[SQUEAKING]
[RUSTLING]
[CLICKING]
SCOTT HUGHES: Last
Thursday, we began
the work of moving
from special relativity
to general relativity,
and we spent
a lot of time unpacking two
formulations of the principle
of equivalence.
So one, which goes
under the name
"weak equivalence principle"--
a simpler way of
saying that is that,
at least over a
sufficiently small region,
if there is nothing
but gravity acting,
I cannot distinguish between
freefall under the influence
of gravity or a
uniform acceleration.
The two things are
equivalent to one another.
Basically, this is a
reflection of the fact
that the gravitational
charge and the inertial mass
are the same thing.
That is the main thing
that really underlies
the weak equivalence principle.
When I gave my tenure talk
a number of years ago here,
I pointed out that there was
this wonderful program called
the Apollo program that was
put together to test this.
And the way it was done was
that they put astronauts
on the moon, and
you actually show
that, if you drop a hammer
and a feather on the moon,
they fall at the same rate.
Of course, the Apollo
program probably
did a few other things as well.
But I'm a general
relativity theorist,
so for me, that was the
outcome of the Apollo program,
was test of the
equivalence principle.
We also have a
different variation
of this we called the Einstein
equivalence principle, which
leads us to a
calculation that we went
through last time, which
states that we can find
a representation
over a sufficiently
small region of spacetime
such that the laws of physics
are reduced to those
of special relativity.
And we did a
calculation to examine
this, where we showed, given
an arbitrary spacetime metric,
I can find a coordinate system
such that this can be written
in the form metric
of flat spacetime
plus terms that are of order
so we have coordinate distance
squared.
Let's put it this way-- so this,
in the vicinity of a point pl--
I'll make that clear
in just a moment.
It ends up looking like of
order of coordinate displacement
squared with corrections
that scale as 1
over second derivative
of the metric.
That sets the scale for what
these end up looking like,
or maybe it's
actually times that.
Sorry.
It's times that.
Why did I divide?
I don't know.
Oh, I know why, because
I wanted to point out
that that is what the scale
of 1 over this thing--
for God's sake, Scott!
Stop putting your
square roots in there!
So it looks like that.
And so this is
what I was saying.
You have a curvature
scale that is
on the order of square root of
1 over the second derivative
of the metric.
Apologies for botching
that as I was writing it
up there quickly.
OK, so we did a calculation
that shows that.
And indeed, what we
did is we went through
and we showed that a general
coordinate transformation has
more than enough
degrees of freedom
to make the metric flat
to get the flat spacetime
metric at a particular point.
And in fact, there are six
degrees of freedom left over,
corresponding to six
rotations and six boosts that
are allowable at that
particular point or event.
Bear in mind we're
working in spacetime.
We have exactly enough degrees
of freedom to cancel out
the first order term, but we
cannot cancel out the second
order term.
And in fact, we find
there are 20 degrees
of freedom left over.
And in a future lecture, we
will derive a geometric object
that characterizes
the curvature that
has indeed 20 degrees
of freedom in it
or 20 independent components
that come out of it.
So this is the foundation of
where we're moving forward.
And so what this
basically tells us
is that, in a
spacetime like this,
we have what we call curvature.
Trajectories that start
out parallel to one another
are not going to remain
parallel as they move forward.
And where we
concluded last time,
we were dealing with
the problem that,
if I want to take
derivatives-- and we're going
to start with vector field.
If I want to
differentiate a vector
field on a curved
manifold, it doesn't work.
If I do the naive thing of just
taking a partial derivative,
it does not work.
So I'll remind you
where we left off.
So we found partial
derivatives of vectors.
Let me put that in there.
Partial derivatives
of vectors, and it
won't take much
work to show it's
true for one forms
or any tensor,
so let's say, partial
derivatives of tensors
do not yield tensors.
We're familiar with
this to some extent
because we already
encountered this
when we began thinking
about the behavior of, even
in flat spacetime, flat
spacetime and curvilinear
coordinates.
So that's the fact that the
basis objects themselves
have some functional dependence
associated with them.
We're interpreting it a
little bit differently now.
And so what we're
doing is we're going
to say that what's
going on here is that,
on my curved
manifold-- and I want
you to visualize
something that's
like a bumpy surface or, if
you like, maybe a sphere.
Think about if you are a
two-dimensional being confined
to the surface of a sphere.
There's only two directions
at any given point.
You can go up, or you can
go on the left-right axis
or the north-south axis, right?
And you would define unit
vectors pointing along there.
But as you move around that
sphere, those of us who
have a three-dimensional life,
and can step back and see this,
we see that these basis objects
point in different directions
at different locations
on the sphere.
The two-dimensional beings
aren't aware of that.
They just know that they're
on a surface that's curved.
And so they would
say that the tangent
space is different
for all these objects.
They've imagined that every
one of these basis objects
lives in a plane that is tangent
to the sphere at any given
point, and that plane is
different at every point
along the sphere.
So we interpret this by saying
that all of our basis objects
live in this tangent space,
and the tangent space
is different at every
point on the surface.
So let me just write
out one equation here.
So when we looked at
the transformation
of a partial
derivative of a vector,
if we looked at just
the components--
so this calculation
is in the notes,
so I would just write down
the result. What we found
was that, if I'm taking,
say, the beta derivative
of component A alpha,
what I found when
I want to go into a
coordinate transformation
is that there's
an extra term that
ruins the tensoriality of this.
So this goes over to
something that looks like--
So the first term is what we'd
expect if this were a tensor
relationship.
That's exactly the matrix
of the Jacobian between two
different coordinate
representations
that we expect to
describe how components
change if they indeed obey
a tensorial relationship.
This extra thing here I
wrote on the second line--
that is spoiling it for us.
So I began to give you the
physical notion of what we
were going to do to fix this.
So let me just
reset that up again.
So let's imagine I have
a particular curve that
goes along my manifold.
I have a point P
here on the curve,
and I have a point Q over here.
And let's say that P
is at event x alpha.
Q is at x alpha plus dx alpha.
Here's my vector
A at the event Q.
And here's my vector
A at the event P.
So what we discussed last time
is that, to get this thing,
I'm just doing the normal
partial derivative.
I'm basically imagining that
these are close to one another,
that I can just subtract A at
Q from A at P, divide by dx
and take the limit.
That's the definition
of a derivative.
And what this is telling us
is, mathematically, yeah,
it's a derivative, but
it's not a derivative
that yields a tensor quantity.
And so we are beginning
to discuss the fact that,
to compare things that have
different tangent spaces, that
live in different points
in my curve manifold,
I need a notion of transport
to take one from the other
in order to compare them.
So there are two
notions of transport
that we're going
to talk about here.
The first one is called
parallel transport.
So transport notion one--
we call this parallel transport.
I'm going to actually focus a
little bit on the math first
and then come back to what is
parallel about this afterwards.
So what I essentially need to
do is say, what I want to do
is find some kind of
a way of imagining
that I take the vector at
P and transport it over
to the point Q, and I will
compare the transported object
rather than the object
originally at point P.
Abstractly, what I'm
essentially going
to do is I'm going to do
what we always do in this.
I'm going to imagine
that there is
some kind of an
operation that is
linear in the separation
between the two of them
that allows me to
define this transport.
So let's do the following.
So I'm going to assume that
we can define an object, which
I will call Pi, capital
Pi, alpha, beta, mu, which
is going to do the following.
So what I'm going to do is
say that A alpha transported--
and to make it even clearer,
how it's being transported.
Let's say it's being
transported from P to Q.
This is given by
alpha at P, and what
I'm going to do is say that,
whatever this object is,
it is linear in both the
coordinate separation
of those two events
and the vector field.
So far, I've said nothing
about physics, by the way.
I'm just laying out some
mathematical definitions.
I'm going to bring it all
together in a few moments.
I'm then going to
say, OK, I know
that I had trouble with my
standard partial derivative.
Let's define a derivative
operator in the following way.
Define a derivative by
comparing the transported vector
to the field at Q.
So what I'm going
to do for the moment is
just denote this notion
of a new kind of derivative
with a capital D. Right now,
it's just another
symbol that we would
pronounce with a "duh" sound so
that it sounds like derivative.
So don't read too much
into that for a moment.
So I'm going to define
this as A at Q minus A
transported from P to Q and
then divide by the separation.
Take the limit-- usual thing.
And when you do
this, you're going
to get something that looks
like the partial derivative
plus an additional term on here.
It looks like this.
So I've said nothing
about what properties
I'm going to demand
of this thing.
And in fact, there
are many ways that one
could define a transport
of an object like this.
In general, when you do this,
this thing I'm calling Pi here
is known as the connection.
It is the object that
connects point P to point Q.
So let's make a couple
demands on its properties.
So now I'll start to put
a bit of physics in this.
So demand one is I'm going to
require that, when I change
my coordinate representation,
when I evaluate this
in my new coordinate
system, that I get something
that looks like this.
If I do this, I'm going
to find that, when
I change coordinates and apply
it to the entire derivative
I've defined over here, that
little extra bit of schmutz
that's on the second line
there is exactly what you need
to cancel out this annoying
bugger so that you have
a nice tensor relationship left.
So I'm going to demand that a
key part of whatever this guy
turns out to be is something
that cancels out the irritating
garbage that came along
with the partial derivative.
Combine that with what I'm
about to say in just a moment--
that pins things
down significantly.
I'm going to make
one more demand,
and this demand is going
to then be connected
to the physical picture
that I'm going to introduce
in about five minutes.
My final demand is I'm
going to require that
whatever this derivative is--
when I apply it to
the metric, I get 0.
If I do that, then it turns out
that the connection is exactly
the Christoffel symbol
that we worked out earlier.
If I put in this demand, the
connection is the Christoffel,
and this derivative
is nothing more
than the covariant derivative.
Der-iv-a-tive.
This is just the
covariant derivative
we worked out earlier.
So much for mathematics.
The key thing which
I want to emphasize
at this point in
the conversation--
I hope we can see the logic
behind the first demand.
That's just something which
I'm going to introduce in order
to clean this guy up.
This is my choice.
Not just my choice--
it's the choice
of a lot of people who have
helped to develop this subject.
But it's a good one,
because, when I do this,
there is a particularly good
physical interpretation to what
this notion of transport means.
And bear with me
for a second while I
gather my notes together.
So let's do the top one first.
OK, so let's make
that curve again.
And actually, let's just go
ahead and redraw my vector
field.
I do want to have a
different copy of this.
Let me introduce
one other object.
So I'm going to give
this curve a name.
I'm going to call
this curve gamma.
And this is a notion
that I'm going
to make much more precise
in a future lecture,
but imagine that there is
some kind of a tape measure
that reads out along the
curve gamma that is uniformly
ticked in a way that
we will make precise
in a future lecture.
I will call the
parameter that is uniform
that denotes these
uniform tick marks--
I will call that lambda.
If you want to read a
little bit about this,
this is what's known
as an affine parameter.
We will make this a little
bit more precise very soon.
With this in mind, you should
be able to convince yourself
that this field U defines the
tangent vector to this curve.
So let's say that my original
point here, that this
is at lambda equals 2--
and let's say this is
at lambda equals 7.
And what I want to do
is transport the vector
from 2 to 7.
Well, a way that I can
do this is by saying,
OK, I now know
that the derivative
that goes over here-- this is
just the covariant derivative.
So I forgot to write this
down but from now on,
this derivative I wrote down--
I'm going to go back to
the gradient symbol we used
for the covariant derivative.
What I can do is take
the covariant derivative
of my vector field, contract
it with this tangent vector.
And what this does--
I'm going to define this as
capital D alpha D lambda.
This is a covariant derivative
with respect to the parameter
lambda as I move along this
constrained trajectory.
What this does is this
tells me how A changes as it
is transported along the curve.
I'm now going to argue--
what parallel
transport comes down to
is when you require that, as
you slide this thing along here,
the covariant total derivative
of this thing as you
move along the curve,
that it's equal to 0.
OK, now let me motivate
where that's coming from.
Why is that?
So let's put all of our
definitions back in.
Let's apply the definition
of covariant derivative
that we discussed in
a previous lecture
and we've all come
to know and love.
Now, this is the bit where
we begin to introduce
a little bit of physics.
Let's imagine that
points P and Q
are sufficiently close
to each other that they
fit within a single,
freely falling frame.
They can go into the
same local Lorentz frame.
So remember, when I go into
my local Lorentz frame,
the metric becomes the
metric of flat spacetime.
G goes over to eta.
There is a second order
correction to that.
So the second derivatives--
I cannot find a [INAUDIBLE]
that gets rid of them.
So there'll be a little bit
of that there, which tells me
how large this
Lorentz frame can be.
But I can get rid of all
the first derivatives.
And if you get rid of all
the first derivatives,
you zero out the Christoffel
symbols in that frame.
So in that frame, there
are no Christoffel symbols.
So what this means is
that, in this frame,
this just becomes the idea
that a simple total derivative
of this object-- you don't need
to include all the garbage that
comes along with the
covariant derivative--
this is equal to 0.
And that's equivalent to saying
that, as I take this vector
and transport it along, I
hold all of its components
constant as I slide it
from one step to the other.
So I start out with
my A at point P here.
And then I slide it
over a little bit,
holding all the components
constant just like this--
slide over again,
slide it over again.
Da, da, da, da da.
Da, da, da, da, da.
Till finally, I get
it over to there.
What this is doing is that--
any two vectors in the middle
of this transport process--
I am holding them
as parallel as it
is possible to hold them,
given that, to be blunt,
you can't even really
define a notion
of two objects being
parallel on a curved surface
if there's a macroscopic
separation between them.
But if you think about
just a little region that's
sufficiently small,
that it's flat
up to quadratic
corrections, then a notion
of these things being parallel
to each other makes sense.
And this idea, that I'm going
to demand that, upon transport,
the derivative of
the metric equals 0,
thereby yielding my connection
being the Christoffel
and this derivative being
the covariant derivative--
it tells us that this
notion of transport
is one in which
objects are just kept
as parallel as possible
as they slide along here.
And that is why this is
called "parallel transport."
It's as parallel as
it's possible to be,
given the curvature.
So as I switch
gears, I just want
to emphasize I went through
the mathematics of that
with a fair amount
of care because it
is important to keep this stuff
as rigorous as possible here.
This notion of
parallel transport
gets used a lot when we start
talking about things moving
around in a curved spacetime.
In particular, if you
think about an object
that is freely falling,
and it is experiencing
no forces other
than gravity, which
we are going to no longer regard
as a force before too long--
if you just go back to Newtonian
intuition, what does it do?
Well, you give it an initial
velocity or initial momentum,
and it maintains it.
It just continues going
in a straight line.
In spacetime, going
in a straight line
basically means, at
every step, I move
and I take the tangent to my
world line, my four-velocity,
and I move it
parallel to itself.
So this notion of
parallel transport
is going to be the
key thing that we
use to actually define
the kinetics of bodies
in curved spacetime.
There's a tremendous
amount of work
being done by all sorts
of things these days
that's based on studies of
orbits in curved spacetime,
and they all come back to this
notion of parallel transport.
All right.
Bear with me a second.
I just want to take
a sip of water.
And then I'm going to talk about
the other notion of transport,
which we are going to discuss--
I always say, briefly, and
then I spend three pages on it,
so we're going to discuss.
All right.
So parallel transport
is extremely important,
and there's a huge
amount of physics
that is tied up in
this, but one thing
which I really want to emphasize
is that it is not unique.
And there is one
other one which we
are going to really use
to define one particularly
important notion, instead of
quantities, for our class.
So suppose I've
got my curve gamma,
and I'm going to, again,
take advantage of the fact
that I can define a set
of tick marks along it
and make that vector U be
the tangent to this curve.
And I'm going to, again,
have my favorite points,
x alpha at point P plus
dx alpha at point Q.
There's another notion of
transport that is-- basically
what you do is
you cheat, and you
imagine that moving from point
x alpha to x alpha plus dx alpha
is a kind of coordinate
transformation.
So let's do the following.
Let's say that x
alpha plus dx alpha--
we're going to take
advantage of the fact
that, since we have this tangent
notion built into the symbols
we've defined, we'll
just say that it's
going to be the tangent
times the interval of lambda.
And what I'm going
to do is define
this as a new coordinate
system, x prime.
So that's the alpha component
of coordinate system x prime.
It's a little bit weird
because your x prime has
a differential built into it.
Just bear with me.
So what we're going to
do is regard the shift,
or the transport, if
you prefer, from P to Q
as a coordinate transformation.
It's the best eraser, so
I'll just keep using it.
So what I mean by that is
I'm going to regard x alpha,
and I'm going to use a
slightly different symbol.
I will define what the L is.
So this is transported, but
I'm going to put an L in here
for reasons that I will
define in just a moment.
This, from P to Q, is what
I get if I regard the change
from point P to point Q
as a simple coordinate
transformation and do my
usual rule for changing
coordinate representation.
So expand what the definition
of x prime is there,
and what you'll see is that
you get a term that's just
basically dx alpha dx beta.
Then you're going
to get something
that looks like the
partial derivative
of that tangent vector.
And remember, this
is being acted on.
I should've said this is
this thing evaluated at P.
Great.
So we fill this out.
OK.
So that's what I get when I
use this notion to transport
the field from P to Q.
Let's think about
it in another way.
Now, these fields are
all just functions.
So I can also express the field
at Q in terms of the field
at P using a Taylor expansion.
I'm assuming that these are
close enough that everything
is accurate to first
order in small quantities,
so nothing controversial
about this.
I'm assuming dx is small
enough that I can do this.
But now I'm going to
get rid of my dx beta
using the tangent field U.
Now, before I move on, I
just want to emphasize--
these two boards here,
over the way the left--
we're talking about two
rather different quantities.
The one I just
moved to the top--
that actually is the field-- if
you were some kind of a gadget
that managed your field A--
that would tell you
what the value is
that you measure at
point Q. This would tell
you-- what do we
get if you picked P
up and, via this transport
mechanism, moved it over to Q?
They are two potentially
different things.
So this motivates another
kind of derivative.
So suppose I look at A--
value it at a Q--
minus A transported--
whoops, that's supposed to be
transported from P to Q--
defined by D lambda.
I will expand this
out in just a moment.
Now I will, at last,
give this a name.
This is written with
a script L. This
is known as the Lie
derivative of the vector A
along U. Anyone heard of
the Lie derivative before?
Yeah.
So at least in the context where
we're going to be using it,
this is a good way to understand
what's going on with it.
We'll see how it is
used, at least in 8.962
in just a few moments.
Filling in the details-- so
plug in these definitions,
subtract, take limits,
blah, blah, blah.
What you find is
that this turns out
to be U contracted on the
partial derivative of A
minus A contracted on the
partial derivative of U.
Exercise for the reader--
it is actually
really easy to show
that you can promote
these partial derivatives
to covariant derivatives.
And what this means is that,
when you evaluate the Lie
derivative-- so
notice, nowhere in here
did I introduce anything
with a covariant derivative.
There was no connection,
nothing going on there.
If you just go ahead
and work it out,
basically, when you
expand this guy out,
you'll find you have connection
coefficients or Christoffel
symbols that are
equal and opposite
and so they cancel each other.
So you can just go from
partials to covariants.
Give me just a second, Trey.
And this is telling us
that the Lie derivative
is perfectly tensorial.
So the Lie derivative
of the vector field
is also a tensor quantity.
You were asking
a question, Trey.
AUDIENCE: In the second term,
did you miss the D lambda?
SCOTT HUGHES: I did.
Yes, I did.
Thank you.
There should be a D
lambda right here.
Thank you.
Yes.
Yeah.
If you don't have
that, then you get what
is technically called "crap."
So thank you for
pointing that out.
For reasons that I hope you
have probably seen before,
you always compute
the Lie derivative
of some kind of an object
along a vector field.
So when you're computing the
Lie derivative of a vector
field along a vector,
sometimes this
is written using a commutator.
I just throw that out there
because you may encounter
this in some of your readings.
It looks like this.
So let me just do
a few more things
that are essentially fleshing
out the definition of this.
So I'm not going to go through
and apply this definition
very carefully to
higher order objects.
What I will just say
is that, if I repeat
this exercise and, instead
of having a vector field
that I'm transporting from
point to point, suppose
I do it for a scalar field--
well, what you actually get--
pardon me for a second--
is this on the partial, but the
partial derivative of a scalar
is the covariant
derivative because there's
no Christoffel that couples in.
If you do this for a one-form,
where it's a 1 indexed
object in the
downstairs position,
you get something
that looks like this.
And again, when you expand out
your covariant derivatives,
you find that your Christoffel
symbols cancel each other out.
And so, if you
like, you can just
go ahead and replace
these with partials.
And likewise, let me just write
one more out for completeness.
Apply this to a tensor.
So it's a very similar
kind of structure
to what you saw when we did the
covariant derivative in which
every index essentially gets
corrected by a factor that
looks like the covariant
derivative of the field
that you are
differentiating along.
The signs are a
little bit different.
So it's a similar tune, but
it's in a different key.
OK, so that's great.
And if you get your
jollies just understanding
different mathematical
transport operations,
maybe this is
already fun enough.
But we're in a physics
class, and so the question
that should be to
your mind is, is there
a point to all this analysis?
So in fact, the most important
application of the Lie
derivative for our purposes--
in probably the
last lecture or two,
I will describe some stuff
related to modern research
that uses it quite heavily.
But to begin with in our class,
the most important application
will be when we
consider cases where,
when I compute the Lie
derivative of some tensor
along a vector U and I get 0.
I'm just going to leave
it schematic like that.
So L U of the tensor
is equal to 0.
If this is the case, we say that
the tensor is Lie transported.
This is, incidentally,
just a brief aside.
It shows up a lot
in fluid dynamics.
In that case, U often
defines the flow lines
associated with
the velocity field
of some kind of a
fluid that is flowing
through your physical situation.
And you would be
interested in the behavior
of all sorts of quantities that
are embedded in that fluid.
And as we're going
to see, when you
find that those quantities are
Lie transported in this way,
there is a powerful
physical outcome
associated with that,
which we are going
to derive in just a moment.
So suppose I, in fact, have a
tensor that is Lie transported.
So suppose I have some tensor
that is Lie transported.
If that's the
case, what I can do
is define a particular
coordinate system
centered on the curve for
which U is the tangent.
So what I'm going to do is I'm
going to define this curve such
that x0 is equal to lambda,
that parameter that defines
my length along the curve in
a way that, I will admit I've
not made very precise
yet but will soon.
And then I'm going to require
that my other three coordinates
are all constant on that curve.
So if I do that, then my tangent
vector is simply delta x0.
In other words, it's only got
one non-trivial component,
and its value of
that component is 1.
And this is the constant.
So the derivatives
of the tangent field
are all equal to 0.
And when you trace this through
all of our various definitions,
what you find is that
it boils down to just
looking at how the
tensor field varies
with respect to that parameter
along the curve itself.
If it's Lie transported,
then this is equal to 0.
And so this means that,
whatever x0 represents,
it's going to be a constant
along that curve with respect
to this tensor field.
Oh, excuse me.
Screwed that up.
The tensor does not vary with
this parameter along the curve.
This was a lot, so let's
just step back and think
about what this is saying.
One of the most important things
that we do in physics when
we're trying to
analyze systems is
we try to identify
quantities that
are constants of the motion.
This is really tricky
in a curved spacetime
because much of
our intuition gets
garbled by all of the facts
that different points have
different tangent spaces.
You worry about whether
something being true,
and is it just a function
of the coordinate system
that I wrote this out in?
What the hell is going on here?
The Lie derivative is giving us
a covariant, frame-independent
way of identifying things that
are constants in our spacetime.
So we're going to wrap
up this discussion.
Let's suppose that the tensor
that I'm looking at here
is called the metric.
Suppose there exists a vector
C such that the metric is Lie
transported along this thing.
What does this tell us?
So first, it means there
exists some coordinate such
that the metric does not vary.
The metric is constant with
respect to that coordinate.
Essentially, if you go through
what I sketched a moment ago,
this is telling us
that the existence
of this kind of a vector, which
I'm going to give a name to
in just a moment--
the existence of
this thing demands
that my metric is constant with
respect to some coordinate.
I am not going to prove
the following statement.
I will just state it,
because, in some ways,
the converse of that statement
is even more powerful.
If there is a coordinate,
such that dgd,
whatever that coordinate
is, is equal to 0,
then a vector field
of this type exists.
So the second thing I want to
do is expand the Lie derivative.
So if I require that my metric
be transported along the vector
C, well, insert my definition
of the Lie derivative.
Now, what is the main
defining characteristic
of the covariant derivative?
How did I get my connection
in the first place?
In other words, what
is this going to be?
OK, students who took
undergraduate classes with me,
I'll remind you of one
of the key bits of wisdom
I always tell people.
If the professor asks you a
question, 90% of the time,
if you just shout out, "0,"
you are likely to be right.
[LAUGHING]
Usually, there's some
kind of a symmetry
that we want you to understand,
which allows you to go,
oh, it's equal to 0.
By the way, whenever I
point that out to a class,
I then work really hard
to make a non-zero answer
for the next time I ask it.
So the covariant derivative
of g is 0, so this term dies.
Because the covariant
derivative of g is 0,
I can always commute the metric
with covariant derivatives.
So I can take this, move
it inside the derivative.
I can take this, move it
inside the derivative.
So what this means is this
Lie derivative equation,
after all the smoke clears,
can be written like this.
Or, if I recall,
there's this notation
for symmetry of indices,
which I introduced
in a previous lecture.
The symmetric covariant
derivative of this C
is equal to 0.
This equation is known
as Killing's equation,
and C is a Killing vector.
Now, this was a fair
amount of formalism.
I was really laying out a lot of
the details to get this right.
So to give you some context
as to why this matters,
there's a bit more that
needs to come out of this,
but we're going to
get to it very soon.
Suppose I have a
body that is freely
falling through some spacetime.
And you know what?
I'm going to leave this here.
So this is a slightly
advanced tangent,
so I'll start a new board.
So if I have some body
that is freely falling,
what we are going to
show in, probably,
Thursday's lecture is that the
equation of motion that governs
it is-- you can argue
this on physical grounds,
and that's all I
will do for now--
it's a trajectory that parallel
transports its own tangent
factor.
For intuition, go into
the freely falling frame
where it's just the trajectory
from special relativity.
It's a straight
line in that frame,
and parallel transporting
its own tangent vector
basically means it just moves
on whatever course it is going.
So this is a
trajectory for which
I demand that the four-velocity
governing it parallel
transports along itself.
Now, suppose you are
moving in a spacetime that
has a Killing vector.
So this will be
Thursday's lecture.
Suppose the spacetime
has a Killing vector.
Well, so there will be some
goofy C that you know exists,
and you know C
obeys this equation.
By combining these
things, you can
show that there is
some quantity, C,
which is given by
taking the inner product
of the four-velocity of
this freely falling thing
and the Killing vector.
And you can prove that this
is a constant of the motion.
So let's think about where this
goes with some of the physics
that you presumably all
know and love already.
Suppose you look at a spacetime.
So you climb a
really high mountain.
You discover that there's
a spacetime metric carved
into the stone
into the top of it.
You think, OK, this
probably matters.
You look at it and you notice it
depends on, say, time, radius,
and two angles.
Suppose you have a metric
that is time-independent.
Hey, if it's
time-independent, then I
know that the derivative of
that thing with respect to time
is 0.
There must exist
a Killing factor
that is related to the
fact that there is no time
dependence in this metric.
So you go and you calculate it.
So this thing, that C is
a constant of the motion--
I believe that's P set 4.
It's not hard.
You combine that equation
that we're going to derive,
called the geodesic equation,
with Killing's equation.
Math happens.
You got it.
So suppose you've got a
metric that's time-dependent
and you know you've
got this thing.
So you know what?
Let's work it out
and look at it.
It becomes clear,
after studying this
for a little bit, that the C for
this Killing vector is energy.
In the same way that, if
you have a time-independent
Lagrangian, your system
has a conserved energy,
if you have a
time-independent metric,
there is a Killing
vector, which--
the language we like to use is--
we say the motion
of that spacetime
emits a conserved energy.
Suppose you find that the metric
is independent of some angle.
We'll call it phi.
Three guesses what's
going to happen.
In this case-- just
one guess, actually.
Conserved--
AUDIENCE: Angular momentum.
SCOTT HUGHES: Angular momentum
pops out in that case.
So this ends up being the
way in which we, essentially,
make very rigorous
and geometric the idea
that conservation laws are put
into general relativity, OK?
So I realize there's a
lot of abstraction here.
So I want to go on a
bit of an aside just
to tie down where we
are going with this
and why this actually matters.
OK.
Let's see.
So we got about 10 minutes left.
So what we're going to do
at the very end of today--
and we'll pick this up
beginning of next time.
So for the people who walked
in a few minutes late,
the stuff that I'm actually
about to start talking about we
need to get through before
you can do one of the problems
on the P set.
So I'm probably going
to take that problem
and move it on to
P set 4, but I'm
going to start talking
about it right now.
So we've really focused a
lot, so far, on tensors.
We're going to now start talking
about a related quantity called
tensor densities.
There's really only two that
matter for our purposes,
but I want to go
through them carefully.
So I will set up with
one, and then we'll
conclude the other
one at the beginning
of Thursday's lecture.
So let me define this first.
I'm going to give a
definition that I like
but that's actually
kind of stupid.
So these are quantities that
transform almost like tensors--
a little bit lame, but,
as you'll see in a moment,
it's kind of accurate.
What you'll find is that
the transformation law
is off by a factor
that is the determinant
of the coordinate
transformation matrix.
Take it to some power.
So there's is an infinite
number of tensor densities
that one could define.
Two are important
for this class.
So the two that are
most important for us
are the Levi-Civita symbol and
the determinant of the metric.
So we use Levi-Civita already
to talk about volumes.
And it was a tensor
when we were working
in rectilinear coordinates,
where the underlying coordinate
system was essentially
Cartesian plus time.
It's not in general, OK?
And we'll go
through why that is.
That'll probably be the last
thing we can fit in today.
So let me remind you--
Levi-Civita-- I'm going to
write it with a tilde on it
to emphasize that
it is not tensorial.
So this is equal to
plus 1 if the indices
are 0, 1, 2, 3 and
even permutations
of that equals minus 1 for
odd permutations of that.
And it's 0 for any
index repeated.
Now, this symbol has a
really nice property when
you apply it to any matrix.
In fact, this is a theorem.
So I'm working in
four-dimensional space.
So let's say I've got a 4-by-4
matrix, which I will call m.
Write a new [INAUDIBLE]
notation, m alpha mu.
If I evaluate Levi-Civita,
contract it on these guys,
I get Levi-Civita
back, multiply it
by the determinant
of the matrix m.
Now, suppose what I
choose for my matrix
m is my coordinate
transformation matrix.
So I'm just going to
write down this result,
and I'll leave it since we're
running a little short on time.
You can just double check
that I've moved things
from one side of the
equation to the other,
and you can just double-check
I did that correctly.
What that tells me
is that Levi-Civita
and a new set of
prime coordinates
is equal to this guy
in the old, unprimed
coordinates with all my usual
factors of transformation
matrices and then
an extra bit that
is the determinant of the
coordinate transformation
matrix.
If it were just
the top line, this
is exactly what you would
need for Levi-Civita
to be a tensor in the way
that we have defined tensors.
It's not.
So the extra factor pushes away
from a tensor relationship.
And so what we would say
is, because this is off
by a factor of what's
sometimes called the Jacobian,
we call this a tensor
density of weight 1.
So in order to do
this properly--
I don't want to rush--
at the beginning of
the next lecture,
we're going to look at how
the determinant of the metric
behaves.
And what we'll see is that,
although the metric is
a tensor, its
determinant is a tensor
density of weight negative 2.
And so what that tells
us is that I can actually
put together a combination
of the Levi-Civita
and the determinant of
the metric in such a way
that their product is tensorial.
And that turns out to
be real useful because I
can use this to define,
in a curved spacetime,
covariant volume elements, OK?
With this as written,
my volume elements--
if I just use this
like I did when we're
taught about special relativity,
my volume elements won't
be elements of a tensor,
and a lot of the framework
that we've developed
goes to hell.
So an extra factor of the
determining of the metric
will allow us to correct this.
And this seems kind of abstract.
So let me just, as a really
brief aside, before we conclude
today's class--
suppose I'm just in
Euclidean three-space
and I'm working in
spherical coordinates.
So here's my line element.
My metric is the
diagonal of 1r squared
r squared sine squared theta.
The determinant of the
metric, which I will write g--
it's r to the fourth
sine squared theta.
What we're going
to learn when we do
this is that the metric is a
tensor density of weight 2.
And so to correct it to
get something of weight 1,
we take a square root.
If you're working in
circle coordinates,
does that look familiar?
This is, in fact,
exactly what allows
us to convert differentials
of our coordinates.
Remember, we're working
in a coordinate basis.
And so we think of
our little element
of just the coordinates.
It's just dr, d theta, d phi.
This ends up being
the quantity that
allows us to convert the little
triple of our coordinates
into something that has
the proper dimensions
and form to actually be
a real volume element.
And so dr, de theta, d phi--
that ain't enough volume.
But r squared sine theta,
dr, d theta, d phi--
that's a volume element, OK?
So basically, that's all
that we're doing right now,
is we're making that
precise and careful.
And that's where I
will pick things up.
We'll finish that
up on Thursday.
