Consider a vector field F(x,y) = (P(x,y),Q(x,y)). Denoting partial derivatives with subscripts as usual, the divergence of F(x,y), div(F(x,y)), is given by
div(F(x,y)) = Px(x,y) + Qy(x,y)
and the curl of F(x,y), curl(F(x,y)), is given by
div(F(x,y)) = Qx(x,y) - Py(x,y)
The questions we will answer in this setion of the notes are:
All of these questions are closely related. We will adress them in four subsections below. The last two are rather deep; they adress the uniqueness and existence of the following question:
To what extent do the divergence and curl of a vector field determine that vector
field?
This question isn't easy, but amazingly, it can be answered just
using the vector calculus that we know.
To see why it is interesting,
recall that Maxwell's equations for electrostatics specify the electric
field E(x,y) and the magnetic field
B(x,y) exactly by specifying their divergence and curl.
For example, in a static situation, the curl of the electric field is zero, and
the divergence of the electric field is a multiple of the electric charge
density c(x,y):
div(E(x,y)) = c(x,y) and
curl(E(x,y)) = 0
Likewise, the electrostatic equations for hte magentic field
are
div(B(x,y)) = 0 and
curl(B(x,y)) = j(x,y)
where j(x,y) is a current density.
Of course, one usually sees these in three dimensions. So here consider a situation in which neiter the charge density nor the current density depends on z. Then the third dimension frops out, as we have it here: planar electrstatics. Moreover, once we have understood the planar case thoroughly, it is very easy to move on to the full three dimensional case. So we will
stay in this section of the notes.
Looking at the elctrostatic eqautions, it looks as if they are only going to specify the electric and magnetic fields the divergence and curl uniquely specify a vector field.
But it can't be that the divergence and curl are
enough to specify a vector field
F(x,y) = (P(x,y),Q(x,y)), because if
we add any
constants to P or Q, these constants will drop out
when we take the partial derivatives in computing the divergence
and curl. Thus
F(x,y) = (xy, x+y) and
G(x,y) = (xy+1,x+y-3)
have the same divergence and the same curl.
Why isn't this a problem for Mr. Maxwell's equations? Because of "boundry conditions"!
For instance, if the total amount of charge is finite, then the electric field
E(x,y) must tend to zero as (x,y)
tends to infinity. This "boundary condition" picks out the physical solution
of Maxwell's equations from among all the others.
If you take a solution of Maxwell's
eqautions that do satisfy this boundary condition, and
you add a constant vector field to the solution,
the equations
will still be satisfied, but not the boundary condition -- the new limiting
value at infinity is clearly the constant vector field that you just added on.
We shall see below that among all solutions to Maxwell's equations, as
above, there is at most one that satisfies the boundary condition.
In fact, we will prove the more general fact: any vector field whose
divergecne and curl both vanish tthroughout the plane is either constant or unbounded.
Here are some vector fields
that have both zero curl and zero divergence, and yet are not constant:
F(x,y) = (x,-y)
G(x,y) = (excos(y),-exsin(y))
As indicated above, they are unbounded -- their magnitudes take on arbitrarily large values.
But now we are getting a bit ahead of ourselves. That's O.K.; I just wanted to
make clear that the considerations we are undertaking here are motivated by
applications. It was partly because Mr. Maxwell understood the meaning of
the divergence and curl, and the extent that they determine the vector field
they came from,
better than his predecessors that he was able to formulate his equations in the
first place. Now, back to the beginning.
Links to the Subsections