[SQUEAKING]
[RUSTLING]
[CLICKING]
SCOTT HUGHES: We're in for a
uncomfortable couple of weeks,
but we will do our damnedest
to make sure that--
I'm not going to say
life won't be disrupted.
Life is going to be
bloody well disrupted.
No question of that.
But number one goal is
making sure everyone
remains healthy, both
physically and mentally.
When we're forced to
isolate a little bit,
we lose the social contact
that makes life worth living,
and so we're going to
be working really hard.
Look to your social group
as well, and to your peers
and and mentors and others to
try to find a way to remain--
if you can't meet in
person quite as much,
there's the phone, there's
Skype, there's FaceTime.
Better than nothing,
and that's sort of what
we're looking at these days.
And we're, of course,
within the department,
very committed to figuring
out a way to make sure
that the education that
we're sort of here to do,
we can deliver it to you
in some form or another.
There may be a few bumps in the
road while we work this out,
but we're getting there.
Hopefully, this will-- if
there is a disruption coming,
it will be short-lived.
If not, let's just
let's focus on what
the important things are.
And today, the important things
are the geodesic equation.
At least that's what we're
going to start things,
and then we're going to take
it into the next major concept
that describes manifolds
with curvature,
I think we were going
to take advantage of.
So just a quick recap, where I
ended things last time was we
described geodesic trajectories.
These are trajectories
which parallel transport
the tangent--
as they move along
their world line,
they parallel transport
the tangent vector
to that world line.
And I forgot to put
the physics up here.
The key reason why
this is interesting
is that this corresponds
to a free fall trajectory.
Free fall basically
means you are
moving under the influence
of nothing but gravity.
And if you want to
understand gravity
in a relativistic theory, well,
that's all you care about.
So these are very
important trajectories.
And you know, it's
not an exaggeration
to say that solving
this equation, OK-- so
in this thing, I've kind of left
agnostic what the spacetime is
that you use to compute
those covariant derivatives
and to write your
Christoffel symbols gamma,
but it could be
any spacetime that
solves the relativistic
field equations,
which we haven't derived
yet but we shall soon.
This is what describes
sort of small bodies moving
through that kind
of a spacetime.
That is the starting point to
a tremendous amount of analyses
in general relativity.
I made a crack last
time that I think
65% of my published
research work
is essentially based on
solving this equation.
I actually went and checked.
It's probably more like 75%, OK.
That wasn't actually
an exaggeration.
It shows up a lot.
All right, so this,
as we went through,
and it's basically
just saying that I'm
going to take the
covariant derivative
of the tangent vector u
and contract it with u.
OK, and there's a
couple of other ways
of writing this which
I've written out here
in just sort of notation.
But if you expand it
out, it looks like this.
So what you're saying
is that the vector
u, the four-vector components
u, are parameterized
by some quantity which if
it's a time-like trajectory,
you can think of
it as essentially
the proper time as you
move along that trajectory.
This is just describing
how this thing
behaves as a function
of that parameter, OK.
So we're going to do
some more with that.
To begin with, there's
a couple of cool results
that we can derive.
So there's a nice side note.
We can rewrite this
in terms of momentum.
So if you imagine, let's
focus on the version
where I'm doing it
per units proper time.
And I'm going to take advantage
of the fact that for a body
moving on a
time-like trajectory,
so a body with a rest mass
m, I just take this equation,
and basically, you
multiply by m twice,
and it very clearly turns into
something that is pretty much
exactly the same, but I just
replace my u's with p's.
Two comments I want
to make about this.
So first, and you know what,
let me actually expand this out.
I mean, it's quite obvious.
Why don't we write it in
terms of the components
in the Christoffel symbols.
OK, so I'm going to
write it like this.
Let's do the following, OK.
Recall that if I have-- so
this trajectory, assuming, when
you write like this,
your parameter lambda
is called an affine parameter.
And that is a parameter
such that the right hand
side of [INAUDIBLE]
equation is 0 on a free fall
trajectory, OK.
If it's something that's
proportional to you,
it is a valid
parameter, but it's
one that's sort of been
defined in a bad way.
One thing we showed
last time is that I
can shift lambda
by any constant,
OK, with the right units.
That essentially
amounts to just changing
the origin of my clock.
And I can multiply it by any
scalar, which essentially
amounts to changing the units
in which I am measuring time.
And it's still a good
affine parameter.
So here's an example
of an affine parameter
that I could use.
Suppose that I defined
this such that an interval
of affine parameter delta
lambda is an interval
of delta tau divided by m.
If I do this, well,
then, p alpha is just d--
so remember this is
now going to be--
let's write it like this.
So this is my original
definition of this thing.
This then becomes
dx alpha d lambda.
So this is a way of choosing
an affine parameter such
that I'm essentially writing
my tangent along the world line
as the momentum rather than
something like the four
velocity, OK?
But there's something
cool about this,
so let's now go
and just write what
my geodesy equation turns into.
It's basically exactly the same.
It's just that I'm going to
absorb the m on the first term.
Basically, it's exactly the
same geodesy equation as I had
before, but with
u's promoted to p's.
What's kind of
cool about this is
you can take a limit
in which m goes
to zero as long as your
interval of proper time
goes to zero at a rate such
that delta t over m is constant.
So what this allows us to
do is just conceptually
reformulate the
geodesy equation,
so that it's perfectly
well behaved.
Not just for time
like trajectories,
but for null or light
like trajectories.
OK, so that's very
important for us.
A lot of the most important
tests of general relativity
actually come down to looking
at the behavior of light
as it moves in some kind
of a curved space time.
And the geodesy
equation, if you sort of
interpret the way we thought
about it before, you're
kind of like, well, let's
go back, and suppose I'm
running it sort of in this one.
And I'm thinking of my
lambda as proper time.
An interval with
proper time is not
defined along a light
like trajectory, OK?
So that just kind of makes
it clear that that's fine.
What we're going
to do when we're
talking about a
light like trajectory
is we're just going to find
the parameter along the world
line, such that the tangent
vector is the momentum
along that world line.
Mass doesn't even make sense.
So the p that goes into
this-- so when I do this,
this is going to be a p,
such that p alpha p alpha
equals minus m squared.
We still have that
rule that it's always
going to be minus m squared,
which is zero in this case.
So this gives us
a tool that we can
use to study the
motion of light as it
reacts to gravity, for example.
OK, we'll switch
gears, so I want
to do one other trick based on
this momentum form of things.
So I can rewrite the
geodesy equation as follows,
and it's going to
start out where it just
looks like I'm essentially doing
what we call index gymnastics.
I'm just sort of moving
a few indices around.
So let's write this as p
alpha, and what I'm going to do
is contract it on this.
I'm putting the
index and my momentum
in the downstairs position now.
First of all, you should
stop and ask yourself,
am I allowed to do that?
If I do that, do I not
generate some additional term
that should then be moved
to the right hand side?
Well, think about
what I'm doing.
If I am lowering an index, that
essentially means that I am--
let's do the following.
Let's change this to
a gamma for a second.
What I have done here
is I have essentially
taken the equation
I wrote over there,
and I had hit it with
g beta gamma, OK?
I can always multiply
by those things.
The covariant derivative
of the metric is zero.
Because the covariant derivative
of the metric is zero,
it commutes with
that derivative,
so I can just walk it inside
the derivative operator.
So I wanted to go through
that just a little
bit carefully, because
that's actually
a trick that once
you've seen it once,
I want you to know it well.
Because I'm just going to
do it many, many times as we
move forward.
There's going to be
a bunch of times,
where I'm taking the partial--
it's going to be the covariant
derivative is something.
I know it's going be raising
indices willy nilly on whatever
it's operating on.
But I'm taking it
down to the fact
that I'm effectively moving a
metric inside and outside here.
All right, so let's take
that for the geodesy equation
and expand it out.
So I end up with mdp beta d
tau, so I'm here using the fact
that the p alpha
that's on the outside,
I'm writing that as mu alpha.
And I'm using that to
convert the derivative I get,
expanding that
into a d by d tau.
And then I get a
term that basically
corrects the downstairs index.
Because it's a downstairs index,
it enters with a minus sign.
Let's move this to the other
side, and what I'm going to do
is make all the indices be
in the downstairs position.
Let's see.
Hang on a second.
Did I do this right?
Sorry.
Yeah, I'm going
make all the indices
in the downstairs position
on this capital gamma,
so I'm going to write
this as follows.
OK, so I chose to
write it this way
as you'll see in just a moment.
Because this is now symmetric
on exchange of alpha and gamma.
Let's expand that
Christoffel symbol,
so I'm going to have one term
beta derivative alpha gamma.
We'll put this up above.
OK, so take a look at that
last line of that expression.
So as written here,
I've got a term
that is symmetric on
exchange of alpha and gamma.
But inside my parentheses,
bearing in mind
that my metric is itself
a symmetric object,
I've got two terms,
this one and this one,
where if I exchange alpha and
gamma, I get a minus sign.
So I got a symmetric
contracted with anti-symmetric.
Therefore, I can
simplify this whole thing
to something that
only involves--
the only derivative
I need to compute
is one partial
derivative of the metric.
Now that's nice, but
if you think about it,
it might even be nicer
than you realize.
Suppose you're working in
some coordinate system, such
that for a particular
derivative, for a particular,
let's say, it's the
derivative with respect
to your time coordinate.
Suppose the metric vanishes.
Suppose that equals zero
for some coordinate.
Then you've just learned
that a particular component
of the four momentum, component
of the downstairs four momentum
mind you, that is a constant of
the motion along the worldwide.
I sort of said in
words a few things
about this a couple
lectures ago when we were
talking about Killing vectors.
I'm going to actually tie this
to that discussion in just
a moment.
This is often operationally
the simplest way
to deduce that you, in fact,
have a constant motion,
so there is some space
times, very complicated ones
that play huge roles in many
of the kind of analysis we do.
But you sort of look
at them, and you
kind of go, oh, thank god.
It's time independent.
All right, that means I know
p downstairs t is constant.
It's independent of
the actual angle.
P downstairs phi is a
constant, and that ends up
giving us some quantities
that we can exploit.
And later when we start
talking about certain solutions
of the field
equations and looking
at the behavior of
these things, we're
going to see how
we can exploit them
to understand the motion of
bodies in very strong gravity.
Before I do this, let me
connect what you're saying here
to-- what I'm saying right here
to stuff that we did a lecture
or two ago with
the Killing vector.
So we know that, if a metric--
so this is something that we
wrote down a little bit before.
We also know that,
if the metric is
independent of some
particular coordinate,
there exists a Killing
field, or Killing vector,
which I will call c beta.
What I want to do now
is say, OK, how does--
let's look at how.
I'm going to define a
particular scalar, so what
do I get when I take--
oh, bugger.
That made no sense.
There we go.
What would I get if I
take that Killing vector?
I contract it with
my for momentum.
How does this guy behave as
I evolve along a trajectory?
So the way we're going
to solve this is we'll
just look at the
time of evolution
as we move along the
trajectory as we--
and the way we'll do
that, I'll show you
how to construct that time
evolution in just a moment.
We're going to assume p
solves the geodesy equation,
c solves Killing's equation.
Let's see what happens.
So what we're going to
do is look at d by d tau,
so that the proper
covariant derivative
along the trajectory
of this guy.
Well, this, if I take advantage
of Leibniz's rule, so first
of all, I can just
write this as--
you know what?
Let's throw an m into
here just to make
things nice and symmetric.
The reason I did
that is so that I
can write that derivative
in the following form,
so this is what I
want to evaluate.
So one thing I do is
expand out that derivative
using Leibniz's
rule, so this is--
let's first plot
my Killing vector.
That term is like this.
Then I got a term, and
it looks like this.
Well, the first term,
it's going to die.
Because like I said,
I'm going to assume p
solves the geodesy
equation, so p is a geodesy.
I kill that.
What about the second term?
Well, for the second
term, what I'm going to do
is note that whenever you have
some general two index object,
so suppose I have some two
index tensor, m alpha beta.
I can always write this in
the following way, right?
Where remember, the parentheses
denote the symmetric part,
and the braces denote
the anti-symmetric part.
So this is just
a theorem, right?
You add it together.
The 1/2s combine, and
you get this thing back
for the first term.
And they combine the minus
with the other one, OK?
Very simple identity.
So if I do that,
applying it up here,
this is symmetric under
exchange of indices.
This is anti-symmetric.
It dies.
The only thing that
is left is this term.
But if this is a Killing
vector by Killing's equation,
this equals zero by
Killing's equation.
So the importance
of this, you've
just shown that what you
get when you contract
for momentum with
the Killing vector
gives you a constant motion.
You've also shown that the
component of the four momentum,
the downstairs component of
the four momentum associated
with whatever
coordinate the metric
happens to be independent of, is
also a constant of the motion.
The key thing to note is both
are actually very powerful
and important statements.
One depends on the coordinate
system and the representation
you've chosen.
The other does not, OK?
So this is really
true and useful,
if you happen to have chosen
the coordinate system such
that this derivative
is equal to zero.
This is true,
though, independent
of your representation,
so these are just
two different ways of calling
out constants of motion.
And we actually find both
of them to be very useful,
so we're going to take
advantage of them.
There's a variation on
this calculation that
is on the next p set, and
you will come back to this,
again, when we start talking
about motion in certain space
times in the second
half of this course.
So let me just do a couple
really quick examples.
I've already kind
of mentioned these,
but let me give names
to what these are.
So if your space time has
a time coordinate, such
that the time derivative of
any metric element is zero,
then you know that a time
like Killing vector, which
I will call ct, and I'm
leaving the vector sign on it.
This thing exists, and you
also know that p downstairs t
is constant.
Now the name that is given
to this is negative energy.
Why negative?
Well, the main reason
why it's negative
is that we will often--
so let me just caution that
this is not always an identity
that we're going to use.
We're going to use it in
a huge number of problems
that we care about.
Many of the space times that
we are going to work with
are those that when you get
really far away from whatever
source is generating
your gravity,
it looks just like
special relativity.
We call such space times
asymptotically flat.
In other words, as you get
asymptotically far away,
it reduces to the
space time that we
studied in the first
couple weeks of the class
when we were doing geometric
spatial relativity.
And in that case, we knew the
timelike component was energy.
And in a flat space
time, you lower that time
like component's index.
You get minus energy.
It just so happens when
you go through the math
carefully that
negative of the energy
ends up defined in this way.
It's going to be the
quantity that is actually
conserved everywhere.
What's kind of
cool is that we use
this associated with
asymptotic flatness
to give you some intuition.
But this is actually
true, even if you're
right outside of the vicinity of
a rapidly rotating black hole.
It is still the case
that p downstairs
t for the right choice
of t is a constant.
In my notes, I also
show you that there
is an example of an
actual Killing vector that
corresponds to angle momentum.
Again, we'll come back to
that a little bit later,
so let me do one
example geodesy before I
sort of change
topic a little bit.
I'm going to write
down a space time
that we either in
person or on video
are going to derive basically
right after Spring break.
So suppose I hand you
the following space time.
This function phi, I'm not going
to say too much about it quite
yet.
What I will say is that
it is small in the sense
that when you're doing
various calculations with it,
feel free to discard terms of
order five squared or higher.
And it only depends on the
coordinates x, y, and z.
No time dependence.
I want to examine slow
motion in the space time.
So what I'm going
to do is I'm going
to imagine that my four
momentum has the usual form.
And I can think of it as having
an energy, and sort of time
like component, and
momentum in the space like.
The magnitude of
the energy is always
going to be much greater than
the magnitude of the momentum,
and in fact, will
be approximately
equal to the mass, where that
is the mass of whatever body is
actually undergoing this
motion in this space
that I've given you.
OK, so the reason why I'm doing
this is what we're going to do
is you want to say,
what does free fall look
like in this space?
Well, I look at geodesics, so
there is my geodesy equation.
This slow motion
condition that I've
applied over here, that tells
me that when I expand all
these terms out here, this
is going to be dominated
by the time time term, OK?
So it'll be dominated
because of the fact
that when you just look at the
numerical magnitude of the size
of the components
of the momentum,
those are going to be the ones
that dominate this calculation.
Everything else, if you put
your factors of c back in,
they're going to
be down by factors
that look like v over c.
So my geodesic
equation turns into--
so all I did was say,
it's dominated by this.
I'm going to move it to the
other side of the equation.
So without giving away
the plot, the beta
equals zero component
is going to turn out
to be very uninteresting.
Can you see why?
Beta equals zero is energy.
When I evaluate
that Christoffel,
I'm going to end up
taking a bunch of-- you
know, I'm going to
have all these zero,
zero, zero, time,
time, time components.
It's time independent, though,
so all my time derivatives
are going to be zero.
It's going to vanish,
but we expect that.
Because it's a time
independent metric,
all the crap I went
through a couple months ago
guarantees that the time like
components, energy's conserved,
right?
So that's kind of
what we expect,
so let's just focus
on-- and if we
had infinite time, which
we clearly don't, it
would be fun to talk about.
Let's just move on and look at
the spatial component of this.
So as I look at the spatial
component of this guy,
let's focus on beta equals i.
What you find when you
actually evaluate this guy is
this turns into--
OK, so go ahead and look up
your Christoffel formulas.
Again, these are one of those
things that, eventually, you
get this memorized by
the end of the term,
but don't feel bad if
you keep forgetting it.
No time derivative.
No time derivative.
The only thing that's
left at the end of the day
is essentially a gradient of the
time time piece of the metric.
So taking advantage
of the fact gi
alpha staying in the
upstairs position,
you can write this as--
it looks like this.
Now, if you like-- remember,
phi is a small value.
You can do binomial
expansion on that.
Knock yourself out.
It's not going to be
important in just a second.
Excuse me, minus one
half one minus two,
five to the minus one power.
So let me talk a
little bit about what
I did here in this last line.
My delta i alpha is coupling
to a partial derivative,
so that partial derivative,
the zero component of that
is the time derivative.
Everything's time
independent, so that's done.
Let's just skip
it, so what I did
was I changed my alpha to a j.
Because I only want
spatial derivatives,
so I'm allowed to do that.
Because I'm just acknowledging
the fact that all
the time derivatives
are uninteresting.
When I do
differentiate g00, I am
differentiating negative
quantity one plus two phi.
The one doesn't contribute.
All that is left is I took
the minus on the inside,
and there's my two phi.
So putting all of these
ingredients together,
I at last get my
Christoffel is delta ij.
I'm saying it looks like
spatial gradient of phi.
And for keeping score, there
are higher order terms,
which we're going to
collect under the assumption
that this phi is small.
Plug it back into my
equation of motion.
I end up with this.
Now let's cancel out the
m's that appeared in here.
If we were not doing relativity,
we would write this as--
ignore the fact that this
is per unit proper time.
This is dpdt is minus gradients
of something that sure as hell
looks like a potential.
What we are going to
do in the lecture rate
after Spring break, so
going into Spring break,
depending on which-- as
I said at the beginning,
it's a little unclear
how these lectures
are going to be delivered.
But bear with me.
We're going to essentially
put together-- we're
going to take all
the last ingredients
and develop the
field equations that
describe relativistic gravity.
The first thing
we're going to do
is solve this in a
particular limit that
describes a body that
is weakly gravitating.
This will emerge
from this with phi
being equal to Newtonian
gravitational potential.
What this is showing is
that the geodesic equation,
this equation that
describes a trajectory that
is as straight as possible in
space time when it is given
that particular
space time, it does
give you the Newtonian
equation of motion, OK?
Yeah?
AUDIENCE: Is there
still supposed
to be a factor of m on
the right-hand side?
SCOTT HUGHES: No, so
it's possible I dropped
an m somewhere in there.
Go through that just a
little bit carefully here,
but you know, it's meant to be--
actually, you know what?
I take it back.
Sorry, so I think I
may have messed up.
I remember what it was,
I remember what it was.
There's an m squared here.
Thank you, Alex.
There was an m squared here.
There was an m and
an m squared here.
That's what I screwed
up, and I think
I have that wrong in my
handwritten notes, which is
probably why I messed that up.
Yeah, so this
equation was correct.
And this should have
been here like so.
We'll clear this
out and get this.
Yeah, so module of
that little bobble.
I just want to show
you this is essentially
the Newtonian limit.
Just give you a little
look at where we go ahead,
there are two ways
that we are going
to derive the field equations
of general relativity.
The first one
essentially boils down
to looking for certain tensors
that have the right symmetries
and allow us to have
sort of a quantity that
looks like derivatives on
the field equaling the stress
energy tensor as the source.
That only works up to
an overall constant,
and this is actually the
way that Einstein originally
developed the field equations,
was worked out all this stuff,
and then by insisting that the
solution that emerge from this
reproduced the
Newtonian equation
and the Newtonian motion,
he was able to fix
what that constant actually is.
There's a more sophisticated
way of doing it,
which I'm going to
also go through.
But it's worth noting that is
the way Einstein originally
did it.
I had the privilege a
couple of years ago--
I was at a conference
in Jerusalem,
where the Einstein Papers
archive is located.
And the guy who is the
main curator of this
was allowing those of
us at the conference
to look through them.
And I actually found--
they had not yet quite
categorized that,
but it was the papers very
much related to working
with these peculiar equations.
I don't think he was trying
to fix the coefficient,
but this spacetime can also be
used to compute the perihelion
precession of mercury.
And so it actually showed
Einstein working through that.
And the thing which is really
cool was he screwed up a lot.
The page that I was looking
at was full of errors.
It would say things like--
big things crossed out
and then "Nein nein nein!"
written on the side.
And so it made me feel
better about myself.
All right.
So everything that
we have done so far--
we've been dancing
around this notion
of what is called "curvature."
So I have used this
word several times,
but I haven't made
this precise yet.
Curvature is going to
be the precise idea
of how two initially
parallel trajectories
cease to be parallel.
So there's a couple of ways
that we can quantify this.
The one which I
am going to use is
one that's amendable
to, with relative ease,
developing a particularly
important tensor, which
characterizes curvature.
And so what we're
going to do is look
at the behavior of a
vector that is parallel
transported in a
non-infinitesimal region
of a curved manifold.
So for the purpose
of this sketch,
I'm going to make this
closed figure be a triangle.
When I actually do the
calculation in just a moment,
I'm going to use a
little parallelogram.
So around a closed figure,
I want a curved manifold.
So suppose my
curvature's actually 0,
and I do this for a triangle
that is on the blackboard.
So let's say I start out with a
vector that points from A to B.
And what I'm going to do is
just parallel transport it.
And in the case, goes, doo, doo,
doo, doo, doo, doo, doo, doo,
doo, doo, doo, doo.
This is an experiment
you can do at home.
When it comes back,
it's pointing exactly
the way it was initially.
Let me also just note
that this triangle--
the sum of its internal
angles is 180 degrees.
Hopefully you all know that.
Now, the next one--
if I'd had a little
bit more time,
I would have grabbed one
of my daughter's balls
to demonstrate this.
But hopefully, if you
guys have something
like a soccer ball
or a basketball,
this is a little experiment
you can do by yourself.
Now imagine a triangle
that is embedded
on the surface of a sphere.
So let's say this is the
North Pole of my sphere.
Here's the equator.
So what I'm going
to imagine is--
let's say I start up here.
Let's make the North
Pole be point A.
I move on a trajectory
that is as straight as I
am allowed to be.
And remember, if I'm a
one-dimensional being living
on the surface of this thing,
that's a straight line.
It only looks straight
looks curved to us
because we see a third dimension
that this whole thing is
embedded in.
And this thing's
going to come in,
and it actually hits the
equator at a right angle, OK?
No ifs, ands, or buts about it.
It's a bloody right angle.
And then I'm going to--
let's call this point B--
walk back along the
equator here till I
reach a point which
I will call C.
And then I'm going to go
straight north until I
come back up to the North Pole.
This is a triangle in which all
three angles are 90 degrees.
So here is a great
little experiment
that's very easy for
you to do at home.
Does anyone happen to
have a ball with them?
OK, never mind.
Let me look at my notes
for just a second.
So let's say I start
out here at point A,
and I have my vector pointing
in the south direction.
So this guy goes down here.
And what you'll see is it
goes down to the equator,
and it keeps pointing south.
Then I bring it along over here,
bring it back up to the north.
The vector has been
rotated by 90 degrees
as it goes around that pass.
It's a really,
fun, exciting demo.
If you've got a ball at home,
you can do this over and over
again.
It's endless fun.
I'm being slightly silly here,
but there's an important point
to be made.
When you do this operation,
parallel transport
rotates the vector.
It turns out that, if you are
working on a two-dimensional
manifold-- particularly, I think
it's a two-dimensional manifold
that is--
it may have to be
of what's called
constant curvature--
in other words,
either a surface, a
plane, or a hyperbola.
It actually rotates by
an angle of whatever
is internal angle of the
triangle minus 180 degrees.
So in this case, it would
rotate it by 90 degrees.
If you took this thing
and you actually opened it
up all the way, you
could basically, just
by taking this leg and making
it as long as you want,
you can make it to 0.
You can make it huge.
And when you do so, you'll just
rotate that vector all the more
as it goes around.
This operation, by the
way, is called a holonomy.
I throw that out there
because, last time I looked,
there was a decent
Wikipedia page
on this that has some cool
animated graphics on it.
Also, MathWorld.Wolfram.com
had some good stuff.
So this has good
descriptions that you
can find it all on Google.
All right.
What I want to do is take
some of these somewhat vague
notions--
so hopefully, I made
it intuitively clear
that there's something
very interesting that
happens when I
parallel transport
a vector around these figures,
depending upon the underlying
geometry of the manifold
that they're embedded in.
Let's try to make
it more precise now.
And I'm going to start
all the way over here
because I'm going to want big,
clean boards to illustrate
this.
OK.
So suppose I'm in some
coordinate system,
and this line I've written here
represents a line of constant.
So lambda is one
particular member
of your set of
spacetime coordinates,
so it might be time
or radius or maybe
you work in some
crazy querying system.
But lambda is meant to
represent some particular member
of your coordinate system.
And then there's another track
over here, which is displaced
from it by delta x lambda, OK?
So everywhere along here,
one of your coordinates
is equal to the value, x lambda.
Everywhere along here wanted,
that same coordinate's
equal to x lambda
plus dx lambda.
Along this trajectory,
there's a different coordinate
that is kept constant.
Lambda and sigma
are not the same.
So there is some
coordinate whose value
I will label as sigma that
is constant along there.
And along this one,
it is also constant.
Let me label the four
vertices, A, B, C, and D.
And let me number
these four edges--
one, two, three, and four.
What I am going to
imagine doing is parallel
transporting some vector,
v alpha, around this loop.
So what I'm going
to do is generate
the equations that describe
how it changes as a transport.
I'm going to start at A, so
v is pointing along here.
Transport it to B to C
to D and then back to A.
So let me very carefully
do the first leg.
Once you get the
pattern, the others
can be done a little
bit more quickly.
So the coordinate-- let's see.
Hang on just one moment.
Yeah, so I am going
from A to B first.
So as I move from A to B,
x lambda remains constant,
and the coordinate x
sigma is increasing.
So I am moving in a
direction that points
along the unit vector associated
with the sigma coordinate.
So I'm going to say that
there's a basis vector.
I shouldn't have
said unit vector.
I don't know its magnitude.
I'm pointing along
the direction in which
coordinate sigma is increasing.
And so parallel
transporting this vector
amounts to requiring that
my covariant derivative
along the sigma
basis vector is 0.
This can be written out.
Turn this into index form.
It looks like this.
OK, no surprises.
So now what I'm going
to do is, essentially,
I'm going to write down an
integral that would describe
how v alpha changes as I move
from A to B. When I do this,
I will then get the value
of the vector at point B.
So the way I'm going to
write this is v alpha at B
is equal to v alpha, the
initial value of this thing,
minus what I get when I
integrate along leg one--
gamma alpha sigma
mu phi mu dx sigma.
Everyone happy with that?
So everything I've done
over here, so far, I think,
is probably just fine.
When you've got a differential
equation, integrate it.
Boom.
You integrate it.
You got your new thing.
We're going to actually
solve these integrals
in a few moments, but we'll
just leave it like this for now.
So that's the first step.
I got a couple more to
do, but hopefully you
can now see the pattern.
If I go from B to
C, I am now moving
in the direction
of lambda, and I'm
holding the value of that
coordinate constant at x
sigma plus dx sigma.
So the vector at
C is going to be
equal to this thing
at B minus what
I get when I integrate along
path two, gamma alpha gamma mu.
We got two more to go.
So this one I am,
again, integrating
along the sigma direction.
But notice, I switched the sign.
I switched the sign
because now my coordinate's
going in the direction
where it's decreasing
rather than increasing.
Get some fresh chalk.
So we've taken it
from A to B, B to C, C
to D. Let's take it
all the way around.
So take it all the way
around my second value
at point A. This is
going to be v alpha at D.
And again, this guy is coming
in the other direction.
So I'll enter this
one with a plus sign.
And I get this.
OK, so the way I'm going
to quantify curvature
is buried in all this stuff.
Let's dig it out.
So the first thing
which you're going to do
is I'm going to say, if I take
v alpha final, basically what
I want to do is
write this guy out,
substitute in for v alpha d,
which requires me to substitute
in for v alpha C, [INAUDIBLE].
So I'm going to get
a big, old mess here.
But in the end, the first
term will be v alpha initial.
So let's subtract that off.
That is the change.
When you actually
work this out, it's
going to involve four integrals.
I have chosen to write
this in a way that
highlights a property I'm
going to take advantage
of in just a moment.
OK, so the reason I
wrote it in this way--
so I have the
integral along four
minus that along two
plus integral along three
minus that along one--
is that each one that
I've written here--
they represent
parts that are sort
of parallel to each
other on the figure, just
offset from each other
by a little bit, parallel
but offset paths.
Yeah, let's put this one high.
Hang on just one moment.
I have a thing in my notes that
said I needed to fix something.
Did I actually fix it?
Yeah, I did.
OK.
So schematically, let's
look at that first line.
The integral along four of--
I have a something, dx
lambda, minus the integral
along path two--
of a something, dx lambda.
So it's the same basic
function inside each of these,
but this one is being
evaluated at x sigma.
This one is being evaluated
at x sigma plus dx sigma.
I can combine them.
So this becomes the integral--
let's say it's along two.
Pardon me for a second.
Make that a little bit bigger.
So what I'm doing
is I'm saying that I
have a function
evaluated at x sigma
minus a function evaluated
at x sigma plus dx sigma.
Let's do a little
binomial expansion.
It's equivalent to an
integral along a single path
of essentially what I
get, the first order
Taylor term of that.
Do the same thing
for the other guy.
Integral along three,
I have a something,
dx sigma minus integral along
1, same something, dx sigma.
This guy is being evaled at
x lambda x delta x lambda.
This guy is being
evaluated at x lambda.
And so this whole
thing is approximately
equal to integral along one.
So it looks like this, OK?
So if you want to do this a
little bit more carefully,
knock yourself out.
Part of that-- I probably
should've said this explicitly,
but hopefully the
notation made it clear--
I'm treating these little
deltas as small quantities, OK?
So it makes sense that I can
introduce a little first order
expansion here.
Let's leave the picture up, but
I'm going to clear this board.
With this way of doing things,
let's rewrite my integrals.
So what this gives me is
delta vx alpha equals--
should really be an
approximately equal
because we're truncating
this expansion.
So the integral from x sigma--
x sigma equals dx sigma.
Alpha x lambda [INAUDIBLE] x
lambda of gamma alpha sigma mu
phi mu dx sigma minus--
So this is just taking
what I wrote there.
Schematically, this is what
you get when you actually
expand all those guys out.
All right.
So I've got a couple
derivatives here.
And I'm doing an
infinite test of a couple
of infinitesimal integrals.
When I'm doing
infinitesimal integrals,
they're very simple to evaluate.
So let's just go ahead, evaluate
them, and also expand out
those derivatives.
Doing so, this
cleans up a fair bit.
First one, I'm going to
be able to finally get rid
of those damn integral signs.
So I'm going to wind up with
something that is essentially
quadratic in these things.
It's going to look
as the product
of my little infinitesimal
displacements.
And I'm going to wind
up with a term that
involves a partial derivative
of my connection here.
So we've got one
term looks like this.
We've got another term
that looks like this.
So don't worry about the
index gymnastics a little bit.
If you go through it
carefully, you'll see it.
Pause here for a second.
This expression sucks.
The reason why it sucks--
it's not just because
there's lots of terms.
There's a bajillion
indices on it.
But it's because I've
got one term that
is linear in the vector
and one that's linear
in the derivative of vector.
However, don't forget-- we get
the derivatives of the vector
by parallel transport.
So we parallel
transported this guy,
which tells us that these
derivatives are simply related
to the vectors themselves.
If I move that to the
other side, that's
equivalent to covariant
derivative of v equals 0.
So if I want to get
rid of my derivative
with respect to x lambda,
here's I you'd write that.
Likewise, if I want to
get rid of-- and I do--
my derivative with
respect to x sigma,
just replace lambda with sigma.
So now let's sub these in.
Now, in 1980, the person
who became my PhD supervisor
wrote this giant review article
on gravitational radiation.
And either the last or second
to last section of the paper--
I'm reminded of it right now.
It begins with the
sentences, "The end is near.
Redemption is at hand.
The end is near.
We shall soon be redeemed."
All right.
Let's plug these in.
So we plug these guys in here.
What do we get?
Delta v alpha
equals these things.
We're supposed to put
this in a parenthesis.
And now, what's going
to happen when I get rid
of all those
derivatives is I'm going
to have a bunch of terms that
look like Christoffel squared.
As Scooby Doo
would say, ruh-roh,
but that is just
what we have to have.
Incidentally, what you see when
you do something like this is
I now have terms entering
into this whole thing that
involved derivatives
of the metric times
derivative of the metric.
Many of you may have
heard sort of the slogan
that general relativity is a
nonlinear theory of gravity.
There's where your
nonlinearity is actually
going to turn out
to be entering,
is the fact you add these
squared terms in here that
involve metric times
itself entering
in such a non-trivial
and important way.
What I'm going to
do, finally, on this
is-- so this is
slightly annoying
because I have one term
in v mu, one term in v nu.
But look.
Both mu and nu
are dummy indices.
They're dummy
indices, so what I'm
going to do is-- on the last
term or the last two terms,
I'm just going to
exchange mu for nu.
And what I finally get is that
the change in the vector v
transported along a loop
whose sides are delta x lambda
and delta sigma--
it's a quantity that is linear
in those two displacements.
It's linear in the vector and
involves this four index tensor
whose value depends on
derivatives of the connection
and two nonlinear terms
in the connection.
This quantity is a mathematical
entity known as the Riemann
curvature tensor.
Even though it involves
connection coefficients,
Christoffel symbols,
and we argued before--
and you guys did a homework
exercise where you show this--
that the connection,
the Christoffel,
is not tensorial, this
combination of them,
basically that the terms come
together in such a way that,
when you change
your representation,
the nontensorial bits
cancel each other out
from the terms that are being
subtracted against one another.
So this is, indeed,
a true tensor.
There's an equivalent
definition,
if you are reading Carroll--
so essentially,
what I just walked
through here is an
integral equivalent
of the following
commutator being applied
to the vector v.
Some textbooks simply
state the Riemann
tensor is related
to the commutator of
partial derivatives acting
upon a four-vector like so.
With a little bit of effort,
you can show that what this is
is, essentially, a way of--
what I worked out over
there is a geometric way
of understanding what
that commutator means.
Incidentally, one thing, which
I think is worth calling out--
when you apply
this to a one-form
or a downstairs component, you
get this with a minus sign.
If you are reading the
textbook by Schutz,
Schutz has this sign wrong
in its first edition.
Hopefully all copies
of the first edition
are rare enough now that, if
you are looking at Schutz--
Schutz it's actually
a wonderful textbook
for an early introduction
to this field,
but if you happen to get a
hold of the first edition,
just be aware that there is--
I think it's on page
171 of the textbook,
you'll see this written.
So I've actually
written a couple papers
with Bernard Schutz, and so
I'm allowed to tease him.
Not only did he get it wrong.
He actually came up with
an intuitive argument
that is wrong.
Sometimes you just need to sit
down and bloody well calculate
something because you
can almost always come up
with an argument to convince you
of something that's not true.
And I'm afraid that's what he
did in this particular case.
So I have a couple notes in
there about what is sometimes
called curvature coupling,
which essentially tells us--
I pointed out in
the last lecture
that when we're
dealing with geodesics,
strictly speaking they describe,
completely point-like, almost
just a monopole and no structure
and no shape whatsoever moving
through spacetime.
If you have a larger
body or a body that
has any kind of multipolar
structure associated with it,
those multipoles--
you can think of that additional
structure is essentially
filling up part of
the local Lorentz
frame around the center
of mass of that point,
and they couple the
spacetime and push it away
from the geodesic.
This Riemann tensor
actually describes the way
in which that body couples
to the background spacetime,
that it might be falling in.
So this ends up playing
a really important role.
For instance, when you study
the precession of equinoxes,
we learn how to do this
Newtonian theory using
the action of tides from
the Earth and the moon
on a planet like the Earth.
This ends up being
the quantity that
mathematically encapsulates
tides and general relativity.
So it enters into there.
So I'm going to sketch
through this very,
very quickly, simply because
we don't have a lot of time,
and there's good discussion of
this in various other places.
But let me just point out that--
you look at this thing.
It's a four-index
tensor, and each index
can take four values.
That makes it look like
it has 256 components.
Now, I'm not going to step
through this in detail
right now.
This will either be in the
next lecture that I do this,
or you'll watch me
on a video once this
gets recorded, depending
upon how things
unroll in the next 24 hours.
Riemann has a lot of symmetries.
I will go through those
symmetries carefully,
either in lecture or on a video.
So symmetries-- Riemann--
and you know what?
Let me write it
out in n dimensions
from n to the 4, which is what
you'd expect for a four index
object in n
dimensions, down to n
squared times m squared
minus 1 over 12th.
So where I want
to conclude today
is let's just take
a look at what that
turns into for a couple
of different numbers
and dimensions.
So if you do n
equals 1, you get 0.
So the Riemann tensor
has no components
on a one-dimensional manifold.
There's a simple
reason for that.
Remember the way
we defined it, OK?
We did this by
parallel transporting
around a particular figure.
If you're in one dimension,
this is all you can do.
There's no holonomy
operation in one dimension.
You can't do that.
So no curvature.
If you want to be a real
pedant and someone says,
well, look at a curved
line, you'll go, ah,
but lines can't be curved.
n equals 2.
So you get 2 squared, 2
squared minus 1 over 12--
you get 1.
So if you're working
in two dimensions,
there is a single number
that characterizes
the curvature at every
point, and this is often
thought of as just a
radius of curvature.
Simplest example is
if you have a sphere.
Sphere is completely
characterized by its radius.
But if you imagine that it's
like a sphere that you squash,
well, you can imagine
at every point
that there is a
particular sphere that
is tangent to that
point, and the radius
of curvature of
the tangent sphere
is the one that defines the
curvature at that point.
I'm going to skip three because
it's not all that interesting.
The one that is more important
is, if you do n equals 4,
you'll wind up with 16 times
15 over 12, which is 20.
This is exactly the
number of derivatives
that we could not
cancel out when
we did the exercise a couple
of lectures ago of assessing
how well we can make spacetime
have a flat representation
in the vicinity of some point.
It's the number of leftover
constraints at second order
in a freely falling frame.
All right so I'm going
to stop there for today.
That's a nice place
for us to stop.
Keep watching your emails.
We're in an
interesting situation.
Life at MIT is evolving.
But when we pick it up
in one form or another,
what I'm going to do first
is talk a little bit more
about the symmetry
of this object
because there's a couple
of explicit symmetries
that lead to that reduction
from n to the 4th to n squared
n squared minus 1 over
12, and it's useful for us
to go through them and
see what they look like.
And then I also want to talk
about a couple of variants
on this curvature tensor, OK?
So just to give you a
little bit of a preview--
the curvature tensor--
it's a four index object.
We have argued
already that we're
going to end up
doing things that
look like looking at
derivatives of the metric
being equal to our source.
Our source is a two-index
object, the stress energy
tensor.
We've got gotta get
rid of two indices.
And so what we're
going to do is we're
going to essentially contract
this guy with a couple
of powers of the metric
in order to trace over
certain combinations of indices
and make two index variants
of the curvature tensor.
And we're also going to
look at derivatives of this,
because it turns out that there
is a particular combination
of derivatives at
the Riemann tensor
that has an important
geometrical meaning.
What we're going to find is
that, when we combine these two
notions, there is a
particular divergence
of a particular variant of the
curvature tensor that is 0.
In other words, we can
make a curvature tensor
that is divergence-free.
Our stress energy tensor
is divergence-free.
I wonder if one is
related to the other.
That, in a nutshell,
is how Einstein came up
with general relativity,
by asking that question
and then just seeing
what happened.
So that's what we're
going to go through next.
