that gives
length to the angular sector
is just r squared.
What this means is
that the coordinate r,
it labels nested
spheres that have
surface area 4 pi r squared.
So if you imagine
that in this spacetime
you make a series of
spheres, and each one of them
has an area 4 pi r squared, that
is the r that goes into this.
Bear in mind-- [CLEARING THROAT]
excuse me-- bear in mind,
that does not mean--
let's say I've got one sphere
of radius 4 pi r1 squared,
and then outside of it a sphere
of radius 4 pi r2 squared.
That does not mean that the
distance, the proper distance,
between those two
spheres is r2 minus r1.
So this is unlikely to measure--
to label radial
distance cleanly.
In fact, it would
only be the case
that this cleanly labels
radial distance if the function
lambda were equal to 0.
And as we'll see, that's
generally not the case.
So whenever we do things
in a curved spacetime,
things are always going to
get a little bit messier.
And what's nice about
this is at least
we do have a clean geometric
meaning to the coordinate r.
In fact, before I
move on, this is
called an areal radius, a radius
simply related to surface area.
So in the end, the
spacetime that I
will be working with in this
lecture takes this form.
And so this is a
choice that is known
as Schwarzschild coordinates.
We will come back to the name
Schwarzschild before too long.
As I said, so this is a
particularly good choice
of radial coordinate.
It'll be very convenient
for what we do.
It's not the only one.
Other coordinate
choices are possible.
And another one that is
quite useful for us--
we're not going to use it
too much in this lecture--
is what are called
isotropic coordinates.
So this is a coordinate
system in which
you choose a radial coordinate
which I will call r bar,
such that your line
element looks like this.
What this does is
this emphasizes
the fact that we are working
in a coordinate system--
or excuse me-- we're working
in a spacetime in which all
of your spatial slices are
fundamentally isotropic.
So get far enough away from
this thing, and the three--
it's just emphasizing
the fact that the three
spatial directions are--
there is something
special about it.
There's presumably a body at r
bar equals 0, in its vicinity,
but other than that,
there's nothing
particularly special about--
if I go into a little
freely falling frame,
all three directions
look the same.
This helps emphasize this.
We lose the areal
interpretation of r here.
I'll remind you that our weak
field solution that we derived
a few weeks ago, a few
lectures ago rather,
it looked like this.
And voila, look at that.
It's exactly the same form.
So this is, in fact--
when we derive this,
this is, in fact, actually
in isotropic coordinates.
What we found was that
the mu that appears here
is equal to minus phi with
phi itself being very small.
And the phi that
came from this is
Newton's gravitational--
the Newtonion gravitational
potential.
Let's switch back.
We're going to use
Schwarzschild coordinates.
This is kind of an aside.
We will use Schwarzschild
shield coordinates
for the bulk of our lecture.
Actually, the last thing
I'll say about this
is that there is a problem.
There's a problem
on a problem set--
I might have made it optional.
I don't quite recall--
in which for one of the
spacetimes that you're
going to be working with
in about two more lectures,
you convert between
Schwarzschild coordinates
and isotropic coordinates.
I think it's an optional lec--
it's an optional
problem at this point.
But it's worth looking at.
So let's build a bot.
Let's build the spacetime
of a compact spherical body.
What we need are curvature
tensors and matter terms.
The curvature tensors,
since I've given you--
whoops.
Not there.
No.
The one at the top.
I'm sorry.
Since I've given
you ds squared, you
know all the metric elements.
It's then just a
matter of a little bit
of having the algebraic
stamina to run through and do
the calculations.
It's straightforward
to construct
all of the curvature
tensor components.
The GR tool Mathematica
notebook that
will soon be released
on the 8.962 website,
if it has not been
released already,
that will allow you to do this.
I'm going to just sort of
quote the results, what
the results turn out to be.
So for this spacetime, it's
given in Carroll's textbook,
a slightly different
notation, but you can easily
translate from the functions
Carroll uses to the ones
that I've written here.
So we need to get those
curvature tensor components.
And we need matter.
And so we will do
what we generally
always do in this class.
We will treat our body as
being made of a perfect fluid.
So let's write out our curvature
tensors that arise from this.
I'm going to go straight
to the Ricci tensor.
If you want to work out all
the complements to the Riemann
tensor, knock yourself out.
As I said, they are
listed in Carroll's book.
It's not terribly difficult to
go and work all these guys out,
though.
Certainly not using Mathematica.
So here's what you get for the
rr complement of the Ricci.
Notice we are seeing
nonlinear terms in here.
These are the kind of terms
that in all of our calculations,
when we were looking at bodies
in a weak field spacetime,
we set those to 0.
So the form for the ttp.
So notice it's quite similar
to the form for rrp's.
If you saw me hesitate
while I was writing,
I actually suddenly just
got concerned I was writing
the wrong line down,
and was just verifying
that I was writing
the right thing.
And two more non-trivial
complements, or really
one more non-trivial
component, and then there's
a fourth component
that is simply related.
So here is our Ricci tensor.
We're going to solve it for
a perfect fluid stress energy
tensor.
We will also solve it for--
sorry-- we will use this
to make an Einstein tensor,
equate it to a perfect
fluid stress energy tensor,
and use that to
build the spacetime
on the interior of this object.
The exterior of the object is
going to be treated as vacuum.
That will have 0
stress energy tensor.
And so it'll be a
little bit easier
when we construct the
exterior solution just
to set Ricci equal
to 0 and solve it.
Before I get into that,
let's look at the source
we're going to use to describe
this compact spherical body.
Like I said, we're
going to treat this guy
as a perfect fluid
stress energy.
So by now, you will
have seen me written--
you'll have seen
that I've written
this form down multiple times.
So these, the pressure
and the density,
are both functions
only of radius,
since this is
spherically symmetric.
And u denotes the four-velocity
of fluid elements in this
fluid--
in this body.
I'm going to start
calling this body a star.
It doesn't necessarily
have to be a star,
but it's just a
convenient shorthand.
So I want this star's
fluid to be static.
So the four-velocity will
have a timeline component,
but there is no spatial motion.
I want the star's fluid
to be at spatial rest.
I am going to require that
this be properly normalized.
u dot u equals minus 1.
So if I do that
in this spacetime,
I'm led to the requirement
that u upstairs
t is e to the phi of r.
Pardon me.
e to the minus 5r.
And u downstairs t
is minus e to the 5r.
Finally, we're going to assume--
so as I said, we're going to
treat this as a compact body.
So we're going to require
that it have a surface.
So we will assume rho
of r equals 0, p of r
equals 0 for r greater
than or equal to r star.
r star denotes the
surface of this body.
Let's attack.
So it's a little bit easiest
to do the exterior first.
So we're going to consider
the region outside of r star.
t mu nu equals 0 there.
And so my Einstein equations
become Ricci equals 0.
So I can just take every
one of these things,
set them equal to 0, and
try to assemble solutions
for phi and for lambda.
Now a particularly convenient--
every one of these
things has to equal 0
so I can make various
linear combinations of them
to make things that
are particularly
convenient to work with.
And if you sort of stare at
the tt and the rr equations,
one thing you see is that
if you multiply the tt
by this combination
of metric functions,
add the r r, what that does
is it allows you to cancel out
those nonlinear terms.
And this combination
tells you that--
whoops-- this
combination tells you
that it's an overall
factor of 2/r,
the radial derivative of phi
equals the radial derivative
of lambda.
And so this tells me that
at least in the exterior phi
equals minus lambda,
perhaps up to some constant
of integration.
If we go when we plug this into
our metric, what that k does
is it just gives me a
rescaling, depending upon
whether I want to attach
it to the function phi
or attach it to the
function lambda.
What we see is that K
just essentially rescales.
But let's do the following.
Let's sort of say that
I'm going to plug this
in and say that I'm going
to plug it in for phi.
So if I plug this
into the metric,
I can capture the impact of the
constant of integration k here
by just rescaling
my time coordinate.
So we are free just
to say, OK, there's
no physics in this thing.
So let's just set k equals 0.
So that's already great.
We have found that at
least in the exterior
phi is equal to minus lambda.
We've reduced this metric
from having two free functions
to one.
So we're feeling spunky now.
Let's take a look at a couple
more of these equations,
and see what we can
learn from them.
So we still have some of
these ugly things involving
nonlinear stuff in
the rr and the tt,
but it's all linear in
the theta theta equation.
So let's look at r
theta theta equals 0.
So I have e to the
minus 2 lambda r
dr of lambda minus r dr
of phi minus 1 plus 1.
All this equals 0.
So now you can either substitute
for lambda or substitute
for phi.
I'm going to follow my
notes here and substitute
in lambda equals minus phi.
Stare at this for a
second, and you'll
realize that you can rewrite
this term on the left-hand side
as dr of r e to
the 2 phi equals 1.
And this can be
easily integrated up.
And the solution is
phi of r equals 1/2
log 1 plus a/r, where a is
some constant of integration.
We're going to figure
out a way to fix
that constant of integration.
And you can probably guess
how I'm going to do that.
But let's just hold
that thought for now.
So let's just take this
solution and write down
what our exterior
spacetime looks like.
So plugging this guy in, I get--
OK.
So here's my line element.
And we're going to
do the usual thing.
This is-- well, remember
that this situation describes
gravity.
And gravity has a
Newtonian limit,
which these general relativistic
solutions must capture.
So let's make sure that this
captures my weak field limit.
So let's consider r
gihugically greater than a.
a must be some parameter with
the dimensions of length.
It might turn out
to be negative.
So let's take the absolute
value of this thing.
And let's consider a
purely radial freefall.
So I'm going to imagine--
when I work this
guy out, I'm going
to look at geodesics
in this limit.
I'm going to look at
purely radial freefall.
I'm going to consider the
non-relativistic large r limit.
So here I am imagining in
my non-relativistic limit
that dt d tau is much
greater than the dr d tau.
So when you work out all those
components of the Ricci tensor,
you will have computed all
of the Christoffel symbols.
And you go back to your
handy dandy little table
of these things that
you will have computed.
So here's what you get for that.
Your phi is given by
1/2 log of 1 plus a/r.
So-- taking the exponential
and the proper derivatives,
what this tells you is
that this is negative
a over 2r squared 1 plus a/r.
And again, I'll emphasize,
we are considering this sort
of very large r limit.
So this is approximately
negative a over 2r squared.
Dividing both sides by the
factor of dt d tau squared,
I finally get my
equation of motion
for non-relativistic weak
field radial freefall
to be that this thing's
radial acceleration is a over
2r squared.
Now, we want this to
capture the Newtonian limit.
Newtonian freefall
is given by dt dr
equals the gravitational
acceleration
minus gm over r squared.
They're exactly
the same, provided
I select a equals minus 2gm.
So what emerges out
of this analysis
is this truly lovely solution.
This is known as the
Schwarzschild metric.
This is a good point
to tell a brief story.
Carl Schwarzschild
derived this solution
shortly after the publication
of Einstein's field equation.
So Einstein's field equations
appeared in late November--
excuse me-- late
in the year 1915.
I believe in November
of 1915, but we
should double-check that.
Schwarzschild published
this exact solution,
a few months later, I
believe in early 1916.
Students of history
will note that if you
were in Germany
in 1915 and 1916,
you were likely to be
involved in some rather
all encompassing, non-academic
activities at that time.
And in fact, Carl Schwarzschild
did this calculation
while he was serving
as an artillery
officer in the German Army
on the Western-- excuse me--
on the Eastern Front.
He, in fact, was quite ill
at the time he did this.
I believe it was
a lung infection
that he had accumulated while
serving in the trenches.
And so while recuperating
from this disease,
he received
Einstein's manuscripts
because he was a professor
in his civilian life,
and just wanted to keep
up with the literature.
He was inspired by this
relativistic theory of gravity,
and decided that it would be
worth a little bit of his time
while he was resting
in the hospital
to see if he could do something
with these exciting new
equations that his colleague,
Herr Doktor Professor Einstein,
had worked out.
And he came up with this,
the first exact solution
to the Einstein field equations.
And indeed, a solution
that continues
to be of astrophysical and
observational importance today.
He succumbed to the disease
that he was suffering
and died several weeks after
coming up with this solution
and publishing it.
This was a solution which
stunned Einstein, who did not
think that such a simple
solution would ever be found,
such a simple exact
solution would ever
be found to these
horrendous equations
that he had developed.
I always find this
story to be somewhat--
well, I'm not quite
sure what to make of it.
In these days of
the coronavirus,
where we are all
isolating, we are--
many of us now are
probably beginning
to hear stories of people we
know who may have the disease.
Sometimes it's hard
for me to crawl out
of bed in the morning.
Somehow I do so.
And I sometimes
think to myself, what
would Karl Schwarzschild do?
He didn't let the fact that he
was dying of a lung infection
from--
he did not let the fact he
was dying of a lung infection
prevent him from leaving
an indelible stamp
on the history of science.
I do not aspire
to such greatness.
I do not think I am
capable of such greatness.
But it at least helps me to
get out of bed in the morning.
So like I said,
this is a solution
that is of observational
importance today.
And one of the reasons
it has such significance
is due to a result known
as Birkhoff's theorem.
Birkhoff's theorem teaches
us that the exterior vacuum
of any spherically
symmetric body
is described by the
Schwarzschild metric.
This is true even if the
source is time varying.
So in doing this
derivation, I focused
on a situation in which
my spacetime was static.
There is no time
dependence whatsoever.
It turns out that
if it is not static,
if it is a time-varying
spacetime, as long as the time
variations preserve
spherical symmetry,
the exterior is still
described by Schwarzschild.
It's even stronger than if--
only if.
As long as the time variations
preserve spherical symmetry.
So for instance, it could
be radial pulsations.
If it is not a variation that
preserves spherical symmetry,
you're very likely to have a
time-varying quadrupole moment,
and then you produce
gravitational waves.
But if you have
time variations that
preserve spherical
symmetry, spherical symmetry
does not allow us to have a
time-varying quadrupole moment.
So no gravitational
waves are produced.
And what this tells you
is that as long as you're
in this vacuum region
on the outside of it,
your thing could be down
there going [VOCALIZING],,
as long as it does
so in such a way that
leaves it spherically
symmetric, then that
is the spacetime
that describes it.
You are completely ignorant
of what the radial extent
of the matter is.
It only matters that
you have a mass of m,
and you are distance r
away from its center,
at least in these coordinates.
In that sense, it's kind of
like Birkhoff's theorem plays
a role similar to Gauss's
law in elementary electricity
and magnetism.
If you have a spherically
symmetric distribution
of charges, you are completely
agnostic as to what that--
what the radial distribution
of that spherical symmetry is.
And in fact-- although, this
isn't usually discussed--
if that were to be sort of a
time variable kind of thing,
as long as it was just
sort of homologous
in moving these guys
in and out like that,
you would still just get the
E-field associated with a point
charge at the origin.
This spacetime is kind of
the gravitational equivalent
of a point charge at the origin,
a point I'm going to come back
to in a future lecture.
Let me just give you a brief
proof of Birkhoff's theorem.
So imagine I want to consider
a spacetime in which I
allow my metric functions
to vary with time.
So go ahead and work
all of these guys.
Use that as your metric.
Work out all of your
curvature tensors.
We are focusing on the exterior
vacuum region of the star.
So what we're going to do is
set those curvature tensors
equal to 0.
What you find when
you do this-- so
when you work out your
curvature tensors, in addition
to those four that we had on
the center boards earlier,
the four components
of Ricci, you
find a new component enters.
What you find is that there
is a component r sub tr,
which looks like--
oops, pardon me--
which looks like 2/r times the
time derivative of your lambda.
This you are going to set
equal to 0 in the exterior.
So this tells you that the
time derivative of lambda is 0.
This means that your
assumed lambda of t and r
is just a lambda of r.
So there is no time
dependence in that lambda.
So you just killed that.
You can then look at your
other curvature components.
And in particular, one of
the things that you'll see
is several of those
curvature components.
You'll see that they pick up
a couple of additional terms
involving d by dt as the various
of the lambdas and the phis.
You can now get rid of those.
Where I think it's particularly
clean to look at this
is if you look at r theta theta.
That turns out to be
completely unchanged.
So your analysis of
the r theta theta term,
it could proceed
exactly as before.
But the other
thing you can do is
you can say, hey, guess what.
I'm now going to imagine that
my phi is time dependent.
My lambda is not time dependent.
We just proved that.
But my phi could be.
Let's look at the time
derivative of this Ricci term.
So if our r theta theta
is 0, the time derivative
of r theta theta must be 0.
So when I compute
this, this leads
to the requirement that
the double derivative,
one derivative in r,
one derivative in phi--
excuse me-- one derivative
in r, one derivative
in t of the potential
phi must be equal to 0.
And so this in turn tells us
for this condition to be true,
it must be the case--
pardon me.
Just one second.
How does this follow?
Oh, no.
Yeah, yeah.
So this tells me
it must be the case
that phi can be separated
into a function of r
in a function of t.
If I take the t derivative and
then the r derivative, I get 0.
If I take the r derivative and
then the t derivative, I get 0.
So my line element, my
most general line element
now appears to be something
that looks like this.
But once I have this, I
can actually just redefine
the time coordinate.
I mean, if I look at
this thing I've got here,
I can just absorb a function
of t into my time coordinate,
just absorb it in a
time coordinate like so.
And what happens is that I then
recover the Schwarzschild form
I originally had.
So that's really nice.
What this tells us is that any
spherically symmetric vacuum
spacetime is going to have
a time-like Killing vector.
And if it has a
time-like Killing vector,
that means that I can define
a notion of conserved energy
for objects that are
moving in that spacetime.
It's going to prove
very useful when
we start discussing orbits.
So this punch line
is this totally
describes the exterior of
my compact spherical body.
What about the interior?
What we're going
to need to do is--
the interior has
non-zero stress energy.
So we're going to need to look
at components of the Einstein
equation and equate them to the
source, my perfect fluid stress
energy tensor, and
see what develops.
So what I'm going to do is
write my Einstein equation
in the form of an
upstairs downstairs index.
And I'm going to do it that
way, because t upstairs mu
downstairs nu, it is represented
in the coordinate system I'm
working in here.
But assuming it's a
diagonal of minus rho ppp.
And I had all the Riemann--
excuse me-- all the Ricci
components on the blackboard
a few moments ago.
From them, I can construct
my Einstein tensor.
So there's a couple components
that turn out to be quite nice.
One of them is the tt component.
This turns out to be minus
1 over r squared d by dr r 1
minus e to the minus 2 lambda.
What's nice about this is notice
it only depends on lambda.
This is going to be
equated to 8 pi g times
t stress energy tensor tt, which
is going to give me a minus rho
there.
So before I go and I do this,
let me make a definition.
I'm going to call e
to the minus 2 lambda,
I'm going to call that
1 minus 2gm of r over m.
What I'm effectively doing is
I am mapping my metric function
lambda to a function
m of r, which
is a sort of mass function.
Hopefully, you can kind
of see that by doing this,
remember, that this
component of the metric,
it's going to enter
as e to the 2 lambda.
And so when I
match this, this is
going to give me something
that very nicely matches
to my exterior spacetime.
So plugging that in there,
and valuing the derivatives,
and equating this guy to
the stress energy tensor
is going to give me a very nice
condition that the function
m of r must obey.
So plugging all this stuff
in, I end up with minus 2g
over r squared dm dr equals
minus 8 pi g rho of r.
Turning things around,
I can turn this
into an integral solution, which
tells me that my mass function
m of r is what I get when I
integrate from 0 to r 4 pi
rho of r prime.
Two comments on this.
You sort of look at this and go,
ah, that makes perfect sense.
This is what you
would obviously get
if you integrate up a density
over a spherical volume.
This is, hopefully,
beautifully intuitive.
And you kind of
look at it and go,
how could it be anything else?
Well, it could be anything else.
So two comments to make.
First of all, I
have set this up.
I have imposed a
boundary condition.
I have imposed a
boundary condition here
that the mass enclosed
at radius 0 is 0.
You might look at
that, and, well, duh.
If I have a ball of radius 0,
there can't be any mass in it.
Of course, you're
going to do that.
Well, it's going to turn out
that black holes are actually
solutions that violate
this boundary condition.
Wah-wah.
So there's that.
The other thing to bear
in mind is that 4 pi--
the reason why this probably
makes intuitive sense
is that in spherical
symmetry in Euclidean space,
4 pi r squared dr is the proper
volume element for a Euclidean
three-volume.
We are not in Euclidean space.
And so this is indeed
the definition of m of r,
but it is not what we would
get if we did a proper volume
integration of rho
over the volume that
goes out to radius r.
If I were to define a quantity
that is the proper volume
integrated rho, let's
call this m sub b.
Let's make it lowercase.
I'll define why I'm calling
that b in just a moment.
I could define
this as what I get
when I do a proper volume
integration of this thing
in my curved spacetime.
You get an additional factor.
And what you find is
that this in general--
so remember, this is going
to go into your spacetime
as a function.
It's going to look
like something
like 1 over that factor there.
This in general is
greater than m of r.
So there is a homework exercise
where you guys are going
to explore this a little bit.
m sub b is what I get when
I count up, essentially
including a factor of the
rest mass of every little--
let's call it baryon
that goes into this star.
This counts up the
total number of baryons
present in my star
in something that's
called the baryonic mass.
This m that appears here
defines the amount of mass
that generates gravity.
They are not the same.
The gravitational mass is
generally somewhat less
than the baryonic mass.
You might sort of think,
well, where did it go?
The missing mass
can be regarded as
gravitational binding energy.
It's really not
that it's missing.
It's just that
gravitational energy
holds this thing together.
And whenever you
bind something, you
do that by putting it
in a lower energy state.
Gravity is essentially
this sort of mb with this--
the way to think of my
mb here is bear in mind
that we're including factors--
let's just put factors of
c squared back in here.
This is adding up
the rest energy
of every particle in this star.
But when I put
them all together,
gravity binds it together.
And that actually takes
away some of that.
It puts it into a bound state.
And so the difference between
the baryonic mass, or energy,
and the gravitational
mass, and energy,
tells you something about how
strongly bound this object is.
So that's one
important equation.
It looks like I might
finish this lecture
up a little bit
on the early side,
which is fine, because
I could use a break.
Another one that
is important for us
will be the rr component of
the stress energy tensor.
So this guy, when
you work it out,
it looks like e to the
minus 2 lambda 2 over
r d phi dr plus
1 over r squared.
That's 1 over r squared.
So let's insert our definition.
This prefactor is
minus-- excuse me--
1 minus 2gm of r over r 2 phi--
oh, shoot.
Pardon me.
Try that again--
2 over r d phi dr plus 1 over
p square 1 over r squared.
And we equate this to
the appropriate component
of the stress energy
tensor, which is just
going to be the pressure.
This can be
rearranged and give us
an equation for a differential
equation governing
the metric potential, the
trigonometric function phi.
So what results of
that exercise something
that looks like this over this.
So I would like to do something
with this in just a moment.
But it's worth
commenting on this.
So when one looks at the
gravitational potential
in a spherical fluid body
in Newtonian gravity,
you get gm of r over r squared.
So there's two interesting
things going on here.
First, your m of
r, which appears
in the numerator
of this thing is
corrected by a term that
involves the pressure.
What you are seeing
here is the fact
that pressure in some
sense reflects work that
is being done in this body.
It's being squeezed.
Gravity is sort of
squeezing it down,
or your hands are
pushing it down.
When you give this
body some pressure,
you're doing work on it.
Work adds to the energy
budget of that body.
Energy gravitates.
In the denominator, you're
getting this r minus 2--
instead of just
having an r squared,
you get a correction that
involves this 2gm term in here.
And this is just
correcting for the fact
that you are not working
in a Euclidean geometry.
So now I want to do a
little bit more with this.
And so let me borrow a result
from an old homework exercise.
I believe this was on P set 3.
Although I don't quite recall.
So on this thing
you guys took a look
at the equation of local
stress energy conservation
for a perfect fluid and
hydrostatic equilibrium.
And what you found was that this
turns into a condition relating
the pressure,
density, and what you
can kind of think of as
the four-acceleration
of the fluid elements.
So we see sort of gradients
of the pressure entering here,
the density entering into
here, and some things
related to the behavior
of the fluid elements.
The spacetime that
we are working with
is one where we have constrained
the behavior of that fluid.
So we are going to
require that our fluid--
let's write it this way.
So our fluid only has a-- it's
only the time-like component
of its four-velocity is 0.
And so when I shoot that all
through the various things
involved in there-- oh, and, of
course, phi only depends on r.
So what this equation
then boils down
to is rho plus p times this
Christoffel symbol equals
minus dp dr. When
you expand this out,
this turns into
d phi dr. So what
this tells me is that d
phi dr is simply related
to the pressure gradient.
When you put all of this
together, what you get
is the following equation.
This is sometimes
called the equation
of relativistic
hydrostatic equilibrium.
My pressure gradient must
satisfy this equation.
Let me write down
two more, because I
think I'm going
to actually begin
with this board in
my next lecture.
So I'm going to just repeat
this equation governing phi.
And m is defined
as this integral.
So here are three coupled
differential equations,
which I can now use
to build a model
for the interior of this body.
We integrate this thing out.
Basically, the way it works is
you need to choose an initial--
some parameter
that characterizes
the initial conditions.
You typically choose a
density at the center.
You choose rho at r equals 0.
You need an equation
of state that relates
the pressure and the density.
And then you just
integrate these guys up.
There's a matching
that must be done
at the surface of the star.
Then once you've
done that, that'll
end up giving you
sort of a notice
that by integrating
up this equation,
you determined the
function phi up
to some constant of integration.
When you match at the
surface of the star,
that lets you fix that
constant of integration.
Boom!
You've made yourself
a spacetime.
This is known as the
Tolman-Oppenheimer-Volkoff
equation, universally
abbreviated TOV.
These are the way in
which we make spherically
symmetric fluid bodies
in general relativity,
spherically symmetric
static fluid bodies
in general relativity.
In the next lecture--
so we're going
to end this one
a tiny bit early.
This is actually a very
natural place to stop.
I am going to start with this.
And what we're going to
do is talk a little bit
about how one solves these.
And we're going to
look at a couple--
we're going to look at one
particular special solution,
which is not very physical,
but it's illustrative.
And I'm going to
talk a little bit
about how to solve this
for more realistic setups.
That is in preparation for what
is one of my favorite homework
exercises.
It's one where I give
you an equation of state.
And you guys just
take these equations.
And using a numerical
integrator--
I hope everyone has access to
something like Mathematica,
or WolframAlpha, or
something like this
in your dispersed lives.
All MIT students should be
able to access the student
license for Mathematica.
So as long as you have
the hardware for that,
you should be OK.
And this is-- actually,
it's pretty cool.
You can make relativistic
stellar models.
I've actually had
in the past students
use this as the basis for
some research projects,
because when you do
things like this,
this is how professionals
make models of stars.
So we will stop there.
And those of you who
are watching at home,
you can go on to the next video,
where we'll be doing this.
As for me, I'm going to
take a bit of a break,
and start recording again
in about a half an hour.
