Hello everyone and welcome to lecture 7.
I'm fairly sure we are still in good time.
Although we can't hang about
with the rest of the section,
the plan is to is to cover.
If we spend 9 lectures 910 and
11 on part three, that's good.
So I my hope is to deal with
the rest of of this chapter,
this part in this lecture and and the next.
So unfortunately we are both this is,
this is Scotland, we have sunlight.
Known again from awkward angles and
we don't seem to have blinds on
these on these windows unfortunately,
so you may have to squint but look cool.
It's autumn where we got to last time
when we finished off this section 322.
And these are the things we covered there.
We talked about the essentially we
talked about the covariant derivative in
flat space and how we could define that
and the covariant derivative in flat
space is the answer to the question,
how does this field vary?
As you move across the space,
given that the space is flat and
so the change in the.
We calculate the change in the.
Um.
In the field has to take account of
the fact that the basis vectors will
be changing across will potentially be
changing across the space for the example,
being the motor the motivating example.
Being the vectors the basis vectors in
in the plane in the basis of spherical
Polaris or of of plane polarized.
Where as you as you're aware the
the the the direction of the of the
radial basis vector changes and the
direction and size of the tangential
basis vector changes as you move
across the space.
And we were able to deal with
that with the current derivative.
I said that for every vector V vector
vector field V there's a tensor.
With one rank higher, called Nabla V.
The components of which tell you
how the vector changes.
The vector field changes as
you move across the space.
And we were able to find the
components of that tensor.
With this rather strange notation,
I think that's the last,
basically the last bit
of notational annoyance.
Are we subjecting you to this
VI semi colon J?
Which is the iconology,
which is just the straightforward
derivative of the the.
That's DV,
the derivative of the ith component of
the of the vector with respect to X J
+ a term involving this Christoffel
symbol called the social connection,
which is basically the thing that encodes.
The way in which the basis
vectors change across the space.
And I said that though I didn't
go into Labor line detail,
there are corresponding expressions for
the covariant derivative of function,
which turns out to be just the.
Derivative. That the gradient operator
we learned about a few lectures back.
And there's corresponding expression for
the great derivative of one form and
associated ones for higher rank tensors,
which we're not going to go
into because the the key idea,
there's nothing really new there.
The key idea is in this expression here,
and that's what we're actually
we're most often use.
So that's what we got last time.
What we're going to do now is 1 bit
of extra calculation and then move
on to the apparently much more exotic
question of how do we do the same thing.
In a case where the space is curved so
so the basis vectors are changing in a
curved space rather just a flat space.
And we'll discover it actually less
less hard than you might anticipate.
But unlikely. Rather annoying.
I think I'll have to put in some
sort of request for blinds to be.
Fixed in this room.
OK. Any questions about that?
We've got to OK, that was just a revision.
But Papa. OK, I'll go back to.
This.
Now what would you do now is work
out how to differentiate the metric,
and that will be illuminating.
In a way. It's illuminating because
it shows a calculation happening,
and it'll be illuminating
because the result is 1.
We'll briefly use later on.
Is that impossible to see?
Really hard, OK?
OK.
So we have a. So we have a vector field V.
Now as you will recall the the
the metric gives us a way of
associating with any vector.
I want phone. In the very
straightforward fashion. One form is.
Um.
We. We.
Metric G.
With. This vector what are the holes?
One of the slots filled in with the vector
and leaves another vector shaped hole.
The result is different one form,
so that's the, that's the one
form corresponding which is sort
of dual to the vector V and it's
mediated by the the metric.
And as we saw that had a has the effect.
In component form of saying that
the components of this one form
VIAGIG. Fiji. Where?
The component of the metric
of the metric vector.
Are these the components of
the corresponding one form?
Are these now?
What happens if we differentiate?
This one form.
What we'll do is we'll
differentiate the left hand side,
we'll differentiate the right
hand side and see what we get.
Now this is our geometrical equation.
In other words,
it's a coordinate independent equation.
There's no, there's no,
although we can talk about the
coordinates of this equation.
This by itself is a coordinate
independent equation.
So we can calculate with it in any.
Coordinate system we're like.
So we pick a coordinate system in
which the calculation is easy.
Of course you pick,
and we're not doing it here.
Here, anything that you haven't
been taught to do previous
stages in your physics education,
you pick coordinates so that
the calculation is easy.
Here the the coordinates we pick are
going to be Cartesian coordinates,
so we're going to calculate with
this using Cartesian coordinates
and in Cartesian coordinates.
The special thing with Cartesian
coordinates is that the basis
vectors are constant.
Across the whole space.
That's when we recorded creating coordinates.
The X&amp;Y basis vectors are are what
they are across the whole space,
and what that means is that the
christophel symbols for Cartesian
coordinates are all zero.
The Christoffel symbols pick up
the change in the basis vectors
as you move around the space,
so the basis vectors don't change.
The Christoffel symbols are zero,
so conveying differentiation in
these coordinates is trivial.
Um. What this means is that.
Uh. And in other words.
In these coordinates.
Equivalent derivative of the vector V.
Is just. DVI by DXJ. E.
Aye without the corresponding term
which involves differentiation of the.
Basis vectors because
that derivative is 0. OK.
So so you can either think
of this as being the.
The the the the expression for the
current derivative with the covenant
with the Christoffel symbol zero.
Or you can think of as just
libras rule with the second term
which would be IDE IDE I DXG.
Disputing because DVD I DXG.
Because that's zero in Cartesian coordinates.
If we even had, you didn't have that
Shadow Cross. It would be better.
And what that means?
Is that if we ask what is the?
And what happens if we apply?
That vector. THG.
To the. Metric metric. That's.
By two and DVI by DX JEI.
And because the metric is a tensor,
it's linear in uterus arguments,
so that's a DVI by DX JPG.
OK.
Now. This. Thing here.
Is. And. If you think of it.
Well, it has the property
that it that is dual. To the.
Basis vectors EI. Because
GEI. EJ. Is equal to well delta.
IG's if you could do one
when the statement and.
00 otherwise that that's
we know that to begin with.
In other words,
GEI.
Is equal to.
One of the beaches one forms.
In other words. This expression
here. Um is. DVI by DXJ.
Who written this? Yes.
Omega I is DVI by DX J Omega I.
Summed over I.
Something over I because this is.
The two eyes are both raised to the
instantiation convention doesn't apply,
so I've got to explicitly say
that what we're summing here.
OK, so that looks rather
strange expression. OK.
Now let's look at that.
That's what we've done by.
Especially differentiating
the right hand side of this.
Of of the situation here.
If we now ask how, what happens if
we differentiate the left hand side?
Then what we get. Um.
Is. The level that you.
32. V tilde. So we're looking at this again.
No difference in the left hand side there.
That's differentiating.
The. I what we got what we got I
because V is our our one form so it
will have some components in the
one form in the one form basis.
Since the. Basis vectors.
Are constant in this basis in
the in this coordinate system.
Then the one form is a constant
in this coordinate system,
so the one form is also.
Do not vary as we move across the space,
so again this ends up being DVI.
By the XG. Omega. Aye with, with.
No D Omega by the XJ term.
Because the basis vectors are.
Constant.
But. In these coordinates. The.
Special thing is that.
The components. No.
I think I have somewhere
explained why this is obvious.
I'm you know I'm have to be
recalled but have thought but
the in these coordinates.
The components of base of
vectors in one forms are this
are equal in these coordinates.
Only in creating coordinates
and what that means is this.
And this. Are equal, so DVI.
Raised by the XG and DVI
lowered by the XG are are equal
so this is equal to this.
And what that tells us.
Is that they are equal in these
in this coordinate system.
But if they're equal as components.
In in one coordinate system.
Then they are equal as tensors
in all coordinate systems.
So two things are equal. In.
One basis. Component by component,
and that's telling you that they
point in the same direction.
They are the same vectors.
And so you've gone from doing
the calculation in a nice easy.
And coordinate system Cartesian
vector Cartesian coordinates.
But it would draw geometrical conclusion
that these two things are equal as vectors.
But that means that they're equal,
independent of the coordinate system.
So we've picked a nice coordinate
system to make the calculation easy.
But still ended up with our
geometrical result which is that the.
Um.
This.
Um. Derivative of the?
Of this one form V. Is equal to.
Um, this thing here? She.
As a geometrical result.
So.
Next right.
10.2 here. Look of it.
So where do we go next from there?
The um.
The component form of this
is this expression here.
And.
VI is equal to
GIJVG.
And the. Component form of. This expression.
Umm.
Yep. Is VI semi colon G? Is equal to G. IG.
VK. To him I key. They're calling G.
And you may have to, you know,
see that a bit to reassure
yourself that that's the case.
So.
This isn't trivial. We know that
there is a given this tensor V VK
semi colon G we know there's some. A
tensor, which is what you get
when you lower the indexes.
What this what worked out is the chance
that you get when you lower the top
index in that covariant derivative.
The chance you get is this covariant
derivative of the corresponding one form.
OK. That's good.
So what we didn't do is we didn't get that
expression there by differentiating that.
It looks like this expression
is just the derivative of that,
but it's not.
We got that by a different route
this calculation here.
So what do we get?
Who would differentiate this?
That's just life.
That's the rule, really.
So V. I semi colon.
Ugg. Is equal to
GIK semi colon JVJ. Plus.
GIKVK. semi colon. Gee.
And that's just liveness rule.
Like this rule where you you
you differentiate a product
by differentiating one you
learned about in school.
It's the I think,
I think it's all generally.
So if.
So this we obtained by this argument.
This we obtained by differentiating that.
If these should be equal.
That term is the same.
And this term must be 0.
So what we have done is
discover that this that well
this term is going to be 0,
so this term must be 0.
In other words.
GIK, semi colon G. Is equal to 0.
It could be a derivative of the metric.
0. What does that mean?
If we ask ourselves what? And.
Consider the.
And. Inner product AB. That's GIGA.
BG. And if we differentiate that, so ask.
With the driver of that of that number.
Libraries. Really. Again. GI. G.
So the the the case component of that.
Semi colon key AIBJ difference in that
1 + g I. GA semi colon K. PG plus G.
IGE. IB. G. key. But if this is 0.
Then that tells us.
That as we move around to the,
the as we move around the space. The.
The way that the inner product varies.
Is purely due to the way that
the to the derivatives of A&amp;B.
Is not due to the inner product is,
which you feel like is the.
The size it describes the the the
the size of this of this object.
So it is a dot A for example,
it would be describing the
side side of the of the vector.
What they're telling us is that as that
varies as you move around the space.
The change in that is
purely just changes in the.
The vector and not changes in the
coordinate in the underlying coordinates.
So something.
So in other words this is telling us.
This is in a sense what gives us
license to think of the metric as
being a measure of the size of the
of of a vector at different points.
Because this is telling us that when
the the the metric between 8:00 or E
changes as you move around is not just
an artifact which of the coordinates
changing under you it's it's it's the,
the, the.
The, the, the,
the vector,
the vector field changing rather than the
an artifact of the coordinate change.
So this is.
This in a sense is what gives us
license to talk about G as a metric as as,
as, as a length,
we and and also via the the dot
product as a way of talking about
the angle between two things.
To that angle is at a well defined thing.
The direct question in the sense
that you say that the derivative
of the number is not.
The derivative of somebody?
No, it's not. Yes, so.
So say that again.
We know that a dot B yeah wouldn't
it be vectors is a number yes.
And then you differentiate the number
but you didn't get 0 differentiated it.
Yes so that it was derivative of a
that would be the length of the vector.
So I as you move around the the space
in different parts of the space this
electric fuel see might not only change
in direction might change the length.
So the derivative of of a dot A
would be the rate the derivative
of the length of that vector.
So it went from here to here.
Then that derivative would be that
that that length will be changing.
And what this is telling you is that
that's because the vector has changed.
Length is not just a thing
about the coordinates.
So that's an important thing and we'll
come and and that I think also shows
that there are two things that's.
That was three things.
It shows that what I said about the vector,
the metric being useful being a length,
the fact that the covariant
derivative of the metric is 0,
and we will pick up later.
But there's also shows the way that
you can do calculations in this.
Sort of context by doing things
like picking the right coordinates.
And this sort of trick that if
you can come up with a geometrical
result in coordinates.
That these two vectors were equal,
then that geometrical result
is coordinate independent.
Because it's no longer just a
coordinate dependent number.
OK Umm any more to say about that?
Ohh yeah yeah the last thing which I
won't work out is I won't go through
the steps for because it's just
rather tedious is but it's it's it's
important but but the derivation is
not is not particularly interesting is
that you can work out that the before.
We were introduced the Christoffel
symbols at the beginning of near
the beginning of this chapter.
I showed how you could work them
out by working out the.
By explicitly working out the way that
the basis vectors in plain Pollers moved,
you're changed as you moved around the space.
I then said, oh,
and these coefficients are called
the Christoffel symbols.
Match these two equations up and
you can work out what the symbols
are for playing pollers.
They're just the components of the of
the of the expression that we got.
So I'll just, I'll just jump back just.
Think about for blank looks there.
Yeah.
And we saw that.
It's not nobody terribly easy there
but but you'll see that in your
notes that this is the way that the
basis vectors of pain pollers change
over the over the over the plane,
and we can with engine identify that
this 1 / r is the Christoffel symbol
gamma R Theta Theta. That one's gamma.
Theta, R, Theta and so on.
So we can just match those up
and discover what the shuffle
symbols are by that process.
But it's also possible to do this rather
more mechanically and discover that the
and I will copy this down because I
don't want to necessarily get it wrong.
This equation says 36 that the the.
Ijk the IJK is at half.
Gil.
GLK. comma key plus.
GKL. G. Minus. GGK.
So that given the derivatives of the metric.
You can just turn the handle and.
To churn out the values of
the Christoffel symbols.
And I'm not going to ask you to remember.
Memorize that.
But you probably will end up memorizing it,
given that you do enough of the exercises.
And that's a nice. Obvious.
I mean, I'm making your promises here,
but that is a nice sort of.
It's a dull but quite a a nicely
contained exam question to
get you to turn that handle in
and and avoid falling asleep.
It's not pedagogically terribly interesting,
but the same as far as the Examiner
and Peggy were interested in
getting some that will,
that will.
Produce our result.
So I I heartily encourage you to look
at the exercises which are covering
that sort of that sort of thing.
And there's lots of them in
the IT would refer to lunch,
but there's several of them in
the exercises for this part.
And it just gives us practice.
Um, so before we go to the next bit,
we're making good time here.
I'm. I'm. It's so much easier
to keep keep your time face
to face than it is in zoom.
It's so much nicer.
Because I can see you go or
and and and and I can see
you and you can see smile.
It's just I can work out.
If things are are.
Keep keeping up on it after
being neurotic and any
questions before go on.
No, OK.
Um.
OK, so that's the thing.
I just wrote down the turn
the handle expression for the
truffle symbols in terms of
the derivatives of the metric.
And those are the key points from
this section. The key thing the
metric tensor this tensor that we've.
That, I said, was somewhat arbitrary.
But because of its of of the way in
which we use the the the the metric,
well, it is arbitrary.
In the sense that.
You pick a metric tensor,
and you pick the shape of the
of of the species describing.
But in mathematical terms
it's somewhat arbitrary.
But because the tensor is what we
the metric tensor is what we use to
map to turn vectors into one forms.
It because of that.
The problem it has the property.
That the, the, the, the.
The convent derivative of the metric
metric is 0 and which means it's
legitimate to use it as a measure of length.
And then this expression here,
which I omit simply because it's
tedious to actually calculate
and it's not interesting.
But the details are on shoots.
Or think in Carol's most you know,
most most fat GR textbooks will
have a derivation of that if you're
interested or don't believe me, OK.
Next.
Umm.
How old are you?
Right. What we now want to do, as I said,
we've now talked a bit about the
convenient derivative in flat space.
Not necessarily Cartesian
coordinates, but flat space.
Meaning one where Euclid's
geometry works and so on.
Or when Minkowski geometry works,
so Minkowski space is also a flat space.
And I I'm I'm here giving a rather
hand waving definition of flat spaces.
We will later discover what you know,
I'm more more precise
definition of our flat pieces,
but I hope you have a a fairly intuitive
notion what a flat space is at this point.
So what we want to do is also
differentiate things in non flat spaces,
for example on the surface of a sphere or.
In the cosmos or something.
So these are not flat spaces.
The nucleus, the parallel axiom nucleus,
doesn't work on a sphere,
and it doesn't work in the cosmos.
So how do we?
Port what we have learned
about conveying differentiation
in flat space into covering
differentiation on a curved space.
What we do is we jump up and down.
Because I said that it's easy. Well,
it's important to pick your coordinates.
If you do a calculation in the
right coordinates, it's easy.
So what you do is you've picked up well.
So, so so how do you change coordinates?
You change coordinates by picking
a Lambda that, that, that,
that this coordinate transformation matrix
that we learned about back in Part 2.
And you have 4 by 4.
By the case of,
you have N by N numbers there,
and as long as the matrix is invertible,
and as long as.
Uh. It's not singular.
And I think the same thing.
And then you have a number of
degrees of freedom so you can pick.
A transformation from whatever
chords you're starting off with.
On your sphere or whatever into
a flat into flat coordinates.
And you can do better.
You can you can pick coordinates
where not only is the.
Is the metric in that new?
Coordinate system diagonal
meaning it's it's flat,
but the derivatives of those of
those components are also zero.
In other words, you can pick coordinates.
In which the metric? Is.
The metric of special activity,
that's the minus plus,
plus plus or plus, minus, minus, minus,
depending what your convention is.
Plus some of which is a
second order in the. Um.
In in, in, in in the in
the component functions.
And that's very easy to do.
You just do that.
You're in freefall, your metric diagonal.
Diagonalizing your personal metric is easy.
You do it every time you jump up and down,
given you jump up and down with the
late more often than once a day,
which is very important to do.
Um. And what that means is we
can start to talk about um.
So I I said we can't do this.
In a moment I'm going to explain
why that's useful to do.
Because now we're going to talk
about differentiation and why
it's hard in a curved space.
Ohh, that, that's, this is called
the local flatness theorem.
The theorem that you can do this is
called the local flatness theorem.
And the the the frame that you get,
the coordinate system that you get when you
do this is called the local and national
frame because it's a it's a natural frame.
It's in freefall.
It corresponds to the natural
frames of special relativity.
So we know what that is.
We the local national frame of
freefall is something we understand,
and it's the freefall frame that
we're talking about back in part one.
OK, so it's a good, nice,
it's a nice good frame.
And this in this sense is one
reason why we made a fuss about the
local national frame in part one,
because it is the framework
which has these nice properties.
That it's it's just special relativity.
No. You know how to differentiate things.
You learned that in school?
That I trust that looks completely familiar,
right? What's happening here? Is that you?
Take up your difference your function.
You ask what's the difference
between the function at the point X,
the function a little bit
further along the opposite.
Divide that by how far you've gone
and take the limit as that goes to 0.
Now that's nice and well defined.
But it depends. On. A minus sign.
And it depends on division.
In other words, that depends on
you being able to say what a
function at this point minus a
function at different point means.
It depends on it being possible
to divide that by a number.
And neither of those things have we
got yet when we're differentiating our
vector field as we move across a space.
So what we have to do is define
what subtraction means for our
vector moving around space,
and define what this divided
by each could correspond to.
So that's what we're doing.
OK, it's just that those two
steps are a little bit hard.
What the the second step is quite easy
once you've got the first step right.
And I know there are there are multiple ways
of talking about the derivative of our.
Of of a vector.
And the we were going to do it
is the the Cuban derivative,
in particular the derivative with
the metric connection so-called.
There are things called leader of this,
there are things called flow derivatives,
are things called 2 forms, and so on.
So there's more than one way of doing this,
but this is the way that's useful in in
GR most useful, most immediately useful,
most commonly useful in GR.
And it relies on the notion
of parallel transport.
And parallel transport.
Age where? I take a vector.
And I see. May I borrow this?
Another vector is parallel to it.
If we put them nearby together,
the fairly obvious parallel
mean that's parallel.
Those two vectors are parallel.
Thank you.
Now if we see that red parallel there.
And to there, and to there,
and to there, and to there,
to there and to there.
Incrementally we can infinitesimally.
Rather, we can end up with a
definition thereby of how you
transport a vector from here.
To a collector somewhere else which is
parallel in the sense I think it is
parallel at each infinitesimal separation.
And what that means?
Is that?
So something like this.
So we can move a vector along a path,
keeping it parallel at each point,
and we end up with these two
vectors being parallel even though
they're separated.
And that matters because.
If you remember when we defined
vectors on the manifold.
I said that the death the definition
of the tangent vector in the manifold.
The tangent vector space on the
manifold was in a space which was
attached to the manifold at one point.
And that space? That's the tangent plane.
TP at M is the tangent plane of the
of the manifold M at the point P.
And the tangent plane TQM which is
tangent plane over the manifold am at
the point Q has nothing to do with this.
Which are different species,
so you can't subtract that vector
there from that vector there.
Which is what we want to do. Super.
What we can do is take a vector here.
And parallel transported along this
curve Lambda. Until we get back,
until we get it into for some Lambda
that goes through both points until
we get it into the same space here.
At that point there are two
vectors in the same space.
So we can subtract.
So we've done the first part of
what we wanted to do with our
definition, differentiation.
We've discovered a way of subtracting
a vector here. Of tracting.
The vector here from the vector here,
which is a little bit further along the path.
And we get a vector as the answer.
We know how far we've had to transport it,
what the difference is in in T so
we've got our H in the bottom line.
We can divide that that vector there by
the amount we've had to do to do this.
At that point we are talking
about the derivative. And and.
So that means that we have are able to
to find the the two steps involved in
defining your derivative in a curved space.
OK, because this is only depends on
the presence of this. Curve Lambda.
No. You know, we haven't said very much here.
Because what we have been vague about.
Is. What this? What we mean by parallel,
I mean what I said was I hope persuasive
that thing to parallel if they are, you know,
if when they're nearby they have the,
they're obviously parallel.
But that is, that's that's that statement.
Has an obviously in it,
so it's not really a mathematical statement.
But we can be precise about what
we mean by parallel by saying, OK,
two things are parallel if in the
local national frame. They are.
Have the same coordinates, so two to so,
so 2 vectors in the local national frame.
Are parallel if the other thing components.
That gives us a definition of parallelism,
a place definition of parallelism.
Which is enough to let us define this.
This. The the the difference.
You let us go from here to here,
and that, and and.
That means that we end up defining.
Great differentiation. In.
And. In the coordinate system
of the local national frame.
But the local metal frame is flat.
And that means that we know how to do that.
So we've gone from defining this
in the recovery space to doing the
same parallel transport thing in.
Our flat space. Which we can do.
So we just have to import what we learned
in the previous section to this case.
And that means that the.
100 hundred freeze this,
and that means that. Well,
that means we're actually finished.
Because we don't have to do
anything else. We can just.
Do I have a slide for that?
The the the definition of the the the.
Greater TV? In. Approach piece
is again just fee I semi colon G.
Umm.
Now, I've I haven't gone through
every step in that whole derivation
because there's more than one way of
of you creeping up on that conclusion.
But I think that I hope
that the the point is clear.
First of all, the parallel transport is
key and in order to give a definition
of what we mean by parallel transport,
we can use the local national frame.
Those two steps are the steps that
allow us to instance step back to the
flat space that we understand already.
But I think it's important to do you know
that you appreciate where those what
what the inputs to that conclusion are.
Um. And and I'm not gonna go through it, but.
You know there are two
further remarks there that.
In the local national frame.
The is it. It's flat,
so the console symbols are zero,
so in the local national frame.
The item Icon G is just VI comma G.
In the local initial. But that's.
That's true for anything,
so it's true for IgG semi colon key.
Well, in this case be IgG IgG comma. Key.
Which because of the what we worked out.
Yep, which because of the definition of the.
Local of the local flatness theorem.
We were able to see not only that
that the vector that the metric
in the new coordinates was zero,
but the derivatives of the metric
in these coordinates was also zero.
So we discovered that this is 0.
In the in the local national frame.
But. We've got tensor equation here
we've got GIIG semi colon K = 0.
That's true as a component calculation.
But seeing this tensor,
the remainder of this tensor is 0.
Is. A geometrical statement.
So although we worked out in these
coordinates the nice easy coordinates,
it's true in general.
So if again worked out.
That the.
And that GGK.
Key is 0. In.
In any coordinates.
So this is one of two comical,
cynical rules.
The other one is insensible.
Interesting. This is a mathematical trick.
We've done the calculation in A-frame,
in which the calculation is easy.
Discovered that by doing so we
have a geometrical statement.
Realize that that is therefore true
in not a coordinate dependent thing,
but our geometrical thing,
and so we're able to go from that
in one frame a special frame,
to the same statement in any frame
as a geometrical statement.
Umm. Right. In the last
couple of moments, I'll just.
No, but it's more sensible if
we start thinking 2.4 next time,
and I'll aim to get through the the
the remainder of this part in the
next lecture which we like to eat,
and that will give us lectures
910 and 11 for Part 4, so.