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The Flux and Divergence of a Vector

Consider a vector that passes through a specified surface. The vector could be ${\bf v}$ the velocity of a fluid, then the flux of ${\bf v}$ would be the amount of fluid passing through that surface in unit time. The flux, $dF$, of a vector ${\bf A}$ through a surface element ${\bf dS}$ is defined as
\begin{displaymath}
dF = {\bf A}\cdot {\bf dS} = {\bf A}\cdot \hat{{\bf n}} dS
\end{displaymath} (1.15)

where ${\bf dS} = \hat{{\bf n}}dS$ is normal to the surface element as shown in Figure 1.10

Figure 1.10: The normal to the surface element and the vector ${\bf A}$.
\includegraphics {fundfig10.eps}

Hence,
\begin{displaymath}
dF = {\bf A}\cdot \hat{{\bf n}} dS = A \cos \theta dS,
\end{displaymath} (1.16)

and $A\cos \theta$ is the component of ${\bf A}$ normal to the surface. Note that the largest value of $dF$ occurs when $\theta= 0$ and $dF = 0$ if $\theta = \pi/2$ so that in the latter case ${\bf A}$ does not cross the surface at all but is instead tangent to the surface. If the vector is tangent to the surface, then the flux through the surface is obviously zero.

Example 1. .6Consider ${\bf A} = \rho {\bf v}$. This is the MASS FLUX of a fluid or plasma and $dF$ is the RATE at which fluid crosses the surface $dS$. Let us consider the units of $dF$ in this example.

\begin{displaymath}
\hbox{units of }dF = \rho v dS = {kg\over m^{3}}{m\over s} m^{2} =
{kg\over s} = \hbox{mass rate of change w.r.t. time}.
\end{displaymath}

Consider ${\bf A} = {\bf B}$, the magnetic field, then $dF$ is the amount of magnetic flux crossing $dS$.

The total flux of a vector across a surface (rather than just the surface element) is

\begin{displaymath}
F = \int {\bf A}\cdot {\bf dS}.
\end{displaymath} (1.17)

There are two types of surfaces that we consider, namely open surfaces and closed surfaces. A closed surface encloses a volume $V$, as shown in Figure 1.11.

Figure 1.11: Two type of surface with an open surface on the left and a closed surface, enclosing a volume, on the right.
\includegraphics [scale=0.7]{fundfig11a.eps} \includegraphics [scale=0.7]{fundfig11b.eps}

If $S$ bounds a finite volume, as in the closed surface case, then ${\bf dS}$ points outwards from the volume. In this case the Divergence Theorem of Gauss (Carl Gauss, 1777-1855) gives
\begin{displaymath}
\int_{S} {\bf A}\cdot {\bf dS} = \int_{V} \nabla \cdot {\bf A} dV,
\end{displaymath} (1.18)

where
\begin{displaymath}
\nabla \cdot {\bf A} = {\partial A_{x}\over \partial x}+ {\...
...l
A_{y}\over \partial y} + {\partial A_{z}\over \partial z},
\end{displaymath} (1.19)

is the divergence of ${\bf A}$. Hence, $\nabla \cdot {\bf A}$ represents the outgoing flux per unit volume and we have two important deductions, namely
  1. If $\nabla \cdot {\bf v} > 0$, a flow is diverging (fluid is flowing out of a volume).

  2. If $\nabla \cdot {\bf v} < 0$, a flow is converging (fluid is flowing into a volume).

Example 1. .7

  1. Consider a fluid of uniform density, $\rho_{0}$, flowing with a velocity given by ${\bf v} = (y,1,1)$. What is the mass flux across the square $0 \le x \le 1$, $0 \le y \le 1$, $z=0$? The situation is shown in Figure 1.12.

    Figure 1.12:
    \includegraphics [bb=20 90 495 320,clip]{fundfig12.ps}

    Here ${\bf dS} = {\bf k} dx dy$ and so

    \begin{eqnarray*}
F & = & \int \rho {\bf v}\cdot {\bf k} dx dy \\
& = & \int_{x=0}^{1}\int_{y=0}^{1} \rho_{0} 1 dx dy \\
& = & \rho_{0}.
\end{eqnarray*}



  2. As in the above example but with a density given by $\rho_{0} = \rho_{0}(2 + \cos \pi x \cos \pi y)$. Hence,

    \begin{eqnarray*}
F & = & \int_{x=0}^{1}\int_{y=0}^{1} \rho_{0}(2 + \cos \pi x ...
...=0}^{y=1} dx \\
& = & \rho_{0}\int_{0}^{1} 2 dx = 2 \rho_{0}.
\end{eqnarray*}



Magnetic Flux Tubes are the building blocks of a magnetic field. A magnetic flux tube (for the magnetic field ${\bf B}$) or a flow tube (for the mass flux $\rho {\bf v}$) is the surface generated by a set of field lines that intersect a simple closed curve, $S_{1}$ as shown in Figure 1.13.

Figure 1.13: A magnetic flux tube.
\includegraphics {fundfig13.eps}

The magnetic flux crossing ${\bf ds}$ is ${\bf B}\cdot
{\bf dS}$ and the total magnetic flux is $\int {\bf B}\cdot {\bf dS}$. Since the flux tube is defined by the magnetic field lines, the walls of the flux tube are parallel to ${\bf B}$. Hence, if there is no flux created inside the tube, then we may state that

\begin{displaymath}
\hbox{flux entering at }S_{1} = \hbox{ flux leaving at }S_{2}.
\end{displaymath}

So the magnetic flux is constant along the tube. To prove this, consider a closed surface defined by $S_{1}$, $S_{2}$ and the walls. Hence, the flux through this closed surface is

\begin{displaymath}
\int {\bf B}\cdot {\bf dS} = - \int_{S_{1}}{\bf B}\cdot {\b...
...{2}}{\bf B}\cdot {\bf dS} + \int_{walls}{\bf B}\cdot {\bf dS}.
\end{displaymath}

The last integral is zero since ${\bf B}$ is parallel to the walls. The first integral on the right hand side has a negative sign. This occurs because the flux through $S_{1}$ into the tube has a direction that is opposite to the direction of the unit normal that is out of the closed surface. Now we make use of Guass's divergence theorem so that

\begin{displaymath}
\int {\bf B}\cdot {\bf dS} = \int \nabla \cdot {\bf b} dV.
\end{displaymath}

However, the basic equation for magnetic fields, see later, is

\begin{displaymath}
\nabla \cdot {\bf B} = 0.
\end{displaymath}

Hence, we have

\begin{displaymath}
\int {\bf B}\cdot {\bf dS} = 0
\end{displaymath}

and so
\begin{displaymath}
\int_{S_{1}}{\bf B}\cdot {\bf dS} = \int_{S_{2}}{\bf B}\cdot {\bf dS}.
\end{displaymath} (1.20)

If the cross-sectional area is small, so that ${\bf B}$ is approximately constant, then the flux is approximately

\begin{displaymath}
F \approx BA \approx \hbox{ constant},
\end{displaymath}

where $A$ is the cross-sectional area and $B$ is the field strength. Thus, if $B$ increases, the area decreases such that $BA$ remains constant.

Example 1. .8For the magnetic field given by ${\bf B} = (1, 2x, 0)$, the field lines are given by

\begin{displaymath}
{dy\over dx} = {B_{y}\over B_{x}} = {2x\over 1}.
\end{displaymath}

Therefore,

\begin{eqnarray*}
dy & = & 2x dx \\
y & = & x^{2} + \hbox{const.}
\end{eqnarray*}



The flux tube passing through $0 \le y \le 1$, $0 \le z \le 1$ and $x=0$ has flux

\begin{displaymath}
F = \int {\bf B}\cdot {\bf dS} = \int {\bf B}\cdot {\bf i} dy dz =
\int_{z=0}^{1}\int_{y=0}^{1} 1 dy dz = 1.
\end{displaymath}

We should check the flux passing through $x=1$. We need to know where the field lines at $x=0$ have gone to at $x=1$.

Table 1.1:
\begin{table}
\centering $\begin{array}{ccc}
Position at $x=0$ & Constant & Po...
... z=1$ & 0 & $y=1, z=1$ \\
$y=1, z=1$ & 1 & $y=2, z=1$
\end{array}$\end{table}


Therefore,

\begin{displaymath}
F = \int_{z=0}^{1}\int_{y=1}^{2} 1 dy dz = 1.
\end{displaymath}

Thus, the flux through $x=0$ equals the flux through $x=1$.
next up previous
Next: Curl , , Triple Up: Introduction Previous: The Gradient Operator,
Prof. Alan Hood
2000-11-06