[SQUEAKING]
[RUSTLING] [CLICKING]
SCOTT HUGHES: So this is
unfortunately a rather sad
lecture, [LAUGHS] since this
will be the last one that
I'm giving to a
live audience here,
although any of you
who are grad students,
if you want to come and keep
me company while I'm recording
my next couple of lectures--
I'm actually not joking.
It's really weird
talking to an empty room,
and so it might be kind of
nice to at least have a face
to react to.
So I will be recording these,
and making them available.
And per the announcements
that I sent out today,
there are some
additional assignments
that I've begun posting.
I've begun putting
things like that up.
But for the time being,
ignore due dates.
In fact, I would like
to say, especially
if you are getting ready
to leave town, just
ignore the assignments.
There will be updates
posted fairly soon.
You know, [CHUCKLES]
sometimes-- there's
a saying that Martin
Schmidt, our provost,
said one time in respect
to a particular initiative
as it was unwinding, and I
feel like it applies here:
we're basically sewing together
the parachute after we've
jumped out of the airplane.
[CHUCKLES] And so that's kind
of where we are right now,
and hopefully it'll be assembled
before we hit the ground.
But we're all kind of figuring
out the way this goes.
So there will be updates
with respect to assignments
and things like
that, and you know,
we're just trying
to figure out--
I'll just be
blunt-- how the hell
we're going to
actually pull this off.
[LAUGHS] It's not curved enough.
All right, so let me get
back to where we were.
I want to begin--
so what we worked
with yesterday was
we derived this mathematical
object called the Riemann
curvature tensor.
And the way we discussed it was
by thinking about some vector
that I parallel transported
around a closed loop.
I chose a parallelogram because
it allowed me to sort of nicely
formulate the mathematics.
So if I imagine a
parallelogram who's
got sides that are
sort of close enough,
that parallel is a
well-defined term,
and the sides are in two
different directions,
one's of length delta x
in the lambda direction,
one is a delta x in
the sigma direction,
when I take this vector all
the way around the loop,
I find the vector has changed.
It's pointing in a
different direction,
and this describes the way
in which it has sort of moved
as I go around that loop.
That object is-- even
though it's created--
it's assembled from the
Christoffel symbols, which
are not tensor, but
it's done in such a way
that their nontensorialness
cancels out,
and this four-index
object R is a tensor.
As you can see,
it involves terms
like derivatives of the
Christoffels and Christoffel
times Christoffel.
So that's where nonlinearity
is going to enter.
I want to sort of
fairly quickly--
because I do want to get into
some other stuff that we do
with this--
talk a bit about the
symmetry of this thing.
So I discussed last time
how you look at this,
and naively, you
think you've got--
in four dimensions, four
dimensions, four [INAUDIBLE]
states.
It looks like you
have 256 components.
But in fact, there are
quite a few symmetries
associated with this which
reduces it significantly.
I kind of gave you
the answer last time,
but let me talk about
where that comes from.
I'm going to describe a way
to see these symmetries, which
there's a couple
ways you can do it,
and I'm going to
pick the one that's
essentially the simplest.
It's perhaps not as--
there's a few ways to do it
that a purist might prefer,
but in the interest
of time, I think
this is sort of the best one.
So the first thing is that
just by inspecting this answer,
you should see that this
object is antisymmetric if you
exchange the final two indices.
OK, just look at that
formula, switch the indices
sigma and lambda, and
you get a minus sign.
Physically, that actually
hopefully makes sense,
because what that is basically
telling you-- remember the way
we did this, this operation
of parallel transporting
around a parallelogram.
We sort of went along, say,
the lambda direction, then
the sigma, then the
lambda, than the sigma.
If I exchange
those indices, it's
kind of like I actually go in
the opposite direction, OK?
I go in the other
direction first.
I go on the sigma
first, then the lambda.
And so this corresponds to
doing this operation, which
I'll remind you is
called a holonomy.
It corresponds to
reversing direction.
OK?
So that's a fairly
easy one to say.
The others are a little bit
tougher, and like I said,
what I'm going to do is follow
kind of a shortcut to do this.
One can do it using sort of
the full glory of the Riemann
tensor here, but life gets
a little bit easier for us
if we do the following.
So what you want to do first
is lower the first index.
So what I'm going
to do is look at R,
alpha in the downstairs,
mu lambda sigma.
This is what I get when
I contract, like so.
And then what I'm
going to do is I'm
going to do all of my analysis
that follows in a local Lorentz
frame.
When I go into the
local Lorentz frame,
the metric at a
particular point is
the metric of flat spacetime.
It's the A to mu nu.
My Christoffel
symbols all vanish,
but you have to be a
little bit careful,
because the derivatives of
the Christoffel do not vanish,
right?
So one of the key
points is that when
you go into those
local Lorentz frame,
there is a little bit of
curvature associated with it.
So the derivatives do not--
OK, so the reason I'm
doing this-- like I said,
you actually could do
the little counting
exercise I'm about to go
through with this whole thing.
It's just messy, OK?
And so this is sort of a
quick way to see the way
it all comes out.
Because this is a
tensor relationship,
you are guaranteed
that something
you conclude in a
convenient reference frame
will hold in all of them, OK?
So it's a nice way to do this.
Those of you who are
sort of, you know,
particularly purist,
knock yourselves out
trying to figure out how to
do this sort of in general.
This is adequate for
our purposes here.
So when I go and
I do this, here's
what my all downstairs
Riemann tensor turns into.
I'm going to lose the
Christoffel squared
terms because
Christoffel vanishes,
and the only thing
that is left are
the terms that involve the
derivative of the Christoffel,
OK?
Let's do something a
little bit further.
So let's now plug in the
definition of the Christoffel,
and we'll use the metric before
we actually go into the frame,
so that we can sort of
get the form that we
would get with this.
I should have made myself be a
little bit more careful here.
So I'm going to put LLF on this,
to make it clear that I'm doing
this in a local Lorentz frame.
So I do this in a
local Lorentz frame,
and what this turns into--
so there's a lot of algebra
that I'm not writing out.
This is an alpha, g sigma mu
minus d sigma d mu g alpha
lambda plus d sigma d
alpha g lambda mu, OK?
So notice, it is only the
second derivative of the metric
that is appearing here, OK?
If I go into the local
Lorentz frame, that's
the one degree of
freedom we were not
able to transform away.
Anything involving the metric
in the first derivative
dies in this particular frame.
But this is good
enough for us to begin
to count up symmetries,
and to see what
things are going to look like.
And so I don't have any great
guidance for how to do this.
This is one of those things
where, literally, you
just sort of stare at
it for a little while,
and you see what happens when
you play around with exchanging
various pairs of indices.
So stare at this for a while.
OK, so you can see right away--
just for counting
purposes, let me write down
the one that I argued
was kind of obvious even
in the full form.
It's the one you get
when you exchange
the last pairs of indices.
We'll call that symmetry one.
Oh, bugger, thank you.
[CHUCKLES]
Yeah.
[CHUCKLES] That's a
good way to-- that's
a very complicated way of
writing zero, otherwise.
[LAUGHING]
All right, if you
exchange the first ones,
you also-- if you
look at this formula,
you can see when you
exchange alpha and mu, you
get a minus sign.
So one that might be--
this one takes a little bit of
creativity to sort of see it,
where it comes out.
Suppose what I do is I
exchange the first two indices
for the second two indices.
Well, it turns out, if
you do that, you just
get the expression back.
OK?
So if I just swap alpha
mu, make them my last two,
make lambda sigma my
first two, [CLICKS TONGUE]
I get Riemann right back.
So that's another symmetry.
And then finally,
someday, when I've
got a little bit more bandwidth
in my class to do this,
I'd really like
to sort of justify
where you can derive this one
a little bit more rigorously.
But for now, it's
sufficient just
to sort of say, look
at that formula,
and you'll see that it's true.
If you take-- and
then you cyclically
permute the second,
third, and fourth indices,
you get zero, OK?
There is a variant of this
particular symmetry, which
is that if you take this and you
antisymmetrize on these things,
you get zero, OK?
And I've got some
words in the notes
that, in the
interest of time, I'm
not going to go through
them, but they are posted,
demonstrating why
those two are actually
equivalent to one another.
What it boils down to is, expand
out this antisymmetrization--
and I'm going to do that
again for a three-index object
a little bit later
in this lecture--
and then occasionally invoke
one of these other ones,
and you'll see that these
two, four and four prime, are
saying the same thing.
So go through all of these
different symmetries,
and you'll find n to the
4 independent components
goes over to n squared times
n squared minus 1 over 12, OK?
Not terribly hard to do that.
It's a little exercise
in combinatorics.
And when you do that for n
equals 20, as I mentioned,
you end up with--
excuse me, not n equals 20.
When you do this with n
equals 4, you end up with 20.
And just for completeness,
let me actually write out
that if you actually-- if you
go into the local Lorentz frame,
your spacetime metric is
in fact minus 1, 1, 1,
1 on the diagonal.
But you, of course, have
these quadratic corrections,
and one can in fact write
them out explicitly.
So that is what the time
piece ends up looking like.
There is an off-diagonal
piece, which
only enters at quadratic order.
It looks like this, and move
this down a little bit lower.
That's what your space-based
piece will look like.
So this explicitly constructs
the coordinate system
used in a freely falling frame,
including these second order
corrections.
So this particular form is known
as Riemann normal coordinates.
So if you are--
this is discussed in a
little bit more detail in one
of the optional
readings in the textbook
by Eric Poisson, A
Relativist's Toolkit.
Very nice discussion.
All right, so now that we have
the curvature tensor in hand,
we really have
essentially every tool
that matters for 8.962, OK?
There's a little
bit more analysis
we need to do with this guy,
but we now have all the pieces.
OK, there's no major
new mathematical object
or no object
described in geometry
that I need to build
in order for us
to make a relativistic
theory of gravity.
I do want to talk about the
curvature tensor a little bit
more, because there are a
couple of properties about it
that are very important for us.
And I think we'll have
just enough time today
to sort of get to them.
That will allow us to set
up, and in the first lecture
that I will record
in an empty room,
actually derive a
relativistic gravity equation.
So let me talk
about a few variants
of this curvature tensor.
So first, suppose you take the
trace on some pair of indices.
So what I mean by the
trace is, essentially,
imagine all the indices are
in the downstairs position,
and I contract it
with the metric,
so that I'm summing over them.
Well, first of all,
you'll note that when
you do this, if you were to take
the trace on indices 1 and 2,
they are antisymmetric.
The metric is symmetric, so
contracting metric with indices
1 and 2 is going to give me 0.
If I can track symmetric with
indices 3 and 4, I get 0, OK?
So the only trace
that makes sense
is do it on either indices 1
and 3 or on indices 2 and 4.
Let's do it on indices 1 and 3.
So what I'm going to do is
evaluate R, alpha mu alpha nu--
which if you like, you can
write this as R alpha beta of R.
Here, I'll turn it like this.
Beta mu alpha nu--
we're going to define this
as R with two indices, OK?
Just little r and mu, nu.
This actually shows up enough
that it's given a name.
This is called the
Ricci curvature tensor.
If you were to like, I said, if
you do your trace on indices 1
and 2, you get 0.
If you do it on 3
and 4, you get 0.
If you do it on 2 and
4, it's exactly the same
as doing it on 1 and 3, because
of all these other symmetries.
So in fact, this ends
up, when you count up
all your different
symmetries, this
is the only trace that
is meaningful, OK?
There are a few other
pairs where you get a minus
sign, but still the same thing.
So this is the only trace
that ends up being meaningful.
This is actually,
in fact, it's not
too hard to show that
this is symmetric.
So even though,
you know, Riemann
had all these crazy
antisymmetries and symmetries,
this one is simpler, OK?
One way to do this is just
to take the exact expression
that we wrote down
for the Riemann tensor
in terms of derivatives of
Christoffel and Christoffel
squared, and just write it
out in this traced over form.
So when you do
this, you get this.
This guy is symmetric on
the bottom two indices.
Now, this isn't
obviously symmetric,
but recall, a
lecture or two ago, I
worked through a couple of
identities that involved
the determinant of the metric.
It turns out, if you go back and
you look up these identities,
you can rewrite this term as
d nu of d mu of the log of j--
excuse me, log of
square root of j.
Symmetric when you
exchange mu and nu.
And then you get two
terms that look like this.
OK, this one, pretty
obviously symmetric
on exchange of mu and nu.
When you do this one, you
might think to yourself, hmm,
that one doesn't
quite look symmetric.
But remember, alpha and
beta are both dummy indices.
And so in fact, when
you exchange mu and nu,
this last term just
turns back into itself.
So that's kind of
interesting, because R mu
nu-- the reason I went
through that little exercise--
this is a symmetric
tensor, and we're
going to work in four
dimensions of spacetime here.
So symmetric 4 by 4 has
10 independent components.
Riemann had 20.
Somehow, this Ricci has--
in some sense, you can
of it as having 10 of--
the 20 components
associated with Riemann
are encoded in this guy.
Where are the other 10?
I'm going to talk about that
in just a moment or two.
Before I do that,
I will just note
that we are going
to want to also know
about the trace of
the Ricci tensor.
So if you compute this, this
is often just abbreviated R
with no indices whatsoever.
This is called the Ricci
scalar or the curvature scalar.
So there's another variant
of curvature which we're not
going to use very much, but I
wanted to just talk about very,
very briefly.
The derivation of this
is highly non-obvious,
so let me just write
it out, and I'm
going to talk about it briefly.
So suppose I define
a four-index tensor,
C alpha mu lambda sigma,
to be the Riemann tensor--
minus-- so working
in n dimensions--
2 over n minus 2, g alpha--
I'm going to
antisymmetrize here.
So antisymmetrizing
on lambda and sigma--
OK, so in the famous words
of Rabi, who ordered that?
So the way that this has been
constructed, if you go through
and you carefully look
at the way this thing
behaves under exchange
of any pairs of indices,
it has exactly the same
symmetries as Riemann.
But if you take the trace of
this, turns out to be zero.
In fact, when you go
through, when you count up
how many independent
components it has,
because it has no trace, it
has 10 independent components.
So this tensor has a name.
It was first formulated by the
mathematician Hermann Weyl.
And so although
it's a German name
and it's not spelled that way,
it is appropriately the "vile"
tensor.
It is a pretty vile
thing to look at,
but it plays an important role.
So in keeping with
the idea that it's
got 10 independent
components, Ricci
has 10 independent
components, Riemann has 20.
Heuristically, you can sort
of think and put big quotes
around all these objects.
This doesn't mean approximately.
Let's just put it this way.
This is, got all the same
information as Ricci plus Weyl,
OK?
One can, in fact-- by
bringing in appropriate powers
in the metric and
things like that,
one can actually write down a
real equation relating these.
It's a bit of a mess.
It's not that interesting.
The key thing I want
you to be aware of
is that all of the
curvature content of Riemann
is sort of split
into Ricci and Weyl.
Looking ahead a little bit-- and
because I'm going to be doing
this in an empty room, I just
want to make this point now--
we are soon going to see
that Ricci is very closely
related to sources of gravity.
So we're going to-- when we
formulate a theory of gravity,
working with things,
we're going to find
that there's a
tensor that is just
a slight modification of
the Ricci tensor, that
is equal to the
stress energy tensor
that we use as our source.
Stress energy tensor is a
symmetric 4 by 4 object.
Those 10 degrees of freedom
in the stress energy tensor
essentially determine
the 10 curvature degrees
of freedom encoded in Ricci.
But what that tells
you is that if you're
in a region of vacuum, where
there is no stress energy,
there's no Ricci, OK?
But there is curvature.
We measure tides.
We see gravitational effects.
Weyl ends up being the quantity
that describes behavior
of gravity in vacuum regions.
Ricci ends up very closely
related to how it describes it
in regions with matter.
One reason I mentioned
this-- so there's only--
[CHUCKLES] only one
LIGO student here now.
Hi, Sylvia.
[CHUCKLES] But in fact,
when one is describing
gravitational radiation, the
behavior of the Weyl tensor
ends up being very
important for characterizing
the degrees of freedom
associated with radiation
and general relativity.
So that's a little bit ahead.
There's a few other
things you can
do with it which are
related to what are called
conformal transformations.
I have a few notes on
them, but they're not
that important for our class.
It's discussed a
little bit in Carroll,
so I'll leave that as a
reading if you are interested,
but we don't need to
go through it here.
So there are two
other things that
are much more important for
us to discuss with respect
to Riemann first, and
I'm going to focus
the time that we have on them.
So one of the really
important aspects
of curvature that I've
emphasized a few times now
is the idea that initially
parallel geodesics
become no longer parallel
when they're moving
on a manifold that is curved.
We're going to use
the Riemann curvature
tensor to quantify what the
breakdown of parallelism
actually means.
There's a couple of
different discussions of this
that you'll see
in various places.
Carroll's discussion
is very brief.
It's rigorous.
It's very brief.
I'm doing something
that's a little bit--
to my mind, it's
a little bit more
physically motivated
OK so I'm going
to do it in a slightly
different way from the way
it's done in Carroll.
It's a little close to the way
it's done in Schutz's textbook.
So what we want to do is
imagine we have initially
parallel geodesics,
and what we want to do
is characterize how
they become nonparallel.
Well, we'll put it this
way, how they deviate
as one moves along
their world lines--
around these world lines.
OK, so here's what I want to do.
That's the idea of the
calculation that I want to do.
So what I want to
start out with is--
let's consider two
nearby geodesics.
And what I'm going
to mean by nearby
is that they are close enough
that they're essentially
in the same local
Lorentz frame, OK?
So they're going to
have the same metric.
I'm going to be able
to choose coordinates
such that the Christoffel
symbol is zero for both of them.
I will not be able to get
rid of the second derivative,
though, OK?
So that will be where a bit of
a difference begins to enter.
Two nearby geodesics--
and I'm going
to use just lambda as the
affine parameter along them.
So here's my first one.
I'm going to make a few
definitions on the next board.
I'm going to call this gamma
sub v. And here's my next one.
I'm going to call
it gamma sub u.
Define two points here.
OK, I'm going to make
a couple of definitions
on the next board.
OK, so definitions--
I'm going to call u the tangent
vector to the curve gamma sub
u.
OK?
So this is equal to dx
d lambda on that curve.
v is the tangent
vector to gamma sub
v. The point A is at lambda
0 on curve gamma sub u.
A prime is at lambda
sub 0 on gamma v.
So they're both parameterized
by a parameter lambda,
and I'm going to set the--
they're basically both
synchronizing their clocks
at the same time.
Like, at those
points, they're going
to find their starting
points as A and A prime.
What I'm going to
do now is I'm going
to define what's called
a geodesic displacement
factor that points from
lambda on the u curve
to lambda on the v curve.
And I'm going to use my
favorite Greek letter, xi.
So this points from
lambda on gamma u
to lambda that is basically
to the same value.
Let me make this a
little bit more precise.
Points from the event
at lambda on gamma u
to the event at
lambda on gamma v--
OK?
Apologies for being
somewhat didactic there,
but we need to define
things carefully.
So on this-- so
here's my initial--
so that's what xi looks
like at parameter lambda 0.
And what we want
to do is examine
how xi evolves as one moves
along these geodesics.
That's going to be our goal.
So let's make things a little
bit more quantitative here.
Xi is going to be equal
to x gamma v at lambda
minus x on gamma u at lambda.
Finally, I'm going to assume
that the curves begin parallel
to each other.
That's a statement that u of
lambda 0 equals v of lambda 0.
And it also tells
me that I can use--
that this must equal 0
at the initial point.
Not going to equal 0 everywhere.
In fact, it won't.
But I want to use this
as a boundary condition
in the calculation
I'm about to do.
OK.
So what we're going to do is
essentially say, you know,
as we move along these two
curves, these are geodesics.
We're going to use the
geodesic equation to slide
along these two curves.
We know what the
equation is that governs
u as it moves along gamma u.
We know what the
equation is that governs
v it moves along
gamma v. We're just
going to take the
difference between them,
and we're going to kind of use
that to develop an acceleration
equation that governs
that displacement xi.
Here's the bit where I differ
a little bit from Carroll.
So Carroll, like I said, Carroll
has a very brief discussion,
which is--
it's absolutely right.
But I want to give a little
bit of physical insight,
and I think you can do
that by choosing to work
in the local Lorentz frame.
So we're going to work in
the local Lorentz frame,
and I'm going to center
this local Lorentz
frame on the event A. The reason
why I'm doing that is this then
allows me to say, g mu nu
at the event is A mu nu.
Pardon me a second.
My Christoffel symbols at event
A are all going to be zero.
g mu nu at A prime is
also going to be A mu nu.
We have to be a
little bit careful.
Our Christoffel symbols do
not vanish at point A prime
because there's a
little bit of curvature.
They are in fact related to the
fact that I'm sort of slightly
far away, and I'm picking up
that second order correction,
OK?
This is sort of where all the
important bits of the analysis
are going to come from.
This is the fact that
they're close to each other.
I can set up a
local Lorentz frame,
but it's got that little
bit of sort of schmutz
at second order
that's coming in,
and kind of pushing me away
from a simple local Lorentz
frame there--
the simple mathematical
form right there.
OK, let's put that up.
OK.
So let's look at the
equations that govern
motion along these two curves.
So the geodesic equation
along curve gamma u--
and we'll just look at it
at A. It's a second order--
I'm going to write it in
terms of the coordinates.
It looks like this.
We're evaluating this at A. At
A, my Christoffel symbols are
all zero, so this is just zero.
Let's look at it along the
other curve, at A prime.
OK?
So in just a second,
I am going to--
wait a second.
I'm going to substitute in
that derivative of Christoffel
that's going to go in there.
Before I do that--
so notice, this also
depends on the velocity as
I move along at that point,
right?
Typical geodesic--
So I'm going to
substitute in the fact
that both the x mu and
the x nu velocity here,
this is defined as v mu.
But remember, these
guys are defined
as being initially parallel.
OK?
So what this tells me is--
OK, that's an alpha.
So this is interesting.
What I'm seeing is I
end up with an equation
where the derivative of
the Christoffel symbol
is coupling to sort of my
motion along that world
line and the displacement.
Now, let me just remind you,
what I really wanted to do
was get an equation that governs
how this guy changes, OK?
But this guy is just
defined as the difference
between the position along
curve v minus the position
along curve u.
So if I take the difference
of these two things--
pardon me, forgot to label
which curve this one is on.
So I'm doing this at A
prime, doing this one at A.
This is an equation
governing the acceleration
of this displacement.
OK?
So this is interesting, OK?
So what you're sort of seeing is
that the geodesic displacement
looks something
like two derivatives
of the metric coupling
in the forward velocity
and the displacement itself.
Now, as written, this equation
is fine if all you are doing
is living life in the
local Lorentz frame, OK?
We want to do a little
better than that, though.
So I'm going to do a little
bit more massaging of this,
but I kind of want to emphasize
that this already brings out
the key physical point, OK?
I can't yet-- you know, you're
sort of looking at this,
and you're thinking
to yourself, that
looks like a piece of Riemann,
but it ain't Riemann, right?
It's a derivative of
a Christoffel symbol,
and that's not a tensor.
So this isn't quite the
kind of thing we want yet.
We need to do a
little bit more work.
And so what I would
sort of say is,
everything I did
up to here, this
is like the key
important physics.
Now I'm going to do
a little bit of stuff
to sort of put the suit and
tie that a tensor is supposed
to wear.
I'm going to dress it
up a little bit, so
that it's wearing the
clothes that all quantities
in this class are
supposed to wear.
So we want to make
this tensorial.
And so for guidance of
that, these time derivatives
or derivatives with
respect to the parameter,
it's not really
nicely formulated, OK?
The thing which
we should note is,
d by d lambda, that's
what I get when
I contract the forward velocity
with a partial derivative.
What we should do is
replace this with something
like what I'm going to call
capital D by d lambda, which
is what I'm going to get
when I do derivatives using--
when I use forward
velocity contracted
on a covariant derivative, OK?
So--
Suppose I just--
I'm agnostic about
what xi actually is.
I just know it's a
vector field, and I
want to compute this thing.
Well, we learned how to do
that a couple lectures ago.
I'm going to work this guy out.
And-- [EXCLAIMS] I couple in
a term that looks like this.
You might be tempted
at this point
to go, oh, I'm in a
local Lorentz frame.
I can get rid of
that Christoffel.
Don't do that quite
yet, because we're
going to want to take
one more derivative.
Christoffel does vanish,
but its derivative does not.
Wait till you've done
all of your derivatives
before you insert
that relationship.
Just getting another
piece of chalk.
So my covariant derivative
along the trajectory
with my parameter is the
usual total derivative plus--
that, OK?
Now, I'm going to want to
take one more derivative.
OK?
So there's a lot
of junk in here.
I'm going to get one term.
It just looks like this.
But now I'm going to get a
whole bunch of other terms
that involve--
so this is what I get when
I expand this guy out,
and I just have--
well, I should hang on a second.
So first, I'm going
to get one that
involves essentially the
first term, u on the partial,
hitting both of these terms.
OK, so when I do that--
pardon me just one moment.
OK, never mind.
Let me just write
it down, and I'll
describe where it comes from.
OK, so this first term, this is
basically the connection term
coupling to this, OK?
Associated with this
covariant derivative.
Now I'm going to have
a whole bunch of terms
that involve this guy operating
on these terms over here.
And my apologies that the board
is not clearing as well as I
would like today.
And you know what?
I'm going to just
write it out like this,
and move to a different
board in a second.
OK?
All right, so I'm going to put
this over on the other side
here.
What I have there-- so the
first term, like I said,
I'm basically just
operating that thing
on the two different terms.
So it's a little easy when
I first hit it on the d xi d
lambda, and it's going to
be messy when I expand out
all the derivatives that
operate on here, which
is why I'm writing
it with some care,
because we're about to make
a lot of mess on the board.
OK.
OK, so first, I'm going to do
what happens when this guy hits
the Christoffel symbol.
Then I'm going to
get a term that
involves this
combination of guys
acting on the four velocity.
And then I get these guys
acting on my displacement.
OK, [BLOWS AIR] now
I'm ready to simplify.
OK, so let me just emphasize,
everything I have here,
all I did was expand
out those derivatives
and write them in a form where
I want to basically call out
different terms and
see how they behave.
Two things to bear
in mind, I am going
to do this in the
local Lorenz frame
and in the vicinity
of the point A.
That's going to allow me to
get rid of a lot of terms.
I'll use the-- here's
an open box of chalk.
OK, so earlier, I didn't want
to get rid of my Christoffel
symbols because I was going
to take another derivative.
I'm done with that.
So anything that's just
a Christoffel on its own,
I go into local Lorentz form--
Christoffel, you die!
Die!
Can't really do much with this.
Let's set this aside.
Let's look at this guy.
What do we know about
the forward vector u?
It's a geodesic.
It obeys the geodesic equation.
The geodesic equation is, this
thing in parentheses equals 0.
You die!
Finally, I have this condition
here at the beginning.
That is essentially
the velocity at point A
minus the velocity at point
B. Our initial condition
is that these things start
out parallel to each other.
These die because these
are initially parallel.
So initially parallel geodesic--
local Lorentz frame--
So the only derivative I'm
going to need to expand out
is this guy, and because I'm
working in a local Lorentz
frame, that's
actually not that bad.
So what I finally get
after all the smoke clears
is that my covariant
acceleration
of this displacement
vector is related
to my noncovariant
acceleration, with a term
that basically ends
up just being--
oops-- a derivative of
the Christoffel symbol.
It couples in two
powers of the four
velocity and the displacement.
But we actually already have--
we worked out earlier
what this guy was, OK?
So my noncovariant acceleration
was a different derivative
of the Christoffel symbol.
So if I go and I plug this in--
My apologies for pausing here.
There are a lot of
indices on this page.
Look at this, I've got a
derivative of Christoffel
minus a derivative
of a Christoffel.
That is exactly what
the Riemann curvature
tensor looks like in
the local Lorentz frame.
Now, if you actually go back and
you look up your definitions,
you'll see it's not quite right.
There's a couple of indices that
are sort of a little bit off,
but they turn out to
all be dummy indices.
So what you should
do at this point
is just relabel a couple
of your dummy indices.
So if on the second term, you
take beta to mu, mu to gamma,
and nu to beta, what you
finally wind up with is--
We'll write it first in
the local Lorentz frame.
This is nothing but the Riemann
tensor in the local Lorentz
frame.
And from the principle
that a tensor equation that
holds in one frame
must hold in all,
we can deduce from
this that this
is the way the
displacement behaves.
So I start out
with two geodesics
that are perfectly
parallel to one another.
If they are in a
spacetime that is curved,
the Riemann curvature tensor
tells how those initially
parallel trajectories evolve as
I move along those geodesics.
So I want to make two remarks.
One is just sort of a way of
thinking about the calculation
I just did.
It may not-- you
might be slightly
dissatisfied with
the fact that I
decided to do the
whole calculation
in a whole special frame.
If that's-- you would prefer to
see something a little bit more
rigorous, feel free to explore
some of the other textbooks
that we have for this class.
They do treat this a little
bit more rigorously than this.
One reason why I
wanted to do this is I
wanted to really
emphasize this idea
that where this
effect comes from is
that if I think about the freely
falling frame describing two
nearby geodesics, it
is that derivative,
it's that secondary derivative
of the metric, that really
plays a fundamental role in
driving these two things apart.
What we get out of it is a
completely frame-independent
equation, and this
ends up really
being the key equation
describing the behavior
of tides in general relativity.
We're going to-- when you go
into a freely falling frame,
we've basically
at that point said
there's no longer such a thing
as gravitational acceleration.
And if there's no
gravitational acceleration
that's coming, well, what
the hell does gravity do?
General relativity, tides
become the key thing
that really defines what
the action of gravity is.
One of the ones I am personally
very interested in applying
this to is that this
equation gives you
a very rigorous and
frame-independent way
to describe how a gravitational
wave detector responds
to the impact of a
gravitational wave.
If you imagine you have
two test masses that
are moving through spacetime,
they just follow geodesics, OK?
They feel gravity.
They are free falling.
They don't do anything.
If you look at them, if
you're falling with them,
you don't see any
effect whatsoever.
But if they're sufficiently
far away from one another
that their displacement
kind of takes the curvature
to that freely falling
frame, then this equation
governs how the separation
between the two of them
actually evolves with time.
And so for instance,
if you're a student who
works in LIGO, where some of
the many important analyses
describing how gravitational
wave detectors respond
to an [INAUDIBLE]
radiation field,
they are based on this equation.
If you're interested
in understanding
how an extended
object is tidally
distorted due to the fact
that it's sort of big enough
that it doesn't all sort
of live at a single point,
but its shape spills across
the local Lorentz frame,
this equation
governs-- it allows
you to set up some
of the stresses that
act upon that body and change
it from a simple point mass.
So I'm waxing slightly
rhapsodic here,
because this is one of the more
important physical equations
that come out of the
class at this point.
All right, in our
last 12 or 13 minutes,
I would like to do one
other little identity which
is extremely important,
but rather messy.
In fact, I think I'm going to
just-- if I have the time--
yeah, you know what?
I think I will have time.
I'm going to set up the first
page of the next lecture.
So let's move away
from-- oh, by the way,
I forgot to give
the name of this.
This is the equation
of geodesic deviation.
OK, so recall when we first
talked about the Riemann
tensor.
There's a very
mathematical action
that the Riemann tensor does.
You can think of
it as what happens
when you have a commutator of
covariant derivatives acting
on a vector.
So if I evaluate the
commentator, sigma lambda--
excuse me-- covariant of
lambda covariant with sigma
acting on v alpha, OK, it
ends up looking like this.
And I argue that this definition
is kind of the differential
version of that little holonomy
operation by which I derived
Riemann in the first place.
A more generalized
form of this is,
suppose I act on some
object with two indices, one
in the upstairs, one
in the downstairs.
Oops.
OK?
So these are just some reminders
of a couple definitions.
I want to use this.
I'm going to apply these
things to two relationships
that I'm going to
write down, and I'm
going to do something
kind of crazy,
then I'm going to do something
a little bit crazier,
and something kind of cool
is going to emerge from this.
And this is something that
deserves a clean board.
So I'm going to write down
relationship A. That's
what I get when I
do the commutator
of the alpha beta derivatives
on the gamma derivative
of some one form, OK?
Now, if I apply those little
definitions I worked out,
this is Riemann mu
gamma alpha beta--
OK?
So this is relationship A.
Relationship B,
I'm going to take
the covariant derivative
along alpha, of the beta gamma
commutator on p delta, OK?
So notice the difference here.
One, I'm doing the
commutator acting
on this particular derivative.
Next one, I'm doing
the derivative
of commutator acting on this.
And so I'm going to skip a line
or two that are in my notes,
but this ends up turning
into minus p mu--
beta alpha delta minus
R mu and delta gamma.
OK?
So the way I went to get
this line, so the two lines
that I skipped over, one is I
just straightforwardly applied
the commutator derivative,
and then took a derivative.
And then I took
advantage of the fact
that I can commute the metric
with covariant derivatives,
and sort of raising
the next, move things--
move an index up on one
side and down on the other,
and take advantage of some
symmetries of the Riemann
tensor to slough
my indices around.
OK, so this board's
kind of a disaster,
so I'm going to go
to a cleaner board
here for what I want to do next.
I'm just going to clean
this so you're not
distracted by its content.
So two equations,
what the hell do they
have to do with each other?
Here is where I'm going to do
something which you might--
let's do the other board.
I'm going through something
that you might legitimately
think is crazy.
What I want to do
is take equation A--
actually, I'm going to look
at both of these equations.
And I am going to antisymmetrize
on the indices alpha, beta,
and gamma.
OK?
So just bear with me a second.
So the way I'm
going to do this--
here's what I'm doing now.
Here's my commutator.
So if I look at equation A--
OK?
So I need to expand
this guy out.
Let me just write out a
step of this analysis.
The way you do this is you
add up every even permutation
of those indices.
That's a beta.
And then you subtract
off every odd permutation
of these indices.
Oh, and don't forget--
exactly on that one form, OK?
So now we can--
if you expand these guys
out, and gather terms exactly
correctly, it's not hard
to show that this turns out
to be what you get if you just
slide those commutators over
by one.
OK.
You know, if you're really
feeling motivated, just try it,
OK?
What this is is, the
right hand side--
sorry, this is the
left hand side--
we'll put this way.
So when I apply this to the
left hand side of equation A,
what emerges is the
antisymmetrized left hand
side of equation B.
So when I antisymmetrize on
the indices alpha, beta, gamma,
these two relationships become--
they say the same thing.
So that means the right hand
side must be the same as well.
So let's apply this
and see what happens
if I require the antisymmetrized
right hand side of A
to equal the antisymmetrized
right hand side of B.
OK, so first, if I remind myself
how I order these equations--
OK, yeah.
So let's do-- here's
A, and here is B.
OK, so a few things to notice.
This term, we
actually showed when
we looked at the different
symmetries of the Riemann
tensor.
This is-- I didn't actually
show this, but it's in the notes
and I stated it.
This is one of those symmetries.
If I take this thing,
and I just add up
what I get when I permute
the three final indices,
I get zero.
And that's equivalent
to antisymmetrizing
on these things.
So this is zero by
Riemann symmetry.
This term here and this term
here are exactly the same, OK?
Because the only
indices that are
in a slightly different order
are alpha, beta, and gamma.
This one goes
alpha, beta, gamma.
Where I wrote it here, it just
became beta, gamma, alpha.
That's a cyclic permutation
I've antisymmetrized.
They are exactly the same.
So these are on the same
side-- or on opposite sides
of the equation.
So they cancel each other out.
And so what arises
from all this is p mu--
pardon me, I missed
the derivative.
I've set no properties
on p, so the only way
this can always hold is if in
fact, this is equal to zero.
This is a result known
as the Bianchi identity.
Let me just write
another form of it,
and then I'm fairly
pressed for time,
so I think I'm just
going to write down
the result of the
next thing, and I
will put the
details of this into
the first prerecorded lecture.
If you expand out
that antisymmetry
and take advantage of
the Riemann symmetry--
the various Riemann
symmetry relationships,
this is an equivalent form.
So these two things both
are an important geometric
relationship the curvature
tensor must go by.
Now, in some notes that
I am going to-- well,
they're actually already
scanned and on the web.
What I do is take the
Bianchi identity and contract
on some of the indices to
convert my Riemann tensor--
my Riemann curvature,
into a Ricci tensor--
Ricci curvature.
So in particular, if I use
this second form of it,
it's just covariant
derivatives, right?
And the metric commutes
with covariant derivatives,
so I can just sort
of walk it through.
What you find when you do
this is that you can write
this relationship in
the following form,
and the derivation of this
will be scanned and posted,
and I will step through it in
the first recorded lecture.
The divergence of a
particular combination
of the Ricci curvature and the
Ricci scalar is equal to zero.
This is a sufficiently
interesting tensor
that it is now given a name.
We write this G, and we call
this the Einstein curvature
tensor.
I sort of mentioned that one
of our governing principles
here is, we're going to change
the source of gravitation
from just matter density to
the stress energy tensor.
That is a two-index
divergence-free tensor.
We want our gravitational--
the left hand side
of the equation
to look like two derivatives
in the metric, which
is a curvature.
So we need a divergence-free
two-index curvature.
Ta-da!
It's got Einstein's
name on it, so you
know it's got to be good.
All right, so that is where,
unfortunately, the COVID virus
is requiring us to
stop for the semester,
at least as far
as in-person goes.
So look for notes to be
posted, videos to be posted,
and I, of course,
will be in contact,
behind a sick wall or something.
But anyway,
[CHUCKLES] certainly,
I am not going out of contact.
And so those of you who are
scattering to parts unknown--
known to you, that
is, unknown to me--
good luck with your travels
and getting yourselves settled.
I really feel--
everyone feels terrible
that you're going through
this crap right now,
but hopefully, this will
flatten the curve, as they say,
and it will be the
right thing to do.
But stay in touch, OK?
This campus is going to be weird
and sad without the students
here, and so we want
to hear from you.
