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SCOTT HUGHES: All
right, good afternoon.
So today's lecture is a
particularly important one.
We are going to introduce the
main physical principle that
underlies general relativity.
We'll be spending a bunch of
time connecting that principle
to the mathematics as the
rest of the term unwinds.
But today is where we're
going to really lay out
where the physics
is in what is known
as the principle of equivalence.
So before I get into
that, just a quick recap.
So in terms of technical
stuff we did last time,
the most important
thing we did was
to introduce these
mathematical objects called
Christoffel symbols.
Christoffel symbols are
those capital gammas.
We began by thinking
of them as just what
you get when you look at
the derivative of your basis
objects.
Pardon me a second.
There was a lot of
chalk on the chalkboard,
and I think I inhaled half of
it while I was cleaning it.
So take derivatives
of your basis vectors.
And there, you get something
that's linearly related
to your basis objects.
And the gamma is the set
of mathematical quantities
that sets that
linear relationship
between the derivatives
and the objects themselves.
One can build under the demand--
so recall how we did this.
We have a physical
argument that tells us
that a quantity we call the
covariant derivative, which
is a way of using these
Christoffel symbols in order
to get tensorial derivatives.
So if I look at just the
partial derivative of a tensor,
if I'm working in a general
curvilinear coordinate system,
the set of partial
derivatives do not
constitute components
of a tensor.
OK?
I posted some notes online,
demonstrating explicitly
why that is.
But in a nutshell,
the main reason
is that, when you
try to work this out
or when you go and
you work this out,
you will find that there
are additional terms.
It was like the derivative of
the transformation matrix that
come in and spoil
the tensoriality
of this relationship.
And what the
Christoffel symbol does,
as you guys are going
to prove on P set three,
it introduces terms
that are exactly
the same but with
the opposite sign
and cancel those things out.
So that the covariant
derivative is
what I get when I
correct each index
with an appropriate factor
of the Christoffel symbol.
It is something that is
a component of a tensor.
In other words, it obeys
the transformation laws
that tensors must have.
We have a physical
argument that essentially
is something that is
tensorial, in other words, both
sides of the equations are
good tensor quantities in one
representation,
have to be tensorial
in any other representation.
And from that, we deduced
that the covariant derivative
of the metric must always be 0.
And by taking appropriate
covariant derivatives
of the metric, sort of doing a
bit of gymnastics with indices
and sort of wiggling
things around a little bit,
we found that the
Christoffel symbol can itself
be built out of partial
derivatives of the metric.
And the result looks like this.
So that's a little bit
more of an in-depth recap
than I typically do.
But I really want to hammer
that home because this
is an important point.
We then began switching gears.
And we're going to do something.
So I introduced, at the end of
the previous lecture, a fairly
silly argument,
which, nonetheless,
has an important
physical concept to it.
So where we were going
with was the fact that,
when we have gravity, we cannot
cover all of spacetime with
inertial frames,
or I should say--
I should watch my wording here--
with an inertial frame.
OK?
We're, in fact,
going to see that you
can have a sequence of them.
And we're going to find
ways of linking them up.
But for now, if you think
of special relativity
as the theory of
physics in which we
can have Lorentz coordinates
that cover the entire universe,
that's out the window.
OK, and so I'm going to
skip my silly derivation.
Part one of it is
essentially just a motivation
that a gravitational
redshift exists.
So I gave a silly
little demonstration
where you imagined dropping
a rock off of a tower.
You had some kind of a magical
device that converts it
into a photon that
shoots it back up
and then converts
it back into a rock.
And you showed that,
as it climbed out
of the gravitational
potential, it
had to lose some of its energy.
Or else, you would get
more energy at the top
than you put in when you
initially dropped it.
It's a way of preventing you
from making perpetual motion
machines.
That is, without a
doubt, a flaky argument.
It can be made a little
bit more rigorous.
But I didn't want to.
And I will justify the reason
I can do this is that this is,
in fact, a highly tested
experimental fact.
Every single one
of you has probably
used this if you've
ever used GPS
to find your way
to some location
where you're supposed
to meet a friend.
And what this basically
tells me is that,
if I imagine a
tower of height h,
if I shoot light with
frequency at the bottom, omega
b, when it reaches
the top, it'll
have a different
frequency, omega t,
where these are
related by omega t
equals, and I'm going to put my
c squareds back in there, just
to keep it all complete.
OK?
So this is, I emphasize,
a very precisely
calibrated experimental
result. There
are many ways of doing it,
some silly, some serious.
The silly one was sort of fun.
But it does actually capture
the experimental fact.
OK?
The key experimental fact is
that it's necessary for light
to lose energy as it climbs
out of the gravitational field,
in essence, because it's
possible, thanks to quantum
field theory, to convert photons
into particles and particles
into photons.
This is a way of
conserving energy.
OK, so this was just part one
of this argument that we cannot
cover all of spacetime
with an inertial frame.
So here's part two
of the argument.
Suppose there was, in fact, a
very large region of space--
now let's say it's near
the Earth's surface--
that could be covered by
a single Lorentz frame.
OK?
So I want you to
imagine that you're
sending your beam of light up.
And we're going to take
advantage of the fact
that light is wave like.
So what I want
you to do is think
of the spacetime trajectory of
successive crests of this wave
as it climbs out of the
gravitational potential.
So I"m going to make my
spacetime diagram the way we
like to do it with
time running up.
And so this direction
represents height.
So down here is the
bottom of the tower.
So let's just make
the drawing clearer.
Let's let this tick be
the bottom of the tower.
This tick over here
will be the top.
Now if we were just
in special relativity,
there was no gravity, we know
light would just move on a 45
degree angle on this thing.
We don't know yet what gravity
is going to do to this light.
OK?
But you can imagine that
whatever it's going to do,
it's going to bend it away in
some way from the trajectory
that special relativity
would predict.
So let's just imagine that
crest one of this light,
it follows some
trajectory in spacetime
that maybe looks like this.
So here is the world
line of crest one.
So what about the world
line of credit two?
So let's think about what
the world line of credit two
must be under this assumption
that we can, in fact, cover
all of spacetime with
a single Lorentz frame.
OK?
If that is the case,
so if that is the case,
no position and no
moment is special.
OK?
Everything is actually
translation invariant
in both time and space.
So the second crest
is going to be
emitted one wave period later.
But there's absolutely nothing
special about spacetime one
wave period later.
And as it moves
along, there can't
be anything special about this.
If it is to be a global
Lorentz frame, whatever
the trajectory is that the crest
two follows through spacetime,
it's got to be congruent with
the trajectory of crest one.
I've got to be able to
just simply slide them
on top of each other.
OK?
That is what this
assumption demands.
So this is going to look
something like this.
OK?
Artwork a little bit off.
The key thing which
I want to emphasize,
though, is that if
this guy is congruent,
the spacing between crest one
and crest two at the bottom,
let's call it delta tb, there'll
be some spacing between the two
at the top.
We'll call it delta tt.
They must be the same.
But that's [INAUDIBLE].
That is just a
period of the wave.
And that's just up
to a factor of 2 pi.
That's 1 over the frequency.
So this implies delta tb is not
equal to delta t at the top.
So a frequency at
the top is smaller.
So the period will actually
be a little bit larger.
So when you think about
this argument carefully,
the weak point is
this assumption
that I can actually cover the
region with a large Lorentz
frame.
Remember what this is telling
me is that, essentially, there
is no special direction, right?
I can do translation,
variance in time and space,
and they're all the same.
But that's actually completely
contradicted by common sense.
If there were no special
direction, why did it go down?
There's obviously something
special about down, right?
So we conclude we cannot have
a global Lorentz frame when
we've got gravity.
And in case you want to read
a little bit more about it,
I'll just give a highlight that
this argument was originally
developed by a gentleman
named Alfred Shield.
So that's unfortunate,
OK, from sort
of a philosophical
perspective because the fact
that we don't have a global
Lorentz frame makes you think,
oh, crap.
What are we going to do
with that framework we
spent all this time developing?
OK?
The existence of a Lorentz frame
and many of the mathematics
we've been developing for
the past few lectures,
they all center around things
you can do in Lorentz frames.
So it's kind of like, OK, so
do we just throw all that out?
Well, here is where Einstein's
key physical insight
came in and gave us
the physical tools
that then needed to be coupled
to some mathematics in order
to turn it into something
that can be worked with
but, nonetheless,
really save the day.
So the key insight has
there's a few steps in it.
So first thing, I was very
careful to use the word global
when I ruled out the
existence of Lorentz frames.
But I'm allowed to have
local Lorentz frames.
Let's see.
What does that actually mean?
So I'll remind that a Lorentz
frame is a tool that we use
to describe inertial observers.
In fact, we often call
them inertial frames
because they're sort of
the constant coordinates.
They represent the coordinates
of an inertial observer
who happens to be at rest in
that Lorentz coordinate system.
So let's switch over.
Let's start calling them
inertial frames for just
a moment.
So an inertial frame means that
there is nothing accelerating,
so no accelerations on
observers or objects at rest
in that frame.
In other words, no
forces are acting.
Einstein's insight
was to recognize
that the next best thing
is a freely falling frame.
OK?
If you are in a freely falling
frame, to you, it sure as hell
seems like there's
a force acting on it
and there's some
accelerations acting upon it.
But let's imagine you're in
this freely falling frame.
And you have a bunch
of small objects
that you release near you.
Let's see if I can find
some bits of broken chalk.
So when you're in this
really falling frame--
this was a lot easier
when I was younger.
All right, when you're in
this really falling frame, OK,
and these objects
are all falling,
there is no acceleration of
these objects relative to you.
The reason for this is that
the gravitational force
is proportional to the mass.
And so the fact that
we have f equals ma.
And fg proportional
to mass means
that all objects in that
freely falling frame--
this is a term that I'm
going to start using a lot.
So I'm just going to
abbreviate it FFF.
If I get really lazy, I
might call it an F cubed.
Anyway, l objects that are
in that freely falling frame,
they're experiencing
the same acceleration.
And so they experience
0 relative acceleration
at least in the absence
of other forces, OK?
Perhaps one of them is charging.
There's an electric force.
Well, then it sort of suggests
that the interesting force
is the extra force provided
to that relative to what
we call the gravitational force
that is driving this freefall.
So I urge you to start getting
comfortable with this idea
because we're going to find this
to be-- so one thing, which I'm
going to do in a
couple of minutes,
is actually a little
calculation that shows me
I can, in fact,
always find a Lorentz
frame in the vicinity of
any point in spacetime.
OK?
And so we're going to actually
regard that Lorentz frame
as being the preferred
coordinate system of a freely
falling observer, one
who is not accelerated
relative to freefall
at that point.
We are not in
freefall right now.
The damn floor is pushing
us out of the way.
But in this way of
doing things, we
would actually regard us as
being the complicated people,
OK?
Someone who is merrily
plummeting to their death,
they're actually the
simple observers,
who are doing what they should
be doing, in some sense.
So the key thing, a
way of saying this,
so coming back to this
and just one more point
about that, because
there are no relative
accelerations, within
this frame, objects
maintain their
relative velocities.
That's basically the
definition of something
that is an inertial.
If I have a frame
where everything is--
two objects are moving with
respect to each other at 1
meter per second, they're
always moving at 1 meter
per second, that's inertial.
And so this sort of demands that
we do all of our experiments
in plummeting elevators
or in space, right?
OK, obviously,
you can't do that.
So there's some
complications we're
going to have to learn
the math to describe.
But this is the way we're
going to from now on think
about things is that the
freely falling frame is
the most natural generalization
of an inertial frame
that we describe using
Lorentz coordinates.
Now before I discuss a
few details about this,
an important thing
that is worth noting
is that a very important
aspect of any realistic source
of gravity is that
it is not uniform.
OK?
Gravity here is a
little bit stronger
than gravity on the ceiling, OK?
And it's a lot stronger than
gravity in geostationary orbit.
We call this variation
in gravity tides.
So tides break down the notion
of uniform freely falling
frames.
What this is telling
us is that this idea,
if I'm going to create
this inertial frame that
is essentially a
freely falling frame,
it will have to
be a finite size.
Hence, it's local
and not global.
OK, physically,
this is telling me
that, if I have a really
tall freely falling elevator,
and I'm here, I have one friend
here and one friend down there,
even if we start absolutely at
rest, let's say we're at rest.
We measure reference back to
the walls of this elevator.
It's a very stiff elevator.
We will see the three
of us diverge away
from each other with time, OK?
Because this person, let's
say earth is down here,
this person is feeling the--
and using Newtonian intuition,
gravity is a little stronger.
Gravity is medium.
Gravity is weak.
And so as viewed in that
freely falling frame,
let's say we center it
on me here in the middle,
I will see the person in the top
go up, the person at the bottom
go down.
Another way of
saying this, so let
me just say that, so
a very tall elevator,
we'll see a
separation of freefall
since gravity is not uniform.
But there's another
way of saying this.
And what this is
telling me is imagine
I made a spacetime diagram that
shows trajectories of the three
of us, so the three of
us that are plummeting
in this elevator, OK?
What we find, if we
make our trajectories
as we fall-- in fact, let's
just go ahead and make
a little sketch of this.
OK, so let's say
here is the middle.
Here is the top.
Here is the bottom.
I'm going to draw this in
coordinates that are fixed
on the person in the middle.
So the person the middle, in
their freely falling frame,
they think they're
standing still.
They see the person at the
top moving up and away,
the person at the bottom
moving up and away.
It's a very similar story to
the little parable I sketched
of that light pole moving up.
These are not
congruent trajectories.
Another way to say this is
that, in any moment, if I think
of these as world lines,
I can, at any moment
along the world line, I can
draw its tangent vector.
The tangents do not
remain parallel.
Now I'm going to take
you back to probably,
I don't know when you
guys all learned this,
but when you first started
learning about geometry.
For me, it was in middle school.
And when you first started
learning about geometry,
you were usually doing
geometry on the plane.
And they gave you various axioms
about the way things behaved.
One of them is now known as
Euclid's parallelism axiom.
And it states that if I have
two lines that start parallel,
they remain parallel.
If you dig into the
history of mathematics,
this axiom bugged the crap out
of people for many, many years,
OK?
Because when you look at
pictures on a piece of paper
or on a chalkboard or something
like that, it seems right, OK?
But it really can't be
justified quite as rigorously
as many other axioms
that Euclid wrote down.
And it turns out, there's
an underlying assumption.
The assumption is that
you are drawing your lines
on a flat manifold.
There's a ready counterexample.
Go to the Earth's equator.
You start, say,
somewhere in Brazil.
Your friend starts
somewhere in Africa.
Both of you stand on the
equator and go due north.
You are moving on
parallel trajectories.
You will intersect
at the North Pole.
OK?
Well, guess what?
You ain't moving
on a flat manifold.
You're moving on the
surface of a sphere.
OK, curvature causes initially
parallel trajectories
to become non-parallel.
What this is telling us is that
the manifold I'm going to want
to use to describe events when
I have gravity cannot be flat.
I'm going to have to
introduce curvature into it.
Take us a little while to unpack
precisely mathematically what
that means.
But this is sort of a sign that
things got more complicated.
And tides are the way in
which that complication
is being introduced.
So switch pages here.
So what we have basically danced
around and put together here
is one formulation
of what is known
as the principle of equivalence.
The principle of equivalence,
or one formulation of it,
tells me that over
sufficiently small regions,
the motion of freely falling
particles due to gravity cannot
be distinguished from
uniform acceleration.
The physical
manifestation of this
is that if you're
an observer who
is in that freely falling
frame moving due to gravity,
you are also experiencing
that uniform acceleration.
So you see no net acceleration.
This particular
formulation of it
is known as the weak
equivalence principle.
I'm going to give you a
couple variations on this
in just a second.
One thing that's
important about it is,
essentially, it's a
statement that, if you
think about Newton's laws,
the idea that f equals ma
and the force of gravity
is proportional to m,
it's a fairly precise
statement that you
think of that m as the
gravitational charge.
It's saying that the
gravitational charge,
the gravitational mass
and the inertial mass
are the same thing.
That's actually a
testable statement.
What you can do is look
at materials that have
very different compositions.
In particular,
what you want to do
is look at things
that maybe have
different ratios of
neutrons to protons
or have highly bound nuclei
with lots of gluons in them
or various kinds
of fields that you
can imagine perhaps couple
to gravity differently
than the quarks do.
And so there are what are called
free fall experiments to test
this, which really
basically just boils down
to dropping lots of different
elements of the periodic table
and making sure they all
fall at the same rate.
And the answer is that they
fall at exactly the same rate
within experimental precision.
And the last time I looked
it up, well, what they do
is they demonstrate the WEP,
the Weak Equivalence Principle,
is valid to about, I believe,
it's about a part in 10
to the minus 13.
Might be a little
bit better now.
So bear with me just
one second here.
OK, so good.
We've got this notion here.
And you might now
think, OK, great.
I have a way of
generalizing what
the notion of an
inertial frame is.
Is this enough for
me to move forward?
Can we now do all of
physics by just applying
the laws of special relativity
in freely falling frames
or, putting it this way, our
new notion of an inertial frame?
For two reasons that
are closely related,
this doesn't quite work.
And it comes down to the fact
that because of tides, we
can't do that.
What this basically means is
that the transformation that
puts you into a freely falling
frame here is not the same
as the transformation
that puts you
into a freely falling frame
10,000 miles over the Earth's
surface, OK?
There are different
transformations
at different locations
in spacetime.
This comes down the idea that
it's not a global Lorentz
frame, OK?
We had a sequence of
local Lorentz frames
that we have to link up.
So I'm going to do a calculation
in just a moment where
we explicitly show that we can
always go into a frame that
is locally Lorentz, but that
there's a term that I'm going
to very strongly
argue is essentially
the curvature associated with
the spacetime metric that
sets the size of what
that region actually is.
So what we are
going to do, what is
going to guide our physics
as we move forward is,
essentially, we're going to take
advantage of a reformulation--
there we go-- of the
equivalence principle, which
I'm going to word as follows.
In sufficiently small
regions of spacetime,
we can find a representation or
a coordinate system such that--
I don't want to cram
this into the margin.
So I'll switch boards--
the laws of physics reduce to
those of special relativity.
This is known as the Einstein
equivalence principle.
At least, that's the name that
Carroll uses in his textbook.
A couple books use
different ones.
But this is a very nice
one because it really
is the principle by which
Einstein guided us to rewriting
the laws of physics.
So we've got a weak equivalence
principle, an Einstein
equivalence principle.
Just as an aside, there
is something called
a strong equivalence principle.
We're not going to talk
about this very much,
but I'm going to give it to you
because it's just kind of cool.
It tells me that gravity
falls in a gravitational field
in a way that is
indistinguishable from mass.
That's a little weird.
I'm not going to explain
it very much right now.
A way to think about
it will become clearer
in some future problem sets.
Basically what it's
telling us is that when
you make very strong--
when you make any kind
of a bound object,
some of the mass of that
object, in the sense
of the mass that is
measured by orbits,
can be thought of as
gravitational energy, OK?
Energy and mass are
equivalent to one another.
So that gravitational energy
should respond to gravity.
And in fact, it falls just
like any other mass, OK?
This is another one.
This is actually very--
well, prior to
September 2015, this
was very difficult to test.
There were some very
precise measurements
of the moon's orbit that
were done to test this.
And there were observations
of binary pulsar systems
that were done to test this.
Now every time LIGO measures
a pair of binary black holes,
which are nothing but
gravitational energy,
and our theoretical models
match the wave forms,
sort of like, boom,
equivalence principle.
Drop mic.
Leave room.
All right, so let's get precise.
What I want to do is
show that we can always
find a local Lorentz frame.
And what I mean
by that is that I
want to be able to
show that I can always
do a change of my coordinates,
a change of my representation
such that, at some point, I can
convert the metric of spacetime
into the metric we used
in special relativity,
at least over some
finite region.
So let me define a
couple of quantities
and then let me formulate
the way this calculation is
going to work.
Let us let coordinates
with unbarred Greek indices
be the coordinate
system that we start in.
And let's say that, in
this representation,
the metric is G alpha beta.
Let's demand that
there exists a set of
coordinates that
I will represent
with bars on the indices.
When we transform to
these coordinates,
I want spacetime to be Lorentz
at least in the vicinity
of some point or some event.
We'll call that event p.
So we will assume that
there is some mapping
between these
coordinates, so there's
no nasty singularities there.
And it's nothing that
we can't deal with.
So we can do it in
either direction,
but let's say x alpha
is x alpha of p written
as some function of these guys.
And so there's a transformation
matrix between the two of them.
It takes the usual form, OK?
So mathematically, the way
I'm going to show this is I
want to show that we can
find a coordinate system such
that, if I compute
the spacetime metric
and the barred
representation, this is always
going to be given by doing
this coordinate transformation.
Then I get the metric
of flat spacetime
over as large a
region as possible.
OK, we know we're not going to
be able to do it everywhere.
What I want to
show is that I can
make that happen at the point
p and that the functional
behavior of this thing
is sufficiently flat
that it remains at
this over some region.
So let me just sketch what
the logic of this calculation
is going to be.
So what we're going
to do is expand.
So remember g is a function.
All of these l's are functions.
So what I'm going
to do is I'm going
to think of this
whole thing as itself
being some kind of a function.
I'm going to expand
in the Taylor series.
There's something
really cool that we're
going to have to do with this.
I mean, if I do this
in general, you're
going to get something
pretty vomitous It's
a giant, giant mess.
OK?
You might worry, what the hell
are we going to do with this?
OK, but there's something
really cool that we can do.
So we are given the metric g.
But we are free to pick our
coordinate transformation to be
whatever we want it to be.
So what we are going
to do is compare
the degrees of
freedom offered by
the coordinate transformation.
Which again, I emphasize,
we are free to make
that whatever we need it to be.
So the coordinate transformation
and its derivative, right?
We're free to play
around with this thing.
It's under our control.
To the constraints
that are imposed
by the metric and its
derivatives, which
we are given.
So what we're
essentially going to do--
so in a moment or two,
I'm going to write out
very schematically what this
coordinate transformation is
going to look like when I
do the Taylor expansion.
And what I want to
do is, essentially,
just count up how many
degrees of freedom
the coordinate system
offers me, count up
the number of
constraints that need
to be matched in order to
effect this transformation,
see whether I've got enough.
Assess what we
learn from that, OK?
We just set up with
a few more things
to help define the logic.
So I'm going to write G
alpha beta as G alpha beta
at the point p plus x gamma.
And you know what?
Let's do the expansion
in the bar coordinate.
Minus x gamma bar at point
p times the derivative
at point p.
And then I'm going
to get something
that involves a
complicated form of x
squared and second derivatives.
OK, this can keep going.
But this is going to be
enough for our purposes.
I am likewise going to expand
my coordinate transformation
matrix.
Make sure I got all
my bars in place.
OK.
OK.
So let me go over
to another board,
just write down a few
more important things.
And then we'll start counting.
So first key thing,
which I want to note
before I dig into
the calculation,
is the metric at this
point, its derivative,
its second derivative.
These have been handed to us.
OK?
So we have no freedom
to play with these.
These are going to
give us constraints
that we need to satisfy.
The coordinate transformation
in its various derivatives,
we are free to specify these.
These are our
degrees of freedom.
OK, so let me just
restate the calculation
that we are going to do.
Basically, we wanted to do this.
Schematically, here's
what this calculation
is going to look like.
And if you look
at this right now,
you might think this is
going to be horrible.
But we don't need to specify
every step of this thing
in absolutely gory detail.
What we want to do
is make this, which
I can write as this guy at
point p, this guy at point p,
this guy at point p plus a
term that is going to be linear
in the distance from the
spacetime displacement away
from point p.
If you want to go and multiply
this out, knock yourself out.
But it suffices to
know that you are
going to get some
new things that
involve first derivatives
of all these quantities.
So this will involve
these two things.
And this will involve
second derivatives
of all these things.
Now what we're
going to do is we're
going to try to make what
we would get multiplying all
this out look as much
like minus 1, 1, 1,
1 of the diagonal as possible.
And like I said, it really just
involves accounting argument.
It's actually one of my
favorite little calculations
we do in this class because
it's my favorite word for this
is it really looks vomitous.
But it's quite elegant.
And I have an
eight-year-old kid.
And I'm constantly
trying to teach her,
math is just all about counting.
And this is a perfect example.
This is really just
about counting.
So let's look at this
thing at 0th order.
So what I'm trying to do is
guarantee that I've got some l,
which I can choose,
which will make
the components of the g at
that point be minus 1, 1, 1, 1.
So my constraints, the
conditions I must satisfy
involve things that hit
these 4 by 4 symmetric tensor
components.
So this guy is something
that is symmetric.
Nature has just handed
me a symmetric tensor
with a symmetric 4 by 4 object.
Those are 10 constraints that
my transformation must satisfy.
What I'm going to use to do this
is this matrix at the point p.
And I'll remind you,
this matrix is just
following set of partial
derivatives at that point p.
This is also 4 by 4.
It is not symmetric.
So this actually gives
me 16 degrees of freedom
to satisfy these constraints.
Ta da!
We can easily make this thing
look Lorentzian at the point p.
In fact, we can do so
and have six degrees
of freedom left over.
Any idea what those six
degrees of freedom are?
Remember you're going
into a Lorentz frame.
Yeah?
Three rotations, three boosts.
Exactly right.
Boom.
So not only does it
work mathematically,
but there's a little
bit of leftover stuff
which hopefully
hits our intuition
as to how things should
behave in special relativity.
OK, that's not really
good enough though, right?
So I just showed that I
can do it at this point.
And if I move a centimeter
away from that point, suppose,
there's some really
steep gradient,
and then it goes
completely to hell.
OK?
Then we're in trouble.
So we have to keep going.
We have to look at
the additional terms
in this expansion.
So when we do things at
0th order, we have now--
so g has been handed to us.
We have now specified
behavior of l at that point.
So l has been
completely soaked up.
I don't have any more freedom
to mess around with it.
When I go to the next
order, OK, so the quantity
that is setting my constraints
is the first derivative
of this.
So these are four derivatives
of my 10 metric functions.
4 by 4 symmetric
by 4 components.
So I have got 40
constraints I must satisfy.
OK, well, the tool
that I have available,
I am free to specify the
derivatives at this point.
Again, remember this is
itself a partial derivative.
So I've got 4 components alpha,
4 derivatives with gamma bar, 4
derivatives with mu bar.
Partial derivatives, it should
not matter what the order is.
So this needs to be
symmetric on mu and gamma.
So I, in fact, have 40 degrees
of freedom at 1st order.
Perfect match.
OK?
So I can make my
coordinate system.
Not only can I make it equal
to the Lorentz form at--
I can make the spacetime
metric Lorentz at the point p,
I can also make it
flat at that point p.
All right, now
we're feeling cocky.
So let's move on.
How far can I go?
I lost the page I need.
OK, so now you have a
flavor for what we're doing.
OK?
I want to count up the
number of degrees of--
sorry-- the number
of constraints
that are in the metric
object, the new metric
object I work with at
this order and compare it
to the degrees of freedom in
the coordinate transformation
at this order.
So when I go to 2nd
order, my new metric thing
that I'm going to be
messing around with
is the second derivative.
So I take two derivatives of
g alpha beta at the point p.
So this again needs to be
symmetric in the derivative.
So I have a symmetric 4
by 4 on these two guys.
And the metric is itself 4
by 4, one alpha and beta.
So there is 100
conditions, 100 constraints
that we must satisfy.
So let's look at the
second derivative of this.
OK, so I'm going to move
this down a little bit.
This is now going to look
like the third derivative
when I do this transformation.
So it's going to be
the third derivative
of x alpha with respect to mu
bar, delta bar, and gamma bar.
So I've got 4 degrees
of freedom for my alpha.
And this must be
perfectly symmetric
under the any interchange
of the indices mu bar,
delta bar, and gamma bar.
OK?
This is a little exercise
in combinatorics.
So the number of equivalent
ways of arranging this
turns out to be n
times m plus 1 times n
plus 2 over 3 factorial.
And I'm in 4 dimensions
for n equals 4.
Work that out, and
you will find that you
have 80 degrees of freedom.
So what we can do
is we can always
find a coordinate transformation
that makes it flat.
It makes it
Lorentzian at point p.
It is flat in that region.
In other words, there is
no first derivative there.
Sorry.
Flat's not really
the right word.
There is no slope
right at that point.
But we cannot transfer
away the second derivative.
And so what this tells me is the
coordinate freedom that we have
means that we can always put our
metric into the following form.
I'm just going to write
this schematically.
Second derivative to
the metric and sort
of the quadratic separation
in spacetime coordinate.
So a couple things about
this are pretty interesting.
So this is basically
telling me that we
can make it Lorentz only up to
terms that look, essentially,
like the second
derivative of the metric.
Well, the second
derivative of any function
tells you about the
curvature of that function.
So this word curvature is going
to be coming up over and over
and over again.
We are, in fact, because
this is general relativity,
we can't do everything simply.
We're going to actually find
that there is a rigorous way
to define a notion of curvature
that we're going to play with,
which indeed looks
like two derivatives
of the spacetime metric.
And we are going to find it
has, I mean, it's actually
going to end up looking
like a 4 index tensor.
OK?
So it's going to be an object
that's got 4 indices on it.
Each of those indices goes over
the 4 spacetime coordinates.
And so naively, it looks like
it's got 4 to the 4th power
independent components.
256.
It has certain symmetries
we're going to see, though.
And when you take into
account those symmetries
and you count up
how many of those
components are
actually independent,
any guesses what the number
is going to turn out to be?
20.
Yes, it exactly compensates
for the number that cannot be
zeroed out by this
coordinate transformation.
It's called the Riemann
curvature tensor,
and we will get to that fairly--
actually, really, just
in a couple lectures.
The other thing which
this is useful for
is in all of my discussion
of the equivalence principle
up to now, there's
been a weasel word
that I have inserted
into much of the physics.
I always said over
sufficiently small regions.
OK?
I say that trajectories begin
to deviate from one another
when they get sufficiently
far away from one another.
OK?
And you should be
going, what the hell
does sufficiently small,
sufficiently far away,
what do all these things mean?
OK?
And the issue is you
need to have a scale.
Well, this scale is going to
be set by the second derivative
of the metric.
So the size of the region
over which spacetime
is inertial in this
coordinate system,
in this representation
is approximately so
imagine it's 1 over--
we can think of d2g
as being 1 over
a length squared.
So 1 over the
square root of that
gives me a rough
idea of how long
the curvature scale
associated with your spacetime
actually is.
And it tells you how big your
inertial region actually is.
So we're going to make this
notion of curvature regress
very, very soon.
As a prelude to
this, we now need
to start thinking about how
we do mathematics and physics
on a curved manifold.
So I'm going to start to
set up some of the issues
that we need to face.
So we're going to need
to define what I mean
by a manifold that is curved.
So a curved manifold
is going to simply
be one in which initially
parallel trajectories do not
remain parallel.
So an example is the surface
of a sphere, as I illustrated.
You start somewhere on
the equator in Brazil.
Your friend starts somewhere
on the equator in Africa.
The two of you
start walking north.
You are exactly
parallel when you take
that first step of the equator.
But your trajectories
cross at the North Pole.
Interestingly, an
example that looks
curved but is not, the
surface of a cylinder.
OK?
If I take two
lines, again, I need
to imagine that this region
continues all the way up here.
I make two lines that are
parallel to each other here
and I have them extend
around this thing,
they would remain parallel
the entire way out, OK?
Another way of stating
that is that you can always
take a cylinder and,
with an appropriate cut,
but without tearing it, you
can flatten it out and make it
into a simple sheet, a
perfectly flat sheet.
You cannot do that with a sphere
without tearing it in some
places.
So what is going to begin
to complicate things
is that we want to work with
vectors and tensors that
live in this curved manifold.
We haven't really thought too
carefully yet about the space.
And let's just focus
on vectors for now.
We haven't thought too carefully
about the space in which
the vectors actually live.
So implicit to
everything we have
talked about up until now is
that we often regard vectors
as objects that themselves
reside in a tangent space.
OK, if I'm working in a
manifold that is flat,
say it's the surface of
this board, OK, and just two
dimensions, every
point on this board
has the same tangent, all right?
So if I draw a vector here and
I draw another vector over here,
it's really easy for
me to compare them
because they actually live
in the same space that
is tangent to this board.
If I'm on the
surface of a sphere,
points that are tangent to
the sphere at the North Pole
are very different
from points that
are tangent to the
sphere on the equator.
So it becomes difficult for
me to actually compare fields
when they are defined on a
curved surface like this.
So this makes it a little
bit-- and so I'm just
going to set up this problem,
and I will sketch the issue.
And then we will resolve it
in our lecture on Tuesday.
This makes it a little
bit complicated for me
to take derivatives
of things like vectors
when I'm working in
a curved manifold.
So let's consider the
following situation.
So I'm going to define some
curve, which I would just
call gamma.
It lives in a curved
space of some sort.
I'm going to draw it
here on the board.
But imagine that this is
on the surface of a sphere.
OK?
So here's the curve gamma.
And let's say this point p
here has coordinates x alpha.
And this point q
here has coordinates
x alpha plus dx alpha.
Suppose there's some
vector field that
fills all this manifold.
OK?
And so let's say
the vector a looks
like this here at the point p.
And it looks like this
here at the point q.
How do I take the derivative
of the vector field
as I go from point p to point q?
Well, your first guess
should basically just
be do what you always
do when you first
learn how to take a derivative.
OK?
So that would be a
notion of a derivative.
There's nothing mathematically
wrong with that.
OK?
But we're going to
find that it gives us
problems for the same reason
that we ran into problems when
we began working
with vector spaces
and curvilinear coordinates.
OK?
If I want this to be a
tensorial object in the same way
that we have been defining
tensors all along here,
I'm going to run into problems.
So let me just actually
demonstrate what
happens if I try to do this.
So the key bit, the way
we want to think about it
is that the points p and q
don't have the same tangent
space, which is a
fancy way of saying
that, as I move from
point p to point q,
the basis vectors are moving.
They're starting to point
in different directions.
So if this were to be--
like I said,
mathematically, if you just
want to get that
derivative, that
is a quantity which has a
mathematical meaning, OK?
But it's not the component
of a tensor, which
we have called out as having a
particularly important meaning
in this geometric construction
of physics that we are doing.
And so if this were
to be tensorial,
then I should be able to switch
to new coordinates, which
I'll designate with primes, such
that the following was true.
The reason why this
doesn't actually work
is I'm going to
demand that alpha--
excuse me-- that a is, in fact,
actually already tensorial.
OK?
It's the compound
of a vector, which
is a particular kind of tensor.
So I'm going to demand
that the following be true.
And I'm going to demand
that my derivatives,
they are just derivatives.
They do the usual rule
Jacobian rule when
I switch coordinate systems.
And so skipping a
line of algebra,
which you can get
in my notes, this
is actually very similar to
the notes I've already posted.
When you work this
out, you're going
to find you get one
term that's correct,
but you get another
term that involves
a derivative of your coordinate
transformation matrix.
And this is an actual term,
spoils the tensorialness
of this quantity.
The way we are going
to cure this is we
are going to demand that if
we want our derivatives to be
derivatives that comport with a
notion of taking tensor objects
and getting other tensor objects
out of them, before we take
the derivative, we have to
have some way of transporting
the objects to the same
location in the manifold,
where I can then compare them.
So here's one example
of a notion of transport
that we could use.
So here's a at point q.
There's a at point p.
Notion one that we'll talk about
is known as parallel transport.
What that's essentially going
to mean is I'm going to say,
well, let's take one of
these guys, either q or p.
The way it's drawn in
my notes, I've used q.
But it could be either.
And let's imagine
I slide it parallel
to itself until this gives me a
alpha from q transported to p.
And then I define a derivative
with that transported notion
of the vector.
Now notice I've called
this notion one.
You can take from
that this idea of how
I transport the vector
from one place to the other
to do this comparison.
I cannot uniquely define it.
There are, in fact, multiple
ways you can do this.
We're going to talk
about two that are
useful at the level of 8.962.
In principle, I imagine
you could probably
come up with a whole butt
load of these things.
These are two that
the physics picks out
as being particularly
useful for us
for the analysis that
we are going to do.
OK?
So we'll pick it up
there on Tuesday.
We're going to
start by coming back
to this notion of I want
to differentiate a vector
field in a curved manifold.
And let me just state
before we conclude,
now that I've transported
a from q to p,
they share the
same tangent space.
Since they share the
same tangent space,
I can compare them more easily.
That allows me to make a
sensible derivative that
respects all the notions
of what a tensor should be.
We'll pick it up from
there on Tuesday.
