We can do the following problem.
You give me any number of
charges which are fixed, anywhere you want.
So here they are.
Here's one guy
q_1, here's q_2,
here's q_3. And instead of these discrete
charges, you may have a line charge too, maybe λ
coulombs per meter. That guy is here.
They are fixed.
They are not dynamical.
The object of interest to us is
this charge q, which can actually move,
allowed to move. The question we ask is,
do we know what this charge will do?
And the answer is,
yes, we know what it will do, because if it's sitting here
and we have already computed the electric field everywhere--
you know how to compute electric field.
Add the field due to this one,
that one, that one, add that.
That is some electric field
sitting here. Then we know,
the force on this charge we're interested in,
this q times the electric field at the location
of the charge. So at every instant,
we know what force is acting on it.
Then from Newton's law,
that will give me the rate of change of velocity or the
acceleration. You can write it either way.
That's all I need,
because using that I can find out the new position,
new velocity, right?
The acceleration tells you,
if you wait 1 nanosecond, you go to this place,
because the initial velocity, I can predict where it will be,
and because you know the acceleration,
I know what velocity it will have.
Then I keep on updating it and
I get my trajectory. Student: I think it
should be dt squared. Prof: Yes, thank you.
So everyone understands this is
how you proceed. This charge will of course push
those guys, but we're not worried, because we don't let
them move. They are just fixed.
And if there are much such
charges, we can find what they all do in this environment,
which is the background electric field.
That's all we can do right now,
this is electrostatics. In the real world,
everybody is moving and all kinds of stuff happens,
but you cannot use electrostatics for that.
This asymmetry between the
background and the test charge that we are studying is in the
static situation. Guys that move and guys that
don't. Now this is something you could
have done on the first day, because I think on the very
first day I told you about Coulomb's law and the principle
of superposition. If you put them together,
you know the force on a charge q.
Then you use F = ma,
you predict its trajectory. But in principle,
that's all you need to do electrostatic problems,
but in practice you want to clean up the machinery a little
bit, make it more efficient.
For example,
we found Gauss's law. Gauss's law follows from
Coulomb's law, but we found it's quite useful.
For example,
if we give you a ball of charge, spherical charge,
going back to Coulomb's law and first principles is very
difficult. But if you use Gauss's law and
the symmetries, you're able to find the
electric field around the sphere,
or an infinitely long cylinder, or an infinite plane or simple
geometries like that. So it's not enough to say
"I've got the basic equations."
You always want to see if there
are ways to speed up the computation to make it more
efficient. So let me remind you what we
did in the first part of the course.
There, if you have the
following question - here's a spring and here's a mass,
the spring likes to have that length and I pulled it to way
over there and I released it, and right now it happens to be
here, going at some speed.
I want to know what velocity it
will have when it is here, so this position is
x_1, this speed is
v_1, this position is
x_2. I want to know
v_2. That's a reasonable thing you
can ask. Unlike the electric force,
the force on this mass is due to the spring.
Apart from that,
it's the same situation. So one way to do that is to
say, I know F = ma, so the instant you start the
problem here, I know the acceleration because
the force of the spring is -k times x_1.
Once I know the force,
I know the acceleration. I can find the new velocity,
new position, keep on doing it,
or I can solve the equation, d^(2)x/dt^(2) is
−kx/m, and I can eventually end up
here. Then I can find the velocity
here. But we all know that in this
problem, there's a shortcut, right?
Yes?
Anybody in the last row want to
tell me what the shortcut is? Yes?
Student: You can use
conservation of energy. Prof: Conservation of
energy. So conservation of energy is a
consequence of Newton's laws and certain properties of the force.
I'm going to remind you how
it's done. One attitude I could take is,
you've already done this before, but usually it's better
to make sure that you're all on the same page when we do this.
So let me remind you how it
works there. You go to Newton's laws and we
write m dv/dt = F.
It's all in one dimension
though. I'm moving just along the line.
Then I multiply both sides by
v. On the left hand side,
if you know basic calculus, this is really
d/dt of mv^(2)/2,
by the chain rule of calculus. The d/dt derivative is
the v derivative, followed by the derivative of
v with respect to t.
v derivative will give
you mv and then there is that.
This becomes F
dx/dt. This means in every tiny
instant of time, the force times the change in
the position is equal to the change in this quantity,
which we know is called kinetic energy.
So if you multiply both sides
by dt, then you find this relation,
and if you add up all the changes,
we get this relation, m--so let's see what we
are integrating from a starting point to an ending point and
some starting time, ending time.
We get
mv_2^(2)/2 - mv_1^(2)/2 =
integral f(x)dx from
x_1 to x_2.
And this is called the work
energy theorem. The work energy theorem tells
you something you may expect, namely, when a force acts on a
body, it's going to accelerate it.
It's going to accelerate it,
it's going to change the velocity.
You're going to ask,
"How much velocity change do I get for a certain action of
the force?" and the statement is,
the force pushes the body from x_1 to
x_2, then this is the change in this
quantity called kinetic energy. This is true no matter what the
force is. It's true for friction,
it's true for spring, it's true for everything,
because Newton's law is always valid.
But if you want to extract from
this a law of conservation of energy, you have to do a certain
manipulation and that's what I'm going to describe.
So let's go back here and try
to extract from it a law of conservation of energy.
So I'm going to write here
k_2 - k_1 is integral
F(x) dx, from x_1 to
x_2. Now it's a fundamental result
in calculus that integral of any function of x from the
point x_1 to the point x_2
can be written as G(x_2) -
G(x_1), where G is any function
whose derivative is the F I'm integrating.
For example,
if F = x, G = x^(2)/2,
because derivative x^(2)/2 is x.
In fact, it's not just
x^(2)/2 . You can add x^(2)/2 a
constant, but the constant is not important,
because they'll cancel between this and this.
So you can pick the solution
without a constant, okay?
Now what does that mean?
Let's rearrange the expression
so I get K_2 −
G(x_2) = K_1 −
G(x_1). That tells you the combination
K - G does not change as the particle moves,
but this doesn't look very nice.
We want to get the old law of
conservation of energy in the standard form.
You define the function
U(x), you simply - this
G(x). It's a very trivial change.
Then you can say
K_2 U(x_2) =
K_1 U(x_1),
and that is the energy that does not change with time.
So the relation between
U, which is called the potential energy,
and the force, is the following.
The force is - the derivative
of the potential energy. The - came because G is
the integral of F, but U is -G,
so that this formula, this means you have this
relation. And you can also say
U(x_2) −
U(x_1) with an integral of
F(x) dx from x_1 to
x_2, there's a - sign here.
Again, the - sign is present
because F is defined to be - the derivative of U.
So you can go back and forth
from potential to force and force to potential.
If you knew the force,
you can integrate it to get the potential.
If you knew the potential,
you can differentiate it to get the force,
and more importantly, you have this great
conservation law, which is very helpful in the
spring problem, because if you know the
potential energy, once and for all,
it's kx^(2)/2, you balance the kinetic and the
potential, you can find the velocity at
one other point. Now why doesn't this work for
friction? Friction is the one dimensional
force. Yes?
Student: It's not
conservative. Prof: It's not
conservative. It's just a definition.
But what's wrong with my
argument? Where in this argument does it
fail? It looks like any function--yes?
Student: Because the
direction of the force depends on the direction that the
particle's moving. Prof: Very good.
The point is that frictional
force is not a function just of your location.
It depends on your velocity.
In other words,
whereas the spring will always exert a force -kx--
if you are displaced to the right by an x,
it exerts a force -kx--whether or not
you're moving to the right, away from the equilibrium or
towards it. Thus given x,
it gives you a given force, but friction is not like that.
Friction does not have a
definite direction. The only thing friction wants
to do is to slow you down. If you want to go right,
it wants to act left. If you want to go left,
it wants to act right. I will resist all political
analogies. I'll just point out to you that
friction does not have a definite sign,
therefore it's not a function of x,
but as she said, it's a function of x and
of x dot, which is the velocity.
So it doesn't have a unique
answer. That does not mean you cannot
find the work due to friction. If you know it's moving to the
right, then the force is to the left.
You can integrate that force,
as long as it is moving to the right.
But if it's going right,
left, right, left and oscillating,
then you cannot integrate it once and for all.
That way doesn't work for
friction. Now here's an even more
powerful example of the law of conservation of energy.
You have this famous roller
coaster and you are here. I say "you are here,"
because you'll never get me on top of one of these guys.
So you are here and you got
some speed, and I want to know your speed when you are here.
Again, it follows from Newton's
laws. You can calculate from Newton's
laws, because at every point,
wherever you may be, there's the force of gravity
acting this way and there is the perpendicular force from the
track acting this way, and the perpendicular force is
such that it cancels this part of gravity,
and this part of gravity will accelerate it and you can find
the velocity here. Keep doing this,
you can find the velocity at the end.
But that's not how we do this.
We use the fact that the
gravity force allows us to define an energy,
and I'll come to that in a moment.
Then we say kinetic potential
is kinetic potential in the beginning and at the end,
and we can simply find the velocity here if you knew the
potential energy here. All right, so the law of
conservation of energy is also a very profound result,
which has survived all the revolutions of quantum
mechanics. After quantum mechanics,
we don't like to think in terms of force, and we don't like to
think in terms of trajectories. You don't even know for sure
where the particle is. But amazingly,
the total energy is always conserved in the problems we
study. So the law of conservation of
energy is very profound and it's also very useful for
computations. Now the next question I have
is, if I go to two dimensions--and it's the same as
in three dimensions. I'm just going to go to two
dimensions--what does it take to get the law of conservation of
energy? So I'm going to mimic this
derivation in 2D and see if I can pull out a law of
conservation of energy. So here's what I will do.
So in two dimensions,
let me start this time with the kinetic energy.
Kinetic energy is
½mv^(2), but
v has got 2 components, x and y,
therefore it's ½mvx^(2)
vy^(2), but I'm going to write this
½m v⋅v.
Then let me ask,
what's the rate of change of this kinetic energy?
You've got to take derivatives.
Now you may be a little queasy
about taking derivatives from this dot product,
but you can check by expanding the component,
just like ordinary product, and you can take derivative of
this guy times that guy, derivative of that guy times
this guy, and they will both be the same,
and you will find it's equal to m dv/dt,
which is the force. I'm sorry.
Let me erase this.
It's not m dv/dt;
it is mdv/dt ⋅v.
So if you integrate this,
cancel the dt and add all the changes,
you will find K_2 -
K_1 is m times dr--
let me see--I'm sorry. I want to keep the
dv/dt here, then I want to write this
v as dr/dt. That's right.
So let me do the following,
excuse me. Let me do it again.
dK/dt = ma
times v and v is dr/dt.
Now we cancel the dts
and we write the change in K as K_2
- K_1 = integral of
F⋅dr from the starting point
r_1 to the ending point
r_2. In other words,
the change in kinetic energy in one dimension is just force
times distance for tiny distances.
In higher dimensions,
like in 2 dimensions, if the particle is moving in
the 2 dimensional plane along some path and it moves a
distance dr, and at that point,
the force is in that direction, you should not simply take the
product of the length of the force times the length of the
distance, but that times the cosine of
the angle. Why should it take that?
Because that's what gives you
the rate of change of kinetic energy.
Just go to the kinetic energy
and it suggests to you that you take this product and define it
to be the work, because if you define it that
way, just like in 1D, the work done by the force will
be the change in kinetic energy. You guys with me now?
In spite of this little mess up
here, this ma is my force. The dts got canceled.
The dK is added up to
K_2 - K_1 and this is
the integral. All right, so now it looks
pretty good, except for the complication.
You have a dot product here.
You have this nice result that
K_2 - K_1 is the
integral F⋅dr from r_1 to
r_2. So let's imagine where
r_1 and r_2 are.
This is say where you started,
and maybe you ended up here at r_2.
You've just got to integrate
this F from r_1 to
r_2. Let's imagine there is no
friction. Let's imagine F depends
only on the position. What's my problem now?
What problem do we have?
Yes?
Student: _______ going
from r_2 to r_1 _______.
Prof: No,
this integral gives me no choice.
I must go from
r_1 to r_2.
Anybody know what's the problem?
Yes?
Student: Your path.
Prof: Your path.
See, in 1 dimension,
if I want to go from here to there, I have no choice,
okay? Just move along that line.
In 2 dimensions,
if you want to go from one point to another,
I can think of many ways. You can do that,
you can take the straight line path, or you can take some other
path. That's an ambiguity we have in
higher dimensions. So even though the work energy
theorem is valid, and no matter which path you
take, the work done by the force will be the change in kinetic
energy, this is not in general equal to
some function at r_1 - some
function at r_2.
Because if it had this form,
we are done. You agree?
If it had this form,
then again you write K_2
U_2 = K_1
U_1. But no one's telling you that
is true. If it was true,
you are in business. But it's a very unreasonable
thing to ask of this force, that its integral from some
point to another point, does not depend on what path
you integrate it on. In other words,
you can take that path, break it up into segments
dr, and on each segment, the force is some variable.
You take F⋅dr
of that little segment, add it up over all the segments
and make the segments vanishingly small.
That's the meaning of this
integral. And you're supposed to get the
same answer, no matter what you do, including going like this,
or anything else. How can they all give the same
answer? In fact, I'll give you a simple
example where it's simply not true, so the answer depends on
the path. In fact, just about any force
you pick randomly will not have the property,
the answer does not depend on the path.
Answers typically will depend
on what path you take. So let's take a simple force,
F = iy.
I can think of more complicated
ones, but I don't have time to do them.
So here's a simple one - force
depends on position. It points in the x
direction only and it grows with y.
And I'm going to do two
calculations. I'm going to find the integral
from the origin to the point (1,1) first by going this way
and then by going that way. They are two possible paths.
And you will see,
when I do the calculation, the answer is different.
Therefore you cannot write the
answer as a difference of something here - something here,
because the answer is not a function only of the starting
and ending points, but how you got there.
And I want to do this just to
remind you how to do these integrals.
So let's take this segment here.
What is the work done?
The work is the force which is
i times y times the distance.
If you move horizontally,
perhaps you guys can see dr for a tiny segment
simply i times dx. And x goes from 0 to 1.
So how much is that?
Can you do this?
I want the first person who
knows the answer. Yes?
Student: Zero
______________. Prof: Right.
On this line, y is 0.
In fact, if you draw a picture
of the force, it looks like this.
It gets bigger and bigger and
bigger as you go further off. And if you go to negative
values, it looks like this. So there is no y,
there is no force. That is 0.
If you go to this segment,
y is definitely non-zero,
but notice, the force is like this and the dr is in the
y direction and they are perpendicular,
so the dot product is 0. Okay?
dr and F are
perpendicular. One is along i and one
is along j dot product to 0.
If you take this path,
the work done is 0. Now let's take the other path,
go like this and then like this.
On this section,
we have the same problem. Force is to the right.
You're moving straight up.
dr and F are
perpendicular. But on the last segment,
from here to here, you get a non-zero
contribution, because force is
iy, displacement is
idx and x goes from 0 to 1,
and y is frozen at the value 1,
because you're on this line. Frozen at the value 1.
Just put 1 for it.
And i⋅i is
1, integral of dx is 1. Therefore the work done by the
force going up and then to the right is 1.
Going to the right and then up
is 0. You can take other paths and
keep getting other answers. So this is a force for which
it's not true, and I bet you that if you just
randomly selected forces in the 2 dimensional plane,
they will never have the property that the integral is
path independent. So it looks like we are onto a
hopeless task, that the answer cannot depend
on the path. So maybe there is no problem in
which you can derive a law of conservation of energy.
Of course, you guys know,
you can derive a law of conservation of energy for a
family of forces. In fact, the answer is,
there's an infinite of numbers of forces you can devise for
which this is true, namely, the answer does not
depend on the path. Before writing those forces,
let me explain to you one other way to write the condition.
The condition,
namely of a conservative force, is that the integral from here
to here is the same as the integral from there to there,
from 1 to 2 along path A and 1 to 2 along path B.
They must be equal for any
starting point and any ending point and any two paths joining
them. So what I'm saying is,
F⋅dr from 1 to 2 along A = integral
F⋅dr from 1 to 2 along B.
Let me rewrite this by putting
a - sign here and equate it to 0.
That's fine.
Shifted it to the left hand
side. But then we use the fact that
when you go from 1 to 2, with a - sign,
this second guy is same as from 2 to 1 along B of
F⋅dr. In other words,
the line integral going this way is - the line integral going
backwards, because at every step,
dr is opposite but F is the same at the point,
F⋅dr changes sign.
Therefore it tells you then
that the line integral from here to there and back is 0.
We write that by saying the
line integral of F⋅dr in a
closed path is 0 for all paths, for all starting points,
all ending points, all paths.
And that is the force that we
want. That's the conservative force.
I've just rewritten.
I've not gotten any closer to
finding conservative force, but I'm saying there are two
ways to write it. Answer doesn't depend on the
path, or the answer on any closed loop, the line integral
on any closed loop, is 0.
That means sometimes it's got
to be positive, sometimes it's got to be
negative. So question is,
where am I going to get a force with these amazing properties?
And I'm just going to give you
the answer. The theorem--if you want,
you can call it a theorem, you can call it whatever you
like. It's a fact--number 1,
here is a recipe for finding conservative forces.
Take any function
U(x,y). I'm just going to do it in 2
dimensions. In 3 dimensions,
you can make it U(x,y,z).
So let's give an example here.
U = x^(3)y^(2).
Take that function.
You can write your own ticket.
Write whatever function you
want, time hyperbolic sinh(y) times
cosh(x), doesn't matter.
Take a function.
Then the force that I want is
the property that F_x =
-dU/dx and F_y =
-dU/dy. In other words,
I'm going to manufacture a force whose x component
is - the x derivative of the function I picked and the
y component is - the y derivative of the
function I picked. So in my example,
F_x will be -x derivative partial
means keep y as a constant,
take the derivative with respect to x.
That's 3x^(2)y and
F_y will be -2x^(3).
3x^(2)y^(2) and
-2x^(3)y. Let me see if I got that right,
because you guys have been catching me too many times.
Okay, I think that's all right.
This force is guaranteed to be
conservative. I will show you why this is
true. Why is this magic working?
But let me write it in another
way that's more compact. So the force I want to get,
written as a vector, is i times
F_x, which is i times
−du/dx j times
−du/dy. That's the force I claim is
conservative. We're going to write that as -
grad U. This is called the grad and it
means gradient, and this symbol is shorthand
for this. There's nothing more to it.
Gradient is a machine.
Derivative is a machine.
You give it a function,
sin x, it gives you an output,
cosine x. you give it x squared,
it gives you 2x. Gradient is a different kind of
machine. You feed into it a function
U of x and y,
and it gives you a vector in all of the xy plane
obtained by taking the two derivatives,
one along x and one along y,
and assembling them into a single vector.
So let's see for this
particular U, the force we have = -i
times 3x^(2)y^(2), -j times 2x^(3)y.
Amazingly, it is claimed that
this force, when you integrate it from one point to another,
answer will not depend on the path;
it will depend only on the end points.
So we have to understand what
makes this work. Before I show that,
I will make another statement. I will show you why this works,
but the second statement, I will not prove,
which is that all conservative forces are of the form - grad
U. In other words,
not only is this an example of conservative force,
that's it. There are no more examples.
Every conservative force you
cook up will always be--yes? Student: Does the
j component ___________? Prof: j component
should be what? Oh, that's negative.
I think I erased it.
How's that?
Thank you.
Is that what you meant?
Student: Yes.
Prof: Okay.
So how are you guys doing with
this? I'm always worried.
You are inscrutable,
and I don't know what's going on in there,
whether silence means "I'm with you"
or silence means, "I'm so far behind,
I don't even know where to start telling you my
problems." I don't know.
But this is your class.
If you don't speak up,
you don't get service. You have to ask.
Don't assume people next to you
know anything. I've met them.
They don't.
So just speak up.
All right.
So it's like one of these
psychotherapy sessions. Turn to the person next to you
and insult him or her, because that's what you want to
do. Don't worry about them, okay.
When you go to physics seminar,
you never assume the speaker has a problem.
I mean, you assume that you
don't have a problem. If you don't follow it,
it's a problem with the speaker.
That's what makes seminars
exciting. It's the closest thing we have
to gladiator fights in the old days.
You bring a speaker from
somewhere, maybe Harvard, then you put them on the stage
and you just roast them for the whole hour.
Now you can do that to me just
fine, because this is not even my discovery.
Somebody else did it hundreds
of years ago. You should relish that
challenge. If this was a small seminar,
you will have time to do it. I don't have that much time,
but I also don't have that little time that if you don't
follow me, you cannot intervene. Okay, you have a question?
No.
Okay.
Now we are going to understand
why a force manufactured in this fashion is conservative,
okay? So let's calculate integral
F⋅dr for this force from some
starting--I'm just going to call it 1 and 2.
It means r_1
and r_2. What is F⋅dr?
It is F_xdx
F_ydy. That's the meaning of the dot
product. I hope you know what I mean by
this. I'm going from here to here,
that vector dr is some amount of dx and some
amount of dy. It's got jdy and
idx. And F is some other
thing, pointing in that direction, so
F⋅dr is this. Just look at this part.
This is what you're integrating.
That becomes,
except for this - sign, dU/dx times
dx dU/dy times dy.
Now what is that?
What is that?
What do you think it stands for?
In 1 dimension,
if I took df/dx times dx,
what am I calculating? Come on guys,
you know what the derivative means.
When you multiply the
derivative by some dx, what do you get?
Pardon me?
Yes, go ahead.
Take a shot.
Yeah.
Student: You're getting
what you originally ___________. Prof: You're finding the
change in the function f, right?
The change in the function
f is df/dx times dx.
There are other changes
proportional to dx^(3) and dx^(2),
but for small dx, that's all it is.
That's the meaning of the rate
of change. If you multiply the rate of
change by the change you get a change in the function.
If you're in 2 dimensions,
if you move from one point to another point,
you've got a function U that depends on x and on
y. Therefore it changes,
because you change x and it changes because you change
y. It's changing because of two
reasons. This is the change due to
change in x; that's the change due to change
in y. For very small,
infinitesimal dx and dy, that is the change.
That is just the change in the
function U. It follows then that if you add
up all these F⋅dr's from
1 to 2, this integral will give me U_1 -
U_2. Therefore integral
F⋅dr from 1 to 2 is U_1 -
U_2. So the magic is the following -
you pick a function U, you can think of the function
U as measured perpendicular to the blackboard.
It's like a height or something
coming out of the blackboard. You cooked up a force which is
related to the rate of change of the function U,
so that if you add all the F⋅dr`s,
you're getting the change in the function between two points.
And that change in the function
is independent of how you got there.
Think of U as a height
of a mountain, sitting on top of the xy
plane. Then F is really
proportional to the rate of change of the height of the
mountain. And F⋅dr is
the change when you move a distance dr,
a vector distance dr. You add all the changes,
what are you adding? You're adding the height,
your elevation. And when you're done,
the total change is the height difference between the final
point and the starting point. It doesn't depend on how you
climb that mountain. You can take a long path,
you can take a short path. As long as you're keeping track
of how many feet have I climbed, you're going to get only one
answer. So not every F has this
property, but an F derived from U by that
trick of taking the gradient has this property.
Therefore for that F,
you might say this is for conservative force.
But we always knew
F⋅dr is K_2 -
K_1, therefore we get
K_1 U_1 =
K_2 U_2.
In other words,
the U that you began with is the potential energy for
that problem. So what you're manufacturing is
pairs of potentials and forces. For every U that you
dream up, there's a force, which is obtained by gradient,
- gradient of U. And in that problem,
where that force is acting on particles,
the energy that is conserved will be the kinetic,
the function U at the starting point,
will be equal to kinetic plus potential at the end point.
And the interesting thing is,
there are no other conservative forces, except forces you can
get by taking a U and taking its gradient.
So all conservative forces fall
under this category, and they all have a conserved
quantity where the potential energy is the U from
which the F is derived. So we can say it as follows -
F = - gradient of U.
The gradient is the combination
of i times the x derivative and j
times the y derivative.
And U at 2 - U at
1 is - integral F⋅dr from 1
to 2. This is the 2 dimensional
generalization of something I wrote in 1 dimension somewhere.
Oh, way on the top.
Here, F = - dU
/dx, U(x_2)
− U(x_1)
is - integral of F. The only difference is the
integrand has become this F⋅dr,
rather than the F_x dx.
So you can get conservation of
energy in higher dimensions for those forces that come from a
potential. U is called a potential.
U is called a potential
energy. So now we can ask the following
question - suppose I give you a force.
I know if it's conservative or
not. How are you going to find out?
Now there are two options.
One is, you are so smart,
you can look at the function and say,
"Hey, this is the gradient of some other function
U." If you do that,
then you are done, because if it's the gradient of
some function, we know it's conservative.
But what if you cannot see that
right away? With simple polynomials,
you can easily guess. Like the example I gave you,
if F_x is -3x^(2)y^(2),
F_y is -2x^(3)y,
maybe you can guess that U is x^(3)y^(2).
But for more complicated
functions, you won't be able to do it.
So there's a process for
testing any force to see if it's conservative.
And the process is the
following. It argues as follows -
F_x is −dU/dx.
F_y is
-dU/dy. Then consider
dF_x / dy, which is
-d^(2)U/dydx. Then consider
dF_y / dx, which is
-d^(2)U/dxdy. And it's the property of
partial derivatives that the cross derivative,
second cross derivative, is independent of whether you
take dydx or dxdy. So let's go to our test case,
U = x^(3)y^(2). dU/dx =
3x^(2)y^(2) and dU/dy =
2x^(3)y. Now I'm saying,
take the x derivative of this y derivative.
That's written as
d^(2)U/dydx. y derivative of the
x derivative gives me 6x^(2)y.
Then I say take the x
derivative of the y derivative.
This gives me 3x^(2).
That means 6x^(2)y.
So cross derivatives are always
equal. Yes?
Student: Is this
_______ if we're going ________ taking __________?
Prof: Yes,
that's correct. So in 2 dimensions,
this is the only test you have to apply.
The test you have to apply is,
are the cross derivatives of the force equal?
Namely, is the y
derivative of the force F_x = to x
derivative of F_y?
You want to see if
dF_x/dy = dF_y/dx.
If that is true,
it is conservative. And it comes from the fact that
if F is the derivative of some function U,
then this mix derivative being equal is a diagnostic for that.
In 3 dimensions,
there are 2 more equations. We're not going to use them,
but I'll just mention what they are.
You get that by moving every
index 1 notch. So this x goes to
y, this y goes to z, this y goes to
z, this x goes to y.
There's one more condition.
And another condition,
z goes back to x, x goes z--let's
see, y goes to z, z goes to x =
dF_x/ dz.
I'm not worried about whether
you write this down or not, but I do want you to know in 2
dimensions, you have to remember this condition.
So this mathematic extends to
all dimensions, but I'm doing it for the case
of 2D. So let's take stock of what has
been done. What has been done is to think
from scratch on how to get the law of conservation of energy in
high dimensions. You start with the kinetic
energy and you ask, why does it change?
And you find it changes due to
the force, and you find the change in kinetic energy is the
line integral of F⋅dr from
start to finish. But that doesn't give you a
work energy-- it gives you a work energy
theorem, but not a law of conservation
of energy, because this integral can
depend on the path. Then you say,
maybe there are some functions for which the answer depends
only on the end points. Then we get a law of
conservation of energy, and we found out that we can
manufacture such functions at will,
by taking any function of U,
taking its gradient, or - the gradient.
Therefore that's a way to
generate conservative forces and they are the only conservative
forces there are. There's no other conservative
forces which are not gradients of something.
And the relation between the
potential and the force, you should know now.
It's what you learned in 1D,
suitably generalized to higher dimensions.
Change in potential is this and
the forces obtained from the potential by taking the
gradient, the change in potential is
obtained by integrating the force.
If you ever get confused about
sign and - sign and so on, go back to the harmonic
oscillator, for which the thing is 1
over--is kx^(2)/2. So let me give you one simple
example that we all know, which is the force of gravity
near the earth. So here is the earth,
so there is some x and y coordinates,
and here is z. So I might as well write down
the answer, because you guys all know this.
U for gravity is
mg times z. And the force of gravity,
which is -idu/dx -
jdu/dy, - kdu/dz,
all you got here is -kmg,
because it doesn't depend on x,
it doesn't depend on y. It depends only on z and
the z derivative is trivial.
Force is -mg pointing
down. So the force is the gradient of
the potential and you can ask, is the potential difference
between two points, U_2 -
U_1, is it equal to integral of
F⋅dr from 1 to 2?
You can easily check that.
If you go from here to here,
straight up, then the integral of -mg
times dz from start to finish, with a - sign is just
mgz - 0. So indeed, it is true.
And if you took two points,
not on top of each other, but something like this,
you can always do the integral from here to here,
where there's no change in potential,
and do the integral there, which will give the same
answer. So gravity's a very trivial
example, a very easy case, but you can see that
gravitation is a conservative force.
And you can see why it's a
conservative force, because if you took a path
where you went like this, let's say, and you went like
this, where you get the same answer as here and here.
Because in this path and this
path, you don't do any work, because F and dr
are perpendicular, right?
F is like this,
and the displacement is like that,
and it's the same thing, whereas here,
if you go like this, F is down and dr
is down, F is down,
dr is down, and it's the same F here
and here; are the same height so you get
the same answer. Okay, but now,
why did we spend quite a bit of time on conservative forces,
because that's not what the primary focus is?
It was electricity and
electrostatics. So we've got to ask ourselves,
does the electrostatic force meet the test of a conservative
force? Well, we know it will,
because otherwise I wouldn't have spent this much time
building up the stuff. But let's just verify that.
It looks like a very formidable
task. Let's see why it's a formidable
task to verify it's conservative.
I'm claiming that you've got
all kinds of these charges producing a field here and I
want to prove that the line integral of this force on any
closed loop is 0. Or I want to prove the integral
on one path is the integral on any other path,
for any 2 paths between any 2 points.
I don't have time to show all
that, right? If the theorem is wrong,
do you agree it may not be so hard to show that?
What will it take to show it is
not conservative? Yes?
Student: Just 2 paths
for which the integral is not equal.
Prof: Yes.
If you pick any 2 paths,
should be the same end points, so always the integral is not
equal, that's it. But the fact that they were
equal doesn't mean that somebody else will not find some other
two paths with some other starting points which they are
not equal. So how are we going to do this?
How are we going to test for
every configuration of electric field, every element of charges
is conservative? So what do you think a strategy
will be? Yes?
Any ideas?
How am I going to take all
possible charges and work it out each time?
Yes?
Student: I guess you
could find the electric field and use that test to see--
Prof: Right, but to find the electric field
and see if it works, you have to calculate the
electric field as a function of x,
y and z for every arrangement of charges.
Then you can ask,
is that true, those derivatives?
Yes?
Student: But what works
for one charge should work for all charges, because of
supersposition. Prof: There we go.
If it works for one charge,
it will work for any arrangement of charges.
So you may not have appreciated
fully the power of superposition,
but you can see now. If you did not realize that,
the problem looks insurmountable.
Who can handle every
arrangement of charges? But if charge by charge it is
true that the electric field due to charge 1 is conservative,
electric field due to charge 2 is conservative.
When I add them up,
the net field is E_1
E_2. The line integral of
E_1 E_2 is the line integral
of E_1_ the line integral of
E_2. E_1 gives 0 on
any closed loop, other will give 0 on any closed
loop. The trick is going to be just
do it for one charge. That's it.
If you can see it for one
charge, you get by superposition that it's true for any charge,
because the field is additive and the integral is additive,
okay. That's the two parts.
If you've got charges,
the net e is E_1
E_2. The integral of
E_1 E_2 is the integral of
E_1 the integral of
E_2. That's the way integrals work.
If each one vanishes on a
closed loop, you're done. So I'm just going to take one
charge. Here's one charge and let me
draw for convenience all the field lines coming out of it.
Then I want to see,
if I do a line integral, say from here to here,
will I get the same answer on some other path?
Let's test that.
So the first integral is from 1
to 2 on this path. Then I'm going to take another
one, which is different from it,
where you can see the answer is the same,
then go on making more and more changes,
and the answer will not change. So I'm going to take another
path where I go tangent to the lines, along a circle.
Then I go here,
till I get to that radius, then I turn around and do that.
I claim the integral will be
the same on this path, this one, that one and that
one. So I want you to think about
why it's going to be true, okay?
Think of it,
why it's okay to do that. You've got a new path but it
doesn't make a difference. Ready?
Tell you the answer now?
How many people don't know why
it's true? So other people know why it's
true. I'm going to ask the other
people. Okay, why is it true?
Student: I raised my
hand. Prof: You raised your
hand. All right.
Yes?
Student: It's true,
because when you're moving on the paths this way,
they're radial, so the radius is constant,
which would mean that there's no _______.
Prof: There's no what?
Student: There's no
change in force because the radius is constant.
Prof: That's not a good
reason. The force is changing.
It's that way there,
it's that way there. Why is it not?
Student: It's always
perpendicular to the force. Prof: The force and the
displacement are perpendicular in this section,
because force is radial, displacement by definition is
tangential along the circle. So dr is perpendicular
to F. Oh, by the way,
I should tell you something. I usually forget this.
We want the electric force to
be conservative, right?
The electric force in any
situation on a charge is q times the electric
field. You will agree that if the
electric field is conservative, multiplying by a number
q is not going to change the fact.
So I'm going to prove the
electric field is conservative, then the electric force on any
charge is just proportional to it.
It will also be conservative.
So really taking force on a
unit charge and seeing what happens.
So this section is for free.
This section--we'll come to
that--this section is also for free, because F looks
like that and dr looks like that.
Here you can pair the sections,
so that that section is paired with that section,
because you're moving the same radial distance with a radial
force. F⋅dr,
the dot product, is the same here,
section by section. Any work you're doing here,
you're doing here, therefore the answer is the
same. Then you can see that you can
go on adding all kinds of changes,
radial and angle, radial or angular,
and when you're done, all the radial integrals will
give you that part, and the angulars won't count.
So as you move around,
it's only when you're moving radially you're even aware that
you're doing work. Going in the angle doesn't
matter. And the net radial change in
going from here to here is that distance, no matter how you
move. Now you might say,
prove to me that that integral is the same as that integral,
because this was made up of radial and angular sections.
This doesn't seem that way,
but we can always make a very fine mesh in which every
displacement is approximated by radial angle or radial angular.
So you can approximate any
contour by sequence of radial and angular moves.
Angular moves are free.
Radial moves keep track of the
change in the distance times the force and you can pair them with
this reference. Every dr we move there
we have a dr here. So that is the pictorial proof
of this. So let's now calculate what is
the potential due to that electric field.
So we're going to do the
calculation. So we use the formula that
U at r_2 - U at
r_1 is - the integral e⋅dr
from 1 to 2. You see, we know it is
conservative. I've shown you with this
argument. So if you want to find the
potential energy between 1 and 2, you can take any path you
like between 1 and 2. So let's pick the points.
Here is point 1 and here is
point 2. You can pick any path you like,
so here is what I'm going to do.
I'm going to go radially out
and then I'm going to go along the arc.
So this is 1,
this is 2, but I'm going to start at some intermediate point
3, then go right. So work is done here;
no work is done there. So let's just find the work in
going from 1 to 3, which is -e_r
(is a vector in the radial direction)
¼Πε _0r^(2).
That is F.
The displacement is
e_r times the little length of
e_r. Do you understand that?
A tiny section here,
dr, has a magnitude equal to the change in the
radius, and a direction equal to the
radius unit rate-- the radial direction.
And you want to do this from
r_1 to r_2.
Okay, e_r
⋅e_r is 1 so I get--I forgot one
thing. You guys know what I forgot
here? One thing's missing.
Yes?
Student: Q.
Prof: When did you
notice that? Just now?
Student: Little bit.
Prof: Don't wait.
Don't wait even one second.
I told you what happens to me,
right? When I go home,
they tie me to a stool and they force me to watch this tape.
And it's just awful when you
think you dropped a symbol, and you're waiting for somebody
to say, "Hey, what about
q?" So don't have any--if you
hesitate, it's only to wait, not to stop me.
All right, so look,
I'll get the q because I know the formula from the time I
was in a sandbox. I know something is wrong,
but I don't want to get to that last stage, then go back and
change everything. So I forgot the q
sitting at the origin here. Okay, so this is
q/4Πε _0.
Integral of -1/r^(2) is just
1/r, so that's 1/r_2 −
1/r_1. Again, sorry,
I made a mistake now. The change in potential energy
is really q times e, but q is the
test charge, so I'm going to call this V.
So it's hard for you guys to do
it on the notebook, because I want to distinguish
between potential energy and potential.
In other words,
the force--now some other test charge q is the electric
field times q. And the force is going to be -
the gradient of the potential energy.
The electric field is going to
be - the gradient of the potential.
Yes?
Student: In that last
line, should that be a negative? Prof: A negative?
Student: Yeah.
Prof: Let's see.
There's a negative here,
and integral of -1/r^(2) is 1/r.
Student: Oh, okay.
Prof: So the upper
limit, then there's lower limit. So let me correct what I said.
I'm trying to find the
difference in what's called the potential.
It's not the potential energy.
Potential difference between
some point and some other point is the work done on a unit
charge. If you've got some other
charge, you should multiply it by the charge.
So the energy of some other
charge in this field, the potential energy,
is q times the potential.
So U is q times
this thing that I'm calling potential.
So we're comparing this
expression to this expression, we can conclude that V
at any point r is q/4Πε
_0(1/r) any constant.
Any constant,
because you cannot immediately say this guy is equal to this
guy and that guy is equal to that guy,
because even if you add a constant to both,
it will still be true. And we can pick the constant
any way we like. It is common to pick the
constant so that V goes to 0 as r goes to
infinity. That's what we will do.
That's the convention.
Then we can write here,
what is the V(r). This is the V due to a
single charge, and the U(r) = to
some q_0. I don't want to call q
again, because q is the guy producing the potential.
q_0 is the guy
experiencing the force. So q_0 times
V is the potential energy of charge q,
of charge q_0 sitting at that point.
Now let's also verify,
if you like, that this V(r)
does produce the electric field.
In other words,
I got the V by integrating electric field.
I want to go back,
just to get you familiar with this.
Getting the electric field from
the potential. Let's try to do that.
So E_x--let me
just do E_x--it's
−dV/dx, right?
-dV /dx =
-q/4Πε _0.
Derivative of 1/r is
-1/r^(2) times dr/dx.
What is dr/dx?
r is (x^(2)
y^(2) z^(2))^(1/2).
dr/dx is just
x divided by this whole thing, is x divided by
r. So this becomes
q/4Πε _0
(1/r^(2)) (x/r).
This is E_x.
So what will be the vector
E? Can you do that in your head?
Multiply this by i,
and do similarly for y and z and add them up.
What will you find?
You will find E will equal
q/4Πε _0
(1/r^(2)), times r over r.
Put r over r,
position vector over r, is the unit vector in the
radial direction. I'm doing this a little fast,
because this is something you can go home and check.
I don't want to spend too much
time doing it. I claim that if you took this
potential and took its x derivative or y
derivative or z derivative,
you get the x, y or z components of
the electric field. Okay, if you didn't follow
this, you should go home and check that this is true.
All right, so this is not that
important for me, this checking,
because we know it's going to work.
What is important for me is for
you to know that we've now found that the potential due to a
single charge leads to a conservative force.
The field due to a single
charge leads to a conservative one,
therefore by superposition, if you add any number of
charges, if you find the potential due
to all of them, then you take its gradient that
will give you a conservative electric field.
So what have you got so far?
Our gain is the following - we
have a law of conservation of energy,
which is the following - ½mv^(2) q
(this q is the q of the body you're interested
in) times V at the point r_1 = ½
mv_2 ^(2) q times v at the
point r_2. And what is V at any
point r? It's the sum
1/4Πε _0.
Let's imagine there are many,
many charges, q_1,
q_2, q_3,
each one at r_1,
r_2, r_3,
etc. This equals
q_i which is charge number I divided by the
length of the vector from r_i to the
r where I'm finding the potential.
In other words,
this is where I want the potential r.
This is
r_1_ and q_1
is sitting there. This is r_2
and q_2_ is sitting there and so
on. And the potential at the point
that I'm interested in is obtained by taking each q
divided by the distance from that point,
and this is the usual 1/4Πε
_0. That is the complete story,
and the charges are not discrete but continuous you can
write an integral. So every problem,
you first find the potential by adding the potential from all
the charges. Then its gradient will give it
the electric field. But we have now a law of
conservation of energy with this as the conserved energy.
The second advantage is that it
is easier to work with V than with E.
So why do you think it's easier
to find V and then find E from it,
by taking derivatives than the other way around?
Yes?
Student: V is a
scalar. Prof: Right.
Student:
> Prof: I'm sorry,
I didn't hear the last part. Student: It's always
easier to take derivatives than to integrate--
Prof: Very good. Yes.
Let me repeat that.
If you want to find the
potential due to many charges, you simply add the numbers
coming from each one. There are no arrows,
there are no vectors. Notice the formula for V
has no vectors in it. Each one contributes a number.
Each charge contributes a
number to the point they're interested in.
Add them all up,
find the number here, there, everywhere else.
Then take derivatives of your
answer to get the field. That's going to be a lot easier
than adding the vectors. So I'm going to illustrate that
with one simple problem and that's the last thing for today.
That's the dipole.
So here is our dipole.
Let's take a charge q,
-q at -a and q at a.
And we want to find the field
there. So let's call that distance
r_ and let's call that distance
r_-. These are not vectors;
these are just lengths. If you want to find the
electric field directly, you know what you have to do.
You have to find the vector
here and you've got to find the vector there due to the second
guy, add the two vectors and find the result.
Of course, you can do it,
but it's going to be very tedious,
because you have to find the direction of the vector here and
the direction of the vector here,
in terms of all these coordinates.
What we will do instead is to
find the potential everywhere, and take the x and
y derivatives of the potential to get the
E_x and E_y.
Well, let's do that.
So let's find the total
potential due to these two guys. The first one is
q/4Πε _0 divided by
1/r_ - 1/r_-. That's
it. Just wanted to simplify the
expression a little bit. So this is true for any
separation, but I want the limit in which
r_ and r_- are much
bigger than the separation between them.
So let's define an r
which is the position vector of the point you're interested in
from the center of the dipole. So we've got to find r_
and r_-.
You get that, you're done.
So vector r_
^( )you can see = r -
ai/2 and vector r_- = vector
r ai/2. Because the vector
ai/2 looks like this, so ai/2
r_ should give you r.
You can check that.
And you can check that
ai/2 r gives you r_-.
so these are the two
expressions. So what is r_ ?
r^( ) is the length of
the vector r_ . That's equal to the square root
of the length squared. The length squared of r_
squared = root of (r - ai/2)
⋅ (r - ai/2).
Student: Why is it
a i/2? Prof: Oh,
I'm sorry, it's not. It is half the distance,
but it's really only a. Thank you.
Yeah, in fact,
one puts the 2a as a separation to avoid all these
factors of 2. So now you take the length
squared. It's equal to r^(2)
a^(2) minus twice r⋅a.
The other one would be r^(2)
a^(2) twice r⋅a.
And that we can approximate as
r times 1 - twice r⋅
a/r^(2)^( )to the 1 half.
I'll tell you what I did here.
I'm going to neglect a
squared compared to r squared, because r is
much bigger than it. But I'm going to keep this
r⋅a term because this has got two
powers of a, where a is a small
number. It's got 1 power of a
and 1 power of r, so it's much stronger than this
term, so I keep that. Keeping the second term,
so I've forgotten the a^(2) term.
In this one,
from the theorem I've been giving you guys all the time,
it's 1 - r⋅ a/r^(2).
Remember 1 x^(n) is
roughly 1 nx dot dot dot if x is small,
and x is indeed small, because it's
r⋅a over r squared.
So now we can see V =
q/4Πε _0{1/[r(1-
r⋅ a/r^(2))]
- 1/[r(1 r⋅ a/r^(2))]} .
This is going to be found in
every textbook in the planet, so don't worry about it.
If you simplify this again by
taking it upstairs, you'll get twice
a⋅r /r^(2).
That's the potential.
Now twice a times q
is the dipole moment, so it's p⋅r
/4Πε _0r^(3).
That's the final piece,
the dipole moment. Dipole moment is a charge times
the separation between the charges, which is 2a
times q. Okay, so what one should do now
is, having taken this V,
one can take its x and y derivatives very easily
to calculate the electric field at each point.
It's very easy to take the
derivative of this, just pick the x or
y. It's one of the homework
problems I've assigned to you, where I show you the dipole
field. So the moral of the story is,
add the potential due to all the charges, then take the
derivative, because derivatives are easier to take.
Yes?
Student: The condition
for that result, is r much bigger than
a? Prof: Yes.
Up till some point,
everything is exact, but then this is long distance
property. Any questions?
Okay, thank you.
