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\begin{document}

{\bf Some comments on the BRRK manuscript -- 2}

\bigskip

1.\\
When defining mean fields by averaging over all $x$ and $y$ we have
\begin{equation}
\bmB = (\mB_x (z), \mB_y (z), \mB_z)
\label{eq01}
\end{equation}
with $\mB_z$ independent of $z$.

It is important that for this $xy$ average the Reynolds rule according to which the average commutes
with spatial derivatives applies only for derivatives with respect to $z$.

\bigskip

2.\\
We have generally
\begin{equation}
\bB = \bnab \x \bA = (\p A_z/\p y - \p A_y/\p z, \,
     \p A_x/\p z - \p A_z/\p x, \,
     \p A_y/\p x - \p A_x/\p y ) \, .
\label{eq03}
\end{equation}
Then
\begin{equation}
\bmB =(\ol{\p A_z/\p y} - \p \mA_y/\p z, \,
     \p \mA_x/\p z - \ol{\p A_z/\p x}, \,
     \ol{\p A_y/\p x} - \ol{\p A_x/\p y} ) \, .
\label{eq05}
\end{equation}
Note that, e.g., $\ol{\p A_z/\p y}$ is in general different from $\p \mA_z/\p y = 0$.

\bigskip

3.\\
A general form $\bA^{\rm m}$ of the vector potential such that $\bmB = \bnab \x \bA^{\rm m}$
reproduces (\ref{eq01}) reads
\begin{equation}
\bA^{\m m} = \big( c_{xy} \!\! \int^z \!\! \mB_y (z') \dd z' + c_{xz} \mB_z y, \,\,
    c_{yx} \!\! \int^z \!\! \mB_x (z') \dd z' + c_{yz} \mB_z x, \,\,
    c_{zx} \mB_x (z) y + c_{zy} \mB_y (z) x \, \big)
\label{eq07}
\end{equation}
with constants $c_{xy}$, $c_{xz}$ ... satisfying
\begin{equation}
c_{xy} - c_{zy} = c_{yz} - c_{xz} = c_{zx} - c_{yx} = 1 \, .
\label{eq09}
\end{equation}
The remaining freedom in the choice of these coefficients corresponds to possibilities of gauging $\bA^{\rm m}$.

We note that, if we average in the sense of BRRK footnote 5,
\begin{equation}
\ol{\bA^{\rm m}} = \big( c_{xy} \!\! \int^z \!\! \mB_y (z') \dd z',  \,\,
    c_{yx} \!\! \int^z \!\! \mB_x (z') \dd z',  \,\, 0 \big)
\label{eq09}
\end{equation}
and that $\bnab \x \ol{\bA^{\rm m}}$ is in general quite different from $\bmB$.

We note further that we may put even $c_{xy} = c_{yx} = 0$ and therefore $\ol{\bA^{\rm m}} = \bzo$.

\bigskip

4.\\
Let us now assume that
\begin{equation}
\bB = \bmB + \bb
\label{eq11}
\end{equation}
where $\bb$ has the property $\ol{\bb} = \bzo$ and is periodic in $x$ and $y$ .
A corresponding vector potential reads
\begin{equation}
\bA = \bA^{\rm m} + \bA^{\rm per}
\label{eq13}
\end{equation}
where $\bA^{\rm per}$ satisfies $\ol{\bA^{\rm per}} = \bzo$ and is periodic in $x$ and $y$.
Then
\begin{equation}
\ba = \bA - \ol{\bA} =  \bA^{\rm m} - \ol{\bA^{\rm m}} + \bA^{\rm per} \, , \quad
    \bA^{\rm m} - \ol{\bA^{\rm m}} = ( c_{xz} \mB_z y, \, c_{yz} \mB_z x, \, c_{zx} \mB_x (z) y + c_{zy} \mB_y x) \, .
\label{eq13}
\end{equation}
Note that $\ba$ in general does not coincide with $\bA^{\rm per}$.

\bigskip

5.\\
Take the average of (BRRK 15) over $x$ and $y$ and find first
\begin{equation}
\p_t \ol{\bA} + S \, \ol{ x \, \p \bA/\p y} = - S \ol{A}_y \hat{\bx} + \bmU \x \bmB + \ol{\bu \x \bb}
    - \eta \bmJ \, .
\label{eq13}
\end{equation}
Split $S \, \ol{ x \, \p \bA/\p y}$ into the two terms
\begin{eqnarray}
S \,\ol{ x \, \p \bA^{\rm m}/\p y} &=& S \, (\ol{c_{xz} \mB_z x},\, 0, \, \ol{c_{zx} \mB_x (z) x} ) \, , \quad
\nonumber\\
S \, \ol{ x \, \p \bA^{\rm per}/\p y} &=& S \, (\ol{\p (A^{\rm per}_x x) / \p y} , \,
    \ol{\p (A^{\rm per}_y x) / \p y}, \, \ol{\p (A^{\rm per}_z x) / \p y} ) \, .
\label{eq13}
\end{eqnarray}
Obviously each of them vanishes with the definition of the average given in BRRK footnote 5.
In this way we confirm (BRRK 16).
The arguments used so far for the justification of (BRRK 16),
which rely, e.g., on $\p \ol{\bA} / \p y = 0$ or on the periodicity of $\ba$ in $x$ and $y$,
are not acceptable.

Taking then the difference between (BRRK 15) and (BRRK 16) we arrive at (BRRK 17).

\bigskip

6.\\
We note that the quantity $\ol{\bA}$ in (BRRK 16) is not uniquely defined.
It is in any case equal to $\ol{\bA^{\rm m}}$ but the latter depends
on the choice of $c_{xy}$ and $c_{yx}$ and may be even equal to zero.
In so far the derivation of equation (BRRK 27) on the basis of (BRRK 16) appears questionable.
This does, however, not mean that (BRRK 27) would be wrong.
It can be derived immediately from the mean--field induction equation, that is, $\p_t \bmB = \cdots$.

\bigskip

7.\\
In this context the following aspect might be of interest:\\
We have derived (BRRK 15) from
\begin{equation}
\p_t \bB = \bnab \x \big( (\bU + \bU^S) \x \bB  - \eta \bJ \big) \, ,
\label{eq15}
\end{equation}
and we may recover this equation by taking the curl of (BRRK 15).
For $\bmB$ we have
\begin{equation}
\p_t \bmB = \bnab \x \big( \bmU \x \bmB + \ol{\bu \x \bb} + \ol{\bU^S \x \bb} - \eta \bmJ \big) \, ,
\label{eq15}
\end{equation}
but we can not find this equation by taking the curl of (BRRK 16).




\end{document}
