Our Energy Sources, Electricity

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.

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