6. Law of Conservation of Energy in Higher Dimensions

Uploaded by YaleCourses on 22.09.2008

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

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

If that is true, then my job is done because
then I have K_1 + U_1 = K_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

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 -

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

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.