1. Gradient
Consider a room in which the temperature is given by a scalar field, T,
so at each point (x,y,z) the temperature is T(x,y,z).
(We will assume that the temperature does not change over time.) At each point
in the room, the gradient of T at that point will show the direction the
temperature rises most quickly. The magnitude of the gradient will determine
how fast the temperature rises in that direction.Consider a surface whose height above sea level at a point (x, y) is H(x, y). The gradient of H at a point is a vector pointing in the direction of the steepest slope or grade at that point. The steepness of the slope at that point is given by the magnitude of the gradient vector.
This observation can be mathematically stated as follows. If the hill height function H is differentiable, then the gradient of H dotted with a unit vector gives the slope of the hill in the direction of the vector. More precisely, when H is differentiable, the dot product of the gradient of H with a given unit vector is equal to the directional derivative of H in the direction of that unit vector.
2. Divergence
Ñ · 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:
òòò Ñ · A dV = òòò [(¶Ax /¶x) + (¶Ay /¶y)
+ (¶Az /¶z)] dx
dy dz =
òòò [(¶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 x-component 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:
òòò Ñ · A dV = òò 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 Ñ · 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)*(4pr2) = 4pkq, and with k = 1/(4peo), we can write this as q/eo . 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, òòò Ñ · E dV = òò closed surface E· dS = qenclosed/eo . If we write òòò Ñ · E dV = qenclosed/eo in differential form, we get Ñ · E = (dqenclosed/eo )dV = r/eo where r is the charge density.
For magnetic fields, since there are no monopoles, the magnetic pole density is always zero, so Ñ · B = 0.
For gravity, we get: Ñ · g = -4pGrm , where rm 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 Ñ 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
Ñ ´ 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:
òò surface
(Ñ ´ A) · dS =
òò [(¶Az/¶y - ¶Ay/¶z) x + (¶Ax/¶z - ¶Az
/¶x) y + (¶Ay /¶x - ¶Ax/¶y) z ]· d[Sxx + Syy
+ Szz]
= òò [(¶Az/¶y - ¶Ay/¶z) dSx
+ (¶Ax/¶z - ¶Az /¶x) dSy
+ (¶Ay /¶x - ¶Ax/¶y) dSz] .
The dSx = dy*dz = dy dz, etc.
However, we must worry about direction since x´ y = z
but y´ x = -z.
After taking this into account, we get:
òò (Ñ ´ A) · dS =
= òò [(¶Az/¶y + ¶Ay/¶z)
dydz + (¶Ax/¶z + ¶Az /¶x)
dxdz + (¶Ay /¶x + ¶Ax/¶y)
dxdy] .
Regrouping gives:
òò (Ñ ´ A) · dS =
òò [¶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:
òò (Ñ ´ A) · dS = òò [dAx
dx + dAy dy + dAz dz] = òclosed 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 Ñ ´ A .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.
Example: Consider the problem we had in the last section:
Fx = ax3 + bxy2 + cz
Fy = ay3 + bx2y
Fz = cx .
We found that this force gave us a work that was independent of path (we tried two different paths and got the same result). Let’s look at the curl of this F. It should equal zero if the Work that this force does is independent of path.
Ñ ´ F = [¶ /¶x x + ¶ /¶y y + ¶ /¶z z] ´ [Fx x + Fy y + Fz z] =
= (¶Fz/¶y - ¶Fy/¶z) x + (¶Fx/¶z - ¶Fz /¶x) y + (¶Fy /¶x - ¶Fx/¶y) z
= (0 – 0) x + (c – c) y + (2bxy – 2bxy) z = 0.
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