The following content is
provided under a Creative Commons license.
Your support will help MIT OpenCourseWare continue to offer
high quality educational resources for free.
To make a donation or to view additional materials from
hundreds of MIT courses, visit MIT OpenCourseWare at
ocw.mit.edu. Today I am going to tell you
about flux of a vector field for a curve.
In case you have seen flux in physics,
probably you have seen flux in space,
and we are going to come to that in a couple of weeks,
but for now we are still doing everything in the plane.
So bear with me if you have seen a more complicated version
of flux. We are going to do the easy one
first. What is flux?
Well, flux is actually another kind of line integral.
Let's say that I have a plane curve and a vector field in the
plane. Then the flux of F across a
curve C is, by definition, a line integral,
but I will use notation F dot n ds.
I have to explain to you what it means, but let me first box
that because that is the important formula to remember.
That is the definition. What does that mean?
First, mostly I have to tell you what this little n is.
The notation suggests it is a normal vector,
so what does that mean? I have a curve in the plane and
I have a vector field. Let's see.
The vector field will be yellow today.
And I will want to integrate along the curve the dot product
of F with the normal vector to the curve, a unit normal vector
to the curve. That means a vector that is at
every point of the curve perpendicular to the curve and
has length one. N everywhere will be the unit
normal vector to the curve C pointing 90 degrees clockwise
from T. What does that mean?
That means I have two normal vectors, one that is pointing
this way, one that is pointing that way.
I have to choose a convention. And the convention is that the
normal vector that I take goes to the right of the curve as I
am traveling along the curve. You mentioned that you were
walking along this curve, then you look to your right,
that is that direction. What we will do is just,
at every point along the curve, the dot product between the
vector field and the normal vector.
And we will sum that along the various pieces of the curve.
What this notation means is that if we actually break C into
small pieces of length delta(s) then the flux will be the limit,
as the pieces become smaller and smaller,
of the sum of F dot n delta S. I take each small piece of my
curve, I do the dot product between F and n and I multiply
by the length of a piece. And then I add these together.
That is what the line integral means.
Of course that is, again, not how I will compute
it. Just to compare this with work,
conceptually it is similar to the line integral we did for
work except the line integral for work -- Work is the line
integral of F dot dr, which is also the line integral
of F dot T ds. That is how we reformulated it.
That means we take our curve and we figure out at each point
how big the tangent component -- I guess I should probably take
the same vector field as before. Let's see.
My field was pointing more like that way.
What I do at any point is project F to the tangent
direction, I figure out how much F is going along my curve and
then I sum these things together.
I am actually summing -- -- the tangential component of my field
F. Roughly-speaking the work
measures, you know, when I move along my curve,
how much I am going with or against F.
Flux, on the other hand, measures, when I go along the
curve, roughly how much the field is going to across the
curve. Counting positively what goes
to the right, negatively what goes to the
left. Flux is integral F dot n ds,
and that one corresponds to summing the normal component of
a vector field. But apart from that
conceptually it is the same kind of thing.
Just the physical interpretations will be very
different, but for a mathematician these
are two line integrals that you set up and compute in pretty
much the same way. Let's see.
I should probably tell you what it means.
Why do we make this definition? What does it correspond to?
Well, the interpretation for work made a lot of sense when F
was representing a force. The line integral was actually
the work done by the force. The interpretation for flux
makes more sense if you think of F as a velocity field.
What is the interpretation? Let's say that for F is a
velocity field. That means I am thinking of
some fluid that is moving, maybe water or something,
and it is moving at a certain speed.
And my vector field represents how things are moving at every
point of the plane. I claim that flux measures how
much fluid passes through -- -- the curve C per unit time.
If you imagine that maybe you have a river and you are somehow
building a damn here, a damn with holes in it so that
the water still passes through, then this measures how much
water passes through your membrane per unit time.
Let's try to figure out why this is true.
Why does this make sense? Let's look at what happens on a
small portion of our curve C. I am zooming in on my curve C.
I guess I need to zoom further. That is a little piece of my
curve, of length delta S, and there is a fluid flow.
On my picture things are flowing to the right.
Here I am drawing a constant vector field because if you zoom
in enough then your vectors will pretty much be the same
everywhere. If you enlarge the picture
enough then things will be pretty much a uniform flow.
Now, how much stuff goes through this little piece of
curve per unit time? Well, what happens over time is
the fluid is moving while my curve is staying the same place
so it corresponds to something like this.
I claim that what goes through C in unit time is actually going
to be a parallelogram. Here is a better picture.
I claim that what will be going through C is this shaded
parallelogram to the left of C. Let's see.
If I move for unit time it works.
That is the stuff that goes through my curve,
for a small portion of curve in unit time.
And, of course, I would need to add all of
these together to get the entire curve.
Let's try to understand how big this parallelogram is.
To know how big this parallelogram is I would like to
use base times height or something like that.
And maybe I want to actually flip my picture so that the base
and the height make more sense to me.
Let me actually turn it this way.
And, in case you have trouble reading the rotated picture,
let me redo it on the board. What passes through a portion
of C in unit time is the contents of a parallelogram
whose base is on C. So it has length delta s.
That is a piece of C. And the other side is going to
be given by my velocity vector F.
And to find the height of this thing, I need to know what
actually the normal component of this vector is.
If I call n the unit normal vector to the curve then the
area is base times height. The base is delta S and the
height is the normal component of F, so it is F dot n.
And so you see that when you sum these things together you
get, what I said, flux.
Now, if you are worried about the fact that actually -- If
your unit time is too long then of course things might start
changing as it flows. You have to take the time unit
and the length unit that are sufficiently small so that
really this approximation where C is a straight line and where
flow is at constant speed are valid.
You want to take maybe a segment here that is a few
micrometers. And the time unit might be a
few nanoseconds or whatever, and then it is a good
approximation. What I mean by per unit time
is, well, actually, that works, but you want to
think of a really, really small time.
And then the amount of matter that passes in that really,
really small time is the flux times the amount of time.
Let's be a tiny bit more careful.
And what I am saying is the amount of stuff that passes
through C depends actually on whether n is going this way or
the opposite way. Actually,
what is implicit in this explanation is that I am
counting positively all the stuff that flows across C in the
direction of n and negatively what flows in the opposite
direction. What flows to the right of C,
well, across C from left to right is counted positively.
While what flows right to left is counted negatively.
So, in fact, it is the net flow through C
per unit time. Any questions about the
definition or the interpretation or things like that?
Yes? Well, you can have both not in
the same small segment. But it could be that,
well, imagine that my vector field accidentally goes in the
opposite direction then this part of the curve,
while things are flowing to the left,
contributes negatively to flux. And here maybe the field is
tangent so the normal component becomes zero.
And then it becomes positive and this part of the curve
contributes positively. For example,
if you imagine that you have a round tank in which the fluid is
rotating and you put your dam just on a diameter across then
things are going one way on one side,
the other way on the other side,
and actually it just evens out. We don't have complete
information. It is just the total net flux.
OK. If there are no other questions
then I guess we will need to figure out how to compute this
guy and how to actually do this line integral.
Well, let's start with a couple of easy examples.
Let's say that C is a circle of radius (a) centered at the
origin going counterclockwise. And let's say that our vector
field is xi yj. What does that look like?
Remember, xi plus yj is a vector field that is pointing
radially away from the origin. Because at every point it is
equal to the vector from the origin to that point.
Now, if we have a circle and let's say we are going
counterclockwise. Actually, I have a nicer
picture. Let me do it here.
That is my curve and my vector field.
And the normal vector, see, when you go counterclockwise in
a closed curve, this convention that a normal
vector points to the right of curve makes it point out.
The usual convention, when you take flux for a closed
curve, is that you are counting the flux going out of the region
enclosed by the curve. And, of course,
if you went clockwise it would be the other way around.
You choose to do it the way you want, but the most common one is
to count flux going out of the region.
Let's see what happens. Well, if I am anywhere on my
circle, see, the normal vector is sticking straight out of the
circle. That is a property of the
circle that the radial direction is perpendicular to the circle.
Actually, let me complete this picture.
If I take a point on the circle, I have my normal vector
that is pointing straight out so it is parallel to F.
Along C we know that F is parallel to n,
so F dot n will be equal to the magnitude of F times,
well, the magnitude of n, but that is one.
Let me put it anywhere, but that is the unit normal
vector. Now, what is the magnitude of
this vector field if I am at a point x, y?
Well, it is square root of x squared plus y squared,
which is the same as the distance from the origin.
So if this distance, if this radius is a then the
magnitude of F will just be a. In fact, F dot n is constant,
always equal to a. So the line integral will be
pretty easy because all I have to do is the integral of F dot n
ds becomes the integral of a ds. (a) is a constant so I can take
it out. And integral ds is just a
length of C which is 2pi a, so I will get 2pi a squared.
And that is positive, as we expected,
because stuff is flowing out of the circle.
Any questions about that? No.
OK. Just out of curiosity,
let's say that we had taken our other favorite vector field.
Let's say that we had the same C, but now the vector field
. Remember, that one goes
counterclockwise around the origin.
If you remember what we did several times,
well, along the circle that vector field now is tangent to
the circle. If it is tangent to the circle
it doesn't have any normal component.
The normal component is zero. Things are not flowing into the
circle or out of it. They are just flowing along the
circle around and around so the flux will be zero.
F now is tangent to C. F dot n is zero and,
therefore, the flux will be zero.
These are examples where you can compute things
geometrically. And I would say,
generally speaking, with flux, well,
if it is a very complicated field then you cannot.
But, if a field is fairly simple,
you should be able to get some general feeling for whether your
answer should be positive, negative or zero just by
thinking about which way is my flow going.
Is it going across the curve one way or the other way?
Still no questions about these examples?
The next thing we need to know is how we will actually compute
these things because here, yeah, it works pretty well,
but what if you don't have a simple geometric interpretation.
What if I give you a really complicated curve and then you
have trouble finding the normal vector?
It is going to be annoying to set up things this way.
Actually, there is a better way to do it in coordinates.
Just as we do work, when we compute this line
integral, usually we don't do it geometrically like this.
Most of the time we just integrate M dx plus N dy in
coordinates. That is a similar way to do it
because it is, again, a line integral so it
should work the same way. Let's try to figure that out. How do we do the calculation in
coordinates, or I should say using components?
That is the general method of calculation when we don't have
something geometric to do. Remember, when we were doing
things for work we said this vector dr, or if you prefer T
ds, we said just becomes symbolically dx and dy.
When you do the line integral of F dot dr you get line
integral of n dx plus n dy. Now let's think for a second
about how we would express n ds. Well, what is n ds compared to
T ds? Well, M is just T rotated by 90
degrees, so n ds is T ds rotated by 90 degrees.
That might sound a little bit outrageous because these are
really symbolic notations but it works.
I am not going to spend too much time trying to convince you
carefully. But if you go back to where we
wrote this and how we tried to justify this and you work your
way through it, you will see that n ds can be
analyzed the same way. N is T rotated 90 degrees
clockwise. That tells us that n ds is --
How do we rotate a vector by 90 degrees?
Well, we swept the two components and we put a minus
sign. You have dy and dx.
And you have to be careful where to put the minus sign.
Well, if you are doing it clockwise, it is in front of dx. Well, actually,
let me just convince you quickly.
Let's say we have a small piece of C.
If we do T delta S, that is also vector delta r.
That is going to be just the vector that goes along the curve
given by this. Its components will be indeed
the change in x, delta x, and the change in y,
delta y. And now, if I want to get n
delta S, well, I claim now that it is
perfectly valid and rigorous to just rotate that by 90 degrees.
If I want to rotate this by 90 degrees clockwise then the x
component will become the same as the old y component.
And the y component will be minus delta x.
Then you take the limit when the segment becomes shorter and
shorter, and that is how you can justify this.
That is the key to computing things in practice.
It means, actually, you already know how to compute
line integrals for flux. Let me just write it explicitly.
Let's say that our vector field has two components.
And let me just confuse you a little bit and not call them M
and N for this time just to stress the fact that we are
doing a different line integral. Let me call them P and Q for
now. Then the line integral of F dot
n ds will be the line integral of
dot product . That will be the integral of -
Q dx P dy. Well, I am running out of space
here. It is integral along C of
negative Q dx plus P dy. And from that point onwards you
just do it the usual way. Remember, here you have two
variables x and y but you are integrating along a curve.
If you are integrating along a curve x and y are related.
They depend on each other or maybe on some other parameter
like T or theta or whatever. You express everything in terms
of a single variable and then you do a usual single integral.
Any questions about that? I see a lot of confused faces
so maybe I shouldn't have called my component P and Q. If you prefer,
if you are really sentimentally attached to M and N then this
new line integral becomes the integral of - N dx M dy.
If a problem tells you compute flux instead of saying compute
work, the only thing you change is
instead of doing M dx plus N dy you do minus N dx plus M dy.
And I am sorry to say that I don't have any good way of
helping you remember which one of the two gets the minus sign,
so you just have to remember this formula by heart.
That is the only way I know. Well, you can try to go through
this argument again, but it is really best if you
just remember that formula. I am not going to do an example
because we already know how to do line integrals.
Hopefully you will get to see one at least in recitation on
Monday. That is all pretty good.
Let me tell you now what if I have to compute flux along a
closed curve and I don't want to compute it?
Well, remember in the case of work we had Green's theorem.
We saw yesterday Green's theorem.
Let's us replace a line integral along a closed curve by
a double integral. Well, here it is the same.
We have a line integral along a curve.
If it is a closed curve, we should be able to replace it
by a double integral. There is a version of Green's
theorem for flux. And you will see it is not
scarier than the other one. It is perhaps less scarier or
perhaps just as scary or just not as scary,
depending on how you feel about it, but it works pretty much the
same way. What does Green's theorem for
flux say? It says if C is a curve that
encloses a region R counterclockwise and if I have a
vector field that is defined everywhere,
not just on C but also inside, so also on R.
Well, maybe I should give names to the components.
If you will forgive me for a second, I will still use P and Q
for now. You will see why.
It is defined and differentiable in R.
Then I can actually -- -- replace the line integral for
flux by a double integral over R of some function.
And that function is called the divergence of F dA.
This is the divergence of F. And I have to define for you
what this guy is. The divergence of a vector
field with components P and Q is just P sub x Q sub y.
This one is actually easier to remember than curl because you
just take the x component, take its partial with respect
to x, take the y component,
take its partial with respect to y and add them together.
No signs. No switching things around.
This one is pretty straightforward.
The picture again is if I have my curve C going
counterclockwise around a region R and I want to find the flux of
some vector field F that is everywhere in here.
Maybe some parts of C will contribute positively and some
parts will contribute negatively.
Just to reiterate what I said, positively here means,
because we are going counterclockwise,
the normal vector points out of the region.
This guy here is the flux out of R through C.
That is the formula. Any questions about what the
statement says or how to use it concretely?
No. OK.
It is pretty similar to Green's theorem for work.
Actually, I should say -- This is called Green's theorem in
normal form also. Not that the other one is
abnormal, but just that the old one for work was,
you could say, in tangential form.
That just means, well, Green's theorem,
as seen yesterday was for the line integral F dot T ds,
integrating the tangent component of F.
The one today is for integrating the normal component
of F. OK. Let's prove this.
Good news. It is much easier to prove than
the one we did yesterday because we are just going to show that
it is the same thing just using different notations. How do I prove it?
Well, maybe actually it would help if first,
before proving it, I actually rewrite what it
means in components. We said the line integral of F
dot n ds is actually the line integral of - Q dx P dy.
And we want to show that this is equal to the double integral
of P sub x Q sub y dA. This is really one of the
features of Green's theorem. No matter which form it is,
it relates a line integral to a double integral.
Let's just try to see if we can reduce it to the one we had
yesterday. Let me forget what these things
mean physically and just focus on the math.
On the math it is a line integral of something dx plus
something dy. Let's call this guy M and let's
call this guy N. Let M equal negative Q and N
equal P. Then this guy here becomes
integral of M dx plus N dy. And I know from yesterday what
this is equal to, namely using the tangential
form of Green's theorem. Green for work.
This is the double integral of curl of this guy.
That is Nx minus My dA. But now let's think about what
this is in terms of M and N. Well, we said that M is
negative Q so this is negative My.
And we said P is the same as N, so this is Nx.
Just by renaming the components, I go from one form
to the other one. So it is really the same
theorem. That's why it is also called
Green's theorem. But the way we think about it
when we use it is different, because one of them computes
the work done by a force along a closed curve,
the other one computes the flux maybe of a velocity field out of
region. Questions?
Yes? That is correct.
If you are trying to compute a line integral for flux,
wait, where did I put it? A line integral for flux just
becomes this. And once you are here you know
how to compute that kind of thing.
The double integral side does not even have any kind of
renaming to do. You know how to compute a
double integral of a function. This is just a particular kind
of function that you get out of a vector field,
but it is like any function. The way you would evaluate
these double integrals is just the usual way.
Namely, you have a function of x and y, you have a region and
you set up the bounds for the isolated integral.
The way you would evaluate the double integrals is really the
usual way, by slicing the region and
setting up the bounds for iterated integrals in dx,
dy or dydx or maybe rd, rd theta or whatever you want.
In fact, in terms of computing integrals, we just have two sets
of skills. One is setting up and
evaluating double integrals. The other one is setting up and
evaluating line integrals. And whether these line
integrals or double integrals are representing work,
flux, integral of a curve, whatever,
the way that we actually compute them is the same.
Let's do an example. Oh, first. Sorry.
This renaming here, see, that is why actually I call my
components P and Q because the argument would have gotten very
messy if I had told you now I call M ,N and I call N minus M
and so on. But, now that we are through
with this, if you still like M and N
better, you know, what this says -- The
formulation of Green's theorem in this language is just
integral of minus N dx plus M dy is the double integral over R of
Mx plus Ny dA. Now let's do an example.
Let's look at this picture again, the flux of xi plus yj
out of the circle of radius A. We did the calculation directly
using geometry, and it wasn't all that bad.
But let's see what Green's theorem does for us here. Example.
Let's take the same example as last time.
F equals xi yj. C equals circle of radius a
counterclockwise. How do we set up Green's
theorem. Well, let's first figure out
the divergence of F. The divergence of this field,
I take the x component, which is x, and I take its
partial respect to x. And then I do the same with the
y component, and I will get one plus one equals two.
So, the divergence of this field is two.
Now, Green's theorem tells us that the flux out of this region
is going to be the double integral of 2 dA.
What is R now? Well, R is the region enclosed
by C. So if C is the circle,
R is the disk of radius A. Of course, we can compute it,
but we don't have to because double integral of 2dA is just
twice the double integral of dA so it is twice the area of R.
And we know the area of a circle of radius A.
That is piA2. So, it is 2piA2.
That is the same answer that we got directly,
which is good news. Now we can even do better.
Let's say that my circle is not at the origin.
Let's say that it is out here. Well, then it becomes harder to
calculate the flux directly. And it is harder even to guess
exactly what will happen because on this side here the vector
field will go into the region so the contribution to flux will be
negative here. Here it will be positive
because it is going out of the region.
There are positive and negative terms.
Well, it looks like positive should win because here the
vector field is much larger than over there.
But, short of computing it, we won't actually know what it
is. If you want to do it by direct
calculation then you have to parametize this circle and
figure out what the line integral will be.
But if you use Green's theorem, well, we never used the fact
that it is the circle of radius A at the origin.
It is true actually for any closed curve that the flux out
of it is going to be twice the area of the region inside.
It still will be 2piA2 even if my circle is anywhere else in
the plane. If I had asked you a trick
question where do you want to place this circle so that that
the flux is the largest? Well, the answer is it doesn't
matter. Now, let's just finish quickly
by answering a question that some of you,
I am sure, must have, which is what does divergence
mean and what does it measure? I mean, we said for curl,
curl measures how much things are rotating somehow.
What does divergence mean? Well, the answer is divergence
measures how much things are diverging.
Let's be more explicit. Interpretation of divergence.
You can think of it, you know, what do I want to say
first? If you take a vector field that
is a constant vector field where everything just translates then
there is no divergence involved because the derivatives will be
zero. If you take the guy that
rotates things around you will also compute and find zero for
divergence. This is not sensitive to
translation motions where everything moves together or to
rotation motions, but instead it is sensitive to
explaining motions. A possible answer is that it
measures how much the flow is expanding areas.
If you imagine this flow that we have here on the picture,
things are moving away from the origin and they fill out the
plane. If we mention this fluid
flowing out there, it is occupying more and more
space. And so that is what it means to
have positive divergence. If you took the opposite vector
field that contracts everything to the origin that will have
negative divergence. That is a good way to think
about it if you are thinking of a gas maybe that can expand to
fill out more volume. If you thinking of water,
well, water doesn't really shrink or expand.
The fact that it is taking more and more space actually means
that there is more and more water.
The other way to think about it is divergence is the source
rate, it is the amount of fluid that
is being inserted into the system,
that is being pumped into the system per unit time per unit
area. What div F equals two here
means is that here you actually have matter being created or
being pumped into the system so that you have more and more
water filling more and more space as it flows.
But, actually, divergence is not two just at
the origin. It is two everywhere.
So, in fact, to have this you need to have a
system of pumps that actually is in something water absolutely
everywhere uniformly. That is the only way to do this.
I mean if you imagine that you just have one spring at the
origin then, sure, water will flow out,
but as you go further and further away it will do so more
and more slowly. Well, here it is flowing away
faster and faster. And that means everywhere you
are still pumping more water into it.
So, that is what divergence measures.
