[SQUEAKING]
[RUSTLING]
[CLICKING]
SCOTT HUGHES: So we'll pick
up where we ended last time.
We're looking at the spacetime
of a compact spherical body,
working in what we call
Schwarzschild coordinates.
We deduce that the line
element describing this body
is of the form ds
squared equals negative e
to the 2 phi dt squared
plus tr squared divided
by 1 minus 2g m of r
over r plus r squared d
omega, where d omega is the
usual solid element, line
element.
And let's see.
Yes, the body is
described, interior,
as a perfect fluid with
particular density profile
rho of r, a pressure
profile p of r.
That describes this thing.
Everywhere inside some radius r
star, which gives its surface.
Sorry, I just distracted myself.
I don't know why I've
been calling this d omega.
Should be d omega squared.
Whatever.
Fluff in that notation.
OK, so in the exterior of
this thing-- so for everywhere
for r greater than our
star, it is vacuum.
There is no density.
There is no pressure.
Once we're outside this
thing, the only mass you see
is the mass of the star.
So another way of saying this
is that when-- you know what?
I'll get to that
in just a moment.
And in the exterior, the e to
the 2, 5 becomes 1 minus 2g,
that same mass over r.
Everywhere in the interior,
the pressure, the function phi,
and the mass r are governed
by these equations.
So the mass, as I
described last time,
it looks like a deceptively
simple spherical integral.
But just be aware
that when you do this,
you are not integrating
over a proper volume.
If you were to integrate
over a proper volume,
you would get a larger mass,
and such a mass, in fact,
does have meaning to it.
And the difference between
that mass and this mass
tells you something
about how gravitationally
bound this object is.
And these two equations
actually have a Newtonian limit
associated with them.
So this is a
relativistic version
of the equation of
hydrostatic equilibrium,
and this has a simple
Newtonian analog.
Did they drop the minus sign?
No.
This has a simple
Newtonian analog,
describing the
gravitational potential
inside a fluid object.
These whole things
taken together
are called the
Tolman-Oppenheimer-Volkoff
equations, or the TOV equations.
So the comment I was making
was that we require m of r star
to equal m, the total
mass of this object.
So that's another
condition on this.
Once you integrate up,
then you switch over
to the mass, the
total mass, that
is used in the
exterior of the star.
So if you wish to
solve these things,
what you basically do is just
choose a central density.
You have to choose
an equation of state,
which allows you to relate
the pressure to the density.
We'll talk about that a little
bit more later in this class.
And then you just
start integrating.
So you basically then
integrate until you find
that the pressure equals 0.
The radius at which this occurs
defines the star surface.
As you integrate along,
this allows you to build up.
You then build the mass profile
of the star, the pressure
profile of the star.
You are building up how the phi
changes as you integrate along
the star.
So note, if you
integrate this up,
this is only defined up to
a constant of integration.
And so what you
then need to do is
once you have reach the surface
of the star, you know m of r.
And so what you're
going to need to do
is adjust the phi that
you found in order
to match with the exterior
solution 1 minus 2gm over r.
This is an exercise you will
do on an upcoming homework
assignment.
This is one of my favorite
assignments in the class.
It's a really good
chance to actually see
the way we solve
these equations.
And this kind of
an exercise, it's
done all the time as
part of modern research,
and you will do this for
a particularly simple kind
of equation of state.
So I want to look at what some
examples of objects like this
actually look like.
So what we're going
to do is consider
an unrealistic but
instructive idealized limit.
So imagine a star that
has rho equal constant.
That is something where no
matter how hard you squeeze it,
you cannot change its density.
The only way that
that can happen
is if you have an object
that is infinitely stiff.
This corresponds to a speed
of sound, which is defined
as dpd rho that is infinite.
Now, of course,
no speed of sound
can actually exceed
the speed of light.
So this is a somewhat
pathological object.
Nonetheless, it's useful
for us for the simple reason
that the mass function
that emerges for this
is quite trivial.
Rho is constant, so m of r
just goes as the volume inside
radius r times rho.
The star then has a
total mass of 4/3 pi r,
to be pi rho r star cubed.
So we know how its density
behaves, it's a constant,
we know how the mass
function behaves.
The challenge is solving
for the pressure profile.
So the pressure is governed
by taking this differential
equation and basically
plug in that mass,
plug in this mass function,
and see what you get.
So if I go and I throw
this guy in here,
this is what you end up getting.
So bear in mind as
you go through this--
yeah, I see what I did here.
So what I do as I just
pulled out my factor of 4/3,
4/3 pi rho r cubed-- excuse
me, my 4/3 pi r cubed.
I factor that out,
and then cancel
it an overall factor of r
squared in the denominator.
So this is simply
what I have over here
with this mass function defined.
We want to solve this to find p.
This is one of
those rare moments
where nature and analysis
conspire and a miracle occurs.
OK, it's a somewhat messy
looking kind of solution,
but nonetheless, it turns
out that the pressure profile
is determined by-- you can
manipulate this equation.
You can integrate it up.
And this ends up describing
what the pressure
profile looks like, p sub c
is the pressure at r equals 0.
We define the
surface of the star
as being the location at
which the pressure goes to 0.
So by using that, so by
exploiting that condition,
we can use this equation
to make a mapping
between the radius of the
star and the central pressure.
So it's a spherically
symmetric thing.
Once we have chosen
what this density is,
then there is essentially a one
parameter family of solutions.
We-- if we choose a central
pressure, that determines
what the radius would be.
Conversely, if we want to
have a particular radius,
that determines what the
central pressure must be.
Let's just do a
little of analysis
that follows from this.
So p equals 0 at
r equals r star.
That defines the surface.
So putting all
that together, you
can manipulate this
equation to find
that the radius of
the star is determined
from the central
pressure, like so.
If you prefer,
you can write this
as an equation for the
central pressure in terms
of the radius of the star.
I've written this
in terms of m tot
but m tot is, of
course, simply related
to this radius-- excuse
me, to this density
and the star's radius cubed.
So what's the
importance of this?
This is, as I've emphasized,
a fairly idealized problem.
Nature will never
give us an object
that has constant density.
Any object, if you give it
a little bit of a squeeze,
the density will change.
Causality requires
that the speed of sound
be less than the speed of light.
So clearly, this is a
somewhat fictional limit.
But we can learn something
very interesting about this.
Notice that this formula
for the central pressure,
it diverges for a particular
compactness of the star.
So p usb c goes to infinity.
Perhaps a little bit
more easily to see,
the denominator goes to 0, for
a certain compactness, m over r.
So let's look at the value
at which the denominator goes
to 0.
So let's see.
Move my 1 to the other side.
Divide by 3.
Square it.
Rearrange terms.
So if the ratio of this star's
total mass to its radius
is such that gm over
r star exceeds 4/9
than the pressure diverges.
What this tells me is that I
cannot construct a physically
allowable static object, even
using this stuff as we can
imagine fluid, a fluid that
has an infinite sound speed.
I cannot make a star
more compact than that.
Making it that compact requires
infinite pressure at the core.
This implies that stars
have a maximum compactness.
We cannot have physically
realizable pressure profiles
if--
and let's turn that around--
if the ratio of the
radius to g times
the total mass is
smaller than 9/4.
Now, this holds for the stiffest
possible fluid that we can even
imagine, one which the laws of
physics actually do not permit.
And so one infers
from this, actually,
that this bound
basically tells me
that given any physical fluid,
any physically realizable star
that I can construct,
I must have
a maximum allowed compactness.
Putting this a little
bit more precisely
brings us to a result that is
known as Buchdahl's theorem.
Buchdahl's theorem
tells me that there
is no stable spherical
fluid configuration
in which the
configuration's radius is
smaller than 9/4 of gm total.
For those of you who like to
put factors of c in there,
divide by c squared, and
this tells you something
that you can convert
to SI units, which
tells you how small you are
allowed to make an object.
So this is clear.
It emerges, and very nicely, in
this idealized but unphysical
fluid limit.
But it can be proven
more generally.
You just have to make a few
assumptions about the way
that the pressure profile is
not singular in any place.
Proof of this can be found in
the beautiful, old textbook
by Weinberg, section 11.6.
Well, when you
hear something like
that, you gotta
think to yourself,
well, suppose I made a star
with some kind of a fluid
and I gave it a radius of
10/4 gm total over c squared.
And I just came along
and squeezed it.
What would happen?
Well, notice the word
"stable" in this definition.
You're free to do
that, and should you
do so, you would simply no
longer have a stable object.
So remember, part of what
went into this analysis
is we were assuming the
spacetime and the fluid that
is the source of
the spacetime, we
were assuming
everything is static.
OK, we're making sure
everything just sits still.
This is telling us you
can't do that if you
want to have a star
as compact as that.
So if you were to do
this, it would collapse.
It would become dynamical
and the spacetime
would transition
into something else.
What that something
else might be
will be a topic that we
get into a little bit more
after we have developed
some additional material.
So let's talk a little bit.
This will help you with the
homework assignment, where
you guys are going to construct
relativistic stellar models.
Let's talk a little bit about
how we describe real objects.
They are not of
constant density,
and they instead
have some p that
is a function of the density.
It's worth noting that in
an even more general case--
this is actually
worth a brief aside--
the equation of state relates
the pressure to local density
and s, where s is the entropy.
For the kind of applications
where general relativity tends
to be important, for
instance, when we're studying
the stellar structure
of a neutron star,
the fluid ends up being so
cold that you don't need
to worry about the entropy.
And where that comes
from is that when
I revisit my first
law of thermodynamics
and I include temperature and
entropy effects, so the term
that I left out in my
earlier accounting, TDS.
If t is small, I can
ignore that term,
and it ends up being
something where
my local energy,
or energy density,
ends up only depending
on the pressure.
But "cold" is a wiggle word.
I have to define a scale
to say whether an object is
cold or hot.
There's a few notes laying
this out right in my notes.
Let me just sketch the key idea.
Cold depends on the
fluids, and it's worth
noting that the kind of fluid
you're playing with here
tend to be made out of fermions.
And so it depends on the
fluids' Fermi temperature.
So the Fermi
temperature is defined
as the Fermi energy normalized
to a Boltzmann factor.
You get the Fermi
energy by looking
at how the energy levels are
filled in your fermion fluid
here.
I have a few additional
notes to lay out
the way in which you can relate
this to the local density,
the mass of each particle that
goes into this Fermi fluid.
The punchline is that
for neutron stars, one
of the cases where
we do, in fact,
make dense general
relativistic fluid stars,
the Fermi temperature tends
to be on the order of 10
to the 13 Kelvin, so
about 10 trillion Kelvin.
When we actually
observe these objects,
they are on the order of 10 to
the 6 to 10 to the 9 Kelvin.
So they're a factor of about
10 of the 7, 10 to the 4 to 10
to the 7 times colder than
the Fermi temperature.
Even though they may be a
billion Kelvin, they are cold.
So we're going to use what are
called cold equations of state
to describe these guys.
So with that out of the
way, let's talk a little bit
about the kinds of equations of
state that we will tend to use.
So people who study the
physics of dense matter,
a lot of their lives is
really down to understanding
what the equation of state of
that cold matter looks like.
Some of them are
concerned about hot matter
as well, in which case they
might be actually worrying
about things at tens
of trillions of Kelvin.
But if you're looking at
astrophysical applications,
you're generally interested
in the cold matter.
And so they end up putting the
other very complicated models
using QCD and effective
field theories to try
to understand how it is
that a particular fluid
of dense matter, how its
pressure and its density
are related.
And what I'm sort of
wheeling around here
is that you generally do not
have a simple analytic form.
You wind up with some kind of
a fairly complicated function
that emerges from a
numerical calculation.
It often ends up being--
if you are a user of
this equation of state--
it ends up being in
the form of a table.
So they might
actually just give you
a file that's got
a bunch of numbers,
which says if the
density is this,
then the pressure is this.
And you can fit little
functions to that
that allow you to look things
up and do your calculations.
But it's not in the
form of a clean thing
that you can write
down on the blackboard.
I want something clean I can
write down the blackboard.
So I'm going to introduce
an approximation, which
is useful for testing
things out, test cases,
and for pedagogy.
What we do is we
take the pressure
to be a power law
of the density.
So what we do is we
write p equals k rho
0 to the gamma, where k
and gamma are constants.
A form that looks like this,
this is called a polytrope.
Now, the thing which I
particularly want to highlight,
and for those of you who
are going to do this highly
recommended homework
exercise, please
pay attention at this point.
This rho 0 is not--
oh, I erased it.
This rho 0 is not the rho--
oh, there it is--
it's not the rho that appears
in the equation of state.
It's slightly different.
Rho 0 is not the rho that
appears, for instance,
in the t of e equations.
Rho 0 is what is called
the rest mass density.
It does not take into account
the fact that if I take a big--
let's say I've got a big
bucket of nuclear fluid.
So I take my bucket here
and I squeeze down on it.
When I squeeze
down, its density is
going to increase,
first of all, because I
have decreased the volume.
So the number of
particles remains fixed,
but I decrease the
space in there.
But I have also done work on
it, because this thing exerts
a pressure that
opposes my squeezing.
And I need to take
into account the fact
that the work I do in
squeezing this fluid
increases the density rho.
So when you write out
your t of e equations,
rho is energy density.
All forms of energy gravitate.
This is just the way
people traditionally
write the equation of state.
This is when one is
doing nuclear physics.
There's good reasons
for doing this,
but it's not the
most convenient form
for the kind of calculations
that we want to do,
and that you are going to want
to do, in the problem set.
Fortunately, it's not too
difficult to convert, so let
me describe to you
how you do that.
So we are going to use the
first law of thermodynamics
in a form in which I've
written now a couple times.
Interestingly, it showed up
in our cosmology lecture.
So my first law tells me du
equals minus pressure dv.
So this is my total
energy in a fluid element
and this is the work
done on a fluid element.
So rho is equal to the amount
of energy in a fiducial volume.
My rest energy-- excuse
me, my rest density--
is the rest energy of
every little body that
goes into this per unit volume.
This means that I can write
du as d rho over rho 0,
provided I throw in an
extra factor of m rest
to get the dimensions right.
And I can write d
volume as d1 over rho 0,
provided I throw in
that factor of m rest
to get the dimensions right.
I know this looks weird
but, it's perfectly valid.
So I'm going to rewrite my
first law of thermodynamics
as d rho over rho 0 equals
minus pd 1 over r 0.
Let's manipulate
that right-hand side.
So I'm going to assume
this polytropic form.
I'm going to use p equals
k rho 0 to the gamma.
But I'm going to
switch that around.
I'm going to write this
as rho 0 equals p over k
to the power of 1 over gamma.
So when I do that, I
get d rho over rho 0
equals kappa 1 over gamma
over kappa pdp over p to the 1
plus 1 over gamma.
Pardon me just one moment.
I did something clever
in my notes here
and I'm just trying to
make sure I understand
what the hell I actually did.
So I'm going to level with you.
I've gone through
this several times.
There's a step in the
calculation that at this point,
I, for some stupid
reason, didn't write down.
I'm going to trust I knew
what I was doing, though,
because I know the
final result was right.
You can integrate
up both sides here.
Oh, I think I see what I did.
OK.
So you integrate
up both sides here.
And what you find
is this becomes rho
equals p over gamma
minus 1 plus a constant.
Yeah, not 100% sure
how I actually did
that, so my apologies on that.
I'm going to assume I
knew what I was doing.
I will try to fix this and
I may post an addendum here.
The next step, actually,
is you want to determine
what that constant is.
So the way you
determine the constant
is you take advantage of the
fact that rho goes to rho 0.
The energy density becomes
the rest energy density
if there is no pressure exerted.
And so this gives us our
final relationship here,
which is that rho equals rho
0 plus p over gamma minus 1.
OK, I will double check how I
went from line 2 line 3 there,
but the final thing that
I have boxed online for
is, indeed, exactly what
you need to do in order
to build a stellar model.
I guess I've been emphasizing
this is something you will
do on an upcoming problem set.
Let me just sketch the recipe.
I've said this verbally, but let
me just write it out explicitly
here.
So I will give you
an equation of state.
You then need to pick
rho 0 at r equals 0.
Using your equation of state
and using that relationship
between rho and rho
0, this will give you
rho at the center,
pressure at the center.
Set m of r at the center to 0.
I emphasize, again, that
this may seem obvious,
but it is somewhat important
that you get it right.
When you do this
homework assignment,
I'll give you a
little hint as to how
to build that in smoothly.
It can be a little bit--
I don't want to say
tricky, but it's worth
thinking about a little bit.
Then what you do is integrate
your equations for the pressure
in the mass from r equals 0.
And this cannot be
done analytically.
You have to use a
numerical integrator.
If you have never used
one of these before,
I will give you a
Mathematica notebook that
demonstrates how to use it.
This is a skill that
is worth knowing.
The plain truth of the matter
is that the class of problems
that are amenable to purely
analytic solutions, those
are interesting.
They're illustrative.
They're good to work with.
But they tend to be
unphysical and they're just
not the ones that are of
interest for many things
that we study in science.
So show while
you're doing this--
this is not necessary
to make your model,
but it's very useful to do this.
You can also integrate,
whoops, d5 er from the center.
So a caution is that you do
not know phi at r equals 0.
So what you should do is just
temporarily set it equal to 0.
And what you're going to be
doing then when you integrate
this up is you will
calculate the delta phi that
describes your model from the
center to the surface, which
brings me to step 4.
When you find p equals 0,
you've hit the surface.
So what we do is we use
the fact that p of r
equals 0 defines the
star's radius r star.
Once you've done that, you
now know the total mass
and the radius.
So you will find, when
you're doing this,
that your numerical integrator
is not super well-behaved
as you approach the surface.
This is a feature, not a bug.
What's going on is that as you
begin to approach the surface,
the gradient and the
pressure gets quite steep.
And so the way one
numerically integrates
a set of couple equations
like this is by, essentially,
if you take advantage of
the fact that an integral,
it's what you get by sort
of dividing things up
into tiny little pieces and
add up like little rectangles.
And when you're solving
a differential equation
like this, you're essentially
taking the continuum solution--
that you guys have learned
how to do in many cases--
and you're approximating
it by a series of smaller
and smaller finite steps.
Because the gradient
in the pressure
gets large as you
approach the surface,
numerical integrators
typically try
taking an infinite number of
infinitesimal steps, which
makes the CPU sad, and so it's
likely to exit with an error
condition.
Generally, when
that has happened,
you've gotten an
answer that's probably
good to within a
part in a million,
or something like that.
Fine for our purposes.
If you need to do something
a little bit more careful,
that's a subject for a
numerical analysis class.
For us, I will
give you some hints
on this when you begin
exploring these solutions.
So you have an additional
boundary condition.
You know by Birkhoff's theorem
that the Schwarzschild metric
describes the exterior.
That means gtt is minus 1 minus,
given by this for everywhere
greater than r star.
This gives us a boundary
condition that phi of r star
must be 1/2 log 1 minus
2m total over r star.
By enforcing this
boundary condition,
you can go back to
your solution for phi
and you can figure out
what the value at r
equals 0 should have been to
give you a continuous function
that matches at the surface.
So that's it for
spherical stars.
I look forward to you
doing these exercises.
My own biases are
perhaps coming out,
but these are a lot of fun.
The one thing which
I will do for you,
and I regret that my notes
didn't really have this,
is I will try to figure out
how on earth I went from line 2
to line 3 in this
calculation over here,
going from the rest mass
density, the rest energy
density to the energy density.
My apologies that
that's not there.
All I can say is that there are
many distractions these days
and I overlooked that when
I was reviewing my notes
and preparing for
today's lectures.
I'd like to take this moment to
take a little bit of a detour.
Let's imagine that we
have a spacetime that
is Schwarzschild everywhere.
In other words, it has this
form for all r, not simply
the exterior of some object.
We already know that this
spacetime is a vacuum solution.
I know that t mu nu equals 0.
Back up for a second.
If I generate the
Einstein tensor for this,
I will get identically
0, which implies
that this corresponds to
a solution, which has t
mu equals 0.
I also know, though,
that if I examine
the behavior of radial
geodesics in the weak field
of this spacetime,
I find that they
fall towards this
like an object that
is falling towards a mass m.
So this spacetime
appears to be something
that is everywhere vacuum.
There is nothing
in this spacetime,
and that nothing
has a mass of m.
I hope that bothers you.
That is among the sillier
things that has been
said in the name of physics.
That sure sounds silly.
But let me remind you that we,
in fact, have seen something
very similar in a much less
complicated theory of physics.
So if I look at the electric
field of a point charge
at the origin--
so that's the three vector
e is just q displacement
factor over r cubed.
If I compute the divergence
of this electric field--
the divergence, of course, tells
me about the charge density--
and I get 0.
So this is an electric field
that has no charge density
anywhere, but that
lack of charge density
has a total charge of q.
This was something that we
easily learn how to resolve.
Usually at the MIT
curriculum, this often
shows up when you take
a course like 8.07.
What we do is we say, oh,
all that's going on here
is that I have a singular
point charge at r equals 0.
So yeah, I've got
no charge density,
but I have a total charge.
Fine.
We were happy with that.
I want you to think of
the Schwarzschild metric
as doing something
similar for gravity.
There is no source
anywhere, but there is mass.
Maybe there's just something
singular and a little
funny going on at r equals 0.
You might be
concerned about what's
happening there at r equal 0.
When I say it plays
a similar role,
it plays a similar role to
the pull on point charge.
So there'll be nothing
there, but perhaps there's
something funny going
on at r equals 0.
And by the way,
the field equations
that govern gravity, my
relativistic theory of gravity,
they're non-linear.
So when I say there's something
funny going on at r equals 0,
it could be really funny.
So we're not going to get
too worked up about that,
but we're just going to bear in
mind this is odd. t mu equals 0
but it has mass.
So let's look at the
spacetime itself.
Just staring at this, we can
see two radii where it appears
something odd is going on.
So you can see right away,
lots of stuff kind of blows up
and behaves badly at r equals 0.
And you can also see that
your gtt and your grr,
they are behaving in a way that
is potentially problematic when
the radius is 2gm.
So you look at that and
think, yeah, there's
two radii there that look sick.
I am worried about
this spacetime.
Well, we should be cautious.
One of the parables that we
learned about when we studied
linearized gravity is
that we can sometimes
put ourselves into a coordinate
system that confuses us.
When we study
linearized gravity,
we found a solution
that looked everywhere.
It looked like the entire
spacetime metric was radiated.
And it turned out only
two of those 10 components
were radiative.
That turned out to
be something that we
were able to cure by introducing
a gauge transformation.
Doing that here's a
little bit trickier,
but we're going to
need to think about,
how can I more clearly call
out the physical content
of this spacetime?
So one of the
lessons that I hope
has been imparted
in this class so far
is that if you really
want understand
the nature of gravity, you
want to go from the metric
to the curvature.
So what I'm going to do is
assemble an invariant scalar
from my curvature.
And I'm going to use
the Riemann tensor
because I know Ricci
vanishes in this spacetime,
so that wouldn't give
me anything interesting.
So what I'm going to do
is assemble an object.
I'm going to call it capital
I. And that's just Riemann
contracted into Riemann.
This actually has a name.
It is known as the
Kretschmann scalar.
And you can go in.
You can work out all
these components.
The gr tool that is posted
to the 8.962 website
is something you
can explore with us.
And this is just a number.
Turns out to be 48 g squared
m squared over r to the sixth.
What does this guy mean?
Well, in an invariant way,
it's kind of Riemann squared.
Riemann tells me to go
back and think about things
like geodesic deviation.
It tells me about the
strength of tides.
So roughly speaking, square
root I is an invariant way
of characterizing tidal forces.
So if you're sitting around
in the Schwarzschild spacetime
and you want to give
yourself an estimate of what
kind of tidal forces are
likely to act on you,
compute the Kretschmann
scalar, take its square root,
and that'll give you an idea
of how strong they typically
tend to be.
So notice, when we look at this,
this tells us r equals 2gm.
If you plug r
equals 2gm in there,
nothing special about it.
It's a radius just
like any other.
As you go from 2.001gm
gm to 1.99999gm,
it increases a little bit.
Of course, it's got the 1
over r to the sixth behavior,
but it's not like there's
a sudden transition,
or anything particularly special
happens right at that radius.
But it is hella
singular at r equals 0.
So sure enough, r equals
0 is a place where
tidal forces blow up.
OK, fine.
We're going to need to do a
little bit more work then,
because I still want understand,
yeah, OK, r equals 2gm.
There's no diverging
tidal forces there,
but that metric still
looks wacky at that point.
So what is going on there?
So let's think about the
geometry of the spacetime
in the vicinity of 2gm.
So let's imagine.
Let's do the following exercise.
Suppose I draw a circle
at some radius r that's
in the theta equals
pi over 2 plane.
So I'm just sweeping
around in phi.
I'm making this like so.
So here is my r cosine phi axis.
Here is my r sine phi axis.
Here is my circle of radius r.
And let's ask,
what is the surface
area that this guy sweeps out
as an advance forward in time?
So as this thing
goes forward in time,
it sort of sweeps out a
cylinder in a spacetime diagram.
Let's compute the proper area
associated with this cylinder
that this circle is sweeping
out as it moves forward in time.
So the surface
area of my tube, I
integrate from some start
time to some end time.
I'm going to integrate
around in phi,
and then the proper
area element that I
need to do this is going
to be gtt g5 phi, the 1/2.
There's actually a minus sign
in there to get the sign right.
Let's write it like this.
To remind you how I do this,
think of this area element
as a 2 volume.
Go back to some of our
earlier discussion of defining
integrals in spacetime, and this
is the proper area associated
with a figure that has some
extent in time and extent
in angle.
So let's compute that guy.
So I take my
Schwarzschild metric.
g5 phi in the theta equals
pi over 2 plane is just r.
gtt is the square root
of 1 minus 2gm over r.
So the area of my
tube is going to be
r, integrate from my start
time to my end time, dt.
So this is easy.
So I get a 2 pi 2 pi r square
root 1 minus 2g m over r,
and let's just say my
interval is delta t.
Notice what happens
as I take the radius
of this thing down to 2gm.
This goes to 0.
This r goes to 2gm.
If I go inside 2gm, I don't
even want to compute that.
Something has gone awry.
But look, I can draw
this thing just fine.
Clearly, there's
a surface there.
It's got to have an
area associated with it.
Why are you telling me
that the area of this thing
is 0 in that limit and is
a nonsense integral if I
go inside this thing?
Well, what's happening
is we have uncovered
a coordinate singularity.
The time coordinate is
badly behaved as we--
not well-- as we approach
this radius, r equals 2gm.
Let me give you an analogy
that describes, essentially--
it's something that is
very similar to that tube,
that world tube
that I just drew.
But let me do it in a,
perhaps, more familiar context.
Suppose I want to
draw a sphere, and all
that I know about a sphere
is that it has got two
coordinates to cover it, an
angle phi and an angle theta.
And so I could say,
OK, here is my sphere.
Here is theta equals
0, phi equals 0.
Here is phi equals pi over 2.
Here's phi equals pi.
Pi equals 3 pi over 2.
Phi equals 2 pi.
Theta equals pi over 2.
Theta equals pi.
There's my sphere.
So this chart that I've
just drawn here, it's true.
This does represent
the coordinate system,
but it's a horrible rendering
of a sphere's geometry.
What I didn't realize when
I wrote this down here is
that, in fact, at theta
equals 0 and theta equals pi,
every phi value should be
collapsed to a single point.
This is reflecting the fact
that if you look at a globe,
all lines of longitude cross the
north pole and the south pole.
Every value of the
azimuthal angle
on the surface of
the earth, they
become singular at the north
pole and the south pole.
This drawing, well,
it's like one of those,
I forget the names of them,
but the various renderings
of a map that try
to take the earth
and write it on a flat
space, and you wind up
with Greenland being three
times the size of Africa,
or something like that.
And it's because there should
be zero area at the top here.
As you approach
the top the area,
it should be getting
much stronger.
And when you do representation
of your map like this,
you're spreading
everything way, way out.
What is going on?
And why this idea of
drawing this world tube that
is swept out by my
circle of radius
r as it advances
forward in time?
That drawing does not
account for the fact
that the Schwarzschild
time coordinate is singular
as you approach r equals 2gm.
It's going to turn out all
times t map to a single sphere,
and r equals 2gm.
To get some insight into
what's going on here,
let's do a little
thought experiment.
What I'm going to do
is imagine I'm at rest
in the Schwarzschild spacetime.
So let's say that I'm
at some finite radius r.
I am not in a weak field.
OK, maybe I am at
something like r equals
4gm, or something like that.
And what I'm going to do is drop
a little rock, drop a particle.
So I'm going to drop a
particle from r equals r0.
I'm going to integrate
the geodesic equation,
and I'm going to parameterize
what its radial motion looks
like as a function of "time."
I put "time" in quotes
here because you
should be saying at
this point, well,
you just told me that time is
doing something kind of funny
here.
What do you mean by that?
I'm actually going to do this
for two different notions
of time.
I'm going to do this for
the coordinate time t,
and I'm also going to
do this for proper time,
tau, as measured along
that world, that infall.
So I'm not going to
go through the details
of this calculation.
It's a straightforward,
moderately tedious exercise.
I would just quote to you
what the result ends up
looking like.
So let's first write down
what the solution looks like,
parameterized by
the proper time.
So this is most easily written
as tau, proper time, and 2gm.
Essentially, I'm
just going to use
it to set a system of units.
I write this thing
as a function of r.
My solution turns out to be--
it looks like this.
So if I were to make a plot of
what this thing's motion looks
like as a function of
time, so here's our 0.
Here is r of tau.
And let's just put in, for
fun, let's say this is 2gm.
Zoom, fallen.
You reach r equals 0
in finite proper time.
The parable of the Kretschmann
scalar is that as you do so,
the tidal forces acting
on you are diverging.
So if you have any last
wishes, send them out
because you're not going
to have a lot of time
to tell people about them.
Let's now write it as a
function of coordinate time t.
This ends up being--
bear with me while I write this
out, this is slightly lengthy.
OK, so what I mean
on this last line
is if you want to get
the complete solution,
just write both of
these functions down.
Again, subtract them off
and place the r's with r0.
When you look at this,
here's what you see.
The motion expressed in times
of the coordinate time t
asymptotically approaches
the radius 2gm,
but it never quite reaches it.
As t goes to infinity,
it eventually reaches--
so r, you can see it
appearing in the behavior
of this natural log.
r gets to 2gm as t
goes to infinity.
So as measured by clocks
on the infalling body,
it rapidly reaches r equals 0.
According to this
coordinate time,
it never even
crosses r equals 2gm.
What the hell is
going on with that?
Well, to give a little
bit of insight into this,
it's useful to stop for a
second and ask ourselves,
what is that coordinate
time t actually measuring?
So let me write down
the Schwarzschild metric
and let's think about this.
So kind of hard to see
what t means in this,
but let's consider a limit.
Suppose I consider observers
who are very far away.
If I look at people who are at
r, much, much larger than 2gm.
For such observers,
spacetime looks like this,
and this is nothing more than
flat spacetime in spherical
coordinates.
This is what we call an
asymptotically flat spacetime.
As you get sufficiently
far away from the source,
it looks just like
flat spacetime.
essentially, special
relativity rules apply.
And that gives us
some insight into what
this coordinate t means.
The t that we are using in
the Schwarzschild coordinate
system, this is time as
measured by distant observers.
Tau is time, as measured
by this infalling observer.
So what we are seeing here
is the infalling observer
crosses 2gm, reaches r equals
0, and has a very short life.
But those who are using
clocks, adapted to things very,
very far away, never
even see it cross 2gm.
Why is that?
Well, we will pick this
up in the next lecture,
but let me remind you
that when we initially
began working on this subject,
one of the very first lectures,
we talked about something called
the Einstein synchronization
procedure, where what we did
was we imagined spacetime
was filled with a conceptual
lattice of measuring
rods and clocks.
And we synchronized
all of those clocks
by requiring that the time
delay between different clocks
is synchronized
according to the time it
takes for light to travel
from one to the other.
This is telling us we
are actually working--
when we use
Schwarzschild time, we
are working in a system
that reflects an underlying
inheritance from
special relativity.
These are clocks that have been
synchronized by the Einstein
synchronization procedure.
And so the pathological
behavior that we
see here, it must ultimately owe
to the behavior of these clocks
that we use to define
our coordinate system,
and the behavior of
those clocks is linked
to the behavior of light.
So in order to get insight as to
what is going on with this, why
is it that if I use a clock
adapted to the infalling body,
I see painful death, but
if I use a clock adopted
to someone very far
away, I don't even
see it approach that dangerous
r equals to a radius.
In order to resolve
that mystery,
I'm going to need to examine
what the motion of light looks
like in this spacetime.
We'll pick that up
in the next lecture.
