% $Id: notes.tex,v 1.1 2008/04/10 11:10:43 brandenb Exp $
\documentclass[twocolumn]{article}
\usepackage{graphicx}
\usepackage{bm}
\usepackage{amsmath}
\graphicspath{{./fig/}{./png/}}
\input apj
\title{Nonlinear magnetic diffusivity and alpha tensors in helical turbulence}
\author{}
\date{\today,~ $ $Revision: 1.1 $ $}
\begin{document}
\maketitle

\section{Introduction}

The magnetic fields in stars and galaxies display large scale order,
and at least in the case of the Sun also long term temporal order,
as manifested by its 22 year cycle.

A final frontier of research in $\alpha^2$ dynamos has been the
quenching of the turbulent diffusivity.

\section{Testfield method}

Testfield method applied to the saturation phase of an $\alpha^2$ dynamo
in a periodic box.
Good fit result (see \Fig{pbm_alpeta_64a}) is given by
\EQ
{\alpha^{\rm test}(\meanBB)\over\alpha_0}\approx
{\eta^{\rm test}_{\rm t}(\meanBB)\over\eta_{\rm t0}}\approx
{u_{\rm rms}(\meanBB)/u_{\rm rms}^{(0)}\over1+\meanBB^2/B_{\rm eq}^2}.
\label{FigFormula}
\EN

This result does not describe the correct saturation values,
because in the steady state we have to have
\EQ
\alpha(\meanBB)-\eta_{\rm T}(\meanBB)k_1=0,
\EN
where $\eta_{\rm T}=\eta_{\rm t}+\eta$ is the total magnetic diffusivity;
see \Fig{pbm_sat_64a}.

\EQ
\Rm=u_{\rm rms}^{(0)}/\eta k_{\rm f}
\EN
defined with respect to kinematic rms velocity, $u_{\rm rms}^{(0)}$.
During saturation the rms velocity drops by typically 30\%.

\begin{figure}[t!]\begin{center}
\includegraphics[width=\columnwidth]{pbm_alpeta_64a}
\end{center}\caption[]{
Saturation of LS field (upper panel) and corresponding evolution of
$\alpha$ and $\eta_{\rm t}$ (dotted lines).
Note that the fit formula \eq{FigFormula} (solid line) describe the
values of $\alpha$ and $\eta_{\rm t}$ quite well.
}\label{pbm_alpeta_64a}\end{figure}

\begin{figure}[t!]\begin{center}
\includegraphics[width=\columnwidth]{pbm_sat_64a}
\end{center}\caption[]{
Incorrect saturation values from testfield method.
[check that kinematic growth rate is ok.]
}\label{pbm_sat_64a}\end{figure}

\section{Varying extent of the box}

\begin{figure}[t!]\begin{center}
\includegraphics[width=\columnwidth]{pbm_16b_rep}
\end{center}\caption[]{
Saturation values for $k_1=1$, with $\Rm=11$.
Fit parameters: $B_{\rm sat}^2=0.037$, $t_{\rm sat}=60$,
$\bra{\meanEMF\cdot\meanBB}=7.9\times10^{-5}$.
}\label{pbm_16b_rep}\end{figure}

\begin{figure}[t!]\begin{center}
\includegraphics[width=\columnwidth]{pbm_elong16b}
\end{center}\caption[]{
Saturation values for $k_1=1/2$, with $\Rm=11$.
Fit parameters: $B_{\rm sat}^2=0.100$, $t_{\rm sat}=100$,
$\bra{\meanEMF\cdot\meanBB}=9.9\times10^{-5}$.
}\label{pbm_elong16b}\end{figure}

\begin{figure}[t!]\begin{center}
\includegraphics[width=\columnwidth]{pm25}
\end{center}\caption[]{
Nonlinear alpha and eta
}\label{pm25}\end{figure}

%\begin{verbatim}
%\end{verbatim}
\onecolumn
\pagebreak
\section{Modification of the test field procedure for a Beltrami mean field}
Assume the mean field in the DNS to be a Beltrami field with an arbitrary phase:
\[
 \meanBB = \overline{B}\big(\cos ( kz + \varphi ), \sin( kz + \varphi ), 0 \big)\;.
\]
Then the $\alpha$ tensor has the shape
\[
 \hat{\alpha} = \alpha_0 + \alpha_1 \left [
 \begin{aligned}
    & \cos^2 ( kz + \varphi )  &\cos ( kz + \varphi )\sin( kz + \varphi ) \\
    & \cos ( kz + \varphi )\sin( kz + \varphi )   &\sin^2( kz + \varphi ) 
  \end{aligned} \right ] \; .
\]
Obviously, 
\begin{align*}
    \langle \alpha_{11} \rangle &=  \langle \alpha_{22} \rangle = \alpha_0 + \alpha_1/2 \,,\\
    \langle \alpha_{12} \rangle &=  \langle \alpha_{21} \rangle = 0  \;,\\
\intertext{but}
    \langle \alpha_{12} \sin(2kz) \rangle& =  \langle \alpha_{21}  \sin(2kz)\rangle = \alpha_1\cos(2\varphi) / 4\,,\\
     \langle \alpha_{12} \cos(2kz) \rangle &=  \langle \alpha_{21}  \cos(2kz)\rangle = \alpha_1\sin(2\varphi)  / 4\; ,  
\end{align*}
where $\langle.\rangle$ denotes the mean over $z$.    
Then 
\[
	|\alpha_1| =4 \sqrt{ \langle \alpha_{12} \sin(2kz) \rangle^2 +  \langle \alpha_{12} \cos(2kz) \rangle^2}  \; 
\]
(analogously with $\alpha_{21}$),
but its sign, as the phase $\varphi$, cannot uniquely be determined because of
\[ 
	\hat{f} \sin( z + \varphi ) = -\hat{f} \sin( z + \varphi + \pi ) \;.
\]
As the sign of course matters, the only resort consists in determining the real phase of the Beltrami field
in the DNS. As this means determining this field itself by averaging of the total field, a simpler way would
be
\[
	\alpha_0 + \alpha_1 = ( \alpha_{11} \overline{B}_x + \alpha_{12} \overline{B}_y ) / \overline{B}_x 
\]
or better 
\[  
	\alpha_0 + \alpha_1 = \big( ( \alpha_{11} \overline{B}_x + \alpha_{12} \overline{B}_y ) \overline{B}_x 
					+  ( \alpha_{21} \overline{B}_x + \alpha_{22} \overline{B}_y ) \overline{B}_y\big)/ \overline{B}^2 \; .
\]	
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\end{document}
