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The Divergence Theorem

Suppose there is ``stuff'' flowing from place to place in three dimensions.

Figure: Illustration of a vector ``flow field'' $\vec{J}$ near a point in three dimensional space. If each vector represents the rate of ``stuff'' flowing per unit area of a plane that is normal to the direction of flow, then the dot product of the flow field integrated over a planar oriented area $\vec{A}$ is the rate of ``stuff'' flowing through that plane. For example, consider the two areas indicated with purple (or dashed) lines. The rate of ``stuff'' flowing through those regions is $\vec{J} \cdot \vec{A_B} = \vec{J} \cdot \hat{k} A_B$ and $\vec{J} \cdot \vec{A_L} = \vec{J} \cdot \hat{k} A_L$ .

If there are no sources or sinks that create or destroy stuff inside a small box surrounding a point, then the change in the amount of stuff in the volume of the box must be related to some integral over the box's surface:

$$\begin{split}\frac{d}{dt} (\text{amount of stuff in box}) & =... ...\mbox{box}}{\mbox{surface}}} \vec{J} \cdot d\vec{A} \end{split}$$ (16-1)

Figure 16-4: Integration of a vector function near a point and its relation to the change in that vector function. The rate of change of stuff is the integral of flux over the outside--and in the limit as the box size goes to zero, the rate of change of the amount of stuff is related to the sum of derivatives of the flux components at that point.

To relate the rate at which ``stuff $M$ '' is flowing into a small box of volume $\delta V = dx dy dz$ located at $(x,y,z)$ due to a flux $\vec{J}$ , note that the amount that $M$ changes in a time $\Delta t$ is:

$$\begin{split}\Delta M(\delta V) & = (M \text{ flowing out of ... ...partial{z}}}) \delta V \Delta t + \mathcal{O}(dx^4) \end{split}$$ (16-2)

If $C(x,y,z) = M(\delta V)/\delta V$ is the concentration (i.e., stuff per volume) at $(x,y,z)$ , then in the limit of small volumes and short times:

$$\ensuremath{\frac{\partial{C}}{\partial{t}}} = -(\ensuremath{\fra... ...} + \ensuremath{\frac{\partial{J_z}}{\partial{z}}}) = -\nabla \cdot \vec{J} = - \vec{J}$$ (16-3)

For an arbitrary closed volume $V$ bounded by an oriented surface $\partial V$ :

$$\ensuremath{\frac{d{M}}{d{t}}} = \ensuremath{\frac{d{}}{d{t}}} \i... ... = -\int_V \nabla \cdot \vec{J} dV = -\int_{\partial V} \vec{J} \cdot d \vec{A}$$ (16-4)

The last equality

$$\int_V \nabla \cdot \vec{J} dV = \int_{\partial V} \vec{J} \cdot d \vec{A}$$ (16-5)

is called the Gauss or the divergence theorem.

MATHEMATICA$^{\text{\scriptsize {\textregistered }}}$ Example

(notebook Lecture-16)

(html Lecture-16)

(xml+mathml Lecture-16)

Hamaker Interaction between a point and Closed Volume Calculating the Van der Walls potential (also called London Dispersion potential) of a point-particle the vicinity of a finite cylinder. The interaction energy due two induced dipoles, one located at $\vec{r} = (\xi,\eta, \zeta)$ and another located at $\vec{x} = (x,y,z)$ goes like

$$\frac{-1}{\norm {\vec{r} - \vec{x}}^6}$$ (16-6)

Integrating this function for $\vec{r}$ ranging over the volume of a cylinder of length $L$ and radius $R$ will give the potential for a point particle located at $\vec{x}$ due to the entire cylinder. This integration has no simple closed form, so a numerical integration is necessary. The following method, using the divergence theorem, makes the numerical integration more efficient by converting a volume integral to a surface integral of a vector potential.