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Let us start with a vector field ${\bf A}$. 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 ${\bf A}$ is the velocity of some fluid, then $\oint_S {\bf A}\cdot
d{\bf S}$ is the rate of flow of material out of $S$.

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

\oint {\bf A}\cdot d{\bf S} = {\bf A}\cdot \oint d{\bf S} = {\bf A} \cdot {\bf S}=0,
\end{displaymath} (119)

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

Figure 18:
\epsfysize =2in
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,
\end{displaymath} (120)

where $dV = dx  dy  dz$ is the volume element (see Fig. 18). 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}{\par...
...ial A_y}{\partial y}+\frac{\partial A_z}{\partial z}\right)dV.
\end{displaymath} (121)

The divergence of a vector field is defined
{\mit div} {\bf A} = \nabla\cdot {\bf A} = \frac{\partial A...
...rac{\partial A_y}{\partial y}+\frac{\partial A_z}{\partial z}.
\end{displaymath} (122)

Divergence is a good scalar (i.e., it is coordinate independent), since it is the dot product of the vector operator $\nabla$ with ${\bf A}$. The formal definition of ${\mit div} {\bf A}$ is
{\mit div} {\bf A} = \lim_{dV\rightarrow 0} \frac{\oint {\bf A}\cdot d{\bf S}}
\end{displaymath} (123)

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

Figure 19:
\epsfysize =2.5in
One of the most important results in vector field theory is the so-called divergence theorem or Gauss' theorem. This states that for any volume $V$ surrounded by a closed surface $S$,
\oint_S {\bf A}\cdot d{\bf S} = \int_V {\mit div} {\bf A}  dV,
\end{displaymath} (124)

where $d{\bf S}$ is an outward pointing volume element. The proof is very straightforward. We divide up the volume into lots of very small 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 Fig. 19). We can use Eq. (121) for each cube individually. This tells us that the summation is equivalent to $\int {\mit div} {\bf A}  dV$ over the whole volume. Thus, the integral of ${\bf A}\cdot d{\bf S}$ over the outer surface is equal to the integral of ${\mit div} {\bf A}$ over the whole volume, which proves the divergence theorem.

Now, for a vector field with ${\mit div} {\bf A} = 0$,

\oint_S {\bf A}\cdot d{\bf S} =0
\end{displaymath} (125)

for any closed surface $S$. So, for two surfaces on the same rim (see Fig. 20),
\int_{S_1} {\bf A}\cdot d{\bf S} = \int_{S_2} {\bf A}\cdot d{\bf S}.
\end{displaymath} (126)

Thus, if ${\mit div} {\bf A} = 0$ then the surface integral depends on the rim but not the nature of the surface which spans it. On the other hand, if ${\mit div} {\bf A}\neq 0 $ then the integral depends on both the rim and the surface.
Figure 20:
\epsfysize =1.75in

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, since what flows into the surface must flow out again. Thus, according to the divergence theorem, $\int {\mit div} {\bf v}  dV = 0$ for any volume. The only way in which this is possible is if ${\mit div}  {\bf v}$ is everywhere zero. Thus, the velocity components of an incompressible fluid satisfy the following differential relation:

\frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y} +
\frac{\partial v_z}
{\partial z}=0.
\end{displaymath} (127)

Consider, now, a compressible fluid of density $\rho$ and velocity ${\bf v}$. The surface integral $\oint_S\rho  {\bf v} \cdot d{\bf S}$ is the net rate of mass flow out of the closed surface $S$. This must be equal to the rate of decrease of mass inside the volume $V$ enclosed by $S$, which is written $-({\partial}/{\partial t})(\int_V \rho dV)$. Thus,

\oint_S \rho {\bf v}\cdot d{\bf S} = -\frac{\partial}{\partial t}\!\left(
\int_V \rho   dV\right)
\end{displaymath} (128)

for any volume. It follows from the divergence theorem that
{\mit div} (\rho {\bf v}) = -\frac{\partial \rho}{\partial t}.
\end{displaymath} (129)

This is called the equation of continuity of the fluid, since it ensures that fluid is neither created nor destroyed as it flows from place to place. If $\rho$ is constant then the equation of continuity reduces to the previous incompressible result, ${\mit div} {\bf v}=0$.

Figure 21:
\epsfysize =1.75in
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 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 $\vert A\vert$. For instance, in Fig. 21, $\vert A\vert$ 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 {\mit div}  {\bf A}  dV.
\end{displaymath} (130)

If ${\mit div} {\bf A} = 0$ 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.

next up previous
Next: The Laplacian Up: Vectors Previous: Gradient
Richard Fitzpatrick 2006-02-02