Divergence

If is constant in space then it is easily demonstrated that the net flux out of is zero. In fact,

(A.129) |

because the vector area of a closed surface is zero.

Suppose, now, that is not uniform in space. Consider a very small rectangular volume over which hardly varies. The contribution to from the two faces normal to the -axis is

(A.130) |

where is the volume element. (See Figure A.21.) There are analogous contributions from the sides normal to the - and -axes, so the total of all the contributions is

The

(A.132) |

Divergence is a good scalar (i.e., it is coordinate independent), because it is the dot product of the vector operator with . The formal definition of is

(A.133) |

This definition is independent of the shape of the infinitesimal volume element.

One of the most important results in vector field theory is the so-called
*divergence theorem*. This states that for any volume
surrounded by a closed surface
,

(A.134) |

where is an outward pointing volume element. The proof is very straightforward. We divide up the volume into very many infinitesimal cubes, and sum over all of the surfaces. The contributions from the interior surfaces cancel out, leaving just the contribution from the outer surface. (See Figure A.22.) We can use Equation (A.131) for each cube individually. This tells us that the summation is equivalent to over the whole volume. Thus, the integral of over the outer surface is equal to the integral of over the whole volume, which proves the divergence theorem.

Now, for a vector field with ,

(A.135) |

for any closed surface . So, for two surfaces, and , on the same rim,

(A.136) |

as illustrated in Figure A.23. (Note that the direction of the surface elements on has been reversed relative to those on the closed surface. Hence, the sign of the associated surface integral is also reversed.) Thus, if then the surface integral depends on the rim, but not on the nature of the surface that spans it. On the other hand, if then the integral depends on both the rim and the surface.

Consider an incompressible fluid whose velocity field is . It is clear that for any closed surface, because what flows into the surface must flow out again. Thus, according to the divergence theorem, for any volume. The only way in which this is possible is if is everywhere zero. Thus, the velocity components of an incompressible fluid satisfy the following differential relation:

(A.137) |

It is sometimes helpful to represent a vector field
by *lines of force*
or *field-lines*.
The direction of a line of force at any point is the same as the local direction of
. The density of lines (i.e., the number of lines crossing a unit surface
perpendicular to
) is equal to
.
For instance, in Figure A.24,
is larger at point 1 than at point 2. The number of lines
crossing a surface element
is
. So, the
net number of lines leaving a closed surface is

(A.138) |

If then there is no net flux of lines out of any surface. Such a field is called a