[SQUEAKING]
[RUSTLING]
[CLICKING]
SCOTT HUGHES: So
we're just picking up
where we stopped last time.
So we are beginning
to discuss how
we are going to sort of
do a geometrical approach
to physics, using a more
general set of coordinates now.
So we began talking about how
things change when I discuss
special relativity,
so for the moment
keeping ourselves just
at special relativity.
We, by the way, are going to
begin lifting our assumptions
that it is simply special
relativity fairly soon.
But to set that
up, I need to start
thinking about how to work
in more general coordinate
systems.
So we're going to do it in the
simplest possible curvilinear
coordinates.
So it's basically just going
from Cartesian coordinates
in the spatial sector to
plane polar coordinates.
One of the things which I
have emphasized a few times,
and I'm going to
continue to hammer on,
is that these are a little bit
different from the curvilinear
coordinates that you are
used to in your past life.
In particular, if I write
out the displacement,
the little vector of
the displacement element
in the usual way,
I am using what's
called a "coordinate basis,"
which means that the vector
dx is related to the
displacement, the differential
of the coordinates, by
just that thing contracted
with all the basis vectors.
And so what that means is I have
a little displacement in time,
which looks normal.
Displacement in radius,
which looks normal.
Displacement in the z
direction, which looks normal,
and a displacement in an
angle, which does not.
In order for this whole thing
to be dimensionally consistent,
that's telling me e phi has to
have the dimensions of length.
And that is a
feature, not a bug.
Last time, we
introduced the matrix
that allows me to convert
between one coordinate system
and another, so just
basically the matrix--
it's sort of a Jacobi matrix.
It's a matrix of
partials between the two
coordinate systems.
And this idea that things are--
they look a little weird.
So the way I did that was I
didn't actually write it out,
but I did the usual mapping
between x, y and r and phi,
worked out all of
my derivatives.
And sure enough,
you've got something
that looks very standard,
with the possible exception
of these r's that are
appearing in here.
So notice the elements
of this matrix.
These do not have
consistent units--
again, feature, not bug.
This guy is basically
just the inverse of that.
This is the matrix that
affects the coordinates
in the opposite direction.
And notice in this case,
you have some elements where
their units are 1 over length.
So let's just continue
to sort of catalog
what some of the things we are
going to be working with look
like in this new
coordinate representation.
And this will lead
us to introduce
one of the mathematical
objects that we
are going to use extensively
as we move forward
in studying this subject.
So what I want to do
next is look at what
my basis vectors look like.
So what I want to do
is characterize what
my e r and my e phi look like.
And these are going to look very
familiar from your intuition,
from having studied
things like E&M
in non-Cartesian coordinates.
So your e r is just related to
the original Cartesian basis
vectors, like so.
And if you like, you
can easily read that out
by performing the
following matrix
multiplication on the original
Cartesian basis vectors.
Your e phi perhaps
looks a little wacky.
So you can see the length
coming into play there.
A good way to think about
this is if your intuition
about basis factors--
I have to be careful with
this language myself--
your intuition
about basis vectors
is typically that
they are unit vectors.
These are not unit vectors.
They do form a nice basis,
but they are not unit vectors.
In particular, the basic idea
we're going to go with here
is that e phi, it's always
going to sort of point
in the tangential direction.
But no matter where
I put it in radius,
I want that vector
to always sort
of subtend the same
amount of angle.
In order to do that, its
length needs to grow with r.
So that's where
that's a little bit
different from your intuition.
And there's a very
good reason for this,
which we will get to, hopefully,
well before the end of today's
class.
So last time, when
we first began
to talk about tensors a
couple of lectures ago,
the first tensor I gave you--
so confining ourselves just
to Cartesian coordinates--
was the metric,
which was originally
introduced as this
mathematical object that
came out of looking at dot
products between basis vectors.
It's essentially a
tensor that allows
me to feed in two
displacements and get
the invariant interval between
those displacements that
comes out of that.
I am going to continue to call
the dot product of two basis
vectors the "metric."
But I'm going to use a slightly
different symbol for this.
I'm going to call
this g alpha beta.
In the coordinate representation
that we are using right now,
so in plane polar
coordinates, this becomes--
you can work it out from what
I've written down right here.
This is just the diagonal
of minus 1, 1 r squared 1.
So this equals dot here.
This is-- I'll put PPC for plane
polar coordinates under that.
And then using that, you
know that you can always
work out the invariant
displacement between two
events.
It's always going to be
the metric contracted
with the differential
displacement element.
And this is going to
be minus dt squared
plus dr squared plus r squared
d phi squared plus dz squared.
That, I hope, makes
a lot of sense.
This is exactly
what you'd expect
if I have two events
that are separated
in plane polar coordinates
by dt, dr, d phi, dz.
This is what the distance
between them should be.
So the fact that
my basis vectors
have this slightly annoying
form associated with them,
it all sort of comes
out in the wash here.
Remember at the end of the day,
if we think about quantities
that are representation
independent--
and that's going to
be the key thing.
When you assemble scalars
out of these things,
the individual
tensor components,
they can be a little
bit confusing sometimes.
They are not things
that we measure.
They are not things that
really characterize what we
are going to be working with.
And we really want to
get into the physics,
unless we're very
careful about it.
This is something
you can measure.
And so sure enough,
it comes out,
and it's got a
good meaning to it.
Let me just wrap
up one last thing
before I talk about sort of
where we're going with this.
So just for completeness, let me
write down the basis one forms.
Just as the basis vectors had a
bit of a funny form associated
with them, you're going to find
the basis one forms likewise
have a bit of a funny
form associated with them.
And the way I'm
going to get these--
and so these are going to be
the Cartesian basis one forms--
basically, I'm not carefully
proving all these relations
at this point, because you
all know how to do that.
I'm just using line
up the indices rule.
And when you do
that, you get this.
And likewise, your basis one
form for the axial direction,
I'll just write down the result.
It's going to look like this.
So the key place where all
of this-- so right now, these
are all just sort
of definitions.
Nothing I've done here
should be anything
that even approaches a
surprise, I hope, just
given the you guys have done--
the key thing
that's probably new
is all this garbage associated
with coordinate bases,
this extra factors of r and
1 over r that are popping up.
But provided you're willing to
sort of swallow your discomfort
and go through
the motions, these
are not difficult calculations.
The key place where all
of this really matters
is going to be when we
calculate derivatives of things.
It'll turn out there is
an important rule when
we talk about integrals as
well a little bit later,
but let's just say that.
So for now, we'll
focus on derivatives.
So all the derivatives that
we've been looking at so far,
we have, indeed, done a couple
of calculations where we've
computed the derivatives
of various vector valued
and tensor valued quantities.
And it was helped by the
fact that all the bases, when
I work in Cartesian
coordinates, are constant.
Well, that's not the case now.
So now, we need to
account for the fact
that the bases all vary
with our coordinates.
So let me just
quickly make a catalog
of all the non-trivial--
there's basically four.
In this one, where I'm just
doing plane polar coordinates,
there are four
non-trivial derivatives
we need to worry about.
One of them actually
turns out to be 0.
So the radial derivative of
the radial unit vector is 0.
But the phi derivative of
the phi unit vector is not.
you go and take the phi
derivative of this guy,
and you basically get--
take the phi derivative
of this, you're
going to get this back,
modulo factor of radius.
So I can write d e r
d phi as e phi over r.
If I take the derivative
of e phi with respect to r,
I get e phi back, divided by r.
So the simplest way to
write this is like so.
And finally, if I take the
phi derivative of the phi unit
vector, I get e r back, with
an extra factor of r thrown in.
And a minus sign.
So we're going to see a way of
doing this that's a little bit
more systematic
later, but I want
to just keep the simple example,
where you can just basically
by hand calculate all the
non-trivial derivatives easily.
Of course, there's also a t
unit vector and a z unit vector.
But they're constants,
so I'm not going
to bother writing them out.
All the derivatives associated
with them are equal to 0.
So let's imagine now that I
have assembled some vector.
So I have some vector field
that lives in this spacetime.
And I'm using this basis.
And so I would write this
vector with components v alpha.
And let's let the--
so this is going to be a
curvilinear coordinate system,
so this will be plane polar
coordinates being used here,
plane polar coordinate
basis vectors.
And what I would like to
do is assemble the tensor
that you can think
of essentially
as the gradient of this vector.
So let's begin by doing this
in a sort of abstract notation.
So the gradient of this guy--
this is sort of ugly notation,
but live with it.
Following what we have
been doing all along, what
you would want to
do is just take
the root of this whole thing.
It's going to have a
downstairs component on it.
So attach to it
the basis one form.
If you prefer, you can write
it using the d notation
like I have there,
but I just want
to stick with the form
I wrote in my notes.
Looking at this the way I've
sort of got this right now,
I can think of, if I
don't include the basis
one forms here, this should be
the components of a one form.
So this should be
a kind of object.
So let's just expand
out that derivative.
Let's write it like this.
So you just-- we haven't
changed calculus.
So when I do this, I'm
going to basically use
the old-fashioned Leibniz
rule for expanding
the derivative
product of two things.
Here's the key thing which
I want to emphasize--
in order for this
whole thing to be--
for this to be a tensorial
object, something that I couple
to this basis one form, the
sum of these two objects
must obey the rules for
transforming tensors.
But the two objects
individually will not.
So this is an
important point which
I'm going to emphasize in
slightly different words
in just a few moments again.
This is one of the key
things I want you to get out
of this lecture, is that
when I'm taking derivatives
of things like this,
you've got to be
a little bit careful about what
you consider to be components
of tensors and what is not.
Now as written like that,
this is kind of annoying.
So my first object has a nice
basis vector attached to it.
My second object involves a
derivative of the basis vector.
However, something
we saw over here
is that derivatives
of basis vectors
are themselves proportional
to basis vectors.
So what I'm going to do is
introduce a bit of notation.
So let me switch
notation slightly here.
So the beta derivative of
e alpha can be written as--
in general, it can be written
as a linear combination
of basis vectors.
So what we're going
to do is define d--
I want to make sure my Greek
letters are legible to everyone
in the room here.
So let me write this
nice and clearly.
d beta of e alpha, I'm going to
write that as capital gamma mu
beta alpha e mu.
This gamma that I've just
introduced here in this context
is known as the
Christoffel symbol.
Fact I'm calling this a symbol,
it's got three indices on it.
You might look at it and go,
ooh, smells like a tensor.
Be a little bit careful.
In much the same way that these
two terms are not individually
components of a tensor,
but their sum is,
this guy individually
is actually
not a component of
a tensor, but when
combined with other things, it
allows us to assemble tensors.
So for our plane
polar coordinates,
there are exactly three
non-zero Christoffel symbols.
So gamma phi r phi is equal
to 1 over r, which is also
equal to gamma phi phi r.
Gamma r phi phi is minus r.
And you can basically just
read that out of that table
that I wrote down over there.
All the others
will be equal to 0.
Now from this
example, this is what
it makes it smell like
every time you introduce
a new coordinate representation.
You're going to need to sit
down for an hour and a half,
or something like
that, and just work out
all the bloody derivatives, and
then go, oh, crap, and read out
all the different
components of this thing,
and assemble them together.
There actually is
an algorithm that we
will get to at the
end of this class that
allows you to easily extract
the Christoffel symbols provided
you know the metric.
But right now, I just want
to illustrate this thing
conceptually.
The key thing which you
should know about it
is that it is essentially the--
I almost said the word "matrix,"
but it's got three indices.
It's a table of
functions that allows
me to relate derivatives
of basis vectors
to the basis vectors.
So before I go on to talk
about some of that stuff,
let's take a look at the
derivative a little bit more
carefully.
So the derivative
of the vector--
so let's basically take what
I've written out up there.
I'm going to write this as
the beta derivative of vector
v. And I can write that as
the beta derivative of e of v
alpha--
so the first term
where the derivative
hits the vector components.
And then I've got
a second term where
the derivative hits the basis.
I'm going to write this like so.
This is sort of annoying.
One term is
proportional to e alpha,
one is proportional to e mu.
But notice, especially in the
second term, both alpha and mu
are dummy indices, so
I'm free to relabel them.
So what I'm going to
do is relabel alpha
and mu by exchanging them.
As long as I do
that consistently,
that is totally kosher.
And when I do that, I can
factor out an overall factor
of the basis object.
This combination
that pops up here--
so we give this a name.
And this is a combination
which, by the time
you have finished this semester,
if you don't have at least
one nightmare in which
this name appears,
I will not have done
my job properly.
This shows up a
lot at this point.
This is called the
"covariant derivative."
And it shows up enough
that we introduce
a whole new notation
for the derivative
to take it into account.
I'm going to call
this combination
of the partial
derivative of v and v
coupled to the
Christoffel symbol--
I'm going to write this using
the, if you're talking LaTeX,
this would be the
nabla operator.
So I made a point
earlier when we
were talking about
derivatives a couple of weeks
ago that we were reserving
the gradient symbol
for a special purpose later.
Here it is.
So whenever I make
a derivative that
involves the gradient
symbol like this,
it is this covariant derivative.
And the covariant derivative
acting on vector components,
it generates tensor components.
Partial derivative does not.
And what I'm going to do,
just in the interest of time--
it's one of those calculations
that's straightforward
but fairly tedious--
I have a set of
notes that I meant
to make live before I headed
over here, but I forgot.
I have a set of notes that I'm
going to show on the website
by this evening which explicitly
works out what happens when you
apply the coordinate
transformation using that--
it's been erased--
when you use that L
matrix to construct
the coordinate transformation
between two representations.
If you try to do it to
partial derivatives of vector
components, basically
what you find
is that there's an extra
term that spoils your ability
to call that-- it spoils the
tensor transformation law,
spoils your ability to call
that a tensor component.
So the partial on its
own doesn't let you.
You get some extra terms that
come along and mess it all up.
On next p set,
you guys are going
to show that if you then try to
apply the tensor transformation
law to the Christoffel
symbols, you
get something that
looks tensorial,
but with an extra term
that spoils your ability
to call it tensorial.
There's a little bit
of extra junk there.
But the two terms
exactly conspire
to cancel each other out so
that the sum is tensorial.
So part one of
this will be notes
that I post to the website
no later than this evening.
Part two, you guys
will do on the p set.
So just saying in
math what I just
said in words, if I
do this, like I said,
you basically will
eventually reach the point
where what I am
writing out right now
will become so automatic
it will haunt your dreams.
Wait a minute, I
screwed that up.
It's so automatic I can't
even write it properly.
Anyhow, something like that
will-- modulo my typo--
that should become automatic.
And the key thing
which I want to note
is that if I take these
guys, and I attach
the appropriate basis
objects to them,
this is an honest-to-god tensor.
And so this derivative is
itself an honest-to-god tensor.
A typical application
of this, so one
that will come up a
fair bit, is how do you
compute a spacetime divergence
in each coordinate system?
So suppose I take the divergence
of some vector field v.
So you're going to
have four terms that
are just the usual, like
when you guys learned
how to do divergence in freshman
E&M in Cartesian coordinates.
You get one term that's just
dv x dx, dv y dy, et cetera.
So you got one term that
looks just like that,
and you're going to have
something that brings
in all of Christoffel symbols.
Notice the symmetry that
we have on this one.
Actually, there is Einstein
summation convention
being imposed here.
But when we look
at this, there's
actually only one
Christoffel symbol
that has repeated indices
in that first position.
So when I put all
this together, you
wind up with something
that looks like this.
So go back and check your
copy of Jackson, or Purcell,
or Griffith, whatever your
favorite E&M textbook is.
And you'll see when you work
in cylindrical coordinates,
you indeed find that there's a
correction to the radial term
that involves 1 over r.
That's popped out exactly
like you think it should.
You have a bit of a wacky
looking thing with your phi
component, of course.
And let me just spend a
second or two making sure.
It's often, especially while
we're developing intuition
about working in a
coordinate basis,
it's not a bad idea to
do a little sanity check.
So here's a sanity check
that I would do with this.
If I take the divergence,
I take a derivative
of a vector field, the
final object that comes out
of that should have the
dimensions of that vector
divided by length.
Remembering c equals 1,
that will clearly have
that vector divided by length.
That will clearly
have that vector
divided by length, vector
divided by length, explicitly
vector divided by length.
That's weird.
But remember, the basis objects
themselves are a little weird.
One of the things we
saw was that e phi
has the dimensions of length.
In order for the vector
to itself be consistent,
v phi must have the dimensions
of v divided by length.
So in fact, when I just
take its phi derivative,
I get something
that looks exactly
like it should if it
is to be a divergence.
Let's move on and
think about how
I take a covariant
derivative of other kinds
of tensorial objects.
This is all you need to know
if you are worried about taking
derivatives of vectors.
But we're going to work with
a lot of different kinds
of tensor objects.
One of the most
important lectures
we're going to do
in about a month
actually involves looking at a
bunch of covariant derivatives
of some four-indexed
objects, so it gets messy.
Let's walk our way there.
So suppose I want to take
the derivative of a scalar.
Scalar have no basis
object attached to them.
There's no basis object.
When I take the
derivative, I don't
have to worry about
anything wiggling around.
No Christoffel symbols come in.
If I want to take the covariant
derivative of some field phi,
it is nothing more than
the partial derivative
of that field phi--
boom.
Happy days.
How about a one form?
The long way to do this would
be to essentially say, well,
the way I started
this was by looking
at how my basis vectors varied
as I took their derivatives.
Let's do the same thing
for the basis one forms,
assemble my table, do a lot
of math, blah, blah, blah.
Knock yourselves out if
that's what you want to do.
There's a shortcut.
Let's use the fact that
when I contract a one form
on a vector, I get a scalar.
So let's say I am looking at the
beta covariant derivative of p
alpha on a alpha.
That's a scalar.
So this is just the
partial derivative.
And a partial derivative of
the product of something I
can expand out really easily.
So using the fact that this
just becomes the partial,
I can write this
as a alpha d beta p
alpha plus p downstairs alpha.
So now what?
Well, let's rewrite this using
the covariant derivative.
Pardon me a second while I
get caught up in my notes.
Here we are.
I can write this as the
covariant derivative
minus the correction that comes
from that Christoffel symbol.
Pardon me just a second.
There's a couple lines here I
want to write out very neatly.
So when I put this in--
oops typo.
That last typo is important,
because I'm now going
to do the relabeling trick.
So what I'm going to
do is take advantage
of the fact that in this
last term, alpha and mu
are both dummy indices.
So on this last term that
I have written down here,
I'm going to swap
out alpha and mu.
When I do that, notice that the
first term and the last term
will both be proportional
to the component a alpha.
Now, let's require that
the covariant derivative
when it acts on two things
that are multiplied together,
it's a derivative, so it should
do what derivatives ordinarily
do.
So what we're going
to do is require
that when I take this
covariant derivative,
I should be able to
write the result like so.
It's a healthy thing that
any derivative should do.
So comparing, I look
at that, and go,
oh, I've got the covariant
derivative of my one form
there.
Just compare forms.
Very, very similar, but
notice the minus sign.
There's a minus sign that's
been introduced there,
and that minus sign guarantees,
if you actually expand out
that combination of
covariant derivatives
I have on the previous line,
there's a nice cancellation
so that the scalar that
I get when I contract p
on a, in fact, doesn't have
anything special going on when
I do the covariant derivative.
So I'm going to
generalize this further,
but let me just make
a quick comment here.
I began this little
calculation by saying,
given how we started
our calculation
of the covariant
derivative of a vector,
we could have begun
by just taking
lots of derivatives of the
basis one forms, and assembling
all these various tables,
and things like that.
If you had done this,
it's simple to find,
based on an analysis
like this, that if you
take a partial
derivative of a one form,
that you get sort of a linear
combination of one forms back.
Looks just like
what you got when
you took a partial derivative
of the basis vector,
but with a minus sign.
And what that minus sign does
is it enforces, if you go back
to a lecture from ages ago, when
I first introduced basis one
forms, it enforces the idea that
when I combine basis one forms
with basis vectors, I get an
identity object out of this,
which is itself a constant.
If you are the
kind of person who
likes that sort of
mathematical rigor,
some textbooks will
start with this,
and then derive other
things from that--
sort of six of one, half
a dozen of the other.
So we could go on at this point.
And I could say,
how do I do this
with a tensor that has two
indices in the upstairs
position?
How do I do this
with a tensor that
has two indices in the
downstairs position?
How do I do it for
a tensor that's
got 17 indices in
the upstairs position
and 38 in the
downstairs position?
The answer is easily
deduced from doing
these kinds of
rules, so I'm just
going to write down a
couple of examples and state
what it turns out to be.
So basically, imagine I want to
take the covariant derivative--
let's do the stress
energy tensor--
covariant derivative of T mu nu.
So remember, the way that the
Christoffel got into there
is that when I looked at
the derivative of a vector,
I was looking at derivatives
of basis objects.
Well, now I'm going
to look at derivatives
of two different basis objects.
So I'm going to wind up with
two Christoffel symbols.
You can kind of think of it
as coming along and correcting
each of these indices.
I can do this with the indices
in the downstairs position.
Guess what?
Comes along and corrects
all them with minus signs.
Just for completeness,
let me just
write down the general rule.
If I am looking at the
covariant derivative of a tensor
with a gajillion
upstairs indices
and a gajillion
downstairs indices,
you get one term that's just a
partial derivative of that guy,
and you get a
Christoffel coupling
for every one of these.
Plus sign for all the
upstairs, minus sign
for all the downstairs.
That was a little tedious.
You basically just, when I
give you a tensor like that,
you just kind of
have to go through.
And it becomes sort
of almost monkey work.
You just have to rotely
go through and correct
every one of the indices using
an algorithm that kind of looks
like this.
Oh, jeez, there's absolutely a
minus sign on the second one.
Thank you.
I appreciate that.
So the way that we have
done things so far,
and I kind of emphasized, it
sort of smells like the way
to do this is you pick your
new coordinate representation,
you throw together all of
your various basis objects,
and then you just
start going whee,
let's start taking
derivatives and see
how all these things vary
with respect to each other,
assemble my table of
the gammas, and then do
my covariant derivative.
If that were, in fact,
the way we did it,
I would not have chosen
my research career
to focus on this field.
That would suck.
Certainly prior to Odin
providing us with Mathematica,
it would have been
absolutely undoable.
Even with it, though, it
would be incredibly tedious.
So there is a better
way to do this,
and it comes via the metric.
Before I derive what the
algorithm actually is,
I want to introduce an extremely
important property of tensor
relationships that we are
going to come back to and use
quite a bit in this course.
So this is something that we
have actually kind of alluded
to repeatedly, but I want to
make it a little more formal
and just clearly state it.
So this relationship
that I'm going to use
is some kind of a tensor
equation, a tensorial equation
that holds in one
representation must
hold in all representations.
Come back to the
intuition when I first
began describing physics in
terms of geometric objects
in spacetime.
One of the key points I
tried to really emphasize
the difference of is that
I can have different--
let's say my arm is a
particular vector in spacetime.
Someone running through
the room at three-quarters
the speed of light will use
different representations
to describe my arm.
They will see
length contractions.
They will see things sort of
spanning different things.
But the geometric
object, the thing
which goes between two events
in spacetime, that does not
change, even though the
representation of those events
might.
This remains true not just
for Lorentz transformations,
but for all classes
of transformations
that we might care to
use in our analysis.
Changing the representation
cannot change the equation.
Written that way, it
sounds like, well, duh,
but as we'll see, it's got
important consequences.
So as a warm-up exercise of
how we might want to use this,
let's think about the
double gradient of a scalar.
So let's define-- let's just say
that this is the object that I
want to compute.
Let's first do this in a
Cartesian representation.
In a Cartesian
representation, I just
take two partial derivatives.
I've got a couple basis
one forms for this.
So I've got something like this.
The thing which I
want to emphasize
is that as written, in
Cartesian coordinates, d
alpha d beta of phi--
those are the components of a
tensor in this representation.
And the key thing is
that they are obviously
symmetric on exchange of
the indices alpha and beta.
If I'm just taking
partial derivatives,
doesn't matter what
order I take them in.
That's got to be symmetric.
Let's now look at
the double gradient
of a scalar in a more
general representation.
So in a general
representation, I'm
going to require
these two derivatives
to be covariant derivatives.
Now, we know one of them can
be very trivially replaced with
a partial, but the
other one cannot.
Hold that thought
for just a second.
If this thing is symmetric in
the Cartesian representation,
I claim it must also be true
in a general representation.
In other words, exchanging the
order of covariant derivatives
when they act on a scalar should
give me the same thing back.
Let's see what this implies.
So if I require the following
to be true, that's saying--
oops.
So let's expand this
out one more level.
So now, I'm correcting that
downstairs index and over here.
So the terms involving
nothing but partials,
they obviously cancel.
I have a common
factor of d mu of phi.
So let's move one of these
over to the other side.
What we've learned is
that this requirement,
that this combination of
derivatives be symmetric,
tells me something about the
symmetry of the Christoffel
symbols itself.
If you go back to
that little table
that I wrote down for
plane polar coordinates,
that was one where
I just calculated
only three non-trivial
components,
but there was a
symmetry in there.
And if you go and
you check it, you
will see it's consistent with
what I just found right here.
Pardon me for just a second.
I want to organize
a few of my notes.
These have gotten
all out of order.
Here it is.
So let me just use
this as an opportunity
to introduce a bit of notation.
Whenever I give you a tensor
that's got two indices,
if I write parentheses
around those indices,
this is going to
mean that I do what
is called "symmetrization."
We're going to use
this from time to time.
If I write square braces,
this is what we call
"anti-symmetrization."
And so what we just learned
is that gamma mu alpha beta
is equal to gamma mu alpha
beta with symmetrization
on those last two indices.
We have likewise learned
that if I contract
this against some object, if
these were anti-symmetric,
I must get a 0 out of it.
So that's a brief aside, but
these are important things,
and I want to make sure you
have a chance to see them.
So trying to make a decision
here about where we're going to
want to carry things forward.
We're approaching the
end of one set of notes.
There's still one more
thing I want to do.
So I set this whole
thing up by saying
that I wanted to give you
guys an algorithm for how
to generate the
Christoffel symbols.
The way I'm going to
do this is by examining
the gradient of the metric.
So suppose I want to
compute the following tensor
quantity-- let's say
is g the metric tensor,
written here in the
fairly abstract notation.
And this is my full-on tensor
gradient of this thing.
So if you want to write
this out in its full glory,
I might write this as
something like this.
But if you stop and think
about this for just a second,
let's go back to this principle.
An equation that is tensorial
in one representation
must be tensorial in all.
Suppose I choose the Cartesian
representation of this thing.
Well, then here's what
it looks like there.
But this is a constant.
So if I do this in Cartesian
coordinates, it has to be 0.
The only way that I can
make this sort of comport
with this principle
that an equation that
is tensorial in
one representation
holds in all representations--
this leads me to say,
I need to require
that the covariant derivative
of the metric be equal to 0.
We're going to use this.
And I think this will be the
last detailed calculation
I do in today's lecture.
We're going to use
this to find a way
to get the Christoffel symbol
from partial derivatives
of the metric.
There's a lot of
terms here and there's
a lot of little indices.
So I'm going to do my best
to make my handwriting neat.
I'm going to write
down a relationship
that I call "Roman numeral I."
The covariant derivative in the
gamma direction, G alpha beta--
you know what, let me put
this down a little bit lower,
so I can get these two
terms on the same line.
So I get this thing that
involves two Christoffel
symbols correcting
those two indices.
This is going to equal 0.
I don't really seem to
have gotten very far.
This is true, but I now
have two bloody Christoffel
symbols that I've somehow
managed to work into this.
What I'm trying to
do is find a way
to get one, and equate it to
things involving derivatives
of the metric.
So this is sort of a
ruh-roh kind of moment.
But there's nothing
special about this order
of the indices.
So with the audacity
that only comes
from knowing the answer in
advance, what I'm going to do
is permute the indices.
Then go, oh, let's permute
the indices once more.
So I'll give you guys a
moment to catch up with me.
Don't forget, these notes
will be scanned and added
to the web page.
So if you don't want to
follow along writing down
every little detail,
I understand,
although personally, I
find that these things gel
a little bit better
when you actually
write them out yourself.
So those are three ways that I
can assert that the metric has
no covariant derivative.
They all are
basically expressing
that same physical fact.
I'm just permuting the indices.
Now there's no better
way to describe this
than you sort of just stare
at this for a few moments,
and then go, gee, I wonder
what would happen if--
so stare at this
for a little while.
And then construct-- you
know I have three things that
are equal to 0.
So I can add them together,
I can subtract one
from the other.
I can add two and
subtract one, whatever.
They should all give me 0.
And the particular
combination I want to look at
is what I get when I
take relationship one
and I subtract from
it two and three.
So I'm going to
get one term that
are just these three
combinations of derivatives,
gamma.
And I get something
that looks like--
let me write this out and
then pause and make a comment.
So I sort of made some lame
jokes a few moments ago
that essentially, the only
reason I was able to get
this was by knowing the answer
in the back of the book,
essentially.
And to be perfectly blunt,
for me personally, that's
probably true.
When I first wrote
this down, I probably
did need to follow an algorithm.
But if I was doing this ab
initio, if I was sitting down
to first do this, what's
really going on here
is the reason I wrote out all
these different combinations
of things is that I was trying
to gather terms together
in such a way that I could take
advantage of that symmetry.
A few moments ago, we showed
that the Christoffel symbols
are symmetric on the
lower two indices.
And so by putting out all
these different combinations
of things, I was then able
to combine them in such a way
that certain
terms-- look at this
and go, ah, symmetry on
alpha and gamma means
this whole term dies.
Symmetry on beta and gamma
means this whole term dies.
Symmetry on alpha and beta
means these two guys combine,
and I get a factor of 2.
So let's clean up our algebra.
Move a bunch of our terms
to the other side equation,
since it's a blah,
blah, blah equals 0.
And what we get when we
do this is g mu downstairs
gamma is equal to 1/2.
What we're going to do now
is we will define everything
on the right-hand side--
I've kind of emphasized earlier
that the Christoffels are not
themselves tensors, but we're
going to imagine that we can
nonetheless-- we're
not going to imagine,
we're just going to define--
we're going to say that we're
allowed to raise and lower
their indices using the metric,
in the same way you guys been
doing with vectors and one forms
and other kinds of tensors.
So let's call everything
on the right-hand side
here gamma with all the indices
in the downstairs position,
gamma sub gamma alpha beta.
And then this is
simply what I get
when I click all of these
things together like so.
If you go and you
look up the formulas
for this in various
textbooks that
give these different
kinds of formulas,
you will typically see it
written as 1/2 g upstairs
indices, and then all this
stuff in parentheses after that.
When you look things up,
this is the typical formula
that is given in these books.
This is where it comes from.
So I need to check one thing
because it appears my notes are
a little bit out of order here.
But nonetheless,
this is a good point,
since we've just finished
a pretty long calculation,
this is a good
point to introduce
an important physical point.
We're going to
come back to this.
We're going to start
this on Thursday.
But I want to begin making
some physical points that
are going to take us
from special relativity
to general relativity.
So despite the fact
that I've introduced
this new mathematical
framework, everything
that I have done so
far is in the context
of special relativity.
I'm going to make a
more precise definition
of special relativity right now.
So special relativity--
we are going
to think of this moving forward
as the theory which allows us
to cover the entire
spacetime manifold
using inertial reference frames.
So we use inertial reference
frames or essentially,
Lorenz reference frames, and
saying that Lorentz coordinates
are good everywhere.
We know we can go between
different Lorentz reference
frames using Lorentz
transformations.
But the key thing is that
if special relativity were
correct, the entire universe
would be accurately described
by any inertial reference
frame you care to write down.
And I will probably only be able
to do about half of this right
now.
We'll pick it up next time,
if I cannot finish this.
The key thing which
I want to emphasize
is, gravity breaks this.
As soon as you put gravity
into your theory of relativity,
you cannot have--
so we will call this a
global inertial frame,
an inertial frame that
is good everywhere,
so "global" in the
mathematical sense,
not "global" in the geographic
sense-- not just earth,
the whole universe, essentially.
As soon as we put in
gravity, we no longer
have global reference frames
and global inertial reference
frames.
That word "inertial"
is important.
But we are going to be allowed
to have local inertial frames.
I have not precisely
defined the difference
what "local" means in this case,
and I won't for a few lectures.
But to give you a preview
as to what that means,
it's essentially
going to say that we
can define an
inertial coordinate
system that is good over a
particular region of spacetime.
And we're going to have
to discuss and come up
with ways of understanding what
the boundaries of that region
are, and how to
make this precise.
So the statement
that gravity breaks
the existence of global
Lorentz frames, like I said,
it's a two-part thing.
I'm going to use a very
handwavy argument which
can be made quite
rigorous later,
but I want to keep it to this
handwaving level, because first
of all, it actually
was first done
by a very high-level
mathematical physicist named
Alfred Schild, who worked in
the early theory of relativity.
It's sort of like he
was so mathematical,
if it was good enough for him,
that's good enough for me.
And I think even though it's
a little handwavy, and kind
of goofy in at
least one place, it
gives a good physical sense
as to why it is gravity
begins to mess things up.
So part one is the
fact that there exists
a gravitational redshift.
So here's where I'm going
to be particularly silly,
but I will back up my silliness
by the fact that everything
silly that I say
here has actually
been experimentally
verified, or at least
the key physical output of this.
So imagine you are
on top of a tower
and you drop a rock of rest mass
m off the top of this tower.
So here you are.
Here's your rock.
The rock falls.
There's a wonderful device down
here which I label with a p.
It's called a photonulater.
And what the
photonulater does, it
converts the rock
into a single photon,
and it does so
conserving energy.
So when this rock
falls, the instant
before it goes into
your photonulater,
just use Newtonian physics
plus the notion of rest energy.
So it's got an energy of m--
mC squared, if you prefer,
its rest energy--
plus what it acquired
after falling--
pardon me, I forgot to
give you a distance here--
after falling a distance h.
So that means that the
photon that I shoot up
from this thing--
let me put a few
things on this board.
So the instant that
I create this photon,
this thing goes
out, and it's going
to have a frequency omega
bottom, which is simply
related to that energy.
This photon immediately is
shot back up to the top,
where clever you,
you happen to have
in your hands a rerockulater.
The rerockulater, as the
name obviously implies,
converts the photon
back into a rock.
Now, suppose it does so--
both the photonulater
and the rerockulater are
fine MIT engineering.
There are no losses
anywhere in this thing.
So there's no friction.
There's no extra heat generated.
It does it conserving
energy, 100%.
What is the energy at the top?
Well, you might naively
say, ah, it's just
going to go up to the top.
It's going to have
that same energy.
It might just have it in the
form of a photon and omega b.
There will be some
frequency at the top.
And your initial
guess might be it's
going to be the same as the
frequency at the bottom.
But if you do that, you're
going to suddenly discover
that your rock has more energy
than it started out with,
and you can redirect it
back down, send it back up.
Next thing you know, you've
got yourself a perpetual motion
machine.
So all you need to do
is get your photonulater
and your rerockulater
going, and you've
got yourself a perpetual
motion machine here.
I will grant that's
probably not the weakest
part of this argument.
Suppose you had this.
I mean, you look at this.
If technology allowed you
to make these goofy devices,
you would instantly look
at this and say, look,
if I am not to have--
let's just say I
live in a universe
where I'm fine
with photonulaters.
I'm fine with
rerockulaters, but damn it,
energy has to be conserved.
I am not fine with
perpetual motion machines.
If that's the case, we always
fight perpetual motion.
We must have that
the energy at the top
is equal to the energy
this guy started with.
When it sort of gets back into--
imagine that your
rerockulater is shaped
like a baseball catcher's mitt.
You want that thing to just
land gently in your mitt,
and just be a perfectly
gentle, little landing there.
And when you put all this
together, going through this,
taking advantage of the fact
that if you work in units where
you've put your
c's back in, there
will be a factor of g h over c
squared appearing in here, what
you find is that the
frequency at the top
is less than the
frequency at the bottom.
In other words, the light has
gotten a little bit redder.
now I fully confess, I did
this via the silliest argument
possible.
But I want to
emphasize that this
is one of the most precisely
verified predictions of gravity
and relativity theory.
This was first done, actually,
up the street at Harvard,
by what's called the
Pound-Rebka experiment.
And the basic principles of
what is going on with this right
now--
I just took this out to
make sure my alarm is not
about to go off, but
I want to emphasize
it's actually built
into the workings
of the global
positioning system.
Because this fact that
light signals may travel out
of a gravitational
potential, they
get redshifted, needs to be
taken into account in order
to do the precise
metrology that GPS allows.
Now, this is part one,
this idea that light
gets redder as it climbs out
of a gravitational field.
Part two, which I
will do on Thursday,
is to show that if there
is a global inertial frame,
there is no way for light to
get redder as it climbs out
of a gravitational potential.
You cannot have both gravity
and a global inertial reference
frame.
That's where I will
pick it up on Thursday.
So we'll do that.
And we will then begin
talking about how
we can try to put the principles
of relativity and gravity
together.
And in some sense, this is when
our study of general relativity
will begin in earnest.
All right, so let us stop there.
