Divergence

Let us start with a vector field ${\bf A}({\bf r})$. Consider $\oint_S {\bf A}\cdot d{\bf S}$ over some closed surface $S$, where $d{\bf S}$ denotes an outward pointing surface element. This surface integral is usually called the flux of ${\bf A}$ out of $S$. If ${\bf A}$ represents the velocity of some fluid then $\oint_S {\bf A}\cdot d{\bf S}$ is the rate of fluid flow out of $S$.

If ${\bf A}$ is constant in space then it is easily demonstrated that the net flux out of $S$ is zero. In fact,

$$\oint {\bf A}\cdot d{\bf S} = {\bf A}\cdot \oint d{\bf S} = {\bf A} \cdot {\bf S}=0,$$ (1.129)

because the vector area ${\bf S}$ of a closed surface is zero.

Figure A.21: Flux of a vector field out of a small box.

Suppose, now, that ${\bf A}$ is not uniform in space. Consider a very small rectangular volume over which ${\bf A}$ hardly varies. The contribution to $\oint {\bf A}\cdot d{\bf S}$ from the two faces normal to the $x$-axis is

$$A_x(x+dx) \,dy\,dz - A_x(x)\, dy\,dz = \frac{\partial A_x}{\partial x}\,dx\,dy\,dz = \frac{\partial A_x}{\partial x}\,dV,$$ (1.130)

where $dV=dx\,dy\,dz$ is the volume element. (See Figure A.21.) There are analogous contributions from the sides normal to the $y$- and $z$-axes, so the total of all the contributions is

$$\oint {\bf A}\cdot d{\bf S} = \left(\frac{\partial A_x}{\partial x}+ \frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}\right)dV.$$ (1.131)

The divergence of a vector field is defined

$${\mit div}\,{\bf A} = \nabla\cdot {\bf A} = \frac{\partial A_x}{\partial x} +\frac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}.$$ (1.132)

Divergence is a good scalar (i.e., it is coordinate independent), because it is the dot product of the vector operator $\nabla$ with ${\bf A}$. The formal definition of $\nabla\cdot{\bf A}$ is

$$\nabla\cdot{\bf A} = \lim_{dV\rightarrow 0} \frac{\oint {\bf A}\cdot d{\bf S}} {dV}.$$ (1.133)

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 $V$ surrounded by a closed surface $S$,

$$\oint_S {\bf A}\cdot d{\bf S} = \int_V \nabla\cdot{\bf A}\,\,dV,$$ (1.134)

where $d{\bf S}$ is an outward pointing volume element. The proof is very straightforward. We divide up the volume into very many infinitesimal cubes, and sum $\int {\bf A}\cdot d{\bf S}$ 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 $\int \nabla\cdot{\bf A}\,\,dV$ over the whole volume. Thus, the integral of ${\bf A}\cdot d{\bf S}$ over the outer surface is equal to the integral of $\nabla\cdot{\bf A}$ over the whole volume, which proves the divergence theorem.

Figure A.22: The divergence theorem.

Now, for a vector field with $\nabla\cdot {\bf A}=0$,

$$\oint_S {\bf A}\cdot d{\bf S} =0$$ (1.135)

for any closed surface $S$. So, for two surfaces, $S_1$ and $S_2$, on the same rim,

$$\int_{S_1} {\bf A}\cdot d{\bf S} = \int_{S_2} {\bf A}\cdot d{\bf S},$$ (1.136)

as illustrated in Figure A.23. (Note that the direction of the surface elements on $S_1$ has been reversed relative to those on the closed surface. Hence, the sign of the associated surface integral is also reversed.) Thus, if $\nabla\cdot {\bf A}=0$ then the surface integral depends on the rim, but not on the nature of the surface that spans it. On the other hand, if $\nabla\cdot{\bf A}\neq 0$ then the integral depends on both the rim and the surface.

Figure A.23: Two surfaces spanning the same rim (right), and the equivalent closed surface (left).

Consider an incompressible fluid whose velocity field is ${\bf v}$. It is clear that $\oint {\bf v}\cdot d{\bf S} = 0$ for any closed surface, because what flows into the surface must flow out again. Thus, according to the divergence theorem, $\int \nabla\cdot{\bf v}\,\,dV = 0$ for any volume. The only way in which this is possible is if $\nabla\cdot{\bf v}$ is everywhere zero. Thus, the velocity components of an incompressible fluid satisfy the following differential relation:

$$\frac{\partial \varv_x}{\partial x} + \frac{\partial \varv_y}{\partial y} + \frac{\partial \varv_z} {\partial z}=0.$$ (1.137)

It is sometimes helpful to represent a vector field ${\bf A}$ 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 ${\bf A}$. The density of lines (i.e., the number of lines crossing a unit surface perpendicular to ${\bf A}$) is equal to $\vert{\bf A}\vert$. For instance, in Figure A.24, $\vert{\bf A}\vert$ is larger at point 1 than at point 2. The number of lines crossing a surface element $d{\bf S}$ is ${\bf A}\cdot d{\bf S}$. So, the net number of lines leaving a closed surface is

$$\oint_S {\bf A}\cdot d{\bf S} = \int_V \nabla\cdot{\bf A}\,\,dV.$$ (1.138)

If $\nabla\cdot {\bf A}=0$ then there is no net flux of lines out of any surface. Such a field is called a solenoidal vector field. The simplest example of a solenoidal vector field is one in which the lines of force all form closed loops.

Figure A.24: Divergent lines of force.