sources and sinks of vector field

Let the vector field $\overrightarrow{U}$ of ${ℝ}^3$ be interpreted, as in the remark of the parent entry (http://planetmath.org/Flux), as the velocity of a liquid.  Then the flux

$${\oint }_a\overrightarrow{U}\cdot 𝑑\overrightarrow{a}$$

of $\overrightarrow{U}$ through a closed surface $a$ expresses how much more liquid per time-unit it comes from inside of $a$ to outside than contrarily.  Since for a usual incompressible liquid, the outwards flow and the inwards flow are equal, we must think in the case that the flux differs from 0 either that the flowing liquid is suitably compressible or that there are inside the surface some sources creating liquid and sinks annihilating liquid.  Ordinarily, one uses the latter idea.  Both the sources and the sinks may be called sources, when the sinks are negative sources.  The flux of the vector $\overrightarrow{U}$ through $a$ is called the productivity or the strength of the sources inside $a$.

For example, the sources and sinks of an electric field ($\overrightarrow{E}$) are the locations containing positive and negative charges, respectively.  The gravitational field has only sinks, which are the locations containing .

The expression

$$\frac{1}{\mathrm{\Delta }v}{\oint }_{\partial \mathrm{\Delta }v}\overrightarrow{U}\cdot 𝑑\overrightarrow{a},$$

where $\mathrm{\Delta }v$ means a region in the vector field and also its volume, is the productivity of the sources in $\mathrm{\Delta }v$ per a volume-unit.  When we let $\mathrm{\Delta }v$ to shrink towards a point $P$ in it, to an infinitesimal volume-element $dv$, we get the limiting value

$\varrho :={\displaystyle \frac{1}{dv}}{\displaystyle {\oint }_{\partial dv}}\overrightarrow{U}\cdot 𝑑\overrightarrow{a},$ (1)

called the source density in $P$.  Thus the productivity of the source in $P$ is $\varrho dv$.  If  $\varrho =0$, there is in $P$ neither a source, nor a sink.

The Gauss’s theorem

$${\int }_v\nabla \cdot \overrightarrow{U}𝑑v={\oint }_a\overrightarrow{U}\cdot 𝑑\overrightarrow{a}$$

applied to $dv$ says that

$\nabla \cdot \overrightarrow{U}={\displaystyle \frac{1}{dv}}{\displaystyle {\oint }_{\partial dv}}\overrightarrow{U}\cdot 𝑑\overrightarrow{a}.$ (2)

Accordingly,

$\varrho =\nabla \cdot \overrightarrow{U}$ (3)

and

${\displaystyle {\oint }_a}\overrightarrow{U}\cdot 𝑑\overrightarrow{a}={\displaystyle {\int }_v}\varrho 𝑑v.$ (4)

This last can be read that  the flux of the vector through a closed surface equals to the total productivity of the sources inside the surface.  For example, if $\overrightarrow{U}$ is the electric flux density $\overrightarrow{D}$, (4) means that the electric flux through a closed surface equals to the total charge inside.

Title	sources and sinks of vector field
Canonical name	SourcesAndSinksOfVectorField
Date of creation	2013-03-22 18:47:03
Last modified on	2013-03-22 18:47:03
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	12
Author	pahio (2872)
Entry type	Definition
Classification	msc 26B15
Classification	msc 26B12
Related topic	Divergence
Related topic	SolenoidalField
Related topic	CirculationAndVorticity
Defines	source
Defines	sink
Defines	source of vector field
Defines	productivity
Defines	strength
Defines	source density