Well, this is exciting.
A whole new, a new, whole new place.
Are you all. I'm not going to
be galloping around the place,
so I think it's really important that
you can see me and see a screen so
we adjust yourself as appropriate.
This is lecture 8 and is basically going
to be the last of the of of the the,
the, the maths bits.
We're going to finish off Part 3 today
and move on to physics next time.
Hooray.
And now there's quite a lot to get
through and this is quite an intricate one.
I don't think it's terribly deep,
but it is quite intricate,
so I propose to get on with
things without too much I do.
I can't think of any announcements,
but are there any questions to do with?
Stuff.
I thought they were in here from now on.
To have left other room behind and can you
hear me OK at that at the back end there?
No, I'm seeing thumbs up in a sort of
middle I I should aim to project suitably.
I think that I think this is
actually a designed lecture theatre.
So I think this will probably have
substantially better acoustics
than the rooms we've been in,
than the room we've been in so far.
Actually the poor astronomy too,
because they've been going from room to
room to room somebody with a days notice.
So you have, you know,
whatever the acoustics of,
of OF446 or 466, it could have been worse.
OK, where we got to last time?
Was talking about. Quite often.
We defined differentiation.
And discovered that the.
With this definition of differentiation, the.
The metric tensor had the property that
it's convenient derivative was zero,
and I mentioned that partly
because it's interesting,
partly because it justifies the our
ascription of the idea of the metric
providing a length which we draw true
for granted if that property that makes
that a reasonable thing to do to to take
the metric to be a definition of length,
and also because working that out was a a
useful way of using some of the mathematical.
Technology we had built up up to that point.
So the a couple of different payoffs,
you're distinct payoffs for that?
For that, now, the next thing
we talk about is a crucial idea,
the notion of geodesics.
A geodesic, as we'll discover,
is the next best is not the next best thing.
It is what it is.
A straight line is in a flat space,
and the thing that most natural corresponds
to a straight line in a curved space is
we have to talk about what autism is.
And to do that we have to 1st we
have to go back to the idea of what
a parallel transport was.
And at what happens if you parallel
transport a vector along a curve?
So imagine those.
Let's take first of all.
A vector field which is.
That's what's at each point in space.
There is a vector.
These both these solid arrows
defined that there's there's a AV,
let's say at each point in the space.
Now let's have a curve.
A curve there.
And that curve will have a a a
function Lambda of T or whatever it is.
So as we advertise T,
we move through the space along
the path in a particular way,
parameterized by the particular
functional form of Lambda that we choose.
Remember these things be a path,
a set of points,
and a curve at one of the most
multiple possible ways you could
parameterize that set of of points
as a function of a parameter T.
And as we learned at the beginning of,
at the beginning of this part.
If you have that structure.
Then you can define a vector
as the directional derivative.
Or the derivative as you.
As you vary the curve parameter, let's say T.
So that you can define the
notion of tangent vectors.
And this is a curve where the have
just tangent vectors which vary
as the as the curve moves about.
Now it's clear.
But as you go along this curve.
The vector field. Changes.
Very obviously.
And so,
so these.
Or or or.
More specifically,
if you parallel transport this
initial vector along this curve,
then it's then you you deviate from the
direction that you you you you you start off.
The curve, if you like, turns a corner.
That curve bends.
It bends in the sense that. And.
Um, do I have a diagram of this?
It bands in the sense that.
If you had a curve.
Like a little of that and you.
Parallel transported
along a curve like that.
Could I lock the focus of that?
What you focus.
Which books are good?
So this curve bends in the
sense it deviates from the. The.
Set of parallel transported digital
vectors here with this curve doesn't,
so we already have a sort of intrinsic
notion of what a straight line is.
Just from that picture,
if what we're doing is we are
parallel transporting this
tangent vector along the curve,
and it stays tangent to the
curve in that nice straight
line parallel transporting this
tangent vector along the curve,
it does not remain tangent to the curve.
That curve bends in a in an
entirely coordinate free way.
It depends on our definition of the
Korean derivative and that's and
that's of the parallel transport,
but it but given that it's a,
it's a bending curve.
And a curve which.
Doesn't do that like this.
It's a curve, which when you.
And parallel transport to
that curve along the curve.
So that's the derivative of the
that that that this tensor.
Is the derivative of the.
Field. Of. You.
Along the. Direction of the curve
if that C is parallel. Then that.
Directional derivative will remain zero.
That means a couple of goals to see PIN but.
What I've written down there is
the sort of mathematical version
of this idea of the vector,
not the parallel transport.
The tangent vector being parallel
transported in such a way that
it remains a tangent vector.
Right. No, that's very pretty.
It it it's. Four symbols,
and it means that, and I'm saying I'm
telling you it means a huge amount,
but it's not something you can very
straightforwardly calculate with.
So how do you calculate with that?
How do you find a curve which
satisfies that equation?
And here we get to safely integrate that.
That microphone is annoying me.
But you know we have.
Turning it down, OK. And.
OK, so I'm going to have to refer to this,
so don't make any index mistakes.
We remember that you is equal
to something like UJEJ.
The usual fashion. So the.
Just going to check.
And I'm going to remind ourselves that this.
Tensor. Remember that this nabla
you is a tensor A11 tensor,
in other words, something which takes.
Are. Vector shaped argument.
And one form shaped argument,
but we write it in an odd way.
So that the.
So the vector.
Argument. We rate down.
And here for for notation like for
reasons of notational convenience.
There's nothing deep there,
it's just handy.
So that which means that this.
Expression here nabla U.
You is really just nabla you with. The.
One of the. With the vector
argument prefilled in.
But since that's the argument to a tensor,
it's linear in that argument, so nabla.
You you will be equal to.
Nabla. You KEK. You.
Which that being linear in that argument,
that pops out to be now you. OK. And?
Nabla and remember he K you and remember
the notation shortcut that rather than
writing this double subscript here,
we we we we end up writing that as you.
Just. I think that's you, I said.
Sorry, that's you. You K now blah.
Key. You. Equals 0.
So, so this is just a sequence of.
There's just a bit of notational
trickery happening here.
There's nothing terribly deep. OK.
But we have an expression for this.
This expression here we we we,
we, we, we, we saw what that was
in the last lecture but one.
To that expression there.
And the danger?
So she makes sure we get the equations right.
It's going to be I.
Yeah. Great, you. G Nabla, G.
You will be equal to. Eugene GUI.
J. The. Aye. OK. Just recalling the
component version of the covenant
derivative that we saw last time.
And. That means we would be equal to
recalling the expansion of that UG.
Uh, you I comma. Gee.
Plus UG. You. Key comma
IGK.
Equals 0.
So again, we're just using the expression
for this this set of components that
were on last time, no time before last.
No. And. Well, good.
Well, haven't said any very
much about the, the, the, the,
the the curve that we're that this
curve we're talking about here.
We were rather that this curve here,
but it'll be this curve here.
Will be. Lambda T. For some,
for some suitable universe suitable
carefully some suitable curve Lambda.
And what we learned before is
that this tangent vector is.
The operator. DDT that will end
at the beginning of this part.
That was our definition of vectors
in this context. And so if we then.
Which means that this this UG.
The JTH component of that vector.
Is going to be you applied to. The XG. That.
The gradient of one of the of the.
That's the basis.
One form corresponding to
the basis vectors without
the component basis vectors.
Which equals DXG by DT.
Just the derivative of
that component along the.
So so so this is this is
fairly directly saying.
Given a curve.
You know that, that, that,
that, that line there, the.
The of the fact is just how much that.
The, the, the, the X or the Y
or the whatever component varies
as you move along that curve.
And that means that looking at
this other term, you I comma G.
Is just D by DX J. And.
DXI. ID T.
And uh. That means.
That. This expression here. Is. DX.
JPITT. And.
DX by DX.
G. Yeah. DX I by DT plus gamma I JK.
DX J by DT. We just completed
the XK by DT equals 0.
And that um. Combination.
There is incredibly deep ODT.
DX. I by DT plus gamma
IGKEDXJ by DT DX K by DT equals 0.
So we have turned this.
Elegant, but rather.
Impenetrable equation,
which we jumped out from considering what
the what a straight line consists of.
We've turned that into a second order
differential equation in the functions
in the coordinate functions X of of T.
And that's a second order
of differential equation.
So from the theory of 2nd
order differential equations,
you can discover that that
will have a solution.
And that solution is the the an
expression for the path followed
by that geodesic or by by the
by our our path which which
satisfies this property.
In the coordinates X of TXI of T.
I don't hear you worried faces
particularly, but OK. Umm.
And a bit bit bit bit a remark.
I, I, I, I'd, I'd said that this curve
Lambda was a suitable curve and it's not
an arbitrary curve that will do that.
Subset of curves which will form
suitable solutions to that,
but once you have a solution Lambda of T.
If Lambda of T. And.
If Lambda 2 is a geodesic then.
That means that Lambda of a T + B. Is.
It's very hard to write to that
person's angle. And that is.
Up and our parameter T which has that
property is known as an affine parameter.
And what that affine parameter
is doing it well what we've seen
that it is affine saying that
this this set of rescaling of that
parameter what that's that's.
Telling you is that you can rescale and
and shift the curve the parameters of
your geodesics more or less at will.
Which makes sense because if
you measure time in seconds.
And and and you have a a duty
that goes through that that that
describes the fall of a of a ball.
You can also whatever that that
that equation of motion is.
You can also rescale it and talk about it
not in in seconds since since midnight,
but in ours since 2:00 o'clock for example.
So you can change the,
you can rescale the the the parameter,
in this case time,
and you can shift the origin of arbitrarily.
And and that is and.
And one way of seeing that well
known we have think of talking about
that is that affine parameters are
defined so that motion looks simple.
If you were to decide that.
Part of the talk about seconds,
I'm going to write down my geodesic
equation in terms of second squared.
You silly,
because that would make the your
duties a really complicated.
Your expression for the parabola,
it would make you the parameter
rewritten in terms of of of
second squared from midnight.
It's going to be a mess.
Motion looks simple in the when
you're using the when you pick up a.
A form for the that your geodesic which.
It's not an affine parameter.
I'm saying to go around circles here,
but but there there's a.
There are more upper stigmatic
way of of saying about that,
but the the point,
the point of saying it is just to
mention this word affine and to trying
to link the that mathematical property,
that of those are a family of
solutions to the duties equation
to something more more physical.
Umm.
That is what it will have to see,
but for the moment about geodesics
is that if anything we can add.
Or. OK, then let's move on,
because now we come to the as were the
main event talking about coverture. Uhm.
And I think it's important
to have a clear idea of.
OK, another parameter is the time coordinate,
some inertial system.
And remember I said by national
system I meant as a system in which,
well for example you jumping up and down
a system in which Newton's laws work.
A system in freefall is international
system and then I think parameter,
the parameter of a geodesic is
a time parameter in in that and
we'll come back to that notion
of of that particular statement
of this implicitly later on.
Um given key points,
we can divide duties equation
as so the process of asking.
Where does if I if I throw
something in in a given space,
where does it go that,
that, that, that, that is?
You're solving the duties equation
to find an equation of motion.
So the answer to the duties equation,
the solution to the Nudestix equation,
is essentially an equation of motion.
Is the creation of a A line
in your space that's the.
Sort of vector version
of the of the equation.
That's simply the the component
version of of the same equation
with with with a a tangent vector
which is parallel transported along.
To remain being attention vector. Anyway.
Moving on talking about. Coverture.
I've drawn there.
The surface of a sphere,
for example the earth.
And if you imagine starting at the equator.
And pointing north.
Your point north.
What would have?
And you you, you, you,
you keep walking in a straight line,
in other words along a geodesic.
So a geodesic is,
I think I said, this area.
A geodesic is a straight
line in Euclidean space,
A geodesic is a straight line.
In the conventional fashion it's with the
with the signature of Euclidean space.
A straight line is the shortest
distance between two points.
In Minkowski space, or something with
the signature of special relativity,
a straight line is the longest distance.
June 2 points all of the alternative
versions of going from HB in
Minkowski space are shorter than the
in the straight than the straight
line and surface of a sphere.
A straight line is a great circle.
It's it's that that has the same
signature as you clean space and
so it's the shortest distance.
So I started the equator,
I point north and I walked NI.
Just keep walking in a straight line and
eventually I will get to the North Pole.
And at that point,
if I started at zero latitude
at that point I'm pointing
toward the 100 degree latitude.
I then start walking sideways.
And head down back down to the equator,
still pointing at the original
direction so I don't turn around.
I end up back at the equator,
this time pointing along the equator,
and then walk backwards through 90
degrees and end up back where I started,
but this time still pointing
along the equator.
So I haven't changed the direction
I'm pointing at any point.
And I've been walking in
straight lines in each case,
but of course when I get
back to where I started,
I'm pointing in a different direction.
If the angular went through
the the North Pole was smaller,
then this deflection would be smaller,
but it would still be still be reflection.
In other words, going for a walk.
And you're saying pointing the same
direction allows me to pick up some
information about the the curved surface.
If it's the fact that the earth is
curved that I can tell the story.
I go up North Pole going to
come down and back and clean,
I have to go all the way to the North Pole.
I could I could clearly do this in any
but any any any circular circular route,
but so,
so the the deflection of this.
This this vector.
Through a circuit is telling us
something about the curvature of.
It's telling us something obviously
telling something about the the
curvature of that surface in a way
which is completely independent of.
Components. So we haven't talked
about components or coordinates or
anything like that at this point.
So there's a, there's a,
a geometrical.
This is a geometrical thing we're
picking up by by this process.
So what we need to do now is find our way of.
Capturing that intuition in a
mathematical form to get some expression,
some something we can calculate with that,
we'll talk about the curvature.
So that's what we're talking
about of a space.
So the way we do that is a rather intricate,
but not fundamentally deep process.
With this poor like you.
So I'm going to set up a path
to go around. Let's have two.
Coordinates here.
We'll call this first one.
X Sigma and that's the,
for example X versus Y or R versus
Theta would it whatever you like
and that line there is the line
where X Sigma is equal to a.
I would draw another line through
the SpaceX Sigma equals a. Plus.
Delta E So that's just just moving
along a bit. And we'll have.
Another period of lines of
constant coordinate lines where X.
Lambda equals B.
Next line equals B plus.
Delta B well and and as you can
guess these this delta and Delta
B I'm going to make small later.
And so at this. Is the.
Tangent vector corresponding to the.
Sigma coordinate.
So that's the the direction in which
the Sigma coordinate changes and this.
Is the. Direction in which the.
X Lambda coordinate changes.
And we'll start off with a with a vector.
V at this point. 8.
And we'll do what we do,
what we described for.
In this case, we'll take this
vector for a walk around this path.
We'll take it to. To be.
We do get to see what you did
and we'll bring it back to.
A and discover that.
Quite unfortunate location of these.
There's a change in the vector which is
picked up as it goes around that circuit.
And we want to, and it's,
it's picked that up by virtue of
going round that curved space.
And what we want to do is work out.
The the the size of that
vector there and how it?
How how it picks up information
about the space as we go around it.
Now, sorry, by transporting the vector
I mean parallel transporting it.
Remember I said we've got in this
case I'm parallel transporting
that back up to there, transport,
parallel transport down here and
parallel transporting it back.
So parallel transport reporting in each case.
OK, in other words,
in such a way that the derivative is 0.
So we know how to do that.
So by parallel transporting.
The vector V. Um.
Along this this fresh leg.
That is as seen.
That the that it will be parallel
transport in such a way that the.
The vector of V.
As parallel transport
it along EE Sigma this.
Right here will be 0.
And what that means is that um again, V.
I comma Sigma.
Will be equal to minus gamma.
IK Sigma V. Key, and that's just the
the the expression for the equate,
derivative and component form
rearranged for the case where it's 0.
Right. Now what? What? What was
the result of that going to be?
What is the? The. The vector.
Let's call that vector at a.
What's the vector? The value
of the vector V would be at B.
What that is going to be the vector.
Well, it's listing with, so we're
going to go with individual components,
include the, the, the, the,
the component I component at a. Plus.
The changes in the the that that
that component as we move along.
So it would be the integral from A to
B of DV. Yep, I by DX Sigma, DX Sigma.
So I'm just I'm just integrating the
derivative along along the curve to get the.
The results would be IA minus.
I Sigma. The. KBTX Sigma.
Um.
And. This A to B is moving from X = A
to X segment equals A+ Delta A. VI.
At a. My integral of a A plus. Dot E.
DX Sigma. Evaluated. At.
Along this line X. Lambda equals B.
And.
That that's the the the neat
version of what I'm growing here.
That's what I've just written down
and it's clear that we can buy
this beam means get the value of
the component of V at B. From E.
We can get to the value of the item point
of C from B in the same way and so on.
And end up. With.
I I ask you to think what that might be.
That is one of the exercises just
I encourage you to think through.
One of the quick exercises at the
at the at the end and I can't
remember off the top of my head,
but this is an exercise in in
keeping track of of signs really.
But the. End result.
When you when you get
all the signs right.
And go through step by
step is this pattern of.
Of of plus minuses and this set
of of of of specific intervals.
So this integrating from A to B,
that's integrating from point B
to Point C, Point C to point D,
and point D back to point A.
Messy, fiddly, but not, but nothing more
exotic is happening than than here.
Now we can take advantage.
We had at this point taking
advantage of the fact that
just A and delta B are small.
So we can do that.
The way we do that is by.
And. Saying that.
The expression such as this.
For. Thank you.
Have a question I'm picking here.
Alright.
And. Yes, I think I'm, I'm,
I'm, I'm picking specifically.
This one here to illustrate.
The.
Showing OK.
The. IG Lambda
VGAT evaluated at X Sigma equals
E Plus delta east and we can just
use Hello theorem to discover
that is going to be IG Lambda V.
G. At X, Sigma equals A+ Delta A.
The body X Sigma gamma I.
G Lambda VG at X,
Sigma equals A+ order. And.
All I'm doing is Taylor theorem to
work out what this. This one is.
In terms of of this. And you can
see that now work out that so the.
The that that particular expression
evaluated A+ Delta E is going to
be that expression plus a bit.
So that when I subtract these two things.
What I'm left with? Is this? OK.
Um.
I end up with a simpler expression
for which I'm not going to put
which is is a numbered in the notes,
but between three 47348,
which I I encourage you to set through,
but we end up.
I I'm not going to go through the
the the index manipulations here
because they're not terribly edifying
that just you to watch me try to
copy indexes from from my nose,
but you but you have the notes and
you go through them very carefully.
But the point is that we end up.
With more things cancelling.
And an expression. For.
This the change in the ith
component of this vector,
in other words the.
The ith component of this change vector here.
In terms of the size?
Of this. The size of this?
The vector we started off with and this.
Rather fiddly expression involving the
christophel symbols, which as you recall,
you tell us information about the way that
the coordinate changes as we move around.
But now we'll look at this and
stare at it a bit and we realize.
This. Component here. Is a number.
I mean is, is, is a component of a vector.
But is a is a number. What number is it?
Is the number we get by applying our.
But by taking a vector and
applying A1 form to it.
So so this this one of the
basis one forms which is.
And one of the this is the ith.
Component of that vector.
We obtained it by applying that vector.
So V. Is equal to. Delta V.
Applied to the.
XI.
So this depends on A1 form.
It also clearly linearly
depends on this number DXL.
But this?
This. To write that. This displacement here.
Is going to be delta A. He Sigma.
So this displacement vector.
Is something which.
Is a vector which has as
its as its size delta E,
so that vector displacement.
There is also something that's
gone in to a thing to get this.
So this is also linearly dependent
on the size that size delta E.
The size there's a B which was written.
I put it here and it's
literally dependent on this,
the size of this vector we started off with.
In other words, this number,
this real number Delta VI is a
number which depends on one,
one form and three vectors.
In other words, this is the.
Would have written like this.
This number is therefore the
the what you get when you.
Take a a tensor which we'll call R
in different rieman, and plug into.
It's a 1/3.
To answer,
we plug in our basis one form vector,
started off with and there's two sizes
of the two vectors which describe
the shape and size of this detour.
And this vector here is
called the Riemann tensor.
It has components this.
And it picks up it.
It encodes information about the,
the, the, the, the, the,
the way that the vector changes as
we move it around around the circuit
in a way which picks up from the.
Christophel symbols.
And we put pictures value from the
console symbols in other words,
which we know which you already
know contain information,
encode information about the way
that coordinates change as you
move around this space question.
So you can create a full path, yes.
So and there will be because what we've
done is these. We'll set this up.
We've constructed this so that these
curves here that we're we're moving along
are the curves of constant coordinates,
so given a coordinate system.
Given a coordinate system and then
you will always be able to to
describe curves of constant cost.
Curves cost X and covers cost
Y or covers of constant R.
And covered, of course,
theatre or or or or whatever or latitude,
longitude.
So you'll be able to set up that grid.
And so, but but of course.
What coordinates you pick?
What coordinate functions you pick?
Will affect the number you get here,
but that's fair enough.
But this is clearly the.
This is clearly the the components
of the Riemann tensor in a
particular coordinate system.
You know this is the basis one
formed in that coordinate system.
These are the basis vectors
in that coordinate system.
So the this number is clearly
a coordinate dependent thing.
But what we've argued here is
that it has nonetheless what we've
indicated that we nonetheless
it's our a geometrical object.
Which just has coordinate
dependent components.
And.
Uh.
And I think that's. Yeah.
And and and and that's a very important,
very important important tensor which is
the includes the information about the,
the, the, the, the curvature.
Now there's another way we can
define that Mark 43 annoying me.
There's another way we can define the.
The Event Center,
which I'm going to just look.
And which I'm going to mention because we
come back to it briefly in a in a moment.
Which is that the human tensor can
also be defined in such a way.
I'm going to not go through
this in detail. It's not it.
It is. It's somewhat peripheral.
We'll write this down and then
explain what it is. And uh.
This is station here. Is a commutator.
Defined so that a B is equal to a B -, B.
E. And I'm not going to you know go
through and calculate it with this,
but the I I mentioned that just in order
to show that there is another way of of of
getting to the same point which in a sense.
Manifestly, it doesn't depend on
coordinates that this this process here,
you know, uses court,
the truffle symbols it was done in terms of
coordinate functions and so on. It's yeah,
it feels like it's somehow codependent,
but I I will assert that there's
another way of getting to the the
the the renter in that way.
And more or less parenthetically.
Um, no, if we stick with this version here.
Then you will recall, I hope,
that in a particular.
And.
In particular coordinate system.
You can calculate the values of the grateful
symbols using the driving from from the
components of the metric. So we can.
If this would have on the next slide.
It's not annoying.
Which I wish I had a slight seeing this,
but I don't really want to write
this whole thing down but. Uh.
OK, I'll do it. Is this is
355 RIJKL is half G. IL comma,
JK minus and and and other stuff.
No write the whole thing the whole day.
I should have. I always had a
slide of it and now that involves.
Is that a question there?
And I think I'll can we
should auto focus back on to.
So the improve that I'm not sure.
So this is the. Um.
The quick confusing this.
So if we recall the the the expression
we found for the Christoffel symbol in
terms of the metric and its derivatives
and do this fairly calculation,
we end up with an expression like
this and if we do this in a.
A local national frame,
which makes the whole thing simpler.
We end up with an expression for
the components of the of the
Riemann tensor in in that frame.
Now we can't do very much
with that in that frame,
and we can't use the comical semi
colon rule that we found last time,
which said that in our local natural
frame we could turn simple derivatives
into covariant derivatives that
works in with single derivatives.
Not, but not with second derivatives.
But what we can do is look at this state
different while and discover a number of.
Symmetries of the Riemann. Tensor.
If you swap the first the the the
the the first pair of of indexes,
you change the sign of the of
this component if you swap the.
Last pier.
That you get you the saying if
you swap 22 like that you get.
An expression which is that that that
that component is is the same so that
although there are N by N by N by
N components in the Riemann tensor,
there are lots of that a lot
of them are equal.
And you can also discover that
if you permute the GKLL GKG if
you promote the the 3rd 3.
Components.
You end up with something which is 0 again.
By staring at this and the point is,
although this is we calculate calculate
this in the local national frame where that
calculation relatively relatively easy.
It's merely tedious and error prone.
That's not a tensor equation
because of the second derivatives,
but we can workout.
On the basis of that,
we can work out these symmetries because
they don't involve any derivatives,
particularly don't involve
any second derivatives.
These are tensor equations.
So these,
although we calculate them using
this in the local national frame,
these are not specific to that frame
and are and correspond to to tensor.
Identities, right?
We are almost. Out of time.
In order to not run over I could I I
do really want to finish this chapter.
Today I'll go through the next part
fairly quickly and I don't think
there are any objectives which depend
on the details of this section,
but do you want to just talk
through them quickly? Um.
If you think back to chapter one,
we have I'm going to this in one minute.
Actually, I'm not. I'm.
I I will pick this up again
again next time I want to try.
At least I have get to go to
the end of the of the chapter.
We remember that diagram from chapter one,
but we were things which were in freefall.
A significant difference apart
from each other if freefall
toward the center of the Earth.
They are distance separation between
them decreased as time moved on,
and the 2nd derivative of that
separation was nonzero even though
the things weren't accelerating.
These as we'll come to discover,
we haven't.
We don't know this yet because we haven't
talked about about general relativity yet.
These paths are geodesics.
The the path that's something
follows as it falls it geodesic and
so the fact that the separation
between the UCS changes as the as
the objects move through space-time.
Is is is telling us something about the
coverage of the space they're moving through?
And So what we discover is through another
similar construct we can talk about.
Appears of geodesics and talk about
the vector. Of course you know joining.
A family of duties.
I'll talk about the vector joining
points on neighboring eugenics
which have the same parameter,
and ask how does that vector change.
We can say that what we have here is also.
A circuit in the space so the remain
coverage tensor is going to tell us how
this connecting vector here changes as
we move through that I'm I'm making.
That's a rather handwaving
remark I at this point.
But the end point is that.
Jumping to the end,
is that the the second derivative of?
That connecting vector joining
points in two neighbouring geodesics.
Ends up. Depending on the.
The change of vector to the geodesic the X it
depends on the on how far apart they start,
off the side of the of the connecting
vector at the beginning and on the
Riemann curvature tensor.
So the recovery tensor is not just
telling you about the shape of the well,
it is telling you the the curvature,
the shape of the.
Space you're moving through.
And one of the ways it does
so is by saying how?
How quickly.
These two.
Two objects.
Two things moving along along geodesics,
how quickly that connecting vector changes.
That's a tidal force.
That's things like these two things falling
towards the center of the Earth and getting
closer together without accelerating.
I think there's basically no objective
depend on the details of that.
I, I, I encourage you to look through
the corresponding part of the notes,
but I I think that is all I'll see with that.
So I think we can therefore declare part
three finished or go into Part 4 next time.
I think the overview video
is up on the stream site.
If it's not,
I'll check and I'll make sure that
the notes are up promptly and I'll