[SQUEAKING]
[RUSTLING]
[CLICKING]
SCOTT HUGHES: All right.
Good morning, 8.962.
This is a very weird experience.
I am standing in here talking
to an empty classroom.
I have some experience
talking to myself,
because like many of us, I
am probably a little weirder
than the average.
But that does not
change the fact
that this is awkward
and a little strange,
and we already miss
having you around here.
So I hope we all get through
MIT's current weirdness
in a healthy and quick
fashion so we can get back
to doing this work we
love with the people we
love to have here.
All right, so all
that being said,
it's time for us to get back
to the business of 8.962, which
is learning about
general relativity.
And today, the lecture
that I am recording
is one in which we will
take all the tools that
have been developed
and we will turn this
into a theory of gravity.
Let me go over a quick
recap of some of the things
that we talked about in
our previous lecture,
and I want to emphasize
this because this quantity
that we derived about
two lectures ago,
the Riemann curvature
tensor, is going
to play an extremely
important role in things
that we do moving forward.
So I'll just quickly remind you
that in our previous lecture,
we counted up the symmetries
that this tensor has.
And so the four most--
the four that are important for
understanding its properties,
its four main symmetries
are first of all,
if you exchange indices 3 and 4,
it comes in with a minus sign,
so it's anti-symmetric under
exchange of indices 3 and 4.
If you lower that
first and next so
that they are all in
the downstairs position
and you exchange
indices 1 and 2,
you likewise pick
up a minus sign.
Again, keeping everything
in the downstairs position,
if you just wholesale
swap indices 1 and 2
for indices 3 and 4 like
so, that's symmetric,
and so you get the whole
thing back with a plus sign.
And finally, one that's a little
bit non-obvious but can be seen
if you're sort of pigheaded
enough to sort of stare at this
thing and kind of pound on
the algebra a little bit,
if you take the Riemann
curvature tensor--
and this can be at the
end of-- the first index
can be either upstairs
or downstairs,
but if you cyclically
permute indices 2, 3, and 4
and add them up,
they sum to 0, OK?
So that tells you that it's a
constraint on this thing when
I look at the behavior,
this thing with respect
to indices 2, 3, and 4.
We introduced a variant of
the Riemann curvature tensor
called the Ricci curvature.
So the way I do this is if I
take the trace on the Riemann
curvature tensor on
indices 1 and 3, which
is equivalent to taking
it with the indices
on the downstairs position
and hitting it with the metric
like so, I get this quantity,
which I forgot to write down,
is symmetric.
One of the major important
physical applications
of the Riemann
curvature tensor is
that it allows us to
describe the way in which two
neighboring geodesics-- if
I have two geodesics that
are separated by
a four vector c,
and I look at how that
separation evolves as they move
forward along their geodesic
paths, this differential
equation describes
how it behaves.
And the key thing
is that what we see
is that the rate of separation
is proportional to the Riemann
curvature.
It ends up playing
the role-- when
we think about-- what this tells
us is this ends up, remember,
geodesics describe free fall.
And so what this is telling me
is a way in which two nearby
but somewhat separated--
separated by a distance c--
nearby but slightly
separated geodesics-- both
are in free fall, but their
free fall trajectories
are diverging from
one other and perhaps
being focused
towards one another
depending upon how R
is actually behaving,
this is the behavior of tides.
Free fall is
gravity, and this is
saying that the free
fall trajectory's
change of separation is
governed by the Riemann tensor,
and that's telling me about the
action of gravitational tides.
The last thing that we did
in Tuesday's-- excuse me--
in Thursday's lecture was I went
through and I developed this
proof of what is known
as the Bianchi identity,
which is an identity on the--
it's an identity on the
covariant derivative
of the Riemann tensor.
And so notice what's
going on here.
I'm leaving indices 3 and
4 on the Riemann tensor--
oh shoot.
I'm actually changing my
notation halfway through,
let me fix that.
My apologies.
OK.
Apologies for that.
Are you leave indices 3 and
4 unchanged, and what you do
is you cyclically permute
the index against--
in the direction which
you're taking a derivative
with indices 1 and 2.
So my first term, it goes alpha,
beta, gamma; then beta, gamma,
alpha; gamma, alpha, beta.
OK, notice the way they are
cyclically permuting like that.
Sum them up and you get 0.
So let's take it from here.
We're going to start with
this Bianchi identity.
What I want to do now is
contract the Bianchi identity
in the following way.
So let's take this form
that I've written out here--
and let me just make sure I've
left it now in a form that
comports with my notes.
I did, good.
So what I'm going to do is
multiply the entire thing.
Using the metric, I'm
going to contract it
on indices beta and mu.
So remember, the metric commutes
with the covariant derivative.
So unless the derivative
itself is with respect
to either the beta
or the mu index,
that g just sort of waltzes
right in top of there.
So when I do this,
it's going to beta--
g beta mu nu upstairs is going
to walk right through this,
it's going to raise
the beta index,
and what I wind up with
here is this first term
becomes the covariant derivative
of the Ricci tensor, OK?
I have contracted on
indices beta and mu.
When it hits this
one, is just going
to raise the index on
that covariant derivative.
So I've got a term here
now that looks something
like the divergence of the
Riemann tensor, divergence
with respect to index 3.
When I hit this term,
it walks through
the covariant derivative
again, and you
see what I'm doing is a
trace on indexes 2 and 3.
Now I can take advantage
of the anti-symmetry
here-- let's reverse this.
And so it's like doing a--
throwing in a minus
sign and then doing
a trace on indices 1 and 2--
excuse me, doing a trace on--
rewind, back up
for just a second.
I'm going to take advantage
that anti-symmetry, I'll
exchange indices 1
and 2, and then I
am doing a trace on indices
1 and 3, which is going
to give me the Ricci tensor.
But because I have used that--
or that anti-symmetry, I
will do so with a minus sign.
So what I get here is this.
I'm going to probably bobble
more than once in this lecture,
because again, doing this in an
empty room is a little weird.
All right.
So I really want to get a
relationship that simplifies
the Riemann tensor, OK?
A Riemann tensor's got
four indices on it.
I'm not scared, but
I don't like it, OK?
So we're going to do one
more contraction operation
to try to simplify this.
Let's now contract once
more using the metric
on indices g and nu, OK?
So when I do it with the first
one, it walks right through--
right through that
covariant derivative,
and I get the trace of the
Riemann tensor, the Ricci--
excuse me-- I get the
trace of the Ricci tensor,
the Ricci scalar.
When I do it on the
second term, OK?
I am now tracing
on indices 1 and 4.
I will invoke anti-symmetry
to change that
into a trace on indices 1 in
3, and I get the Ricci tenser
with a minus sign.
And then the next one,
I just trace on this
and I wind up with something
that looks like this.
OK.
Now these two terms are
both divergences on the--
they are both divergences
on the second index.
The second index
is a dummy index,
so I can put these two together.
So this is equivalent to--
or dividing out a
factor of minus 2,
the way this is
more often written.
You can also factor
out that derivative.
Let's write it like this.
OK?
So what I'm doing here
is I divided by a minus 2
so that I can put this guy in
front and I get a minus 1/2
in front of my Ricci term.
And because I want to factor
out my covariant derivative,
I need to throw in a
factor of the metric
there so that the
indices line up right.
So what we do at this point is
we do what-- whenever you reach
a certain point in
your calculation
where you've got something good,
you do what every mathematician
or physicist would do,
you give this guy a name.
So switching my
indices a tiny bit,
we define g mu nu to be
the Ricci tensor minus 1/2
metric Ricci scalar,
and this is an entity
known as the Einstein tensor.
This is a course on
Einstein's gravity,
so the name alone
should tell you,
this guy is going to matter.
One quick side note.
So suppose I take the trace
of the Einstein tensor.
When we took the trace
of the Ricci tensor,
I didn't write it down, but
if I take a trace of this guy,
I just get the Ricci scalar
R, which I used over here.
So when I do this here--
oops.
Suppose I just want to call
this the Einstein scalar g.
Well, applying that
to its definition,
this is going to be equal to
the Ricci scalar minus 1/2 trace
of the metric times R.
And it's a general rule
in any theory of spacetime
that the trace of the metric
is equal to the number
of dimensions in
your spacetime, OK?
You can easily work it
out in special relativity,
you're basically just raising
one index, and as we'll see,
it holds completely generally.
In fact, it follows
directly from the fact
that the upstairs metric
is the matrix inverse
of the downstairs metric.
So this is equal to
4, so this whole thing
is just the negative
of the Ricci scalar.
What this means is
that the Einstein
tensor is the trace-reversed
Ricci tensor, OK?
I just want to
plant that for now.
This is a fact that we're
going to take advantage
of a little bit later,
but for now, it's
just a mathematical fact
that I want to point out,
I want to set aside.
We'll come back to
it when it matters.
OK.
We now have everything
that we need
to take all of the
framework that we
have been developing
all term and turn it
into a theory of gravity.
I just had a nightmare.
Am I being recorded?-- yes, OK.
Sorry, just suddenly thought
I might have forgotten
to turn my microphone on!
So let's turn this into
a theory of gravity.
Ingredient 1 is something that
we have discussed quite a bit
before.
I want to restate it and I
want to sort of remind us.
Several times over
the past couple of
lectures I have implicitly
used this rule already,
but I want to make it a
little bit more explicit now.
We're going to use the
principle of equivalence.
In particular,
we're going to use
what is known as the
minimal coupling principle.
So here's the way-- what
this basically means.
We're going to try to take
laws of physics that are
well-understood from
laboratory experiments,
from special relativity,
everything that we have known
and loved and tested for--
since we started
studying physics,
and we're going to try to see
how that can be carried over
to working in the curved
spacetime that will describe
gravity with as little
additional sort of coupling
to spacetime entities
as is possible.
So here's what
we're going to do.
Take a law of physics that is
valid in inertial coordinates
and flat spacetime, or
equivalently, the local Lorentz
frame, which corresponds
to the local region
of a freely-falling frame or
a freely-falling observer,
take that law of physics
that is good in that form
and rewrite it in a
coordinate-invariant tensorial
form.
This is one the reasons
why throughout this term,
we have been brutally didactic
about insisting on getting all
of our laws of physics expressed
using tensors, quantities which
have exactly the
transformation laws
that we demand in order for
them to be true tensors that
live in the curve manifold that
we use to describe spacetime.
The last time I
actually did something
like this was when I derived--
I've erased it now, but
when I derived the equation
of geodesic deviation, OK?
I first did it using very,
very simple language,
and then I sort of
looked at it and said,
well this is fine according
to that local Lorentz frame,
according to that
freely-falling observer.
But this is not
tensorial, it's actually
as written only
good in that frame.
And so what we did was we
took another couple minutes
and said, let's see how I
can change this acceleration
operator that describes my
equation of geodesic deviation,
put in the extra
structure necessary so
that the acceleration
operator is tensorial,
and when we did that, we saw
that the result was actually
exactly what the Riemann tensor
looks like in the local Lorentz
frame.
We said, if it holds
a local Lorentz frame,
I'm going to assert it
holds in all other frames.
And that indeed is the final
step in this procedure.
We assert that the resulting
law holds in curved spacetime.
OK?
So this is the procedure
by which general relativity
takes the laws of physics,
good and flat spacetime,
and rejiggers them so that
they work in curved spacetime.
Ultimately, this is
physics, and so ultimately
the test for these
things are experiments.
And I will simply
say at this point
that this procedure has
passed all experimental tests
that we have thrown
at it so far,
and so we're happy with it.
So let me just
describe one example
of where we did this-- actually,
I'm going to do two examples.
So if I consider the
force-free motion of an object
in the freely-falling frame--
so recall, in the
freely-falling frame,
everything is being acted upon
by gravity in an equal way.
If I am in the
local Lorentz frame,
I can simply say that
my object feels--
my freely-falling object,
freely-flying observer--
feels no acceleration.
That is a perfectly rigorous
expression of the idea
that this observer or object
is undergoing force-free motion
in this frame.
This is not tensorial, though.
And so we look at
this and say, well,
if I want to make this
tensorial, what I'm going to do
is note that the tensor
operator that describes-- now
let me keep my indices
consistent here.
My tensor operator that
describes this equation
is given by taking the covariant
derivative of the four velocity
and contracting it with
the four velocity itself.
These say the exact same thing
in the local Lorentz frame.
This one is tensorial,
though, that one is not.
And so we then
say, OK, well this
is the version
that is tensorial,
I'm going to assert that
it holds in general.
Another example in flat
spacetime, local conservation
of energy and momentum
was expressed by the idea
that my stress energy
tensor had no divergence
in the local Lorentz frame.
Well, if I want to
make this tensorial,
all I do is I promote that
partial derivative I used
to define the divergence
to a covariant derivative.
This is how we are going
to define conservation--
we're going to define local
conservation of energy
and momentum in a
general spacetime theory.
So that's step 1.
We need-- or sorry,
ingredient 1.
Ingredient 2 is-- well, let's
just step back for a second.
We have done a lot of work
to describe the behavior
of curved spacetimes, OK?
Spacetimes that are
not just the spacetimes
of special
relativity, spacetimes
when my basis objects are
functional, where the Riemann
curvature tensor is non-zero.
We've done a lot to do that,
but I haven't said anything
about where that curved
spacetime actually comes from.
So the next thing which I
need is a field equation
which connects my spacetime to
sources of matter and energy.
That's a tall order.
The way we're going
to do this, we're
actually going to do
it two different ways.
So in this current
lecture, I'm going
to do it using a
method that parallels
how Einstein originally
did it when he derived--
what resulted out
of this is what
we call the field equation
of general relativity
or the Einstein field equation,
and in this first presentation
of this material, I'm going to
do it the way Einstein did it.
So what we are going to do is
we will require that whatever
emerges from this procedure, it
must recover Newtonian gravity
in an appropriate limit.
This is a philosophical
point about physics.
When you come up
with a new theory,
you may conceptually
overturn what came before.
You may have an entirely new
way of thinking about it.
You may go from saying that
there is a potential that
is sourced by mass
that fills all of space
and that objects react to
to saying something like,
we now decide that
the manifold of events
has a curvature
that is determined
by the distribution of matter
and energy in spacetime.
It's very different
philosophical and ultimately
mathematical ways
of formulating this,
but they have to give
consistent predictions,
because at the end of the
day, Newtonian gravity
works pretty damn well, OK?
We can't just throw that away.
So what we're going
to do is demand
that in an appropriate limit,
both the field equation
for Newtonian gravity--
so this is the
Laplace operator now,
which I'm going to write in a
semi-coordinate-invariant form,
as the chronic or delta
contracted-- basically
it's the trace on a matrix
of partial derivatives acting
on a potential.
This equals 4 pi rho, and I call
this semi-coordinate invariant
because part of
what goes into this
is this Newtonian limit
only works if everything
is sufficiently slowly
varying in time,
that things having to
do with time derivatives
can be neglected, OK?
It's never really been--
prior to some of the more
modern experiments that we've
had to do, time-varying
sources of gravity
are very hard to work with.
And so Newton was never
really tested in that way.
Nonetheless, whatever
emerges from Einstein
had best agree with this.
And we are also going to require
that the equation of motion
in this framework agree
with Newtonian gravity.
We actually went through
this-- this was a little bit
of a preview of this
lecture, we did this in our--
we concluded our
discussion of geodesics.
Let me just recap the result
that came out of this.
So our equation of
motion was that--
you can write it as the
acceleration of an observer
is related to the
gradient of the potential.
All right.
So let's follow in the footsteps
of Einstein and do this.
So what we're going to do--
let's do the equation of motion
first.
I've already gone
through this briefly,
but I want to go
over it again and I
want to update the
notation slightly.
So let's do the
equation of motion
by beginning with the
geodesic equation.
We will start with
the acceleration
coupled to the four velocity
by the Christoffel symbols.
All tests of Newtonian
gravity, especially
those that Einstein had
available at the time
that he was formulating
this, were slow motion ones.
We were considering
objects moving
at best in our solar system.
And so things there on a human
scale certainly move quickly,
but they're slow compared
to the speed of light.
And so let's impose the slow
motion limit, which tells us
that the 0th component
of the four velocity
is much larger than the
spatial components of the four
velocity, OK?
Remember working in units
where the speed of light
is equal to 1.
And so if this is being measured
in human units, kilometers
per second, and
things like that,
this is on the order
of the speed of light.
So when we throw this in, we
see that we expand this out,
that the contributions
from the dt
d tau terms here are going to be
vastly larger than any others.
And so we can
simplify our equation
to a form that looks like this.
OK?
In the spirit of
being uber complete,
let's write out that
Christoffel symbol.
So dig back into your
previous lectures' notes,
remind yourself what the formula
for the Christoffel symbol is.
OK.
Notice, two of the terms
here are time derivatives.
The Newtonian limit-- all
the tests that were available
when Einstein was formulating
this, the limit that we care
about here, the
gravitational field,
the gravitational potentials
that he was studying,
what the Newtonian limit
emerges from, they are static.
So we're going to do is neglect
time derivatives to recover
this limit.
And when we do this, what we
find is that the component--
the Christoffel
component that we
care about looks like
one derivative of the 0 0
piece of the spacetime metric.
It's not too hard
to convince yourself
that this, in fact, reduces--
oops, pardon me.
I skipped a step.
Pardon me just one moment.
Just one moment, my apologies.
I'm going to write
the spacetime metric--
I'm going to work in a
coordinate system such
that spacetime looks
like the flat space
time of special relativity
plus a little bit else, OK?
This is consistent with
the idea that every system
we have studied in
Newtonian gravity
is one where the predictions
of special relativity
actually work really,
really well, OK?
Gravity is new,
it's special, it's
why we have a whole other
course describing it.
But clearly it can't
be too far different
from special relativity
or we wouldn't
have been able to formulate
special relativity
in the first place.
So my apologies, I sort of
jumped ahead here for a second.
We're going to treat
the g mu nu that
goes into this as the metric of
flat spacetime plus something
else where I'm going
to imagine that all
the different
components of this--
so a typical component
of this h mu nu
has an absolute value that
is much smaller than 1.
It's not too hard to prove that
when you invert this, what you
wind up with is a form that
looks like so where this h mu
nu with the indices in
the upstairs position
is given by raising
h's indices using
the metric of flat spacetime.
We're going to talk about this
in a little bit more detail
in a future lecture,
it doesn't really
matter too much right now.
I just want to point out
that the inverse g, which
we need to use here,
also has this form that
looks like flat spacetime metric
and this h coupling into it.
Now the reason I'm
going through all this
is that in order to work
out this Christoffel symbol,
I need to take a derivative.
The derivative of eta is 0, OK?
So the only thing that
gets differentiated is h.
So when you work out
this Christoffel symbol,
what you get is this.
If you're being-- keeping
score, there are corrections
of order h squared,
and pardon me,
I should have
actually noted, there
are corrections
of order h squared
that go into this inverse.
Let me move this
over so I can fit
that in a little bit better.
But in keeping with
the idea that--
in keeping with the idea
that for the Newtonian limit,
the h squared could--
h is small, we're going to
treat the h squared corrections
as negligible and
we will drop them.
OK.
So let's look at what motion
in this limit turns into, then.
We now have enough pieces
to compute all the bits
of the equation of motion.
So in keeping with the idea that
I am going to neglect all time
derivatives, this tells
me that the gamma 0,
00 term is equal to 0.
And from this, we find that
there is a simple equation
describing time.
In our equation
describing space, OK?
So what I've done there is just
taken this geodesic equation,
plugged in that result for
the Christoffel symbol,
and expanded this guy out.
So what results, I can
divide both sides now
by two powers of dt d tau.
All right.
If you take a look at
what we've got here,
this prediction of the--
no even a prediction, this
result from the
geodesic equation
is identical to our
Newtonian equation of motion
provided we make the
following identification.
h00 must be minus
2 phi where phi
was Newtonian
gravitational potential.
Or equivalently, g00 is the
negative of 1 plus 2 phi.
All right.
So that's step 1.
We have made for
ourselves a correspondence
between what the metric should
be and the equation of motion.
We still have to do
the field equation,
so let's talk about that.
So very helpfully I've already
got the Newtonian field
equation right above me here.
Let me rewrite it
because I'm going
to want to tweak my
notation a tiny bit.
I don't want to think about
what this is telling me.
So eta ij is the same
thing as delta ij,
I just want to put
it in this form
so that it looks like a
piece of a spacetime tensor.
This is manifestly not
a tensorial equation.
I have a bunch of derivatives
on my potential being
set equal to--
OK, there's a couple
constants, but this, OK?
When we learned about
quantities like this
in undergraduate
physics, usually we're
told that this is a--
excuse me, this is a scalar.
But we now know,
rho is not a scalar,
it is the mass density,
which up to a factor of c
squared, is the same thing
as the energy density.
And when we examined
how this behaves
as we change between
inertial reference frames,
we found this transforms
like a particular component
of a tensor.
And as I sort of emphasize,
not that long ago we
have been in something of
a didactic fury insisting
that everything be formulated
in terms of tensors.
Pulling out a particular
component of a tensor
is bad math and bad physics.
So we want to promote this
to something tensorial.
So s on the
right-hand side, we've
got one component of the
stress energy tensor.
We would like whatever
is going to be
on the right-hand
side of this equation
to be the stress
energy tensor, OK?
We can sort of
imagine that what's
going on in Newton's gravity is
that there is one particular--
maybe there's one
component of this equation
that in all the
analyses that were done
that led to our formation
of Newtonian gravity,
there may be one component
that was dominant, which
is how it was that Newton
and everyone since then
was able to sort of pick
out a particular component
of this equation
as being important.
Over here on the left-hand
side, we saw earlier that
the equation of motion we're
going to look for corresponds
to the Newtonian limit if the
metric plays the same role--
up to factors of 2 and offsets
by 1 and things like that--
the metric must play the
same role as the Newtonian
gravitational potential.
So if I look at--
if I look at the
Newtonian field equation,
I see two derivatives
acting on the potential.
So I want my metric to
stand in for the potential,
we expect there to be
two derivatives of metric
entering this relationship.
So now two derivatives
of the metric
is going to give me something
that smells like a curvature.
So we want to put a curvature
tensor on the left-hand side
of this equation.
We have several to
choose from, OK?
It clearly can't be
the Riemann tensor.
There's too many indices,
it just doesn't fit.
It could be the
Ricci curvature, OK?
The Ricci curvature
has two indices.
That has two indices,
that's a candidate.
But it's worth stopping
and reminding ourselves,
wait a minute, this
guy has some properties
that I already know about.
t mu nu tells me
about the properties
of energy and momentum
in my spacetime,
and as such, conservation--
local conservation
of energy and momentum requires
that it be divergence-free.
So whatever this
curvature tensor is here
on the left-hand
side, we need it
to be a divergence-free
2-index mathematical object.
At the beginning
of today's lecture,
I showed how by contracting
on the Bianchi identity,
you can, in fact, deduce that
there exists exactly such
a mathematical object.
So let us suppose
that our equation that
relates the properties
of the spacetime
to the sources of energy
and momentum of my spacetime
is essentially that that
Einstein tensor, g mu nu,
be equal to the
stress energy tensor.
Now in fact, they don't have the
same dimensions as each other,
so let's throw in a kappa, some
kind of a constant to make sure
that we get the right
units, the right dimensions,
and that we recover
the Newtonian limit.
The way we're going to
deduce how well this works is
see whether an
equation of this form
gives me something that looks
like the Newtonian limit
when I go to what I'm going to
call the weak gravity limit,
and I'm going to then use,
assuming it does work--
not to give away
the plot, it does--
we'll use that to figure out
what this constant kappa must
be.
So if we do, in fact, have a
field equation of the form g
mu nu is some
constant t mu nu, it's
not too hard to figure out
that an equivalent form of this
is to say that the Ricci tensor
is k times t mu nu minus 1/2 g
mu nu t where this t is just
the trace of the stress energy
tensor.
Remember, I spent a few moments
after we derived the Einstein
tensor pointing out
that it's essentially
the same thing as Ricci but
with the trace reversed.
This is just a trace-reversed
equivalent to that equation.
This step that I'm
introducing here,
basically it just
makes the algebra
for the next
calculation I'm going
to do a little bit easier, OK?
So I just want to emphasize
that this and that are exactly
the same content.
All right.
So to make some headway, we need
to choose a form for a stress
energy tensor.
Our goal is to recover
the Newtonian limit,
and so what we want to do is
make the stress energy tensor
of a body that corresponds to
the sort of sources of gravity
that are used in studies
of Newtonian gravity.
So let's do something
very simple for us.
Let's pick a static--
in other words,
no time variation,
a static perfect fluid
as our source of gravity.
So I'm going to choose
for my t mu nu--
dial yourself back to lectures
where we talk about this, OK?
So this is the perfect
fluid stress energy tensor.
We're working in
the Newtonian limit,
and we are working in units
where the speed of light
is equal to 1.
If you put speed of light
back into these things,
you explicitly include
it, this is actually a rho
c squared that appears here.
And so what this tells me
is that if I'm studying sort
of Newtonian limit
problems, rho is much, much,
much greater than P in the
limit that we care about.
Furthermore, I am treating
this fluid as being static.
So that means that
my four velocity only
has one component, OK?
The fluid is not flowing.
You might be tempted to say,
oh, OK, I can just put a 1
in for this.
Not so fast, OK?
Let's be a little bit
more careful about that.
One of the key governing
properties of a four velocity
is that it is
properly normalized.
So this equals g mu nu
u mu u nu is minus 1.
We know that the only
components of this that matter,
so to speak, are the
mu and nu equal 0.
So this becomes g00 mu 0
squared equals minus 1.
But g00 is-- well, let's
write it this way--
Negative 1 plus h00.
Go through this algebra,
and what it tells you is u0
equals 1 plus 1/2 h00.
Again, I'm doing my algebra
at leading order in h here.
We raise and lower indices.
So in my calculation,
I'm going to want to know
the downstairs version of this.
And if I, again,
treat this thing--
treat this thing
consistently, OK.
What I'll find is I just
pick up a minus sign there.
OK.
OK.
Let's now put all
the pieces together.
The only component of
my stress energy tensor
that's going to
now really matter
is rho u0 u0, which, putting
all these ingredients back
together, is rho 1 plus h00.
The trace of this guy, putting
all these pieces together,
is just equal to negative rho.
Since I only have
one component that's
going to end up
mattering, let's just
focus on one component of
my proposed field equation.
OK?
So this is the guy
that I want to solve.
I'll let you digest that
and set up the calculation.
We've got T00 minus 1/2
T00 T. This is going to be,
plugging in these bits that I
worked out on the other board,
here's my T00.
Just make sure I did
that right earlier--
I did.
OK.
So this is my right-hand
side of my field equation.
It will actually be
sufficient for our purposes
to neglect this term, OK?
We'll see why in just a moment.
So plugging that in, I need
to work out the 00 component
of my Ricci.
So I go back to its
foundational definition.
This is what I get when I
take the trace on indices 1
and 3 of the Riemann tensor.
I can simplify that to
just doing the trace over
the spatial indices, because the
term I'm leaving out is the one
that is of the form 00 here,
which by the anti-symmetry,
on exchange of those
indices, must vanish.
Plugging in my
definition, what I find
is it is going to
look like this here.
So I'm just going to neglect
the order of gamma squared term
because I'm working in
a limit where I assume
that all these h's are small.
This is going to vanish because
of my assumption of everything
being static in this limit.
So this, I then go and
plug in my definitions.
OK.
Again, I'm going to lose
these two derivatives
by the assumption of
things being static.
And pardon me just a second--
yeah, so I'm going to lose these
two because of the assumption
of things being static.
The only derivative--
the only term
that's going to matter,
the derivative here is h.
And so when I hit it
with the inverse metric,
this becomes simply the
derivative of the h00 piece,
OK?
I can go from g straight to eta
because the correction to this
is of order h squared, which
as I've repeatedly emphasized,
we're going to neglect.
All right, we're almost there.
Let me put this board
up, I want to keep this.
OK, where was I?
So I've got it down
to here, let me just
simplify this one step more.
Eta is-- if mu is not
spatial, then this is just 0.
So I can neatly change my
mu derivative into a j.
I can just focus on the
spatial piece of it.
So this tells me R00 is minus
1/2 Kronecker delta delta i
delta j acting on h00.
This operator is
nothing more than--
it's a Laplace operator.
So this is minus 1/2, our
old-fashioned, happily,
well-known from
undergrad studies
Laplace operator on h00.
So putting all this
together, my field
equation, which I
wrote in this form,
reduces down to del squared
h00 equals minus kappa rho.
The Newtonian limit that we
did for the equation of motion,
the fact that we showed that
geodesics correspond to this,
that already led me
to deduce that h00
was equal to minus 2 phi.
My Newtonian field
equation requires
me to have the Laplace operator
acting on the potential phi,
giving me 4 pi g rho.
Put all these pieces
together, and what
we see is this
proposed field equation
works perfectly provided we
choose for that constant.
Kappa equals 8 pi j.
And so we finally get g mu
nu equals 8 pi g t mu nu.
This is known as the
Einstein field equation.
So before I do a few
more things with it,
let us pause and
just sort of take
stock of what went
into this calculation.
We have a ton of
mathematical tools
that we have developed
that allow us
to just to describe the
behavior of curved manifolds
and the motion of bodies in
a moving curved manifolds.
We didn't yet have
a tool telling us
how the spacetime metric
can be specified, OK?
We didn't have the equivalent
of the Newtonian fuel equation
that told me how gravity
arises from a source.
So what we did was we looked
at the geodesic equation,
we went into a limit
where things deviated
just a little bit
from flat spacetime,
and we required objects be
moving non-relativistically
so that their spatial four
velocity components were all
small.
That told us that we were able
to reproduce the Newtonian
equation of motion if h00, the
little deviation of spacetime
from flat spacetime
in the 00 piece,
was equal to negative 2 times
the Newtonian potential.
We then said, well, the
Newtonian field equation
is sort of sick from a
relativistic perspective
because it is working with
a particular component
of a tensor rather
than with a tensor.
So let's just ask
ourselves, how can we
promote this to a properly
constructed tensorial equation?
So we insisted the
right-hand side be t mu nu.
And then we looked
for something that
looks like two derivatives
of the potential,
or, more properly, two
derivatives of the metric
which is going to give
me a curvature tensor,
and say, OK, I want a
two-index curvature tensor
on the left-hand side.
Since stress energy
tensor is divergence-free,
I am forced to choose
a character tensor that
is divergence-free, and that's
what leads me to this object,
and there's the
Einstein curvature.
And then insisting that
that procedure reproduce
the Newtonian limit when things
sort of deviate very slightly
from flat spacetime,
that insisted
the constant proportionality
between the two sides
be 8 pi j.
This, in a nutshell, is how
Einstein derived this equation
originally when it
was published in 1915.
When I first went
through this exercise
and really appreciated
this, I was struck
by what a clever guy he was.
And it is worth noting that
the mathematics for doing this
was very foreign to
Einstein at that time.
There's a reason
there's a 10-year gap
between his papers
on special relativity
and his presentation
of the field equations
of general relativity.
Special relativity was 1905,
field equation was 1915.
He was spending most of those
intervening 10 years learning
all the math that we have
been studying for the past six
or seven weeks, OK?
So we kind of have the luxury
of knowing what path to take.
And so we were able to sort
of pick out the most important
bits so that we could sort of--
we knew where we wanted to go.
He had to learn all
this stuff from scratch,
and he worked with quite
a few mathematicians
to learn all these pieces.
Having said that,
though, it did strike me
this is a somewhat ad
hoc kind of a derivation.
When you look at this, you
might sort of think, well,
could we not--
might there not be
other things I could put on
either the left-hand side
or the right-hand
side that would still
respect the Newtonian limit?
And indeed, we can add any
divergence-free tensor onto--
depending how you count it--
either the left-hand side
or the right-hand side--
let's say the left-hand side--
and we would still have
a good field equation.
Einstein himself was the
first one to note this.
Here's an example of such
a divergence-free tensor.
The metric itself, OK?
The metric is compatible with
the covariant derivative.
Any covariant derivative
of the metric is 0.
And so I can just put
the metric over here,
that's perfectly fine.
Now the dimensions
are a little bit off,
so we have to insert a
constant of proportionality
to make everything
come out right.
This lambda is known as
the cosmological constant.
Now what's kind
of interesting is
that one can write
down the Einstein field
equations in this way, but
you could just as easily
take that lambda g
mu nu and move it
onto the right-hand side and
think of this additional term
as a particularly special
source of stress energy.
Let's do that.
So let's define t
mu nu lambda equal
negative lambda over
8 pi g times g mu nu.
If we do that, we
then just have g
mu nu equals 8 pi gt mu nu
with a particular contribution
to our t mu nu being this
cosmological constant term.
When we do this, what
you see is that t mu
nu is nothing more than
a perfect fluid with rho
equals 8 pi g in the
freely-falling frame, pressure
of negative lambda over 8 pi j.
Such a stress energy
tensor actually
arises in quantum
field theories.
This represents a form
of zero-point energy
in the vacuum.
You basically need
to look for something
that is a stress energy tensor
that is isotopic and invariant
to Lorentz transformations
and the local Lorentz frame,
and that uniquely picks out
a stress energy tensor that
is proportional to the metric
in the freely-falling frame.
So this is an argument
that was originally
noted by Yakov Zeldovich.
Whoops.
And much of this
stuff was considered
to be kind of a
curiosity for years
until cosmological
observations--
we haven't done cosmology yet.
We will do this in
a couple of weeks--
a couple of lectures,
I should say.
And it turns out that
the large-scale structure
of our universe seems to support
the existence potentially
of there being a
cosmological constant.
So the behavior of all these
things is a lot more relevant,
it's been a lot more
relevant over the past,
say, 15 or 20 years than
it was when I originally
learned the subject in 1993.
So I want to just
conclude this lecture
with a couple of remarks
about things that are commonly
set equal to 1 when we are doing
calculations of this point.
So one often sets G equal to
1 as well as c equal to 1.
Carroll's textbook does not--
several other modern
textbooks do not--
I personally like for
pedagogical purposes leaving
the G in there, because it
is very useful for calling
out-- helping to understand the
way in which different terms
sort of couple in.
It can-- if nothing else, it
serves as a very useful order
counting parameter,
something that we'll
see in some of the future
calculations that we do.
But there's a
reason why one often
works with G equal to 1 in
many relativity analyses.
Fundamentally, this is because
gravity is a very weak force.
G is the most
poorly known of all
of the fundamental
constants of nature.
I think it's only known--
I forget the number
right now, but it's
known to about
five or six digits.
Contrast this with things like
the intrinsic magnetic moment
of the electron, which is known
to something like 13 digits.
What this sort of means is that
because G is so poorly known--
well, let me just write
that out in words first.
So G is itself poorly known.
And so when we
measure the properties
of various large
objects using gravity,
we typically find that something
like G times an object's mass
is measured much
better than M alone.
Basically, the observable
that one is probing
is G times M. To get M out
of that, you take G times M
and you divide by the
value of G that you
have determined independently.
If you only know this guy
to five or six digits,
you're only going to know this
guy to five or six digits.
Whereas, for
instance, for our sun,
GM is known to about nine
digits, maybe even 10 digits
now.
When you set both G and
c to 1, what you find
is that mass, time,
and length all come out
having the same dimension.
And what that means is that
certain factors of G and c
can be combined to become
convergent factors.
So in particular, the
combination over c squared,
it converts a normal mass--
let's say an SI mass--
into a length.
A very useful one is G times the
mass of the sun over c squared
is 1.47 kilometers.
GM over c cubed, OK?
You're going to divide by
another factor of velocity.
This takes mass to time.
So a similar one, G
mass sun over c cubed,
this is 4.92 times 10
to minus 6 seconds.
One more before I
conclude this lecture.
(I can't erase this equation,
it's too beautiful!)
If I do g over c to the fourth,
this converts energy to length.
I'm not going to give the
numeric value of this,
but I'm going to make
a comment about this.
So bear in mind that your
typical component of T mu nu
has the dimensional form
energy per unit volume--
i.e., energy over length cubed.
So if I take G over c to
the fourth times T mu nu,
that is going to give me a
length over a length cubed--
in other words, 1
over length squared,
which is exactly what
you get for curvature.
So when one writes the
Einstein field equations,
if you leave your G's
and your c's in there,
the correct coupling factor
between your Einstein tensor
and your stress energy
tensor is actually
8 pi G over c to the fourth.
And I just want to leave
you with the observation
that G is a pretty
small constant,
c to the fourth is a
rather large constant,
and so we are getting a
tiny amount of curvature
from a tremendous
amount of stress energy.
Spacetime is hard to bend.
