Professor Dave again, let’s check out vector fields.
We spent a lot of time with vectors when we learned about linear algebra, and not surprisingly,
we won’t be done with them any time soon.
We are used to vectors being made of components which are scalars, for example <2,1> which
can also be written as 2i+1j.
We then got used to being able to move these vectors around however we wanted.
But now we are going to introduce vector fields, which can be thought of as functions that
take in coordinates and give back vectors.
These vector fields will end up having vectors assigned to every point throughout the coordinate
system, as shown here.
Of course, we are not limited to just two dimensions, vector fields can exist in any
number of dimensions, and an example of a three dimensional field is shown here.
These vector fields can have a variety of practical interpretations, such as a force
being exerted through space, or as the representation of motion.
But for now we will just focus on the general concept of a vector field and cover some basics.
As we mentioned, these vector fields can be thought of as functions, F of x and y, where
the function has vector components that depend on the coordinates x and y.
The most general way we could express this is F(x,y) = P(x,y)i + Q(x,y)j, where P and
Q are ordinary scalar functions that will depend on the coordinates x and y.
These scalar functions, also sometimes called scalar fields, determine the components of
the vectors at each point in the coordinate system.
In three dimensions, our vector field could look like this.
F(x,y,z) = P(x,y,z)i + Q(x,y,z)j + R(x,y,z)k.
For a quick example, consider the vector field F(x,y) = -yi + xj.
To get an idea of how this vector field looks, let’s plug in a few coordinates and plot
our resultant vectors.
We will take a few simple coordinates of the form (x,y) and plug them into the vector field.
To start, consider (1,0).
Here x = 1 and y = 0, so we get back the vector 0i + 1j, or <0,1>.
We will now plot this vector, originating from the coordinate we plugged in, (1,0).
We can quickly go through several more points and plot the resulting vectors as we go.
For (0,1) we get <-1,0>.
For (-1,0) we get <0,-1>.
For (0,-1) we get <1,0>.
For (1,1), <-1,1>.
For (-1,1), <-1,-1>.
For (-1,-1), <1,-1>.
And for (1,-1), we get <1,1>.
We are starting to see a clear circular pattern here.
If we continued with more points we would see this continue, and that vectors increase
in size as we move away from the origin.
This is the general approach that can be used when wanting to see what a vector field looks like.
Obviously we can’t plot out the vectors at all the infinite points of the coordinate
system, but we can see the general trend quite quickly.
Now let’s bring back the del operator we introduced in the previous tutorial regarding
gradient.
Recall that del is a vector made of the differential operators or if in three dimensions
.
The gradient of some function f, del f, is itself a vector field with components being
the partial derivatives of the function, so .
To get more specific, if we have a vector field F, and it can be written as the gradient
of a function f, the vector field is called a “conservative vector field” and the
function f is called a “potential function” for the vector field F.
There are two more uses for the del operator with vector fields, and these involve the
vector dot product and cross product.
We can take the del operator and dot it into a vector field F.
This is referred to as the “divergence” of F. Similarly, we can take del cross F,
and this is referred to as the “curl” of F. Let’s take a closer look at what these
expressions are.
Recall that we can write a vector field F in the form
so for now we’ll just focus on three dimensions, while P, Q, and R will be ordinary scalar
functions that depend on x, y, and z.
While we can take the divergence in any number of dimensions, the curl can only be taken
in three dimensions.
First let’s look at the divergence, del dot F. Recalling from a previous tutorial
how the dot product works, we will simply find the product of the corresponding elements
of the vectors, and then add those up.
So del dot F ends up giving us d/dx P + d/dy Q + d/dz R, which means we take the partial
derivative of P with respect to x, plus the partial derivative of Q with respect to y,
plus the partial derivative of R with respect to z.
So the divergence of a vector field will no longer be a vector field, rather it will be
a scalar field.
Consider the vector field F = <(x^2)y, -x/y, xyz>.
To find the divergence we will take del dot this expression, which gives d/dx[(x^2)y]
+ d/dy(-x/y) + d/dz(xyz).
All of these derivatives follow the power rule that we already know, so we end up with
2xy + x/(y^2) + xy, which simplifies to 3xy + x/(y^2).
We will get more into the application of divergence in a future tutorial, but for now you can
think of it as a measure of how much of the vector field is going out of some point.
Next let’s go over the curl.
As we mentioned, the curl will be the cross product between del and a vector field F.
Remembering how the cross product works from a previous tutorial, we set up a matrix with
i, j, and k on the top row, the elements of del on the second row, and the elements of
F on the last row.
Let’s once again consider the vector field F = <(x^2)y, -x/y, xyz>.
To find the curl we will take the determinant of the matrix i, j, k; d/dx, d/dy, d/dz; (x^2)y,
-x/y, xyz.
Taking the determinant we get i times the quantity d/dy(xyz) - d/dz(-x/y), minus j times
the quantity d/dx(xyz) - d/dz[(x^2)y], plus k times the quantity d/dx(-x/y) - d/dy[(x^2)y].
Now we take all these derivatives.
For the first term, this y becomes one so we get xz, and this expression has no variable,
so it becomes zero.
For the second term, this x becomes one so we get yz, and again here there is no z so
this becomes zero.
For the third term, this x becomes one, leaving us with negative one over y, and this y becomes
one, leaving us with x^2.
Simplifying, we get xzi – yzj + (-1/y + x^2)k, or in vector form <xz, -yz, -1/y +
x^2>.
This is the curl of our vector field.
Notice that the result is also a vector field.
So while the divergence gives a scalar field, the curl gives a vector field.
The curl represents the rotation of a vector field F, the direction of the curl being the
axis of rotation and the magnitude of the curl being the magnitude of rotation.
Before we wrap up, let’s mention some interesting properties of divergence and curl.
First off, for a function that has continuous second-order partial derivatives, the curl
of its gradient is always the zero vector.
This means that the curl of any conservative vector field is zero, because being conservative
means the field can be written as the gradient of some function.
We can go ahead and show this by writing the gradient as and taking
the curl.
To do this, we set up the matrix shown here and take the determinant.
We end up getting i times (d^2)f/dydz – (d^2)f/dzdy, minus j times (d^2)f/dxdz – (d^2)f/dzdx,
plus k times (d^2)f/dxdy – (d^2)f/dydx.
This notation is how we express taking multiple derivatives of the function f, with the order
of differentiation being the variables on the denominator from right to left.
If we took the derivative with respect to the same variable twice, say x, it would be
expressed as (d^2)f/dx^2.
Now here’s where we need the continuous second derivative to come into play.
If these derivatives are all continuous, that means the order in which we take derivatives
does not matter, meaning that (d^2)f/dydz is the same as (d^2)f/dzdy, and so on.
So all of these terms cancel and we are left with a vector that has a zero for all its
components.
This is the zero vector and we have proved this property.
Another property is that if the components of F have continuous second-order partial
derivatives, then the divergence of a curl is always zero.
We can prove this as well.
Jumping ahead and writing out the components of the curl, we get del cross F being equal
to .
Remembering what del represents and taking the dot product of del and this expression
will result in the derivative of the first element with respect to x, plus the derivative
of the second with respect to y, plus the derivative of the third with respect to z.
This gives us [(d^2)R/dxdy – (d^2)Q/dxdz)] + [(d^2)P/dydz – (d^2)R/dydx)] + [(d^2)Q/dzdx
– (d^2)P/dzdy)].
Once again, if these are all continuous, the order of differentiation does not matter,
and since this is all just one long series of terms being added and subtracted anyway,
we can find the corresponding terms and get them all to cancel, giving us zero and therefore
verifying the property.
Vector fields are a very useful concept, and being able to take the divergence and curl
of these fields allows for a surprisingly wide variety of calculations.
There are several important theorems that utilize these operations, so let’s check
out some of those next, but first, let’s check comprehension.