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6. Law of Conservation of Energy in Higher Dimensions

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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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