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MHD Equilibrium Structures

A magnetic arcade in the corona, in equilibrium with no flow, may be modelled by solving,

\begin{displaymath}
{\bf J} = 0,
\end{displaymath}

which satisfies (4.1) when ${\bf v}= 0$ and $p = 0$. Thus, ${\bf J} =
0$ is a solution to ${\bf J}\times {\bf B} = 0$. This is equivalent to
\begin{displaymath}
\nabla \times {\bf B} = 0.
\end{displaymath} (4.12)

In addition, $\nabla \cdot {\bf B} = 0$ and so, taking the curl of (4.12) we obtain

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

Using a vector identity we obtain

\begin{displaymath}
\nabla \times \nabla \times {\bf B} = \nabla \left ( \nabla \cdot
{\bf B}\right ) - \nabla^{2}{\bf B} = 0.
\end{displaymath}

Thus, as the first term in the middle expression is zero, we have
\begin{displaymath}
\nabla^{2}{\bf B} = 0.
\end{displaymath} (4.13)

For ${\bf B} = B_{x}(x,y){\bf i} + B_{y}(x,y){\bf j}$, (4.12) and (4.13) become
\begin{displaymath}
{\partial B_{x}\over \partial y} - {\partial B_{y}\over \partial x} =
0,
\end{displaymath} (4.14)

and
\begin{displaymath}
{\partial^{2}B_{x}\over \partial x^{2}} + {\partial^{2}B_{x}\over
\partial y^{2}} = 0,
\end{displaymath} (4.15)

respectively. To solve these equations we note that they are linear equations and that the coefficients are constants and so we look for seperable solutions of the form

\begin{displaymath}
B_{x} = X(x) Y(y).
\end{displaymath}

Substituting into (4.15) gives

\begin{displaymath}
{d^{2}X\over dx^{2}} Y + X {d^{2}Y\over dy^{2}} = 0.
\end{displaymath}

Dividing by $XY$ and rearranging gives

\begin{displaymath}
{1\over X}{d^{2}X\over dx^{2}} = - {1\over Y}{d^{2}Y\over dy^{2}} =
\hbox{ constant}.
\end{displaymath}

Taking the constant as $-1/L^{2}$ we obtain equations for $X(x)$ and $Y(y)$ as

\begin{displaymath}
{d^{2}X\over dx^{2}} + {1\over L^{2}} X = 0, \qquad {d^{2}Y\over
dy^{2}} - {1\over L^{2}} Y = 0.
\end{displaymath}

These are easily solved in terms of trigonometric functions and expontential functions respectively to give one possible solution as

\begin{displaymath}
B_{x} = B_{0}\cos {x\over L} e^{-y/L},
\end{displaymath}

and

\begin{displaymath}
B_{y} = - B_{0}\sin {x\over L} e^{-y/L}.
\end{displaymath}

The field lines are shown in Figure 4.5.

Figure 4.5: The field lines for the magnetic arcade model.
\includegraphics [scale=0.7]{fundfig43.ps}



Subsections
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Next: Potential Problems leading to Up: Magnetohydrodynamics MHD Previous: Effect of on
Prof. Alan Hood
2000-11-06