Download Gradient, Divergence and Curl: the Basics

Gradient, Divergence and Curl: the Basics
We first consider the position vector, r:
r = xx + yy + zz,
where x, y, and z are rectangular unit vectors. Since the unit vectors for rectangular
coordinates are constants, we have for dr:
dr = dx x + dy y + dz z .
The operator, del:  is defined to be (in rectangular coordinates) as:
 = /x x + /y y + /z z ,
This operator operates as a vector.
1. Gradient
If the del operator,  operates on a scalar function, f(x,y,z), we get the gradient:
f = (f/x) x + (f/y) y + (f/z) z .
We can interpret this gradient as a vector with the magnitude and direction of the
maximum change of the function in space. We can relate the gradient to the differential
change in the function:
df = (f/x) dx + (f/y) dy + (f/z) dz = f  dr = df .
Since the del operator should be treated as a vector, there are two ways for a vector to
multiply another vector: dot product and cross product. We first consider the dot
product:
2. Divergence
The divergence of a vector is defined to be:
A = [/x x + /y y + /z z]  [Ax x + Ay y + Az z]
= (Ax /x) + (Ay /y) + (Az /z) ;
since the rectangular unit vectors are constant, x/x = 0 (etc.). This will not necessarily
be true for other unit vectors in other coordinate systems. We'll see examples of this
soon.
To get some idea of what the divergence of a vector is, we consider Gauss' theorem
(sometimes called the divergence theorem). We start with:
 A dV =  [(Ax /x) + (Ay /y) + (Az /z)] dx dy dz =
 [(Ax /x)dx dydz + (Ay /y)dy dxdz + (Az /z)dz dxdy]
We can see that each term as written in the last expression gives the value of the change
in vector A that cuts perpendicular through the surface. For instance, consider the first
term: (Ax /x)dx dydz . The first part: (Ax /x)dx gives the change in the xcomponent of A and the second part, dydz, gives the yz surface (or x component of the
surface, Sx) where we define the direction of the surface vector as that direction that is
perpendicular to its surface. The other two terms give the change in the component of A
that is perpendicular to the xz (Sy) and xy (Sz) surfaces. We thus can write:
 A dV = closed surface AdS
where the vector S is the surface area vector. Thus we see that the volume integral of the
divergence of vector A is equal to the net amount of A that cuts through (or diverges
from) the closed surface that surrounds the volume over which the volume integral is
taken. Hence the name divergence for A .
Example from electromagnetism: Consider a single point charge, q, and its electric field:
E = kq/r2 which points radially away from the center. Now let’s enclose that charge in a
sphere of radius, r, with the charge initially at the center. The dS vector will also point
radially away from the center. Since the magnitude of the electric field will remain the
same at all points on the surface (since r is constant), the dot product of E dS will be a
constant; and so the integral over the entire surface will simply give (kq/r2)*(4r2) =
4kq, and with k = 1/(4), we can write this as q/ . We recognize that the charge is
the source of all the field, so it really doesn’t matter where the charge is inside the sphere
– it doesn’t have to be at the center of the sphere. If there is no charge inside, then the
integral is equal to zero. Thus,    E dV =  closed surface E dS = qenclosed/If we
write    E dV = qenclosed/in differential form, we get   E = (dqenclosed/dV =
 where  is the charge density

For magnetic fields, since there are no monopoles, the magnetic pole density is
always zero, so   B = 0.
For gravity, we get:   g = -4Gm , where m is the mass density.
Due to the spherical symmetry of the fields from point charges and masses, we
will have to wait until we get expressed in spherical form to show that the divergence
theorem actually works – that is, that both sides of the equation give the same result. We
do this in the section Kinematics in 3-D.
3. Curl
The curl of a vector is defined to be:
A = [/x x + /y y + /z z]  [Ax x + Ay y + Az z] =
(Ay /x)(z) + (Az /x)(-y) + (Ax/y)(-z) + (Az/y)(x) + (Ax/z)(y) + (Ay/z)(-x)
= (Az/y - Ay/z) x + (Ax/z - Az /x) y + (Ay /x - Ax/y) z
where we have used the fact that the unit vectors do not change with position (x/x = 0)
and the fact that (xx=0 and xy=z, etc.). For other coordinate systems, unit vectors may
change with position.
To see what the curl of a vector means, we use Stokes Theorem. We begin with:
surface (A)  dS =
 [(Az/y - Ay/z) x + (Ax/z - Az /x) y + (Ay /x - Ax/y) z ]d[Sxx + Syy + Szz]
=  [(Az/y - Ay/z) dSx + (Ax/z - Az /x) dSy + (Ay /x - Ax/y) dSz] .
The dSx = dx*dy = dx dy, etc. However, we must worry about direction since xy = z but
yx = -z. After taking this into account, we get:
 (A)  dS =
=  [(Az/y + Ay/z) dydz + (Ax/z + Az /x) dxdz + (Ay /x + Ax/y) dxdy] .
Regrouping gives:
 (A)  dS =
 [Ax/z)dz + (Ax/y)dy]dx + [Ayx)dx + (Ayz)dz]dy + [Az/y)dy + (Az/x)dx]dz .
Now we note that dAx = (Ax/x)dx + (Ax/y)dy + (Ax/z)dz . In the above
integration, x was held constant when we integrated over the other variables, so the
(Ax/x)dx term is zero. Thus the above double integral becomes:
 (A)  dS =  [dAx dx + dAy dy + dAz dz] = closed loop Adr .
If the integral around a closed loop is not zero, then that implies that there is some
circulation of the vector field. Note that if the curl of the vector is zero everywhere, then
there cannot be any circulation of the vector field anywhere in space. Hence the name of
curl for A .
Another derivation of Stoke’s Theorem can be found in Volume 2 of the Feynmann
lectures at: http://feynmanlectures.caltech.edu/II_03.html#Ch3-S6 .
We can see an immediate use for the curl if we recall our discussion about work. If the
curl of a force field is zero, then the work done around a closed path must be zero
regardless of the closed path chosen. This means that the work done between any two
points must be path-independent! This then allows a potential energy change to be
defined for this force that depends only on the beginning and ending points.