Uploaded by YaleCourses on 22.09.2008

Transcript:

If you've got a function f(x), you know the value

at some point, and you want to go to a

neighboring point, a distance Δx away.

You ask, "How much does the function change?"

And the answer is, the change in the function

Δf is the derivative of the function at the starting

point times the distance you move.

This is not an equality, unless f happens to be a

straight line; it's an approximation and there

are corrections to this. That's what all the dots mean;

corrections are proportional to Δx^(2) and

Δx^(3) and so on. But if Δx is tiny this

will do; that's it.

But today, we are going to move the whole Work Energy Theorem

and the Law of Conservation of Energy to two dimensions.

So, when you go to two dimensions, you've got to ask

yourself, "What am I looking for?"

Well, in the end, I'm hoping I will get some

relation like K_1 + U_1 = K_2 +

U_2, assuming there is no friction.

U_2 is going to be my new potential energy.

Then, if it's the potential energy and the particle is

moving in two dimensions, it's got to be a function of

two variables, x and y.

So, I have to make sure that you guys know enough about

functions of more than one variable.

So, this is, again, a crash course on

reminding you of the main points.

There are not that many. By the time I get here I'll be

done. So, how do you visualize the

function of two variables? The function of one variable,

you know, you plot x this way and the function along

y. If it's two variables,

you plot x here and y here and the function

itself is shown by drawing some surface on top of the xy

plane, so that if you take a point and

you go right up till you hit the surface, that's the value of the

function, at that point xy.

It's like a canopy on top of the xy plane and how high

you've got to go is the function.

For example, starting with the floor,

I can ask how high you have to go till I hit the ceiling.

That function varies; there are dents and dimples and

so on, so the function varies with xy.

Another example of a function of x and

y--x and y could be coordinates in the

United States and the function could be the temperature at that

point. So, you plot it on top of that,

each point you plot the temperature at that point.

So, once you've got the notion of a function of two variables,

if you're going to do calculus the next thing is what about the

derivatives of the function. How does it change?

Well, in the old days, it was dependent on x

and I changed the x and I found the change in the function

divided by the change in x and took the limit and

that became my derivative. And now, I'm sitting in the

xy plane, so here is x and here is

y; I'm here, the function is

coming out of the blackboard. So, imagine something measured

out but I'm sitting here. Now, I want to move and ask how

the function changes. But now I have a lot of options;

in fact, an infinite number of options.

I can move along x, I can move along y,

I can move at some intermediate angle,

we have to ask what do you want me to do when it comes time to

take derivatives. So, it turns out,

and you will see it proven amply as we go along,

that you just have to think about derivatives on two

principle directions which I will choose to be x and

y. So, we're going to define one

derivative, which is defined as follows.

You start at the point xy, you go to the point

x + Δx, the same y and subtract

the function at the starting point, divide by Δx and

take all the limits, Δx goes to 0.

That means you go from here to here, you move a distance

Δx, you'd find the change in the function,

you take the derivative. As you move horizontally,

you notice you don't do anything to y;

y will be left fixed at whatever y you had;

x will be changed by Δx;

you find the change; you take the derivative and

take the limit that is denoted by the symbol df/dx.

[Reader note: Partial derivatives will be

written as ordinary derivatives to avoid using fonts that may

not be universally available. It should be clear from the

context that a partial derivative is intended here,

since f depends on x and y.]

So, this curly d instead of the straight d tells

you it's called a "partial derivative."

Some people may want to make it very explicit by saying this is

the derivative with subscript y.

That means y is being held constant when x is

varied, but we don't have to write that because we know we've

got two coordinates. If I'm changing one,

the other guy is y so we won't write that;

that's the partial derivative. So, you can also move from here

to here, up and down, and see how the function

changes. I won't write the details,

you can define obviously a df/dy.

So, one tells you how the function changes,

with x I should move along x,

and the other tells you how it changes at y as you move

along y. Let's get some practice.

So, I'm going to write some function, f = x^(3)y^(2),

that's a function of x and y.

You can write any--let's make it a little more interesting,

plus y, or y^(2);

that's some function of x and y.

So, when I say df/dx, the rule is find out how it

varies with x keeping y constant.

That means, really treat it like a constant,

because number five--What will you do if y was equal to

5? This'll be 25 and it's not part

of taking derivatives because it's not changing,

and here it'll be just standing in front doing nothing;

it will just take the x derivative, treating y as

a constant. You're supposed to do that here;

this is a derivative 3x^(2)y;

so that's the x derivative.

Now, you can take the y part, y^(2),

yes, thank you. Then, I can take df/dy,

so then I look for y changes, this is 2y here,

so it's 2x^(3)y + 2y;

so that's the x derivative and the partial

y derivative. Okay, so then you can now take

higher derivatives. We know that from calculus and

one variable, you can take the derivative of

the derivative. So, one thing you can think

about is d^(2)f/dx^(2), that really means take the

d by dx of the d by dx.

That's what it means. First, take a derivative and

take the derivative of the derivative.

So, let's see what I get here. I already took df/dx.

I want to take its derivative, right, the derivative of

x^(2) is 2x, so I get 6xy^(2).

Then, I can take the y derivative of the y

derivative, d^(2)f/dy^(2),

that's d by dy of df/dy,

if you take the y derivative there.

You guys should keep an eye out for me when I do this.

I get this. But now, we have an interesting

possibility you didn't have in one dimension,

which is to take the x derivative of the y

derivative, so I want to take d by dx of

df/dy. That is written as

d^(2)f/dxdy. Let's see what I get.

So, I want the x derivative of the y

derivative. So, I go to this guy and take

his x derivative, I get 3x^(2) from there,

so I get 6x^(2)y and that's it.

So, make sure that I got the proper x derivative of

this, I think that's fine. Then, I can also take

d^(2)f/dydx. That means, take the y

derivative of the x derivative, but here's the

x derivative, take the y derivative of

that, put a 2y there, and I get 6x^(2)y.

So, you're supposed to notice something.

If you already know this, you're not surprised.

If you've never seen this before, you will notice that the

cross derivative, y followed by x

and x followed by y, will come up being

equal. That's a general property of

any reasonable function. By reasonable,

I mean you cannot call the mathematicians to help because

they will always find something where this won't work,

okay? But if you write down any

function that you are capable of writing down with powers of

x and powers of y and sines and cosines,

it'll always be true that you could take the cross derivatives

in either order and get the same answer.

I'd like to give a little bit of a feeling for why that is

true; it's true but it's helpful to

know why it's true. So, let's ask the following

question. Let's take a function and let's

ask how much the function changes when I go from some

point x, y to another point

x + Δx, y + Δy.

I want to find the change in the function.

So, I'm asking you what is f(x + Δx,

y + Δy) - f(x,y) for small values of Δx and

Δy. For neighboring points,

what's the change? So, we're going to do it in two

stages. We introduce an intermediate

point here, whose coordinate is x + Δ x and y,

and I'm going to add and subtract the value of the

function here. Adding and subtracting is free;

it doesn't cost anything, so let's do that.

Then, what do I get? I get f(x + Δx,

y + Δy ) - f(x + Δx, y) + f(x + Δx,

y) - f(x,y).

I'm just saying, the change of that guy minus

this guy is the same as that minus this, plus this minus

that; that's a trivial substitution.

But I write it this way because I look at the first entity here,

it looks like I'm just changing y.

I'm not changing x, you agree?

So what will this be? This is going to be the rate of

change of the function with respect to x times

Δx, and this--I'm sorry,

I got it wrong, with respect to y times

Δy.

This one is df/dx times Δx.

Therefore, the change in the function, if I add it all up,

I get df/dx Δx + df/dy Δy.

But if you are pedantic, you will notice there is

something I have to be a little careful about.

What do you think I'm referring to, in my notation,

yes? Student: [inaudible]

Professor Ramamurti Shankar: Meaning?

Student: [inaudible] Professor Ramamurti

Shankar: Oh, but this is nothing to do with

dimension, you see. It's one number minus another

number. I put another number in between

and in this function; it's a function only of here.

Let me see, x is not changing at all and y is

changing, so it is df/dy Δy.

I meant something else. That is, in fact,

a good approximation but there is one thing you should be

careful about which has to do with where the derivatives are

really taken. For the term here,

f(x + Δx, y) - f(xy),

I took the derivative, at my starting point.

If you want, I will say at the starting

point xy. The second one,

when I came here and I want to move up, I'm taking the

derivative with respect to y at the new point.

The new point is (x + Δx, y), so derivatives are not

quite taken at the same point. So, you've got to fix that.

And how do you fix that part? You argue that the derivative

with respect to y is just another function of y;

f is a function of x and y,

its derivatives are functions of x and y.

Everything is a function of x and y,

and we are saying this derivative has been computed at

x + Δx, instead of x,

so it is going to be df/dy at xy +

d^(2)f/dxdy times Δx.

In other words, I am saying the derivative at

this location is the derivative at that location,

plus the rate of change of the derivative times the change in

x. In other words,

the derivative itself is changing.

So, if you put that together you find it is df/dx Δx +

df/dy Δy, where if I don't put any bars

or anything it means at the starting point xy +

(d^(2)f/dxdy) Δx Δy.

Now, you've got to realize that when you're doing these calculus

problems, Δx is a tiny number, Δy is a tiny number,

Δx Δy is tiny times tiny. So normally,

we don't care about it, or if you want to be more

accurate, of course, you should keep that term.

In the first approximation, where you work to the first

power of everything, this will be your Δf.

But if you're a little more ambitious but you keep track of

the fact the derivative itself is changing, you will keep that

term. But another person comes along

and says, "You know what, I want to go like this.

I want to introduce as my new point, intermediate point,

the one here. It had a different value of

y in the same x and then I went horizontally

when I keep track of the changes."

What do you think that person will calculate for the change in

the function? That person will get exactly

this part, but the extra term that person will get will look

like d^(2)f/dydx times Δy Δx.

If you just do the whole thing in your head,

you can see that I'm just exchanging the x and

y roles. So, everything that happened

with x and y here will come backwards with

y and x. But then, the change between

these two points is the change between these two points and it

doesn't matter whether I introduce an intermediate point

here or an intermediate point there;

therefore, these changes have to be equal.

This part is of course equal; therefore, you want that part

to be equal, Δx Δy is clearly Δy Δx,

so the consequence of that is d^(2)f/dxdy =

d^(2)f/dydx and that's the reason it turns out when you

take cross derivatives you get the same answer.

It comes from the fact -- if you say, where is that result

coming from -- it comes from the fact,

if you start at some point and you go to another point and you

ask for the change in the function,

the change is accumulating as you move.

You can move horizontally and then vertically,

or you can move vertically and then horizontally.

The change in the function is a change in the function.

You've got to get the same answer both ways.

That's the reason; that's the requirement that

leads to this requirement. Now, I will not be keeping

track of functions to this accuracy in everything we do

today. We'll be keeping the leading

powers in Δx and Δy.

So, you should bear in mind that if you make a movement in

the xy plane, which is Δx

horizontally and Δy vertically, then the change in

the function is this [df/dx Δx + df/dy Δy].

Draw a box around that, because that's going to

be--This is just the naive generalization to two dimensions

of what you know in one dimension.

We were saying look, the function is changing

because the independent variables x and y

are changing, and the change in the function

is one part, which I blame on the changing x,

and a second part, which I blame on the changing

y and I add them. So, you're worried about the

fact that we're moving in the plane and there are vectors.

That's all correct, but f is not a vector,

f is just a number and this change has got two parts,

okay. So, this is basically all the

math we will need to do, what I want to do today.

So, let's now go back to our original goal,

which was to derive something like the Law of Conservation of

Energy in two dimensions instead of one.

You remember what I did last time so I will remind you one

more time what the trick was. We found out that the change in

the kinetic energy of some object is equal to the work done

by a force, which was some F times

Δx and then if you add all the changes over a not

infinitesimal displacement, but a macroscopic displacement,

that was given by integral of F times dx and the

integral of F times dx from

x_1 to x_2.

If F is a function only of x, from the rules of

calculus, the integral can be written as a difference of a

function at this limit minus that limit,

and that was U(x_1) -

U(x_2), where U is that function

whose derivative with a minus sign is F.

That's what we did. Then, it's very simple now to

take the U_1 to the left-hand side.

Let me see, K_1 to the

right-hand side and U(x_2) to the

left-hand side, to get K_2 +

U_2 = K_1 + U_1 and that's the

conservation of energy. You want to try the same thing

in two dimensions; that's your goal.

So, the first question is, "What should I use for the work

done?" What expression should I use

for the work done in two dimensions, because the force

now is a vector; force is not one number,

it's got an x part and a y part.

My displacement has also got an x part and a y

part and I can worry about what I should use.

So, I'm going to deduce the quantity I want to use for

ΔW, namely, the tiny work done which is an

extension of this. I'm going to demand that since

I'm looking for a Work Energy Theorem, I'm going to demand

that, remember in one dimension ΔK = F Δx.

If we divide it by the time over which it happens,

I find dK/dt is equal to force [F]

times velocity [v]. I'm going to demand that that's

the power, force times velocity, is what I call the power.

And I'm going to look for dK/dt in two dimensions,

of a body that's moving. What's the rate at which

kinetic energy is changing when a body is moving?

For that, I need a formula for kinetic energy.

A formula for kinetic energy is going to be again ½

mv^(2). I want that to be the same

entity, so I will choose it to be that, but v^(2) has

got now v_x^(2) + v_y^(2),

because you know whenever you take a vector V,

then its length is the square root of the x part

squared plus the y part squared.

Any questions? Okay, so let's take the rate of

change of this, dK/dt.

Again, you have to know your calculus.

What's the time derivative of v_x^(2)?

The rule from calculus says, first take the derivative of

v_x^(2) with respect to v_x,

which is 2v_x, then take the derivative of

v_x with respect to time.

So, you do the same thing for the second term,

2 v_y dv_y/dt,

and that gives me--the 2s cancel, gives me

mdv_x /dtv_x + m

dv_y /dtv_y.

This is the rate at which the kinetic energy of a body is

changing. So, what's my next step, yes?

Student: So what happened to the half?

Professor Ramamurti Shankar: The half got

canceled by the two here. Yes?

Student: [inaudible] Professor Ramamurti

Shankar: That's correct. So, what we want to do is to

recognize this as ma in the x direction,

because m times dv_x/dt is

a_x, and this is m times

a_y. Therefore, they are the forces

in the x and y directions, so I write

F_xv_x + F_yv_y.

So the power, when you go to two dimensions,

is not very different from one dimension.

In one dimension, you had only one force and you

had only one velocity. In two dimensions,

you got an x and y component for each one,

and it becomes this. So, this is what I will define

to be the power. When a body is moving and a

force is acting on it, the force has two components,

the velocity has two components;

this combination shall be called "Power."

But now, let's multiply both sides by Δt and write

the change in kinetic energy is equal to F_x

and you guys think about what happens when I multiply

v_x by Δt;

v_x = dx/dt.

Multiplying by Δt just gives me the distance traveled

in the x direction + F_y times

dy. So, this is the tiny amount of

work done by a force and it generalizes what we had here,

work done is force times Δx.

In two dimensions, it's F_xdx +

F_ydy. It's not hard to guess that but

the beauty is--this combination--Now you can say,

you know what, I didn't have to do all this,

I could have always guessed that in two dimensions,

when you had an x and a y you obviously have to

add them. There's nothing very obvious

about it because this combination is now guaranteed to

have the property that if I call this the work done.

Then, it has the advantage that the work done is in fact the

change in kinetic energy. I want to define work so that

its effect on kinetic energy is the same as in 1D,

namely, work done should be equal to change in kinetic

energy. And I engineered that by taking

the change in kinetic energy, seeing whatever it came out to,

and calling that the work done, I cannot go wrong.

But now, if you notice something repeating itself all

the time, which is that I had a vector F,

which you can write as I times F_x + J

times F_y, I had a vector velocity,

which is I times v_x + J times

v_y. Then, I had a tiny distance

moved by the particle, which is I times dx +

J times dy. So, the particle moves from one

point to another point. The vector describing its

location changes by this tiny vector.

It's just the step in the x direction times

I, plus step in the y direction times

J. So, what you're finding is the

following combination. The x component of

F times the x component of v,

plus the y component of F, times the y

component of v. Or F component of

x times the distance moved in x plus y

component of F times the displacement in y.

So, we are running into the following combination.

We are saying, there seem to be in all these

problems two vectors, I times A_x

+ J times A_y.

A, for example, could be the force that I'm

talking about. There's another vector

B, which is I times B_x + J

times B_y.

Right? For example,

this guy could be standing in for F,

this could be standing in for v.

The combination that seems to appear very naturally is the

combination A_xB_x +

A_yB_y. It appears too many times so I

take it seriously, give that a name.

That name will be called a dot product of A with

B and is written like this.

Whenever something appears all the time you give it a name;

this is A.B. So, for any two vectors A

and B, that will be the definition of

A.B. Then, the work done by a force

F, that displaces a particle by a tiny vector

dr, is F.dr. The particle's moving in the

xy plane. From one instant to the next,

it can move from here to there. That little guy is dr;

it's got a little bit horizontal and it's got a little

bit vertical, that's how you build dr.

The force itself is some force which at that point need not

point at the direction in which you're moving.

It's some direction. At each point,

the force could have whatever value it likes.

So, the dot product, you know, sometimes you learn

the dot product as A_xB_x +

A_yB_y, you can ask,

"Who thought about it?" Why is it a natural quantity?

And here is one way you can understand why somebody would

think of this particular combination.

So, once you've got the dot product, you've got to get a

feeling for what it is. The first thing we realize is

that if you take a dot product of A with itself,

then it's A_xA_x +

A_yA_y, which is A_x^(2) +

A_y^(2), which is the length of the

vector A, which you can either denote

this way or just write it without an arrow.

So A.A is a positive number that measures the length

squared of the vector A, likewise B.B.

Now, we have to ask ourselves, "What is A.B?"

So, somebody know what A.B is?

Yep? Student: [inaudible]

Professor Ramamurti Shankar: Okay,

so how do we know that? How do we know it's length of

A times length of B times cosine of the

angle? You should follow from--Is it

an independent definition or is it a consequence of this?

Yep? Student: Well,

you can derive using the Law of Cosines.

Professor Ramamurti Shankar: Yes.

He said we can derive it using the Law of Cosines and that's

what I will do now. In other words,

that definition which you may have learnt about first is not

independent of this definition; it's a consequence of this

definition. So, let's see why that's true.

Let's draw two vectors here. Here is A and here is

B, it's got a length A, it's got a length

B, this makes an angle,

θ_A with the x axis,

and that makes an angle θ_B for the

x axis. Now, do you guys agree that the

x component of A with the horizontal part of

A is the A cos θ_A?

You must've seen a lot of examples of that when you did

all the force calculation. And A_y is

equal to the length of A times sin θ_A

and likewise for B; I don't feel like writing it.

If you do that, then A.B will be length

of A, length of B times (cos θ_A cos

θ_B + sin θ_A sin

θ_B). You've got to go back to your

good old trig and it'll tell you this cos cos plus sin

sin is cos (θ_A – θ_B).So,

you find it is length of A, length of B,

cos (θ_A – θ_B).

Often, people simply say that is AB cos θ,

where it's understood that θ is the angle between

the two vectors. So, the dot product that you

learnt--I don't know which way it was introduced to you first,

but these are two equivalent definitions of the dot product.

In one of them, if you're thinking more in

terms of the components of A,

a pair of numbers for A and a pair of numbers for

B, this definition of the dot product is very nice.

If you're thinking of them as two little arrows pointing in

different directions, then the other definition,

in which the lengths and the angle between them appear,

that's more natural. But numerically they're equal.

An important property of the dot product, which you can check

either way, is the dot product of A with B + C,

is dot product of A with B plus dot product of

A with C. That just means you can open

all the brackets with dot products as you can with

ordinary products.

Now, that's a very important property of the dot product.

So, maybe I can ask somebody, do you know one property,

significant property of the dot product?

Student: When two vectors are perpendicular the

dot product equals 0. Professor Ramamurti

Shankar: Oh that's interesting, I didn't think

about it. Yes.

One thing is if two vectors are perpendicular the dot product is

0 because the cosine of 90 is 0. But what I had in mind--Of

course it's hard for me, it's not fair I ask you some

question without saying what I'm looking for.

What can you say if you use a different set of axis?

Yep? Student: In this case it

doesn't matter. Professor Ramamurti

Shankar: If you go to the rotated axis,

the components of vector A will change.

We have done that in the homework;

we've done that in the class. The components of B will

also change. Everything will get a prime.

But the combination, A_xB_x +

A_yB_y, when you evaluate it before or

after, will give the same answer because it's not so obvious when

you write it this way. But it's very obvious when you

write it this way because it's clear to us if you stand on your

head or you rotate the whole axis.

What you're looking for is the length of A,

which certainly doesn't change on your orientation,

or the length of B, or the angle between them.

The angle θ_A will change but the angle with

the new x axis won't be the same.

The angle with the new y axis won't be the same.

Likewise for B, but the angle between the

vectors, it's an invariant property,

something intrinsic to the two vectors, doesn't change,

so the dot product is an invariant.

This is a very important notion. When you learn relativity,

you will find you have one observer saying something,

another observer saying something.

They will disagree on a lot of things, but there are few things

they will all agree on. Those few things will be analog

of A.B. So, it's very good to have this

part of it very clear in your head.

This part of elementary vector analysis should be clear in your

head. Okay, so any questions about

this? So, if you want,

geometrically, the work done by a force when

it moves a body a distance dr,

a vector dr, is the length of the force,

the distance traveled, times the cosine of the angle

between the force and the displacement vector.

Okay, I'm almost ready for business because,

what is my goal? I find out that in a tiny

displacement, F.dr,

I start from somewhere, I go to a neighboring place,

a distance dr away. The change in kinetic energy is

F.dr. So, let me make a big trip,

okay, let me make a trip in the xy plane made up of a

whole bunch of little segments in each one of which I calculate

this, and I add them all up.

On the left-hand side, the ΔKs,

this whole thing was defined so that it's equal to ΔK,

right? ΔK was equal to this,

that's equal to all that. So, if you add all the

ΔKs, it's very clear what you will get.

You will get the kinetic energy at the end minus kinetic energy

at the beginning. On the right-hand side,

you are told to add F.dr for every tiny segment.

Well, that is written symbolically as this.

This is the notation we use in calculus.

That just means, if you want to go from A

to B along some path, you chop up the path into tiny

pieces. Each tiny segment,

if it's small enough to be approximated by a tiny vector

dr, then take the dot product of

that little dr, with the force at that point,

which means length of F, times the length of the

segment, times cosine of the angle;

add them all up. Then somebody will say,

"Well, your segments are not short enough for me."

We'll chop it up some more, then chop it up some more,

chop it up till your worst critic has been silenced.

That's the limit in which you can write the answer as this

integral. Just like in calculus when you

integrate a function you take tiny intervals Δx,

multiply F by Δx, but then you make

the intervals more and more numerous but less and less wide

and the limit of that is the area under the graph and that's

called "integral," you do the same thing now in

two dimensions. So, now maybe it'll be true,

just like in one dimension, the integral of this function

will be something that depends on the end points.

I'm just going to call it U(1) - U(2),

of some function U, just like it was in one

dimension.

If that is true, then my job is done because

then I have K_1 + U_1 = K_2 +

U_2.

So, when I'm looking for the Law of Conservation of Energy,

I've got to go to some calculus book and I got to ask the

calculus book, "Look, in one dimension you

told me integral of F from start to end is really the

difference of another function G of the end minus the

start, with G as that function

whose derivative is F." Maybe there is going to be some

other magic function you knew in two dimensions related to

F, in some way,

so that this integral is again given by a difference of

something there minus something here.

If that is the case, then you can rearrange it and

get this. But you will find that it's not

meant to be that simple. So, again, has anybody heard

rumors about why it may not be that simple?

What could go wrong?

Okay so, yes? Student: [inaudible]

Professor Ramamurti Shankar: Okay,

that's probably correct but say it in terms of what we know.

What can go wrong? I'm saying in one end,

I did an integral; the integral was the difference

of two numbers and therefore I got K + U = K + U.

So, something could go wrong somewhere here when I say this

integral from start to finish is the difference only of the

ending point minus the starting point.

Is that reasonable or could you imagine it depending on

something else? Yes?

Student: Well, say you have one of your forces

is friction. If you take a certain path from

point one to point two, you should have the same thing

in potential energy. But if you take a longer path

and there's friction involved, your kinetic energy would be

reduced. Professor Ramamurti

Shankar: Absolutely correct. It is certainly true that if

you've got friction, this is not going to work.

It doesn't even work in 1D. In 1D, if I start here and end

here, and I worry about friction, if it went straight

from here to here there's amount of friction.

If I just went back and forth 97 times and then I ended up

there is 97 times more friction. So, we agreed that if there's

friction, this is not going to work.

But the trouble with friction was, the force was not a

function only of x and y.

It depended on the direction of motion.

But now, I grant you that the force is not a function of

velocity. It's only a function of where

you are. Can something still be wrong?

Well, let me ask you the following question.

Another person does this.

[Shows a different path from 1 to 2]

Do you think that person should do the same amount of work

because the force is now integrated on a longer path?

So, you see, in one dimension there's only

one way to go from here to there.

Just go, right? That way, when you write an

integral you write the lower limit and the upper limit and

you don't say any more because it's only one way in 1D to go

from x_1 to x_2.

In two dimensions, there are thousands of ways to

go from one point to another point.

You can wander all over the place and you end up here.

Therefore, this integral, even if I say the starting

point is r_1 and the ending point is

r_2, that this is

r_1 and that's r_2,

it's not adequate. What do you think I should

attach to this integration? What other information should I

give? What more should I specify?

Yep? Student: Along a closed

path. Professor Ramamurti

Shankar: No, it's not a closed path.

I'm going from 1 to 2. What more should I tell you

before you can even find the work done?

Yes? Student: What path

you're going on. Professor Ramamurti

Shankar: You have to say which path you're going on

because if you only have two points,

1 and 2, then there is a work done but it depends on the path.

If the work depends on the path, then the answer cannot be

a function of U(1) - U(2), cannot just be U(1)

- U(2). U(1) - U(2) says,

tell me what you entered, tell me where you start,

and that difference of some function U between those

two points is the work done. In other words,

I'm asking you to think critically about whether this

equality really could be true. This is some function U

in the xy plane evaluated at one point minus the other

point. This is on a path joining those

two points but I have not told you which path and I can draw

any path I like with those same end points.

And you got to realize that it's very unreasonable to expect

that. No matter which path you take

you will get the same answer. Okay, so you might think that

I'm creating a straw man because it's going to turn out by some

magic, that no matter what force you

take somehow, due to the magic of

mathematics, the integral will depend only

on the end points. But that's not the case.

In general, that won't happen. So, I have to show you that.

So, I'm going to start by asking you, give me a number

from 1 to 3. 2?

Okay, 2. Then, I want a few more

numbers, another number from 1 to 3.

Professor Ramamurti Shankar: From 1 to 3,

[audience laughs]. All right then, two more.

Mark, you pick a number. Student: [inaudible]

Professor Ramamurti Shankar: Good,

thank you. Then I need one more, yes?

1 to 3, a number from 1 to 3. Student: [inaudible]

Professor Ramamurti Shankar: 2,

very good. Okay, so you pick these numbers

randomly, and I'm going to take a force which looks like

I times x^(2)y^(3) + J times xy^(2),

okay? I put the powers based on what

you guys gave me. So, we picked the force in two

dimensions out of the hat. Now, let's ask,

"Is it true for this force that the work done in going from one

point to another depends only on the path,

or does it depend, I mean, it depends only on the

end points or does it depend in detail on how you go from the

end points?" You all have to understand,

before you copy anything down, where we are going with this.

What's the game plan? So, let me tell you one more

time because you can copy this all you want,

it will get you nowhere. You should feel that you know

where I'm going but the details remain to be shown or you should

have an idea what's happening. If you try to generalize the

Work Energy Theorem to two dimensions, this is what

happened so far. You found a definition of work

which has the property that the work done is the change in

kinetic energy. Then, you added up all the

changes of kinetic energy and added up all the work done and

you said K_2 - K_1 is the

integral of F.dr; that's guaranteed to be true;

that's just based on Newton's laws.

What is tricky is the second equality that that integral is

the difference in a function calculated at one end point

minus the other end point. If that was true,

if the integral depended only on the end points,

then it cannot depend on the path that you take.

If it depends on the path, every path you take within the

same two end points will give different numbers.

So, the answer cannot simply be U_1 -

U_2. First, I'm trying to convince

you, through an example that I selected randomly,

that if you took a random force and found the work done along

one path, or another path, you will in fact get two

different answers, okay?

That's the first thing, to appreciate that there's a

problem. So generally,

if we took a random force, not a frictional force,

a force that depends only on location and not velocity,

it will not be possible to define a potential energy nor

will it be possible to define a K + U so that it doesn't

change. So, it's going to take a very

special force for which the answer depends only on the

starting and ending point and not on the path.

To show you that that's a special situation,

I'm taking a generic situation, namely, a force manufactured by

this class, without any prior consultation with me,

and I will show you that for that force the answer is going

to depend on how you go. So, let's take that force and

let's find the work done in going from the origin to the

point 1,1. So, I'm going to take two paths.

One path I'm going to go horizontally till I'm below the

point. Then I'm going straight up,

okay. So, let's find the work done

when I go this way. So again, you should be

thinking all the time. You should say if this guy got

struck by lightning can I do anything, or am I just going to

say "Well, I don't know what he was planning to do."

You've got to have some idea what I'm going to do.

I'm going to integrate F.dr, first on the

horizontal segment, then on the vertical segment.

F is some vector you give me at the point x

and y; I'll plug in some numbers,

I'll get something times I and something times

J. It may be pointing like that at

this point, and that's F, that's given.

Now, when I'm moving horizontally my displacement

dr, has only got a dx part.

I hope you see that. Every step I move,

there's no dy in it, it's all horizontal.

So, F.dr just becomes F_xdx because

there is no dy when you move horizontally.

So, when you do the integral, you have

F_xdx. F_x happens to

be x^(2)y^(3)dx, x going from 0 to 1.

Now, what do I do with y^(3)?

You all know how to integrate a function of x times

dx. What do I do with y^(3)?

What do you think it means? Student: [inaudible]

Professor Ramamurti Shankar: Pardon me?

What should I do with y? Evaluate y on that path

at that point. Well, it turns out,

throughout this horizontal segment y = 0,

so this is gone. Basically, the point is very

simple. When you move horizontally,

you're working against horizontal forces doing work,

but on the x axis when y is 0 there is no

horizontal force; that's why there's nothing to

do. Now, you come to this segment.

I think you all agree, the distance traveled is

J times dy. So, I have to have another

segment, the second part of my trip, which is

F_y times dy,

and y goes from 0 to 1, and the y component is

xy^(2)dy, where y goes from 0 to 1.

But now, on the entire line that I'm moving up and down,

x = 1. Do you see that?

x is a constant on the line so you can replace x

by 1 and integral of y^(2)dy is y^(3)

over 3.The work done is, therefore, 1/3 joules.

So, the work done in going first to the right and then to

the top is 1/3.

You can also go straight up and then horizontally.

By a similar trick, I won't do that now,

because I want to show you something else that's useful for

you, I'm going to pick another path

which is not just made up of x and y segments.

Then it's very easy to do this. So, I'm going to pick another

way to go from 0,0 to 1,1, which is on this curve;

this is the curve y = x^(2).

First of all, you've got to understand the

curve y = x^(2) goes through the two points I'm

interested in. If you took y =

5x^(2), it doesn't work.

But this guy goes through 0,0 and goes through 1,1.

And I'm asking you, if I did the work done by the

force along that segment, what is the integral of

F.dr? So, let's take a tiny portion

of that, looks like this, right, that's dr,

it's got a dx, and it's got a dy.

Now, you notice that as this segment becomes very small

dy/dx is a slope; therefore, dy will be

dy/dx times dx. In other words,

dx and dy are not independent if they're moving in

a particular direction. I hope you understand that.

You want to follow a certain curve.

If you step to the right by some amount, you've got to step

vertically by a certain amount so you're moving on that curve.

That's why, when you calculate the work done,

dx and dy are not independent.

So, what you really want is F_xdx +

F_ydy, but for dy I'm going to

use dy/dx times dx.

In other words, every segment Δy that

you have is related to the Δx you took horizontally

so that you stay on that curve. So, everything depends only on

dx. But what am I putting inside

the integral? Let's take

F_x, the x component is

x^(2)y^(3). On this curve y = x^(2),

so you really write x^(2),

and y = x^(2), that is x^(6).

This x^(3) [should have said x^(6)]

is really y^(3) written in terms of x.

Then, I have to write F_y,

which is x times y^(2),

which is x^(4) times dy/dx which is 2x.

So, I get here, from all of this,

(x^(8) + 2x^(5))dx. Did I make a mistake somewhere?

Pardon me? Student: The second part

of this. Professor Ramamurti

Shankar: Did I, here?

Here? Student: No.

Professor Ramamurti Shankar: Oh here?

Student: [inaudible] Professor Ramamurti

Shankar: Oh, x^(6),

right, thank you. How is that?

Thanks for watching that; you have to watch it.

So, now what do I get? x^(8) integral is

x^(9)/9. That gives me 1,9 because x is

going from 0 to 1. The next thing is

x^(7)/7, which is 2 times that.

Well, I'm not paying too much attention to this because I know

it's not 1/3, okay?

There's no way this guy's going to be 1/3, that's all I care

about. So, I've shown you that if we

took a random force, the work done is dependent on

the path. For this force,

you cannot define a potential energy whereas in one dimension

any force that was not friction allowed you to define a

potential energy. In higher dimensions,

you just cannot do that; that's the main point.

So, if you're looking for a conservative,

this is called a "conservative force."

It's a force for which you can define a potential energy.

It has the property of the work done in going from A to

B, or 1 to 2, is independent of how you got

from 1 to 2. And the one force that the

class generated pretty much randomly is not a conservative

force because the work done was path dependent.

So, what we have learned, in fact I'll keep this portion

here, if you're looking for a conservative force,

a force whose answer does not depend on how you went from

start to finish, then you have to somehow dream

up some force so that if you did this integral the answer does

not depend on the path. You realize,

that looks really miraculous because we just wrote down an

arbitrary force that we all cooked up together with these

exponents, and the answer depends on the

path. And I guarantee you,

if you just arbitrarily write down some force,

it won't work. So, maybe there is no Law of

Conservation of Energy in more than one dimension.

So, how am I going to search for a force that will do the

job? Are there at least some forces

for which this will be true? Yes?

Student: The force is always parallel to the path

along which you have [inaudible] Professor Ramamurti

Shankar: No, but see the point is you've got

to first write down a force in the xy plane.

Then, I should be free to pick any two points and connect them

any way I like and the answer shouldn't depend on how we

connected them. Only then, you can have

conservation of energy. I hope you all understand that

fact. So, I'm saying,

a generic force on the xy plane doesn't do that.

Then we ask, "Can there ever be an answer?"

It is so demanding. Yes?

Student: Well, in the inverse square force

they've got it. Professor Ramamurti

Shankar: Right, so he's saying,

"I know a force." The force of gravity,

in fact, happens to have the property that the work done by

the force of gravity does not depend on the path.

We will see why that is true. But you see,

what I want is to ask, "Is there a machine that'll

manufacture conservative forces?"

and I'm going to tell you there is.

I will show you a machine that'll produce a large number,

an infinite number of conservative forces,

and I'll show you how to produce that.

Here is the trick. The trick is,

instead of taking a force and finding if there's a potential

that will come from it by doing integrals,

let me assume there is a potential.

Then, I will ask what force I can associate with the

potential, and here is the answer.

The answer is, step one, pick any U of

x and y. You pick the function first.

That function is going to be your potential energy;

it's been anointed even before we do anything.

But then, from the potential, I want you to manufacture the

following force. I want the force that I'm going

to find to have an x component, which is [minus]

the derivative of U with respect to x,

and is going to have a y component which is [minus]

the derivative of U with respect to y.

That is step two. In fact, the claim now is this

force is going to be a conservative force.

And in fact, this potential energy

associated with it will be the function you started with.

So, how do I know that? So, for example,

I'm saying suppose U = xy^(3), then

F_x = -dU/dx, which is equal to

-y^(3), and F_y,

which is -dU/dy would be -3xy^(2).

The claim is, if you put I times

F_x and J times

F_y, the answer will not depend on

how you go from start to finish. So, let me prove that to you.

Here is the proof. The change in the function

U, in the xy plane, you guys remember me

telling you, is dU/dx times dx +

dU/dy times dy. That's the whole thing of

mathematical preliminaries which was done somewhere,

I forgot, here. It says, here,

ΔF is dF/dx times Δx + dF/dy times

Δy. I'll apply that to the function

U, but who is this? You notice what's going on here?

dU/dx times dx, this is F_x,

with a minus sign, that's F_y

with the minus sign. Therefore this is equal to

-F.dr.

Right? Because we agree,

the force that I want to manufacture is related to

U in this fashion. Now, if you add all the

changes, the right-hand side becomes the integral of

F.dr, and the left-hand side becomes

the sum of all the ΔUs with a minus sign.

Add all the ΔUs with a minus sign, that'll just become

U at 1 -U at 2.

It is cooked up so that F.dr is actually at the

change in the function U. That's if you say,

what was the trick that you did?

I cooked up a force by design so that F.dr was a change

in a certain function U. If I add all the F.drs,

I'm going to get a change in the function U from start

to finish and it's got to be U_1 -

U_2.

So, I don't know how else I can say this.

Maybe the way to think about it is, why do certain integrals not

depend on how you took the path, right?

Let me ask you a different question;

forget about integrals. You are on top of some hilly

mountain. We have a starting point,

you have an ending point, okay.

I started the starting point, and I walked to the ending

point. At every portion of my walk,

I keep track of how many feet I'm climbing;

that's like my ΔU. I add them all up.

At the end of the day, the height change will be the

top of the mountain minus bottom of the mountain,

the height of the mountain. You go on a different path,

you don't go straight for the summit, you loop around and you

coil and you wind down, you go up, you do this,

and you also end up at the summit.

If you kept track of how long you walked, it won't be the same

as me. But if you also kept track of

how many feet you climbed and you added them all up,

what answer will you get? You'll get the same answer I

got. Therefore, if what you were

keeping track of was the height change in a function,

then the sum of all the height changes will be simply the total

height change, which is the height at the end

minus height at the beginning. Therefore, starting with the

height function, by taking its derivatives,

if you manufacture a force, this will be none other than

the fact of adding up the changes.

And that's why this will be K_2 -

K_1, and then you will get

K_1 + U_1 = K_2 +

U_2. I hope you understand how

conservative forces are not impossible to get.

In fact, for every function U of x and

y you can think of, you can manufacture a

conservative force. So, you may ask the following

question. Maybe there are other ways to

manufacture the conservative force and you just thought of

one, and the answer is "no." Not only is this a machine that

generates conservative forces, my two step algorithm,

pick a U and take its derivatives, every conservative

force you get is necessarily obtained by taking derivatives

with respect to x and y of some function

U and that U will be the potential energy

associated with that force. So, when that force alone acts

on a body the kinetic plus that potential will not change.

Finally, you remember that when the class picked a certain

force, which I wrote here, I went on a limb and I said I'm

going to do the integral of this force along this path and that

path and I'm going to get different answers and show you

we have a problem. What if the force you had given

me was actually a conservative force?

Then, I would be embarrassed because then I'll find,

after all the work, this'll turn out to be again

1/3. So, I have to make sure right

away that the force is not conservative.

How can you tell? One way to say it is,

ask yourself, "Could that be some function

U whose x derivative of this was this,

and whose y derivative was that?"

You can probably convince yourself no function is going to

do it for you because if you took an x derivative you

should've lost a power of x here.

That means, if you just took the y derivative to go

here, you should have more powers of x and less

powers of y but this is just the opposite way.

So, we know this couldn't have come from a U.

But there's a better test. Instead of doing all that,

instead of saying I'm satisfied that this doesn't come from

taking derivatives of the U by looking at possible

Us are not being satisfied because maybe I'm not

clever enough. There is a mechanical way to

tell. The mechanical way to tell is

the following. Maybe I want you guys to think

for a second about what the recipe may be.

If a force came from a function U by taking derivatives,

as written up there, what can you say about the

components of that force, from the fact--yes?

Student: What is the cross derivative [inaudible]

Professor Ramamurti Shankar: Right,

we know that for every function U the cross derivatives

are equal but the ordinary derivatives are just the

F_x and F_y;

therefore, d^(2)U over dydx is really d

by dy of F_x.

And I want that to be equal to d by dx of

F_y because that would then be d^(2)U

over dxdy. In other words,

if the force satisfies this condition, the y

derivative of F_x is the

x derivative F_y.

Then, it has the right pedigree to be a conservative force

because if a force came from a function U by taking

derivatives, the simple requirement of the

cross derivatives are equal for any function U tells me.

See, if I take the y derivative of

F_x, I'm taking d^(2)U over

dxdy. And here, I'm taking

d^(2)U over dydx and they must be equal.

So, that's the diagnostic. If I give you a force and I ask

you, "Is it conservative?" you simply take the y

derivative of F_x and the

x derivative of F_y and if they

match you know it's conservative.

So, I can summarize by saying the following thing.

In two dimensions, there are indeed many,

many forces for which the potential energy can be defined.

But every one of them has an ancestor which is simply a

function, not a vector, but a scalar function,

an ordinary function of x and y.

Then, the force is obtained by taking x and y

derivatives of that function; the x derivative with a

minus sign is called F_x,

and the y derivative is called F_y.

Okay, so let's take the most popular example is the force of

gravity near the surface of the Earth.

The force of gravity we know is -mg times J.

Is this guy conservative? Yes, because the x

derivative of this vanishes and the y derivative of

F_x you don't even have to worry because there

is no F_x; it's clearly conservative.

Then you can ask, "What is the potential U

that led to this?" Well, dU/dy with a minus

sign had to be -mg and dU/dx had to be 0.

So, the function that will do the job is mgy.

You can also have mgy + 96 but we will not add those

constants because in the end, in the Law of Conservation of

Energy, K_1 +U_1 = K_2 +

U_2, adding a 96 to both sides

doesn't do anything. So this means,

when a body's moving in the gravitational field ½

mv^(2) + mgy, before, is the same as ½

mv^(2) + mgy, after.

Now, you knew this already when you're moving up and down the

y direction but what I'm telling you is this is true in

two dimensions. So, I'll give you a final

example so you guys can go home and think about it.

That is a roller coaster. So, here's the roller coaster,

it has a track that looks like this.

This is x, this is the height of the

roller coaster, but at every x there's a

certain height so y is the function of x and it

looks like this. It's just the profile of the

roller coaster. But that is also the potential

energy U, because if you multiply this by

mg, well, you just scale the graph

by mg it looks the same. So, take a photograph of a

roller coaster, multiply the height in meters

by mg, it's going to still look like

the roller coaster. That is my potential energy for

this problem and the claim is that kinetic plus potential will

not change, so K + U is a constant;

let's call the constant E.

So, if a trolley begins here at the top, what is its total

energy? It's got potential energy equal

to the height, it's got no kinetic energy,

so total energy in fact is just its height.

And total energy cannot change as the trolley goes up and down.

So, you draw a line at that height and call it the total

energy. You started this guy off at

that energy, that energy must always be the same.

What that means is, if you are somewhere here,

so that is your potential energy,

that is your kinetic energy, it's very nicely read off from

this graph, to reach the same total.

So, as you oscillate up and down, you gain and lose kinetic

and potential. When you come here your

potential energy is almost your whole energy but you've got a

little bit of kinetic energy. That means, your roller coaster

is still moving when it comes here with the remaining kinetic

energy. You can have a roller coaster

whose energy is like this. This is released from rest here.

It is released from rest here. That's the total energy and

this total energy line looks like this.

That means, if you released it here, it'll come down,

pick up speed, slow down,

pick up speed again, and come to this point,

it must stop and turn around because at that point the

potential energy is equal to total energy and there is no

kinetic energy. That means you've stopped,

that means you're turning around, it'll rattle back and

forth.

By the way, according to laws of quantum mechanics,

it can do something else. Maybe you guys know.

You know what else it can do if it starts here?

Yes? Student: It can tunnel.

Professor Ramamurti Shankar: It can suddenly

find itself here and that's not allowed by classical mechanics

because to do that it has to go over the hump.

Look what's happening at the hump.

I've got more kinetic energy than I have total energy.

I'm sorry, I've got more potential energy than I have

total energy. That means kinetic energy is

negative, that's not possible, because ½ mv^(2) can

never be negative. So, quantum theory allows these

forbidden processes and it's called tunneling.

But for us, the roller coaster problem, you will just turn

around. There's one thing I want you to

think about before you go. You should not have simply

accepted the Law of Conservation of Energy in this problem

because gravity is not the only force acting on this.

What else is acting on this roller coaster?

Student: No friction. Professor Ramamurti

Shankar: Pardon me? No friction, then what?

Student: The normal force.

Professor Ramamurti Shankar: The force of that

track. Look, if I didn't want to have

anything but gravity, here's a roller coaster ride

you guys will love, just push you over the edge.

That conserves energy and it's got no track and it's got the

only gravitational force and you can happily use this formula.

Why am I paying all this money? Because there's another force

acting on it, but that force,

if it's not frictional, is necessarily perpendicular to

the motion of the trolley, and the displacement of the

trolley is along the track, so F.dr vanishes.

So, I will conclude by telling you the correct thing to do

would be to say K_2 - K_1 is the

integral of all the forces, divided into force due to the

track and the force due to gravity.

The force due to the track is 0. I mean, it's not 0,

but that dot product with dr would be 0,

because F and dr are perpendicular.

For that reason, you'd drop that and the force

of gravity is cooked up so then it becomes U_1 -

U_2 and that's what we used.

Okay, one challenge you guys can go home and do this,

take any force in two dimensions, F;

it is parallel to the direction where you are measured from the

origin times any function of the distance from the origin.

Show, convince yourself that this force is a conservative

force by applying the test I gave you and the trick is to use

x and y instead of r.

Take this force, write it now in terms of

x and y, take the cross derivatives and

you can see it's conservative. So, any radial force, yep?

Student: [inaudible] Professor Ramamurti

Shankar: Oh, here.

This is the force to which you should try your thing and

gravity is a special example of this.

Okay, judging from the class reaction, and the stunned and

shocked look, a lot of you people,

maybe this material is new, so you should think about it,

talk about it, go to discussion section,

but this is the level at which you should understand energy

conservation in a course like this.

at some point, and you want to go to a

neighboring point, a distance Δx away.

You ask, "How much does the function change?"

And the answer is, the change in the function

Δf is the derivative of the function at the starting

point times the distance you move.

This is not an equality, unless f happens to be a

straight line; it's an approximation and there

are corrections to this. That's what all the dots mean;

corrections are proportional to Δx^(2) and

Δx^(3) and so on. But if Δx is tiny this

will do; that's it.

But today, we are going to move the whole Work Energy Theorem

and the Law of Conservation of Energy to two dimensions.

So, when you go to two dimensions, you've got to ask

yourself, "What am I looking for?"

Well, in the end, I'm hoping I will get some

relation like K_1 + U_1 = K_2 +

U_2, assuming there is no friction.

U_2 is going to be my new potential energy.

Then, if it's the potential energy and the particle is

moving in two dimensions, it's got to be a function of

two variables, x and y.

So, I have to make sure that you guys know enough about

functions of more than one variable.

So, this is, again, a crash course on

reminding you of the main points.

There are not that many. By the time I get here I'll be

done. So, how do you visualize the

function of two variables? The function of one variable,

you know, you plot x this way and the function along

y. If it's two variables,

you plot x here and y here and the function

itself is shown by drawing some surface on top of the xy

plane, so that if you take a point and

you go right up till you hit the surface, that's the value of the

function, at that point xy.

It's like a canopy on top of the xy plane and how high

you've got to go is the function.

For example, starting with the floor,

I can ask how high you have to go till I hit the ceiling.

That function varies; there are dents and dimples and

so on, so the function varies with xy.

Another example of a function of x and

y--x and y could be coordinates in the

United States and the function could be the temperature at that

point. So, you plot it on top of that,

each point you plot the temperature at that point.

So, once you've got the notion of a function of two variables,

if you're going to do calculus the next thing is what about the

derivatives of the function. How does it change?

Well, in the old days, it was dependent on x

and I changed the x and I found the change in the function

divided by the change in x and took the limit and

that became my derivative. And now, I'm sitting in the

xy plane, so here is x and here is

y; I'm here, the function is

coming out of the blackboard. So, imagine something measured

out but I'm sitting here. Now, I want to move and ask how

the function changes. But now I have a lot of options;

in fact, an infinite number of options.

I can move along x, I can move along y,

I can move at some intermediate angle,

we have to ask what do you want me to do when it comes time to

take derivatives. So, it turns out,

and you will see it proven amply as we go along,

that you just have to think about derivatives on two

principle directions which I will choose to be x and

y. So, we're going to define one

derivative, which is defined as follows.

You start at the point xy, you go to the point

x + Δx, the same y and subtract

the function at the starting point, divide by Δx and

take all the limits, Δx goes to 0.

That means you go from here to here, you move a distance

Δx, you'd find the change in the function,

you take the derivative. As you move horizontally,

you notice you don't do anything to y;

y will be left fixed at whatever y you had;

x will be changed by Δx;

you find the change; you take the derivative and

take the limit that is denoted by the symbol df/dx.

[Reader note: Partial derivatives will be

written as ordinary derivatives to avoid using fonts that may

not be universally available. It should be clear from the

context that a partial derivative is intended here,

since f depends on x and y.]

So, this curly d instead of the straight d tells

you it's called a "partial derivative."

Some people may want to make it very explicit by saying this is

the derivative with subscript y.

That means y is being held constant when x is

varied, but we don't have to write that because we know we've

got two coordinates. If I'm changing one,

the other guy is y so we won't write that;

that's the partial derivative. So, you can also move from here

to here, up and down, and see how the function

changes. I won't write the details,

you can define obviously a df/dy.

So, one tells you how the function changes,

with x I should move along x,

and the other tells you how it changes at y as you move

along y. Let's get some practice.

So, I'm going to write some function, f = x^(3)y^(2),

that's a function of x and y.

You can write any--let's make it a little more interesting,

plus y, or y^(2);

that's some function of x and y.

So, when I say df/dx, the rule is find out how it

varies with x keeping y constant.

That means, really treat it like a constant,

because number five--What will you do if y was equal to

5? This'll be 25 and it's not part

of taking derivatives because it's not changing,

and here it'll be just standing in front doing nothing;

it will just take the x derivative, treating y as

a constant. You're supposed to do that here;

this is a derivative 3x^(2)y;

so that's the x derivative.

Now, you can take the y part, y^(2),

yes, thank you. Then, I can take df/dy,

so then I look for y changes, this is 2y here,

so it's 2x^(3)y + 2y;

so that's the x derivative and the partial

y derivative. Okay, so then you can now take

higher derivatives. We know that from calculus and

one variable, you can take the derivative of

the derivative. So, one thing you can think

about is d^(2)f/dx^(2), that really means take the

d by dx of the d by dx.

That's what it means. First, take a derivative and

take the derivative of the derivative.

So, let's see what I get here. I already took df/dx.

I want to take its derivative, right, the derivative of

x^(2) is 2x, so I get 6xy^(2).

Then, I can take the y derivative of the y

derivative, d^(2)f/dy^(2),

that's d by dy of df/dy,

if you take the y derivative there.

You guys should keep an eye out for me when I do this.

I get this. But now, we have an interesting

possibility you didn't have in one dimension,

which is to take the x derivative of the y

derivative, so I want to take d by dx of

df/dy. That is written as

d^(2)f/dxdy. Let's see what I get.

So, I want the x derivative of the y

derivative. So, I go to this guy and take

his x derivative, I get 3x^(2) from there,

so I get 6x^(2)y and that's it.

So, make sure that I got the proper x derivative of

this, I think that's fine. Then, I can also take

d^(2)f/dydx. That means, take the y

derivative of the x derivative, but here's the

x derivative, take the y derivative of

that, put a 2y there, and I get 6x^(2)y.

So, you're supposed to notice something.

If you already know this, you're not surprised.

If you've never seen this before, you will notice that the

cross derivative, y followed by x

and x followed by y, will come up being

equal. That's a general property of

any reasonable function. By reasonable,

I mean you cannot call the mathematicians to help because

they will always find something where this won't work,

okay? But if you write down any

function that you are capable of writing down with powers of

x and powers of y and sines and cosines,

it'll always be true that you could take the cross derivatives

in either order and get the same answer.

I'd like to give a little bit of a feeling for why that is

true; it's true but it's helpful to

know why it's true. So, let's ask the following

question. Let's take a function and let's

ask how much the function changes when I go from some

point x, y to another point

x + Δx, y + Δy.

I want to find the change in the function.

So, I'm asking you what is f(x + Δx,

y + Δy) - f(x,y) for small values of Δx and

Δy. For neighboring points,

what's the change? So, we're going to do it in two

stages. We introduce an intermediate

point here, whose coordinate is x + Δ x and y,

and I'm going to add and subtract the value of the

function here. Adding and subtracting is free;

it doesn't cost anything, so let's do that.

Then, what do I get? I get f(x + Δx,

y + Δy ) - f(x + Δx, y) + f(x + Δx,

y) - f(x,y).

I'm just saying, the change of that guy minus

this guy is the same as that minus this, plus this minus

that; that's a trivial substitution.

But I write it this way because I look at the first entity here,

it looks like I'm just changing y.

I'm not changing x, you agree?

So what will this be? This is going to be the rate of

change of the function with respect to x times

Δx, and this--I'm sorry,

I got it wrong, with respect to y times

Δy.

This one is df/dx times Δx.

Therefore, the change in the function, if I add it all up,

I get df/dx Δx + df/dy Δy.

But if you are pedantic, you will notice there is

something I have to be a little careful about.

What do you think I'm referring to, in my notation,

yes? Student: [inaudible]

Professor Ramamurti Shankar: Meaning?

Student: [inaudible] Professor Ramamurti

Shankar: Oh, but this is nothing to do with

dimension, you see. It's one number minus another

number. I put another number in between

and in this function; it's a function only of here.

Let me see, x is not changing at all and y is

changing, so it is df/dy Δy.

I meant something else. That is, in fact,

a good approximation but there is one thing you should be

careful about which has to do with where the derivatives are

really taken. For the term here,

f(x + Δx, y) - f(xy),

I took the derivative, at my starting point.

If you want, I will say at the starting

point xy. The second one,

when I came here and I want to move up, I'm taking the

derivative with respect to y at the new point.

The new point is (x + Δx, y), so derivatives are not

quite taken at the same point. So, you've got to fix that.

And how do you fix that part? You argue that the derivative

with respect to y is just another function of y;

f is a function of x and y,

its derivatives are functions of x and y.

Everything is a function of x and y,

and we are saying this derivative has been computed at

x + Δx, instead of x,

so it is going to be df/dy at xy +

d^(2)f/dxdy times Δx.

In other words, I am saying the derivative at

this location is the derivative at that location,

plus the rate of change of the derivative times the change in

x. In other words,

the derivative itself is changing.

So, if you put that together you find it is df/dx Δx +

df/dy Δy, where if I don't put any bars

or anything it means at the starting point xy +

(d^(2)f/dxdy) Δx Δy.

Now, you've got to realize that when you're doing these calculus

problems, Δx is a tiny number, Δy is a tiny number,

Δx Δy is tiny times tiny. So normally,

we don't care about it, or if you want to be more

accurate, of course, you should keep that term.

In the first approximation, where you work to the first

power of everything, this will be your Δf.

But if you're a little more ambitious but you keep track of

the fact the derivative itself is changing, you will keep that

term. But another person comes along

and says, "You know what, I want to go like this.

I want to introduce as my new point, intermediate point,

the one here. It had a different value of

y in the same x and then I went horizontally

when I keep track of the changes."

What do you think that person will calculate for the change in

the function? That person will get exactly

this part, but the extra term that person will get will look

like d^(2)f/dydx times Δy Δx.

If you just do the whole thing in your head,

you can see that I'm just exchanging the x and

y roles. So, everything that happened

with x and y here will come backwards with

y and x. But then, the change between

these two points is the change between these two points and it

doesn't matter whether I introduce an intermediate point

here or an intermediate point there;

therefore, these changes have to be equal.

This part is of course equal; therefore, you want that part

to be equal, Δx Δy is clearly Δy Δx,

so the consequence of that is d^(2)f/dxdy =

d^(2)f/dydx and that's the reason it turns out when you

take cross derivatives you get the same answer.

It comes from the fact -- if you say, where is that result

coming from -- it comes from the fact,

if you start at some point and you go to another point and you

ask for the change in the function,

the change is accumulating as you move.

You can move horizontally and then vertically,

or you can move vertically and then horizontally.

The change in the function is a change in the function.

You've got to get the same answer both ways.

That's the reason; that's the requirement that

leads to this requirement. Now, I will not be keeping

track of functions to this accuracy in everything we do

today. We'll be keeping the leading

powers in Δx and Δy.

So, you should bear in mind that if you make a movement in

the xy plane, which is Δx

horizontally and Δy vertically, then the change in

the function is this [df/dx Δx + df/dy Δy].

Draw a box around that, because that's going to

be--This is just the naive generalization to two dimensions

of what you know in one dimension.

We were saying look, the function is changing

because the independent variables x and y

are changing, and the change in the function

is one part, which I blame on the changing x,

and a second part, which I blame on the changing

y and I add them. So, you're worried about the

fact that we're moving in the plane and there are vectors.

That's all correct, but f is not a vector,

f is just a number and this change has got two parts,

okay. So, this is basically all the

math we will need to do, what I want to do today.

So, let's now go back to our original goal,

which was to derive something like the Law of Conservation of

Energy in two dimensions instead of one.

You remember what I did last time so I will remind you one

more time what the trick was. We found out that the change in

the kinetic energy of some object is equal to the work done

by a force, which was some F times

Δx and then if you add all the changes over a not

infinitesimal displacement, but a macroscopic displacement,

that was given by integral of F times dx and the

integral of F times dx from

x_1 to x_2.

If F is a function only of x, from the rules of

calculus, the integral can be written as a difference of a

function at this limit minus that limit,

and that was U(x_1) -

U(x_2), where U is that function

whose derivative with a minus sign is F.

That's what we did. Then, it's very simple now to

take the U_1 to the left-hand side.

Let me see, K_1 to the

right-hand side and U(x_2) to the

left-hand side, to get K_2 +

U_2 = K_1 + U_1 and that's the

conservation of energy. You want to try the same thing

in two dimensions; that's your goal.

So, the first question is, "What should I use for the work

done?" What expression should I use

for the work done in two dimensions, because the force

now is a vector; force is not one number,

it's got an x part and a y part.

My displacement has also got an x part and a y

part and I can worry about what I should use.

So, I'm going to deduce the quantity I want to use for

ΔW, namely, the tiny work done which is an

extension of this. I'm going to demand that since

I'm looking for a Work Energy Theorem, I'm going to demand

that, remember in one dimension ΔK = F Δx.

If we divide it by the time over which it happens,

I find dK/dt is equal to force [F]

times velocity [v]. I'm going to demand that that's

the power, force times velocity, is what I call the power.

And I'm going to look for dK/dt in two dimensions,

of a body that's moving. What's the rate at which

kinetic energy is changing when a body is moving?

For that, I need a formula for kinetic energy.

A formula for kinetic energy is going to be again ½

mv^(2). I want that to be the same

entity, so I will choose it to be that, but v^(2) has

got now v_x^(2) + v_y^(2),

because you know whenever you take a vector V,

then its length is the square root of the x part

squared plus the y part squared.

Any questions? Okay, so let's take the rate of

change of this, dK/dt.

Again, you have to know your calculus.

What's the time derivative of v_x^(2)?

The rule from calculus says, first take the derivative of

v_x^(2) with respect to v_x,

which is 2v_x, then take the derivative of

v_x with respect to time.

So, you do the same thing for the second term,

2 v_y dv_y/dt,

and that gives me--the 2s cancel, gives me

mdv_x /dtv_x + m

dv_y /dtv_y.

This is the rate at which the kinetic energy of a body is

changing. So, what's my next step, yes?

Student: So what happened to the half?

Professor Ramamurti Shankar: The half got

canceled by the two here. Yes?

Student: [inaudible] Professor Ramamurti

Shankar: That's correct. So, what we want to do is to

recognize this as ma in the x direction,

because m times dv_x/dt is

a_x, and this is m times

a_y. Therefore, they are the forces

in the x and y directions, so I write

F_xv_x + F_yv_y.

So the power, when you go to two dimensions,

is not very different from one dimension.

In one dimension, you had only one force and you

had only one velocity. In two dimensions,

you got an x and y component for each one,

and it becomes this. So, this is what I will define

to be the power. When a body is moving and a

force is acting on it, the force has two components,

the velocity has two components;

this combination shall be called "Power."

But now, let's multiply both sides by Δt and write

the change in kinetic energy is equal to F_x

and you guys think about what happens when I multiply

v_x by Δt;

v_x = dx/dt.

Multiplying by Δt just gives me the distance traveled

in the x direction + F_y times

dy. So, this is the tiny amount of

work done by a force and it generalizes what we had here,

work done is force times Δx.

In two dimensions, it's F_xdx +

F_ydy. It's not hard to guess that but

the beauty is--this combination--Now you can say,

you know what, I didn't have to do all this,

I could have always guessed that in two dimensions,

when you had an x and a y you obviously have to

add them. There's nothing very obvious

about it because this combination is now guaranteed to

have the property that if I call this the work done.

Then, it has the advantage that the work done is in fact the

change in kinetic energy. I want to define work so that

its effect on kinetic energy is the same as in 1D,

namely, work done should be equal to change in kinetic

energy. And I engineered that by taking

the change in kinetic energy, seeing whatever it came out to,

and calling that the work done, I cannot go wrong.

But now, if you notice something repeating itself all

the time, which is that I had a vector F,

which you can write as I times F_x + J

times F_y, I had a vector velocity,

which is I times v_x + J times

v_y. Then, I had a tiny distance

moved by the particle, which is I times dx +

J times dy. So, the particle moves from one

point to another point. The vector describing its

location changes by this tiny vector.

It's just the step in the x direction times

I, plus step in the y direction times

J. So, what you're finding is the

following combination. The x component of

F times the x component of v,

plus the y component of F, times the y

component of v. Or F component of

x times the distance moved in x plus y

component of F times the displacement in y.

So, we are running into the following combination.

We are saying, there seem to be in all these

problems two vectors, I times A_x

+ J times A_y.

A, for example, could be the force that I'm

talking about. There's another vector

B, which is I times B_x + J

times B_y.

Right? For example,

this guy could be standing in for F,

this could be standing in for v.

The combination that seems to appear very naturally is the

combination A_xB_x +

A_yB_y. It appears too many times so I

take it seriously, give that a name.

That name will be called a dot product of A with

B and is written like this.

Whenever something appears all the time you give it a name;

this is A.B. So, for any two vectors A

and B, that will be the definition of

A.B. Then, the work done by a force

F, that displaces a particle by a tiny vector

dr, is F.dr. The particle's moving in the

xy plane. From one instant to the next,

it can move from here to there. That little guy is dr;

it's got a little bit horizontal and it's got a little

bit vertical, that's how you build dr.

The force itself is some force which at that point need not

point at the direction in which you're moving.

It's some direction. At each point,

the force could have whatever value it likes.

So, the dot product, you know, sometimes you learn

the dot product as A_xB_x +

A_yB_y, you can ask,

"Who thought about it?" Why is it a natural quantity?

And here is one way you can understand why somebody would

think of this particular combination.

So, once you've got the dot product, you've got to get a

feeling for what it is. The first thing we realize is

that if you take a dot product of A with itself,

then it's A_xA_x +

A_yA_y, which is A_x^(2) +

A_y^(2), which is the length of the

vector A, which you can either denote

this way or just write it without an arrow.

So A.A is a positive number that measures the length

squared of the vector A, likewise B.B.

Now, we have to ask ourselves, "What is A.B?"

So, somebody know what A.B is?

Yep? Student: [inaudible]

Professor Ramamurti Shankar: Okay,

so how do we know that? How do we know it's length of

A times length of B times cosine of the

angle? You should follow from--Is it

an independent definition or is it a consequence of this?

Yep? Student: Well,

you can derive using the Law of Cosines.

Professor Ramamurti Shankar: Yes.

He said we can derive it using the Law of Cosines and that's

what I will do now. In other words,

that definition which you may have learnt about first is not

independent of this definition; it's a consequence of this

definition. So, let's see why that's true.

Let's draw two vectors here. Here is A and here is

B, it's got a length A, it's got a length

B, this makes an angle,

θ_A with the x axis,

and that makes an angle θ_B for the

x axis. Now, do you guys agree that the

x component of A with the horizontal part of

A is the A cos θ_A?

You must've seen a lot of examples of that when you did

all the force calculation. And A_y is

equal to the length of A times sin θ_A

and likewise for B; I don't feel like writing it.

If you do that, then A.B will be length

of A, length of B times (cos θ_A cos

θ_B + sin θ_A sin

θ_B). You've got to go back to your

good old trig and it'll tell you this cos cos plus sin

sin is cos (θ_A – θ_B).So,

you find it is length of A, length of B,

cos (θ_A – θ_B).

Often, people simply say that is AB cos θ,

where it's understood that θ is the angle between

the two vectors. So, the dot product that you

learnt--I don't know which way it was introduced to you first,

but these are two equivalent definitions of the dot product.

In one of them, if you're thinking more in

terms of the components of A,

a pair of numbers for A and a pair of numbers for

B, this definition of the dot product is very nice.

If you're thinking of them as two little arrows pointing in

different directions, then the other definition,

in which the lengths and the angle between them appear,

that's more natural. But numerically they're equal.

An important property of the dot product, which you can check

either way, is the dot product of A with B + C,

is dot product of A with B plus dot product of

A with C. That just means you can open

all the brackets with dot products as you can with

ordinary products.

Now, that's a very important property of the dot product.

So, maybe I can ask somebody, do you know one property,

significant property of the dot product?

Student: When two vectors are perpendicular the

dot product equals 0. Professor Ramamurti

Shankar: Oh that's interesting, I didn't think

about it. Yes.

One thing is if two vectors are perpendicular the dot product is

0 because the cosine of 90 is 0. But what I had in mind--Of

course it's hard for me, it's not fair I ask you some

question without saying what I'm looking for.

What can you say if you use a different set of axis?

Yep? Student: In this case it

doesn't matter. Professor Ramamurti

Shankar: If you go to the rotated axis,

the components of vector A will change.

We have done that in the homework;

we've done that in the class. The components of B will

also change. Everything will get a prime.

But the combination, A_xB_x +

A_yB_y, when you evaluate it before or

after, will give the same answer because it's not so obvious when

you write it this way. But it's very obvious when you

write it this way because it's clear to us if you stand on your

head or you rotate the whole axis.

What you're looking for is the length of A,

which certainly doesn't change on your orientation,

or the length of B, or the angle between them.

The angle θ_A will change but the angle with

the new x axis won't be the same.

The angle with the new y axis won't be the same.

Likewise for B, but the angle between the

vectors, it's an invariant property,

something intrinsic to the two vectors, doesn't change,

so the dot product is an invariant.

This is a very important notion. When you learn relativity,

you will find you have one observer saying something,

another observer saying something.

They will disagree on a lot of things, but there are few things

they will all agree on. Those few things will be analog

of A.B. So, it's very good to have this

part of it very clear in your head.

This part of elementary vector analysis should be clear in your

head. Okay, so any questions about

this? So, if you want,

geometrically, the work done by a force when

it moves a body a distance dr,

a vector dr, is the length of the force,

the distance traveled, times the cosine of the angle

between the force and the displacement vector.

Okay, I'm almost ready for business because,

what is my goal? I find out that in a tiny

displacement, F.dr,

I start from somewhere, I go to a neighboring place,

a distance dr away. The change in kinetic energy is

F.dr. So, let me make a big trip,

okay, let me make a trip in the xy plane made up of a

whole bunch of little segments in each one of which I calculate

this, and I add them all up.

On the left-hand side, the ΔKs,

this whole thing was defined so that it's equal to ΔK,

right? ΔK was equal to this,

that's equal to all that. So, if you add all the

ΔKs, it's very clear what you will get.

You will get the kinetic energy at the end minus kinetic energy

at the beginning. On the right-hand side,

you are told to add F.dr for every tiny segment.

Well, that is written symbolically as this.

This is the notation we use in calculus.

That just means, if you want to go from A

to B along some path, you chop up the path into tiny

pieces. Each tiny segment,

if it's small enough to be approximated by a tiny vector

dr, then take the dot product of

that little dr, with the force at that point,

which means length of F, times the length of the

segment, times cosine of the angle;

add them all up. Then somebody will say,

"Well, your segments are not short enough for me."

We'll chop it up some more, then chop it up some more,

chop it up till your worst critic has been silenced.

That's the limit in which you can write the answer as this

integral. Just like in calculus when you

integrate a function you take tiny intervals Δx,

multiply F by Δx, but then you make

the intervals more and more numerous but less and less wide

and the limit of that is the area under the graph and that's

called "integral," you do the same thing now in

two dimensions. So, now maybe it'll be true,

just like in one dimension, the integral of this function

will be something that depends on the end points.

I'm just going to call it U(1) - U(2),

of some function U, just like it was in one

dimension.

If that is true, then my job is done because

then I have K_1 + U_1 = K_2 +

U_2.

So, when I'm looking for the Law of Conservation of Energy,

I've got to go to some calculus book and I got to ask the

calculus book, "Look, in one dimension you

told me integral of F from start to end is really the

difference of another function G of the end minus the

start, with G as that function

whose derivative is F." Maybe there is going to be some

other magic function you knew in two dimensions related to

F, in some way,

so that this integral is again given by a difference of

something there minus something here.

If that is the case, then you can rearrange it and

get this. But you will find that it's not

meant to be that simple. So, again, has anybody heard

rumors about why it may not be that simple?

What could go wrong?

Okay so, yes? Student: [inaudible]

Professor Ramamurti Shankar: Okay,

that's probably correct but say it in terms of what we know.

What can go wrong? I'm saying in one end,

I did an integral; the integral was the difference

of two numbers and therefore I got K + U = K + U.

So, something could go wrong somewhere here when I say this

integral from start to finish is the difference only of the

ending point minus the starting point.

Is that reasonable or could you imagine it depending on

something else? Yes?

Student: Well, say you have one of your forces

is friction. If you take a certain path from

point one to point two, you should have the same thing

in potential energy. But if you take a longer path

and there's friction involved, your kinetic energy would be

reduced. Professor Ramamurti

Shankar: Absolutely correct. It is certainly true that if

you've got friction, this is not going to work.

It doesn't even work in 1D. In 1D, if I start here and end

here, and I worry about friction, if it went straight

from here to here there's amount of friction.

If I just went back and forth 97 times and then I ended up

there is 97 times more friction. So, we agreed that if there's

friction, this is not going to work.

But the trouble with friction was, the force was not a

function only of x and y.

It depended on the direction of motion.

But now, I grant you that the force is not a function of

velocity. It's only a function of where

you are. Can something still be wrong?

Well, let me ask you the following question.

Another person does this.

[Shows a different path from 1 to 2]

Do you think that person should do the same amount of work

because the force is now integrated on a longer path?

So, you see, in one dimension there's only

one way to go from here to there.

Just go, right? That way, when you write an

integral you write the lower limit and the upper limit and

you don't say any more because it's only one way in 1D to go

from x_1 to x_2.

In two dimensions, there are thousands of ways to

go from one point to another point.

You can wander all over the place and you end up here.

Therefore, this integral, even if I say the starting

point is r_1 and the ending point is

r_2, that this is

r_1 and that's r_2,

it's not adequate. What do you think I should

attach to this integration? What other information should I

give? What more should I specify?

Yep? Student: Along a closed

path. Professor Ramamurti

Shankar: No, it's not a closed path.

I'm going from 1 to 2. What more should I tell you

before you can even find the work done?

Yes? Student: What path

you're going on. Professor Ramamurti

Shankar: You have to say which path you're going on

because if you only have two points,

1 and 2, then there is a work done but it depends on the path.

If the work depends on the path, then the answer cannot be

a function of U(1) - U(2), cannot just be U(1)

- U(2). U(1) - U(2) says,

tell me what you entered, tell me where you start,

and that difference of some function U between those

two points is the work done. In other words,

I'm asking you to think critically about whether this

equality really could be true. This is some function U

in the xy plane evaluated at one point minus the other

point. This is on a path joining those

two points but I have not told you which path and I can draw

any path I like with those same end points.

And you got to realize that it's very unreasonable to expect

that. No matter which path you take

you will get the same answer. Okay, so you might think that

I'm creating a straw man because it's going to turn out by some

magic, that no matter what force you

take somehow, due to the magic of

mathematics, the integral will depend only

on the end points. But that's not the case.

In general, that won't happen. So, I have to show you that.

So, I'm going to start by asking you, give me a number

from 1 to 3. 2?

Okay, 2. Then, I want a few more

numbers, another number from 1 to 3.

Professor Ramamurti Shankar: From 1 to 3,

[audience laughs]. All right then, two more.

Mark, you pick a number. Student: [inaudible]

Professor Ramamurti Shankar: Good,

thank you. Then I need one more, yes?

1 to 3, a number from 1 to 3. Student: [inaudible]

Professor Ramamurti Shankar: 2,

very good. Okay, so you pick these numbers

randomly, and I'm going to take a force which looks like

I times x^(2)y^(3) + J times xy^(2),

okay? I put the powers based on what

you guys gave me. So, we picked the force in two

dimensions out of the hat. Now, let's ask,

"Is it true for this force that the work done in going from one

point to another depends only on the path,

or does it depend, I mean, it depends only on the

end points or does it depend in detail on how you go from the

end points?" You all have to understand,

before you copy anything down, where we are going with this.

What's the game plan? So, let me tell you one more

time because you can copy this all you want,

it will get you nowhere. You should feel that you know

where I'm going but the details remain to be shown or you should

have an idea what's happening. If you try to generalize the

Work Energy Theorem to two dimensions, this is what

happened so far. You found a definition of work

which has the property that the work done is the change in

kinetic energy. Then, you added up all the

changes of kinetic energy and added up all the work done and

you said K_2 - K_1 is the

integral of F.dr; that's guaranteed to be true;

that's just based on Newton's laws.

What is tricky is the second equality that that integral is

the difference in a function calculated at one end point

minus the other end point. If that was true,

if the integral depended only on the end points,

then it cannot depend on the path that you take.

If it depends on the path, every path you take within the

same two end points will give different numbers.

So, the answer cannot simply be U_1 -

U_2. First, I'm trying to convince

you, through an example that I selected randomly,

that if you took a random force and found the work done along

one path, or another path, you will in fact get two

different answers, okay?

That's the first thing, to appreciate that there's a

problem. So generally,

if we took a random force, not a frictional force,

a force that depends only on location and not velocity,

it will not be possible to define a potential energy nor

will it be possible to define a K + U so that it doesn't

change. So, it's going to take a very

special force for which the answer depends only on the

starting and ending point and not on the path.

To show you that that's a special situation,

I'm taking a generic situation, namely, a force manufactured by

this class, without any prior consultation with me,

and I will show you that for that force the answer is going

to depend on how you go. So, let's take that force and

let's find the work done in going from the origin to the

point 1,1. So, I'm going to take two paths.

One path I'm going to go horizontally till I'm below the

point. Then I'm going straight up,

okay. So, let's find the work done

when I go this way. So again, you should be

thinking all the time. You should say if this guy got

struck by lightning can I do anything, or am I just going to

say "Well, I don't know what he was planning to do."

You've got to have some idea what I'm going to do.

I'm going to integrate F.dr, first on the

horizontal segment, then on the vertical segment.

F is some vector you give me at the point x

and y; I'll plug in some numbers,

I'll get something times I and something times

J. It may be pointing like that at

this point, and that's F, that's given.

Now, when I'm moving horizontally my displacement

dr, has only got a dx part.

I hope you see that. Every step I move,

there's no dy in it, it's all horizontal.

So, F.dr just becomes F_xdx because

there is no dy when you move horizontally.

So, when you do the integral, you have

F_xdx. F_x happens to

be x^(2)y^(3)dx, x going from 0 to 1.

Now, what do I do with y^(3)?

You all know how to integrate a function of x times

dx. What do I do with y^(3)?

What do you think it means? Student: [inaudible]

Professor Ramamurti Shankar: Pardon me?

What should I do with y? Evaluate y on that path

at that point. Well, it turns out,

throughout this horizontal segment y = 0,

so this is gone. Basically, the point is very

simple. When you move horizontally,

you're working against horizontal forces doing work,

but on the x axis when y is 0 there is no

horizontal force; that's why there's nothing to

do. Now, you come to this segment.

I think you all agree, the distance traveled is

J times dy. So, I have to have another

segment, the second part of my trip, which is

F_y times dy,

and y goes from 0 to 1, and the y component is

xy^(2)dy, where y goes from 0 to 1.

But now, on the entire line that I'm moving up and down,

x = 1. Do you see that?

x is a constant on the line so you can replace x

by 1 and integral of y^(2)dy is y^(3)

over 3.The work done is, therefore, 1/3 joules.

So, the work done in going first to the right and then to

the top is 1/3.

You can also go straight up and then horizontally.

By a similar trick, I won't do that now,

because I want to show you something else that's useful for

you, I'm going to pick another path

which is not just made up of x and y segments.

Then it's very easy to do this. So, I'm going to pick another

way to go from 0,0 to 1,1, which is on this curve;

this is the curve y = x^(2).

First of all, you've got to understand the

curve y = x^(2) goes through the two points I'm

interested in. If you took y =

5x^(2), it doesn't work.

But this guy goes through 0,0 and goes through 1,1.

And I'm asking you, if I did the work done by the

force along that segment, what is the integral of

F.dr? So, let's take a tiny portion

of that, looks like this, right, that's dr,

it's got a dx, and it's got a dy.

Now, you notice that as this segment becomes very small

dy/dx is a slope; therefore, dy will be

dy/dx times dx. In other words,

dx and dy are not independent if they're moving in

a particular direction. I hope you understand that.

You want to follow a certain curve.

If you step to the right by some amount, you've got to step

vertically by a certain amount so you're moving on that curve.

That's why, when you calculate the work done,

dx and dy are not independent.

So, what you really want is F_xdx +

F_ydy, but for dy I'm going to

use dy/dx times dx.

In other words, every segment Δy that

you have is related to the Δx you took horizontally

so that you stay on that curve. So, everything depends only on

dx. But what am I putting inside

the integral? Let's take

F_x, the x component is

x^(2)y^(3). On this curve y = x^(2),

so you really write x^(2),

and y = x^(2), that is x^(6).

This x^(3) [should have said x^(6)]

is really y^(3) written in terms of x.

Then, I have to write F_y,

which is x times y^(2),

which is x^(4) times dy/dx which is 2x.

So, I get here, from all of this,

(x^(8) + 2x^(5))dx. Did I make a mistake somewhere?

Pardon me? Student: The second part

of this. Professor Ramamurti

Shankar: Did I, here?

Here? Student: No.

Professor Ramamurti Shankar: Oh here?

Student: [inaudible] Professor Ramamurti

Shankar: Oh, x^(6),

right, thank you. How is that?

Thanks for watching that; you have to watch it.

So, now what do I get? x^(8) integral is

x^(9)/9. That gives me 1,9 because x is

going from 0 to 1. The next thing is

x^(7)/7, which is 2 times that.

Well, I'm not paying too much attention to this because I know

it's not 1/3, okay?

There's no way this guy's going to be 1/3, that's all I care

about. So, I've shown you that if we

took a random force, the work done is dependent on

the path. For this force,

you cannot define a potential energy whereas in one dimension

any force that was not friction allowed you to define a

potential energy. In higher dimensions,

you just cannot do that; that's the main point.

So, if you're looking for a conservative,

this is called a "conservative force."

It's a force for which you can define a potential energy.

It has the property of the work done in going from A to

B, or 1 to 2, is independent of how you got

from 1 to 2. And the one force that the

class generated pretty much randomly is not a conservative

force because the work done was path dependent.

So, what we have learned, in fact I'll keep this portion

here, if you're looking for a conservative force,

a force whose answer does not depend on how you went from

start to finish, then you have to somehow dream

up some force so that if you did this integral the answer does

not depend on the path. You realize,

that looks really miraculous because we just wrote down an

arbitrary force that we all cooked up together with these

exponents, and the answer depends on the

path. And I guarantee you,

if you just arbitrarily write down some force,

it won't work. So, maybe there is no Law of

Conservation of Energy in more than one dimension.

So, how am I going to search for a force that will do the

job? Are there at least some forces

for which this will be true? Yes?

Student: The force is always parallel to the path

along which you have [inaudible] Professor Ramamurti

Shankar: No, but see the point is you've got

to first write down a force in the xy plane.

Then, I should be free to pick any two points and connect them

any way I like and the answer shouldn't depend on how we

connected them. Only then, you can have

conservation of energy. I hope you all understand that

fact. So, I'm saying,

a generic force on the xy plane doesn't do that.

Then we ask, "Can there ever be an answer?"

It is so demanding. Yes?

Student: Well, in the inverse square force

they've got it. Professor Ramamurti

Shankar: Right, so he's saying,

"I know a force." The force of gravity,

in fact, happens to have the property that the work done by

the force of gravity does not depend on the path.

We will see why that is true. But you see,

what I want is to ask, "Is there a machine that'll

manufacture conservative forces?"

and I'm going to tell you there is.

I will show you a machine that'll produce a large number,

an infinite number of conservative forces,

and I'll show you how to produce that.

Here is the trick. The trick is,

instead of taking a force and finding if there's a potential

that will come from it by doing integrals,

let me assume there is a potential.

Then, I will ask what force I can associate with the

potential, and here is the answer.

The answer is, step one, pick any U of

x and y. You pick the function first.

That function is going to be your potential energy;

it's been anointed even before we do anything.

But then, from the potential, I want you to manufacture the

following force. I want the force that I'm going

to find to have an x component, which is [minus]

the derivative of U with respect to x,

and is going to have a y component which is [minus]

the derivative of U with respect to y.

That is step two. In fact, the claim now is this

force is going to be a conservative force.

And in fact, this potential energy

associated with it will be the function you started with.

So, how do I know that? So, for example,

I'm saying suppose U = xy^(3), then

F_x = -dU/dx, which is equal to

-y^(3), and F_y,

which is -dU/dy would be -3xy^(2).

The claim is, if you put I times

F_x and J times

F_y, the answer will not depend on

how you go from start to finish. So, let me prove that to you.

Here is the proof. The change in the function

U, in the xy plane, you guys remember me

telling you, is dU/dx times dx +

dU/dy times dy. That's the whole thing of

mathematical preliminaries which was done somewhere,

I forgot, here. It says, here,

ΔF is dF/dx times Δx + dF/dy times

Δy. I'll apply that to the function

U, but who is this? You notice what's going on here?

dU/dx times dx, this is F_x,

with a minus sign, that's F_y

with the minus sign. Therefore this is equal to

-F.dr.

Right? Because we agree,

the force that I want to manufacture is related to

U in this fashion. Now, if you add all the

changes, the right-hand side becomes the integral of

F.dr, and the left-hand side becomes

the sum of all the ΔUs with a minus sign.

Add all the ΔUs with a minus sign, that'll just become

U at 1 -U at 2.

It is cooked up so that F.dr is actually at the

change in the function U. That's if you say,

what was the trick that you did?

I cooked up a force by design so that F.dr was a change

in a certain function U. If I add all the F.drs,

I'm going to get a change in the function U from start

to finish and it's got to be U_1 -

U_2.

So, I don't know how else I can say this.

Maybe the way to think about it is, why do certain integrals not

depend on how you took the path, right?

Let me ask you a different question;

forget about integrals. You are on top of some hilly

mountain. We have a starting point,

you have an ending point, okay.

I started the starting point, and I walked to the ending

point. At every portion of my walk,

I keep track of how many feet I'm climbing;

that's like my ΔU. I add them all up.

At the end of the day, the height change will be the

top of the mountain minus bottom of the mountain,

the height of the mountain. You go on a different path,

you don't go straight for the summit, you loop around and you

coil and you wind down, you go up, you do this,

and you also end up at the summit.

If you kept track of how long you walked, it won't be the same

as me. But if you also kept track of

how many feet you climbed and you added them all up,

what answer will you get? You'll get the same answer I

got. Therefore, if what you were

keeping track of was the height change in a function,

then the sum of all the height changes will be simply the total

height change, which is the height at the end

minus height at the beginning. Therefore, starting with the

height function, by taking its derivatives,

if you manufacture a force, this will be none other than

the fact of adding up the changes.

And that's why this will be K_2 -

K_1, and then you will get

K_1 + U_1 = K_2 +

U_2. I hope you understand how

conservative forces are not impossible to get.

In fact, for every function U of x and

y you can think of, you can manufacture a

conservative force. So, you may ask the following

question. Maybe there are other ways to

manufacture the conservative force and you just thought of

one, and the answer is "no." Not only is this a machine that

generates conservative forces, my two step algorithm,

pick a U and take its derivatives, every conservative

force you get is necessarily obtained by taking derivatives

with respect to x and y of some function

U and that U will be the potential energy

associated with that force. So, when that force alone acts

on a body the kinetic plus that potential will not change.

Finally, you remember that when the class picked a certain

force, which I wrote here, I went on a limb and I said I'm

going to do the integral of this force along this path and that

path and I'm going to get different answers and show you

we have a problem. What if the force you had given

me was actually a conservative force?

Then, I would be embarrassed because then I'll find,

after all the work, this'll turn out to be again

1/3. So, I have to make sure right

away that the force is not conservative.

How can you tell? One way to say it is,

ask yourself, "Could that be some function

U whose x derivative of this was this,

and whose y derivative was that?"

You can probably convince yourself no function is going to

do it for you because if you took an x derivative you

should've lost a power of x here.

That means, if you just took the y derivative to go

here, you should have more powers of x and less

powers of y but this is just the opposite way.

So, we know this couldn't have come from a U.

But there's a better test. Instead of doing all that,

instead of saying I'm satisfied that this doesn't come from

taking derivatives of the U by looking at possible

Us are not being satisfied because maybe I'm not

clever enough. There is a mechanical way to

tell. The mechanical way to tell is

the following. Maybe I want you guys to think

for a second about what the recipe may be.

If a force came from a function U by taking derivatives,

as written up there, what can you say about the

components of that force, from the fact--yes?

Student: What is the cross derivative [inaudible]

Professor Ramamurti Shankar: Right,

we know that for every function U the cross derivatives

are equal but the ordinary derivatives are just the

F_x and F_y;

therefore, d^(2)U over dydx is really d

by dy of F_x.

And I want that to be equal to d by dx of

F_y because that would then be d^(2)U

over dxdy. In other words,

if the force satisfies this condition, the y

derivative of F_x is the

x derivative F_y.

Then, it has the right pedigree to be a conservative force

because if a force came from a function U by taking

derivatives, the simple requirement of the

cross derivatives are equal for any function U tells me.

See, if I take the y derivative of

F_x, I'm taking d^(2)U over

dxdy. And here, I'm taking

d^(2)U over dydx and they must be equal.

So, that's the diagnostic. If I give you a force and I ask

you, "Is it conservative?" you simply take the y

derivative of F_x and the

x derivative of F_y and if they

match you know it's conservative.

So, I can summarize by saying the following thing.

In two dimensions, there are indeed many,

many forces for which the potential energy can be defined.

But every one of them has an ancestor which is simply a

function, not a vector, but a scalar function,

an ordinary function of x and y.

Then, the force is obtained by taking x and y

derivatives of that function; the x derivative with a

minus sign is called F_x,

and the y derivative is called F_y.

Okay, so let's take the most popular example is the force of

gravity near the surface of the Earth.

The force of gravity we know is -mg times J.

Is this guy conservative? Yes, because the x

derivative of this vanishes and the y derivative of

F_x you don't even have to worry because there

is no F_x; it's clearly conservative.

Then you can ask, "What is the potential U

that led to this?" Well, dU/dy with a minus

sign had to be -mg and dU/dx had to be 0.

So, the function that will do the job is mgy.

You can also have mgy + 96 but we will not add those

constants because in the end, in the Law of Conservation of

Energy, K_1 +U_1 = K_2 +

U_2, adding a 96 to both sides

doesn't do anything. So this means,

when a body's moving in the gravitational field ½

mv^(2) + mgy, before, is the same as ½

mv^(2) + mgy, after.

Now, you knew this already when you're moving up and down the

y direction but what I'm telling you is this is true in

two dimensions. So, I'll give you a final

example so you guys can go home and think about it.

That is a roller coaster. So, here's the roller coaster,

it has a track that looks like this.

This is x, this is the height of the

roller coaster, but at every x there's a

certain height so y is the function of x and it

looks like this. It's just the profile of the

roller coaster. But that is also the potential

energy U, because if you multiply this by

mg, well, you just scale the graph

by mg it looks the same. So, take a photograph of a

roller coaster, multiply the height in meters

by mg, it's going to still look like

the roller coaster. That is my potential energy for

this problem and the claim is that kinetic plus potential will

not change, so K + U is a constant;

let's call the constant E.

So, if a trolley begins here at the top, what is its total

energy? It's got potential energy equal

to the height, it's got no kinetic energy,

so total energy in fact is just its height.

And total energy cannot change as the trolley goes up and down.

So, you draw a line at that height and call it the total

energy. You started this guy off at

that energy, that energy must always be the same.

What that means is, if you are somewhere here,

so that is your potential energy,

that is your kinetic energy, it's very nicely read off from

this graph, to reach the same total.

So, as you oscillate up and down, you gain and lose kinetic

and potential. When you come here your

potential energy is almost your whole energy but you've got a

little bit of kinetic energy. That means, your roller coaster

is still moving when it comes here with the remaining kinetic

energy. You can have a roller coaster

whose energy is like this. This is released from rest here.

It is released from rest here. That's the total energy and

this total energy line looks like this.

That means, if you released it here, it'll come down,

pick up speed, slow down,

pick up speed again, and come to this point,

it must stop and turn around because at that point the

potential energy is equal to total energy and there is no

kinetic energy. That means you've stopped,

that means you're turning around, it'll rattle back and

forth.

By the way, according to laws of quantum mechanics,

it can do something else. Maybe you guys know.

You know what else it can do if it starts here?

Yes? Student: It can tunnel.

Professor Ramamurti Shankar: It can suddenly

find itself here and that's not allowed by classical mechanics

because to do that it has to go over the hump.

Look what's happening at the hump.

I've got more kinetic energy than I have total energy.

I'm sorry, I've got more potential energy than I have

total energy. That means kinetic energy is

negative, that's not possible, because ½ mv^(2) can

never be negative. So, quantum theory allows these

forbidden processes and it's called tunneling.

But for us, the roller coaster problem, you will just turn

around. There's one thing I want you to

think about before you go. You should not have simply

accepted the Law of Conservation of Energy in this problem

because gravity is not the only force acting on this.

What else is acting on this roller coaster?

Student: No friction. Professor Ramamurti

Shankar: Pardon me? No friction, then what?

Student: The normal force.

Professor Ramamurti Shankar: The force of that

track. Look, if I didn't want to have

anything but gravity, here's a roller coaster ride

you guys will love, just push you over the edge.

That conserves energy and it's got no track and it's got the

only gravitational force and you can happily use this formula.

Why am I paying all this money? Because there's another force

acting on it, but that force,

if it's not frictional, is necessarily perpendicular to

the motion of the trolley, and the displacement of the

trolley is along the track, so F.dr vanishes.

So, I will conclude by telling you the correct thing to do

would be to say K_2 - K_1 is the

integral of all the forces, divided into force due to the

track and the force due to gravity.

The force due to the track is 0. I mean, it's not 0,

but that dot product with dr would be 0,

because F and dr are perpendicular.

For that reason, you'd drop that and the force

of gravity is cooked up so then it becomes U_1 -

U_2 and that's what we used.

Okay, one challenge you guys can go home and do this,

take any force in two dimensions, F;

it is parallel to the direction where you are measured from the

origin times any function of the distance from the origin.

Show, convince yourself that this force is a conservative

force by applying the test I gave you and the trick is to use

x and y instead of r.

Take this force, write it now in terms of

x and y, take the cross derivatives and

you can see it's conservative. So, any radial force, yep?

Student: [inaudible] Professor Ramamurti

Shankar: Oh, here.

This is the force to which you should try your thing and

gravity is a special example of this.

Okay, judging from the class reaction, and the stunned and

shocked look, a lot of you people,

maybe this material is new, so you should think about it,

talk about it, go to discussion section,

but this is the level at which you should understand energy

conservation in a course like this.