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\DOI{10.5802/crphys.245}
\datereceived{2024-12-18}
\daterevised{2025-03-07}
\dateaccepted{2025-03-07}
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\def\bU{\mathbi{U}}
\def\bJ{\mathbi{J}}
\def\bB{\mathbi{B}}
\def\be{\mathbi{e}}
\def\bq{\mathbi{q}}
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\dateposted{2025-03-26}
\begin{document}

\begin{noXML}

\CDRsetmeta{articletype}{research-article}

\title{Three-body anisotropic dynamo: the rotor, the gap and the
stator}

\alttitle{Dynamo anisotrope \`a trois corps : le rotor, l'entrefer et
le stator}

\author{\firstname{Franck} \lastname{Plunian}\CDRorcid{0000-0002-4043-057X}\IsCorresp}
\address{Univ. Grenoble Alpes, Univ. Savoie Mont Blanc, CNRS, IRD,
Univ. Gustave Eiffel, ISTerre, 38000 Grenoble, France}
\email[F. Plunian]{Franck.Plunian@univ-grenoble-alpes.fr}

\author{\firstname{Paul} \lastname{Gomez}\CDRorcid{0000-0002-1725-3739}}
\address{Universit\'e Lyon 1, ENS de Lyon, CNRS, Laboratoire de
G\'eologie de Lyon, Lyon 69622, France}
\email[P. Gomez]{paul.gomez@ens-lyon.fr}

\author{\firstname{Thierry} \lastname{Alboussi\`ere}\CDRorcid{0000-0002-3692-899X}}
\addressSameAs{2}{Universit\'e Lyon 1, ENS de Lyon, CNRS, Laboratoire
de G\'eologie de Lyon, Lyon 69622, France}
\email[T. Alboussi\`ere]{thierry.alboussiere@ens-lyon.fr}

\begin{abstract}
Following the success of the experimental Fury dynamo
\cite{Alboussiere2022}, we are studying (i) the possibility of a new
experiment with the same geometry and the same anisotropic electrical
conductivity, but with a wider galinstan gap between the rotor and the
stator than in Fury, in order to allow richer dynamics via Lorentz
forces. Of course, to do this, we must first be able to start the
dynamo with a large gap, which is the first objective of this study.
The second objective (ii) is to design a miniature dynamo experiment,
smaller than Fury, and with a narrow gap comparable to that of Fury.
The use this time of anisotropic magnetic permeability instead of
anisotropic electrical conductivity is the most appropriate solution.
Theory shows that without a gap, using one rather than the other does
not change the threshold of the dynamo, and only the direction of
rotation of the rotor has to be reversed~\cite{Plunian2021}. With a gap
filled with galinstan, even a very narrow one, and unlike Fury whose
rotor and stator were mainly made of copper, here the use of iron leads
to a significant jump in magnetic permeability with galinstan, which
could be detrimental to the dynamo. In order to study these two issues
relating to (i) the thickness of the gap, and (ii) the electromagnetic
properties of the material filling the gap compared with those of the
rotor and stator, we need to solve the problem of the three-body
anisotropic dynamo: the rotor, the gap and the stator.
\vspace*{2pt}
\end{abstract}

\begin{altabstract}
Suite au succ\`es de la dynamo exp\'erimentale Fury
\cite{Alboussiere2022}, nous \'etudions (i) la possibilit\'e d'une
nouvelle exp\'erience avec la m\^eme g\'eom\'etrie et la m\^eme
conductivit\'e \'electrique anisotrope, mais avec un entrefer de
galinstan entre le rotor et le stator plus large que dans Fury, afin de
permettre une dynamique plus riche via les forces de Lorentz. Bien
s\^ur, pour ce faire, nous devons d'abord \^etre en mesure de
d\'emarrer la dynamo avec un large entrefer, ce qui constitue le
premier objectif de cette \'etude. Le deuxi\`eme objectif (ii) est de
concevoir une exp\'erience de dynamo miniature, plus petite que Fury,
et avec un entrefer \'etroit, comparable \`a celui de Fury.
L'utilisation cette fois d'une perm\'eabilit\'e magn\'etique anisotrope
au lieu d'une conductivit\'e \'electrique anisotrope est la solution la
plus appropri\'ee. La th\'eorie montre que sans entrefer, l'utilisation
de l'une plut\^ot que l'autre ne change pas le seuil de la dynamo, et
seul le sens de rotation du rotor doit \^etre invers\'e
\cite{Plunian2021}. Avec un entrefer rempli de galinstan, m\^eme tr\`es
\'etroit et contrairement \`a Fury dont le rotor et le stator \'etaient
principalement constitu\'es de cuivre, ici l'utilisation de fer
entra\^{\i}ne un saut significatif de perm\'eabilit\'e magn\'etique avec
le galinstan, ce qui pourrait \^etre pr\'ejudiciable \`a la dynamo.
Pour \'etudier ces deux questions li\'ees (i) \`a l'\'epaisseur de
l'entrefer, et (ii) aux propri\'et\'es \'electromagn\'etiques du
mat\'eriau remplissant l'entrefer par rapport \`a celles du rotor et du
stator, nous devons r\'esoudre le probl\`eme de la dynamo anisotrope
\`a trois corps : le rotor, l'entrefer et le stator.
\end{altabstract}

\keywords{\kwd{Dynamo effect}\kwd{Anisotropic electromagnetic
properties}\kwd{Dynamo experiment}}

\altkeywords{\kwd{Effet dynamo}\kwd{Propri\'et\'es \'electromagn\'etiques anisotropes}\kwd{Dynamo exp\'erimentale}}

\thanks{French Academy of Sciences and Electricit\'e de France, PEPR Origins}

\shortrunauthors

\maketitle

\end{noXML}

\def\eqref#1{(\ref{#1})}

\section{Introduction}\label{sec:introduction}

The Fury experiment demonstrated that it was possible to produce a
dynamo effect using a material with anisotropic electrical conductivity
\cite{Alboussiere2022}. Such material was obtained by alternating
layers of copper and kapton in order to favour electrical currents
along the copper layers and slow them down in the direction
perpendicular to the layers.  Fury's geometry consists of a cylindrical
rotor rotating inside a cylindrical stator, both made with this
copper/kapton material, and a thin gap between the rotor and stator,
filled with galinstan to ensure good electrical contact between the two
solid parts (see Figure~\ref{fig:anisotopric cylinders}).

In Fury, the gap is so thin that even if the galinstan flow is
turbulent, with a Reynolds number of about $5000$, it plays no other
role than to ensure good electric contact between the rotor and the
stator, which is why it is neglected in the theoretical models
\cite{Plunian2021}.  Above the dynamo threshold, the additional
mechanical power injected is then entirely dissipated in the form of
Joule dissipation, and the magnetic field has no dynamic behaviour
other than  saturation.  This contrasts with fluid dynamo experiments
in which the additional mechanical power is mainly dissipated in the
form of turbulent hydrodynamic dissipation~\cite{Gailitis2001,
Stieglitz2001, Monchaux2007}, and for which the magnetic field
generated can be dynamically unstable, with chaotic or cyclic behaviour
\cite{Ravelet2008}, similar to the magnetic field observed at the
surface of the Earth or the Sun. 

The advantage of Fury is that it is relatively small, inexpensive, and
avoids the use of liquid sodium, which requires an appropriate safety
environment.  Therefore, it could be interesting to study a new
experimental configuration, similar to Fury but with a larger galinstan
gap so that the system benefits from the advantages of Fury as well as
from a larger number of degrees of freedom that could potentially lead
to dynamical instabilities. {Although the role of
small-scale turbulence in the fluid may be crucial in the dynamo
process, presumably leading to additional magnetic dissipation in the
fluid, it is reasonable to start with a simple solid-body rotation for
the fluid motion in the gap. Such simple configuration has also the
advantage to be mathematically tractable. It can even be approximated
to be at rest, at least for some time after the rotor has started to
rotate, with Stewartson boundary layers at the lateral wall of the
rotor, sufficiently thin to be ignored.} Theoretically, we expect the
dynamo threshold to increase with the thickness of the gap. It is
therefore necessary to solve a three-body problem, consisting of two
anisotropic bodies separated by a gap filled with an isotropic
material. This is the first objective of this paper. We note that the
limiting case of an infinite gap (without stator) has already being
studied~\cite{Plunian2023}. It corresponds to a two-body system,
consisting of an anisotropic rotor immersed in an isotropic medium
(e.g.\ a liquid metal). In this case the dynamo threshold is higher than
in Fury, requiring a larger rotor size and also a large quantity of
liquid metal around it. This configuration has been suggested as a good
candidate for starting a dynamo in large installations such as those
built in Maryland, Wisconsin or Dresden~\cite{Lathrop2011,
Stefani2015}. 

The three-body anisotropic dynamo is also crucial to study in the case
where the presence of a galinstan gap has an  effect on the dynamo's
threshold not because of its thickness but because of the difference in
electromagnetic properties between the liquid metal filling the gap and
the materials making up the rotor and stator.  In Fury, although the
electrical conductivity ratio between galinstan and copper is small,
around $6\times 10^{-2}$, the presence of a thin galinstan gap does not
significantly alter the dynamo threshold. On the other hand, if we use
a rotor and stator with an anisotropic magnetic permeability instead of
an anisotropic electric conductivity, the difference in magnetic
permeability with galinstan can be significantly lower and potentially
be much more detrimental to the dynamo. By way of comparison, the
magnetic permeability ratio between galinstan and soft iron is around
$2\times 10^{-4}$, the one between galinstan and $\mu$-metal is around
$6.7\times 10^{-6}$ (see Table~\ref{table:materials}). This is all the
more interesting because, if we neglect the presence of the gap, the
dynamo threshold for a rotor and stator made of soft iron (resp.\ 
$\mu$-metal)  and tin is around a hundred (resp.\ thousand) times
smaller than Fury's, made of copper and kapton, suggesting the
possibility of building a miniature dynamo. In fact, the degree of
miniaturisation will depend on the effect of the jump in magnetic
permeability between galinstan filling the gap and soft iron (resp.\ 
$\mu$-metal). In terms of miniaturisation, Fury's net mass (without
infrastructure) is around 43 kg, while the Maryland and Dresden
experiments exceed 14 tonnes and 6 tonnes respectively. We are aiming
here for an experiment net mass of around\break 1~kg. 

{Last but not least, in order to construct a stator and rotor with an
anisotropic electric conductivity (resp.\ magnetic permeability), an
alternation of two materials is necessary, in order to constrain the
electric currents (resp.\ the magnetic field). Then, this study seems in
line with previous dynamo studies underlining the benefic action of
inhomogeneous materials~\cite{Busse1992, Gallet2012, Gallet2013,
Petrelis2016, Marcotte2021}. However, from a theoretical point of view,
the anisotropic dynamo only relies on anisotropy, and not on
inhomogeneity. To illustrate this point,  the corresponding dynamo
mechanism is illustrated in Figure~\ref{fig:dynamo mechanism} in
cartesian geometry. In an appropriate reference frame, the shear
between two moving parts of a single homogeneous material is
represented by the two opposite velocities $\bU$. We assume that the
electrical conductivity is maximum along vertical planes parallel to
the fine lines drawn at the top horizontal surfaces. These lines are
here only to fix the eye, the material then staying homogeneous, but
with an anisotropic electrical conductivity. We note that these lines
are not perpendicular to $\bU$, implying that the horizontal component
of the electric current density $\bJ$ is not perpendicular to $\bU$
either. Considering a current density loop as represented in red in
Figure~\ref{fig:dynamo mechanism}, the magnetic field $\bB$ which is
induced by $\bJ$ has a component perpendicular to $\bU$, such that $\bU
\times \bB \neq 0$ (Ohm's law) reinforces the vertical component of
$\bJ$. This illustrates the dynamo mechanism, provided $\bU$ is
sufficiently large. On the other hand, if $\bU$ is changed in $-\bU$ ,
or equivalently if the direction of the anisotropy $\alpha$ is changed
in $-\alpha$, then $\bU \times \bB$ will be opposite and then will
weaken the vertical component of $\bJ$, making the dynamo impossible.
If the electrical conductivity is isotropic, then the current loop will
follow the shortest way which is in planes perpendicular to $\bU$. In
that case, the induced magnetic field $\bB$ becomes parallel to $\bU$
and then unable to drive a current density, as $\bU \times \bB=0$.}  

\begin{figure}
\includegraphics{fig01}
\caption{\label{fig:dynamo mechanism}Illustration of the anisotropic
dynamo mechanism. The two moving parts are made of the same homogeneous
material. The maximum electrical conductivity is along vertical planes
parallel to the lines drawn at the top surfaces. The current density
loop $\bJ$ (in red) induces a magnetic field $\bB$ (in green). The
component of $\bB$ which is perpendicular to $\bU$ reinforces the
vertical component of $\bJ$ in each moving part, through the cross
product $\bU \times \bB$ (Ohm's law), making the dynamo possible
(provided a velocity threshold is exceeded  to compensate for Joule
dissipation).} 
\end{figure}

In Section~\ref{sec:General formulation}, the configuration of the
three-body problem is introduced as well as the set of equations to be
solved. In Section~\ref{sec:renormalization and BC}  the boundary
conditions are presented, followed by a renormalisation of the problem
leading to dimensionless variables. In Section~\ref{sec:resolution}, a
dispersion relation is derived, and an explicit expression for the
magnetic Reynolds number is given, followed in section
\ref{sec:Results} by significant results, dimensionless and
dimensional.{\vspace*{-12pt}}

\section{General formulation}\label{sec:General formulation}

As shown in Figure~\ref{fig:anisotopric cylinders}, we consider three
cylindrical bodies, the rotor ($r&lt;R$), the gap  ($R&lt;r&lt;R+e$) and the
stator ($r&gt;R+e$), where $r$ is the radial coordinate in cylindrical
geometry. The rotor rotates with a solid body rotation given by the
angular velocity $\Omega$, while the stator is at rest. The gap in
between can also rotate, with an angular velocity $\omega$ which, for
mathematical convenience, is considered to be also constant in $r$
(solid body rotation), with $\omega \in [0,\Omega ]$.  The velocity
field is then given by
{\begin{equation} \label{eq:infinite shear}
\bU = r\tilde{\Omega} \be_{\theta}, \quad \mbox{with }
\tilde{\Omega} = \begin{cases}
\Omega, &amp; 0&lt;r&lt; R \\
\omega, \quad \mbox{for} &amp; R&lt;r&lt;R+e\\
0,      &amp;r&gt;R+e,
\end{cases}
\end{equation}}\unskip
where $(\be_r,\be_{\theta},\be_z)$ and $(r, \theta, z)$ are unit basis
vectors and coordinates in cylindrical geometry.

The electrical conductivity and magnetic permeability of the rotor and
stator have the same anisotropy. We define the electrical conductivity
and the magnetic permeability by $\sigma^{\parallel}$ and
$\mu^{\parallel}$ in a given direction $\bq$, and by $\sigma^{\perp}$
and $\mu^{\perp}$ in the directions perpendicular to $\bq$, where $\bq$
is a unit vector.  In the direction parallel to $\bq$,  Ohm's law and
the relation between $\bH$ and $\bB$  are written in the form $\bJ\cdot
\bq=\sigma^{\parallel} (\bE\cdot \bq)$ and $\bB\cdot
\bq=\mu^{\parallel}(\bH\cdot \bq)$, while in the directions
perpendicular to $\bq$,  they  are written as
$\bJ-(\bJ\cdot\bq)\bq=\sigma^{\perp} (\bE-(\bE\cdot\bq)\bq)$ and
$\bB-(\bB\cdot\bq)\bq=\mu^{\perp}(\bH-(\bH\cdot\bq)\bq)$. This leads to
two symmetric tensors, $\lbrack\sigma_{ij}\rbrack$ for the electrical
conductivity and $\lbrack\mu_{ij}\rbrack$ for the magnetic
permeability, defined by~\cite{Ruderman1984}
\changeEqNo{2a,b}
{\begin{subequations}\label{eq:conductivity permeability tensors}
\begin{equation}
\lbrack\sigma_{ij}\rbrack = \sigma^{\perp} \delta_{ij} + (\sigma^{\parallel}-\sigma^{\perp})q_i q_j, 
\quad 
\lbrack\mu_{ij}\rbrack=\mu^{\perp} \delta_{ij} + (\mu^{\parallel}-\mu^{\perp})q_i q_j. 
\end{equation}
\end{subequations}}\unskip
\resetEqNo{2}\unskip
\noindent
The expressions (\ref{eq:conductivity permeability tensors}a,b) also encompass the isotropic electromagnetic
properties of the fluid in the gap, $\sigma_g$ and $\mu_g$, replacing
$\sigma^{\perp}$ and $\sigma^{\parallel}$ by  $\sigma_g$, $\mu^{\perp}$
and $\mu^{\parallel}$ by  $\mu_g$. Anticipating a renormalisation of
the electrical conductivity and magnetic permeability by
$\sigma^{\perp}$ and $\mu^{\perp}$, it is then useful to
rewrite (\ref{eq:conductivity permeability tensors}a,b) as following, 
\changeEqNo{3a,b}
{\begin{subequations}\label{eq:conductivity permeability tensors inout}
\begin{equation}
\lbrack\sigma_{ij}\rbrack=\sigma^{\perp}\; \tilde{\varphi}_{\sigma} \left(\delta_{ij} -\dfrac{\tilde{\sigma}}{1+\tilde{\sigma}}q_i q_j\right), 
\quad
\lbrack\mu_{ij}\rbrack=\mu^{\perp}\; \tilde{\varphi}_{\mu} \left(\delta_{ij} -\dfrac{\tilde{\mu}}{1+\tilde{\mu}}q_i q_j\right), 
\end{equation}
\end{subequations}}\unskip
\resetEqNo{3}\unskip
\noindent
with
\changeEqNo{4a\ndash d}
{\begin{subequations}\label{eq:phisigmasigmaphimumu}
\begin{equation}
\tilde{\varphi}_{\sigma} = \begin{cases}
1 \\ \varphi_{\sigma} \\ 1,
\end{cases} \quad
\tilde{\sigma} = \begin{cases}
\sigma \\ 0 \\ \sigma,
\end{cases} \quad
\tilde{\varphi}_{\mu} = \begin{cases}
1  \\ \varphi_{\mu} \\ 1,
\end{cases} \quad 
\tilde{\mu} = \begin{cases}
\mu &amp; r&lt;R \\ 
0 \quad \mbox{for }&amp; R&lt;r&lt;R+e\\  
\mu &amp; r&gt;R+e,
\end{cases}
\end{equation}
\end{subequations}}\unskip
\resetEqNo{4}\unskip
\noindent
and
\changeEqNo{5a\ndash d}
{\begin{subequations}\label{eq:sigmamu}
\begin{equation}
\varphi _{\sigma} = \dfrac{\sigma_g}{\sigma^{\perp}}, \quad 
\sigma = \dfrac{\sigma^{\perp}}{\sigma^{\parallel}}-1, \quad
\varphi _{\mu} = \dfrac{\mu_g}{\mu^{\perp}}, \quad 
\mu = \dfrac{\mu^{\perp}}{\mu^{\parallel}}-1.
\end{equation}
\end{subequations}}\unskip
\resetEqNo{5}\unskip

As in~\cite{Plunian2020, Plunian2021, Plunian2022, Plunian2023}, we
choose $\bq$ as a unit vector in the horizontal plane defined by
{\begin{equation}\label{eq:q}
\bq=c \; \be_r+s \; \be_{\theta},
\end{equation}}\unskip
where $c=\cos\alpha$, $s=\sin\alpha$, with $\alpha$ a prescribed pitch
angle.  The vector $\bq$ follows the direction of logarithmic spirals
which are perpendicular to the ones shown in
Figure~\ref{fig:anisotopric cylinders} (see~\cite{Alboussiere2020} for
an alternative expression in cartesian geometry).

\begin{figure}
\includegraphics{fig02}
\caption{\label{fig:anisotopric cylinders}Left: Exploded view of the
rotor surrounded by the gap, itself surrounded by the stator. Right:
The logarithmic spirals in the rotor and stator correspond to the
direction of the electric current (resp.\ magnetic field) for
$\sigma^{\perp}\gg\sigma^{\parallel}$ (resp.\ 
$\mu^{\perp}\gg\mu^{\parallel}$), in the horizontal plane seen from
above. They are perpendicular to $\bq$.} 
\end{figure}

In the magnetohydrodynamic approximation, Maxwell's equations and Ohm's
law take the following forms
{\begin{subequations}\label{eq:MaxwellOhm}
\begin{eqnarray}
\bH &amp;=&amp; \lbrack\mu_{ij}\rbrack^{-1}\bB,\label{eq:Maxwell1}\Seqnsplit
\nabla \cdot \bB &amp;=&amp; 0, \label{eq:Maxwell2}\Seqnsplit
\bJ &amp;=&amp; \nabla \times \bH, \label{eq:Ampere}\Seqnsplit
\partial_t \bB &amp;=&amp; - \nabla \times \bE\label{eq:Faraday}\Seqnsplit
\bJ &amp;=&amp; \lbrack\sigma_{ij}\rbrack(\bE + \bU \times \bB),\label{eq:Ohm}
\end{eqnarray}
\end{subequations}}\unskip
leading to the equation for the magnetic induction $\bB$,
{\begin{equation}\label{eq:induction equation}
\partial_t \bB = \nabla \times (\bU \times \bB) - \nabla \times (
\lbrack\sigma_{ij}\rbrack^{-1} \nabla \times (
\lbrack\mu_{ij}\rbrack^{-1}\bB ) ),
\end{equation}}\unskip
where
\changeEqNo{9a,b}
{\begin{subequations}\label{eq:resistivity and inv permeability tensors}
\begin{equation}
\lbrack\sigma_{ij}\rbrack^{-1} = 
(\tilde{\varphi}_{\sigma}\sigma^{\perp})^{-1} 
(\delta_{ij} + \tilde{\sigma} q_i q_j ), \quad 
\lbrack\mu_{ij}\rbrack^{-1} = 
(\tilde{\varphi}_{\mu}\mu^{\perp})^{-1} 
(\delta_{ij} + \tilde{\mu} q_i q_j ).
\end{equation}
\end{subequations}}\unskip
\resetEqNo{9}\unskip
\noindent
Equations (\ref{eq:resistivity and inv permeability tensors}a,b)
are derived from (\ref{eq:conductivity permeability tensors inout}a,b)
knowing that, for any unit vector $\bq$ and any scalar quantity
$X \neq -1$, we have
{\begin{equation}
\left[\delta_{ij} -\dfrac{X}{1+X}q_i q_j\right]^{-1} = 
\delta_{ij} + X q_i q_j.
\end{equation}}\unskip

Since the velocity is stationary and independent of $z$, and as we are
considering only axisymmetric solutions, because they are the least
dissipative ones and then likely to be dominant at the dynamo onset, we
look for a magnetic induction in the form
{\begin{equation}\label{eq:magnetic induction}
\bB=\bB(r)\exp(\gamma t + \mathrm{i} kz),
\end{equation}}\unskip
where $\bB(r)$ is the axisymmetric magnetic mode at vertical wave
number $k$. In (\ref{eq:magnetic induction}) a positive value of the
real part of the magnetic growth rate $\gamma$ is the signature of
dynamo action. The dynamo threshold that we will be sought, corresponds
to $\Re\{\gamma\}=0$.

Replacing (\ref{eq:infinite shear}) and (\ref{eq:magnetic induction})
in (\ref{eq:induction equation}), and after some algebra (see
e.g.~\cite{Plunian2023}), one obtains the following equations for
$B_r(r)$ and $B_{\theta}(r)$,
{\begin{subequations}\label{eq:gammaBrBteta}
\begin{eqnarray}
\tilde{\gamma} B_r &amp;=&amp; -(\sigma^{\perp}\mu^{\perp})^{-1} [ \tilde{\mu}
c^2 k^2 B_r + (1+\tilde{\sigma} s^2) D_k(B_r) - cs(\tilde{\sigma}
-\tilde{\mu}) k^2 B_{\theta} ], \label{eq:gammaBr} \Seqnsplit
\tilde{\gamma} B_{\theta}  &amp;=&amp; -(\sigma^{\perp}\mu^{\perp})^{-1} [
\tilde{\sigma} c^2 k^2 B_{\theta} + (1+\tilde{\mu} s^2)D_k (B_{\theta})
- cs(\tilde{\sigma} -\tilde{\mu}) D_k(B_r)], \label{eq:gammaBteta}
\end{eqnarray}
\end{subequations}}\unskip
where
{\begin{equation}\label{eq:gammatilde}
\tilde{\gamma} = \tilde{\varphi}_{\sigma}\tilde{\varphi}_{\mu}\gamma = 
\begin{cases}
\gamma, &amp; r&lt; R  \\
\varphi_{\sigma}\varphi_{\mu}\gamma, \quad \text{for } &amp; R&lt;r&lt; R+e \\
\gamma, &amp; r&gt; R+e
\end{cases}
\end{equation}}\unskip
and 
{\begin{equation}
D_{k}(X) = k^2X - \dfrac{\partial}{\partial r}
\left(\dfrac{1}{r}\dfrac{\partial}{\partial r}(rX)\right). 
\end{equation}}\unskip

\section{Boundary conditions and renormalization}
\label{sec:renormalization and BC}

The system of Equations (\ref{eq:gammaBrBteta}a,b) is completed
by the appropriate boundary conditions for $r=0$ and $r\rightarrow
\infty$,
\changeEqNo{15a\ndash d}
{\begin{subequations}\label{eq:r0infinite}
\begin{equation}
B_r(r=0) = B_{\theta}(r=0) = \lim_{r\rightarrow \infty}B_r = 
\lim_{r\rightarrow \infty}B_{\theta} = 0,
\end{equation}
\end{subequations}}\unskip
\resetEqNo{15}\unskip
\noindent
and by the continuity across $r=R$ and $r=R+e$ of the normal component
of $\bB$ and of the tangential components of $\bH$ and $\bE$,
\changeEqNo{16a\ndash e}
{\begin{subequations}\label{eq:r1}
\begin{equation}
  [B_r]^{R^+}_{R^-} = [H_{\theta}]^{R^+}_{R^-}
= [H_z]^{R^+}_{R^-} = [E_{\theta}]^{R^+}_{R^-}
= [E_z]^{R^+}_{R^-} = 0,
\end{equation}
\end{subequations}}\unskip
\resetEqNo{16}\unskip
{\vspace*{-1pc}}
\changeEqNo{17a\ndash e}
{\begin{subequations}\label{eq:r2}
\begin{equation}
  [B_r]^{(R+e)^+}_{(R+e)^-} =
  [H_{\theta}]^{(R+e)^+}_{(R+e)^-}
= [H_z]^{(R+e)^+}_{(R+e)^-}
= [E_{\theta}]^{(R+e)^+}_{(R+e)^-}
= [E_z]^{(R+e)^+}_{(R+e)^-} =0,
\end{equation}
\end{subequations}}\unskip
\resetEqNo{17}\unskip
\noindent
where $\left[X\right]^{r_2}_{r_1} =X(r=r_2)-X(r=r_1)$.

From (\ref{eq:Maxwell1}) and (\ref{eq:resistivity and inv permeability
tensors}b), the magnetic field $\bH$ takes the form
{\begin{equation}\label{eq:magnetic field}
\bH = (\tilde{\varphi}_{\mu}\mu^{\perp})^{-1}
\begin{pmatrix} 
(1+\tilde{\mu} c^2)B_r+ \tilde{\mu} cs B_{\theta}\\ 
\tilde{\mu} cs B_r + (1+\tilde{\mu} s^2)B_{\theta}\\ 
B_z 
\end{pmatrix},
\end{equation}}\unskip
with, from (\ref{eq:Maxwell2}), 
{\begin{equation}\label{eq:divB3}
B_z = \mathrm{i} k^{-1} \left(\dfrac{B_r}{r}+\dfrac{\partial B_r}{\partial r}\right).
\end{equation}}\unskip
Then (\ref{eq:r1}a--c) and (\ref{eq:r2}a--c) can be rewritten
as
{\begin{eqnarray}
B_r(R^+) &amp;=&amp; B_r(R^-) \label{eq:BCBrR} \Seqnsplit
\varphi_{\mu}^{-1}B_{\theta}(R^+) &amp;=&amp; \mu cs B_r(R^-) + 
(1+\mu s^2)B_{\theta}(R^-) \label{eq:BCHtetaR}\Seqnsplit
\varphi_{\mu}^{-1}\left(\dfrac{B_r(R^+)}{R} + \dfrac{\partial
B_r}{\partial r}(R^+)\right) &amp;=&amp; \dfrac{B_r(R^-)}{R} + 
\dfrac{\partial B_r}{\partial r}(R^-) \label{eq:BCHzR}\Seqnsplit
B_r((R+e)^-) &amp;=&amp; B_r((R+e)^+)\label{eq:BCBrR+e}\Seqnsplit
\varphi_{\mu}^{-1}B_{\theta}((R+e)^-) &amp;=&amp; 
\mu cs B_r((R+e)^+)+(1+\mu s^2)B_{\theta}((R+e)^+) \label{eq:BCHtetaR+e}\Seqnsplit
\varphi_{\mu}^{-1}\left(\dfrac{B_r((R+e)^-)}{(R+e)} + 
\dfrac{\partial B_r}{\partial r}((R+e)^-)\right) 
&amp;=&amp; \dfrac{B_r((R+e)^+)}{(R+e)} + \dfrac{\partial B_r}{\partial r}((R+e)^+).
\label{eq:BCHzR+e}
\end{eqnarray}}\unskip
From (\ref{eq:Faraday}) we have $E_{\theta}=-\mathrm{i} k^{-1}\gamma
B_r$, implying that the two conditions (\ref{eq:r1}a) and
(\ref{eq:r1}d) are redundant, as well as  (\ref{eq:r2}a) and
(\ref{eq:r2}d). As for the last boundary conditions (\ref{eq:r1}e) and 
(\ref{eq:r2}e), using (\ref{eq:Ohm}) we have
{\begin{equation}
\tilde{\varphi}_{\sigma}^{-1}J_z + \sigma^{\perp} r \tilde{\Omega}
B_r = \sigma^{\perp}E_z
\end{equation}}\unskip
with $\tilde{\Omega}$ defined in (\ref{eq:infinite shear}), leading to
{\begin{eqnarray}
&amp; \varphi_{\sigma}^{-1}J_z(R^+) + \sigma^{\perp}\omega R B_r(R^{+}) = 
J_z(R^-)+ \sigma^{\perp} \Omega R B_r(R^{-}),&amp; \label{eq:Ezcont1} 
\Seqnsplit
&amp; J_z((R+e)^+) = \varphi_{\sigma}^{-1} J_z((R+e)^-)+\sigma^{\perp} 
\omega (R+e) B_r((R+e)^{-}).&amp; \label{eq:Ezcont2}
\end{eqnarray}}\unskip
Renormalizing the distance and time by $R$ and
$\mu^{\perp}\sigma^{\perp}R^2$, and the current density $\bJ$ by
$(\mu^{\perp}R)^{-1}$, corresponds to replacing $R$, $\sigma^{\perp}$
and $\mu^{\perp}$ by unity in the system (\ref{eq:gammaBrBteta}a,b) and
in the boundary conditions                     
(\ref{eq:r1}a--e), 
(\ref{eq:r2}a--e),
(\ref{eq:BCBrR})--(\ref{eq:BCHzR+e}) and
(\ref{eq:Ezcont1})--(\ref{eq:Ezcont2}). Then the absolute value of the
dimensionless angular velocity $|\Omega|$ corresponds to the magnetic
Reynolds number. For the sake of simplicity, we will retain the
notation $e$ for the now-normalised thickness of the gap.

\section{Resolution}\label{sec:resolution}
As we are interested in the dynamo threshold, the system
(\ref{eq:gammaBrBteta}a,b) is solved for $\gamma=0$, in the
same line as in~\cite{Plunian2022, Plunian2023}. Introducing
\changeEqNo{29a,b}
{\begin{subequations}\label{eq:ksigma kmu}
\begin{equation}
k_{\tilde{\sigma}} = k\left(\dfrac{1+\tilde{\sigma}}{1+\tilde{\sigma}
s^2}\right)^{1/2}, \quad k_{\tilde{\mu}} = k\left(\dfrac{1 + 
\tilde{\mu}}{1+\tilde{\mu} s^2}\right)^{1/2},
\end{equation}
\end{subequations}}\unskip
\resetEqNo{29}\unskip
\noindent
and with the help of the identity
{\begin{equation}
D_{k_1}(X)=D_{k_2}(X) +(k_1^2-k_2^2)X, \label{eq:identity}
\end{equation}}\unskip
the system (\ref{eq:gammaBrBteta}a,b) takes the following
form
{\begin{subequations}\label{eq:gammaBrBteta3}
\begin{eqnarray}
(1+\tilde{\sigma} s^2)  D_{k_{\tilde{\sigma}}}(B_r) &amp;=&amp;
(\tilde{\sigma} -\tilde{\mu})ck^2 (cB_r + s B_{\theta})
\label{eq:gammaBr3} \Seqnsplit
(1+\tilde{\mu} s^2)D_{k_{\tilde{\mu}}} (B_{\theta})  &amp;=&amp;
(\tilde{\sigma} -\tilde{\mu})c (s D_k(B_r)-ck^2 B_{\theta}).
\label{eq:gammaBteta3}
\end{eqnarray}
\end{subequations}}\unskip
Then we can show that
\changeEqNo{32a,b}
{\begin{subequations}\label{eq:Dkk}
\begin{equation}
D_{k_{\tilde{\mu}}}(cB_r + s B_{\theta}) = 
D_{k_{\tilde{\sigma}}} (s D_k(B_r)-ck^2 B_{\theta}) = 0.
\end{equation}
\end{subequations}}\unskip
\resetEqNo{32}\unskip
\noindent
Then, using (\ref{eq:Dkk}a,b) and (\ref{eq:gammaBrBteta3}a,b) leads to
\changeEqNo{33a,b}
{\begin{subequations}\label{eq:differential equations}
\begin{equation}
(D_{k_{\tilde{\mu}}}\circ D_{k_{\tilde{\sigma}}})(B_r) = 
(D_{k_{\tilde{\sigma}}}\circ D_{k_{\tilde{\mu}}})(B_{\theta}) = 0.
\end{equation}
\end{subequations}}\unskip
\resetEqNo{33}\unskip
\noindent
The two operators $D_{k_{\tilde{\sigma}}}$ and $D_{k_{\tilde{\mu}}}$
being commutative, $B_r$ and $B_{\theta}$ satisfy the same  linear
differential equation of fourth order.   As the solution of
$D_{\nu}(X)=0$ is a linear combination of $I_1(\nu r)$ and $K_1(\nu
r)$, where $I_1$ and $K_1$ are first and second kind modified Bessel
functions of order 1, the solutions of (\ref{eq:differential
equations}a,b) are a linear combination of
$I_1(k_{\tilde{\sigma}} r)$, $K_1(k_{\tilde{\sigma}} r)$,
$I_1(k_{\tilde{\mu}} r)$ and $K_1(k_{\tilde{\mu}} r)$. Applying the
boundary conditions (\ref{eq:r0infinite}a--d), $B_r$ and
$B_{\theta}$ can be written in the following form,
{\begin{eqnarray}\label{eq:Brlambda}
B_r = \begin{cases}
r&lt; 1,   &amp; -s\left(\lambda_{\sigma}^R \dfrac{I_1(k_{\sigma} r)}{I_1(k_{\sigma})}+ \lambda_{\mu}^R \dfrac{I_1(k_{\mu} r)}{I_1(k_{\mu})}\right) \\[.75pc]
1&lt;r&lt;1+e &amp; -s\left(\lambda_{r}^I \dfrac{I_1(k r)}{I_1(k)}+ \lambda_{r}^K \dfrac{K_1(k r)}{K_1(k)}\right) \\[.75pc]
r&gt; 1+e, &amp; -s\left(\lambda_{\sigma}^S \dfrac{K_1(k_{\sigma} r)}{K_1(k_{\sigma}(1+e))}+ \lambda_{\mu}^S \dfrac{K_1(k_{\mu} r)}{K_1(k_{\mu}(1+e))}\right),
\end{cases}
\end{eqnarray}}\unskip
{\vspace*{-1pc}}
{\begin{eqnarray}\label{eq:Btetalambda}
B_{\theta} = \begin{cases}
r&lt; 1,     &amp; c \left(\lambda_{\sigma}^R \dfrac{I_1(k_{\sigma} r)}{I_1(k_{\sigma})}+ \dfrac{\mu s^2}{1+\mu s^2}\lambda_{\mu}^R \dfrac{I_1(k_{\mu} r)}{I_1(k_{\mu})}\right) \\[.75pc]
1&lt;r&lt; 1+e, &amp; c \left(\lambda_{\theta}^I \dfrac{I_1(k r)}{I_1(k)}+ \lambda_{\theta}^K \dfrac{K_1(k r)}{K_1(k)}\right) \\[.75pc]
r&gt; 1+e,   &amp; c \left(\lambda_{\sigma}^S \dfrac{K_1(k_{\sigma} r)}{K_1(k_{\sigma}(1+e))}+ \dfrac{\mu s^2}{1+\mu s^2}\lambda_{\mu}^S \dfrac{K_1(k_{\mu} r)}{K_1(k_{\mu}(1+e))} \right),
\end{cases}
\end{eqnarray}}\unskip
where $B_{\theta}$ has been obtained from $B_r$ by replacing
(\ref{eq:Brlambda}) in (\ref{eq:gammaBr3}). For $1&lt;r&lt;1+e$, the system
(\ref{eq:gammaBrBteta}a,b) leads to
$D_{k}(B_r)=D_{k}(B_{\theta})=0$, and to the solutions
(\ref{eq:Brlambda}b) and (\ref{eq:Btetalambda}b).

In (\ref{eq:Brlambda}) and (\ref{eq:Btetalambda}) there are eight
unknowns: $\lambda_{\sigma}^R$ and $\lambda_{\mu}^R$ related to the
rotor, $ \lambda_{r}^I, \lambda_{r}^K, \lambda_{\theta}^I$ and
$\lambda_{\theta}^K$ related to the gap, $\lambda_{\sigma}^S$ and
$\lambda_{\mu}^S$ related to the stator. Therefore, to solve the
problem we need eight equations. The six first ones are given by the
boundary conditions (\ref{eq:BCBrR})--(\ref{eq:BCHzR+e}), leading to
{\begin{eqnarray}
\lambda_{r}^I + \lambda_{r}^K &amp;=&amp; \lambda_{\sigma}^R + \lambda_{\mu}^R \label{eq:BC1}, \Seqnsplit
\varphi_{\mu}^{-1}\left(\lambda_{\theta}^I+\lambda_{\theta}^K\right) &amp;=&amp; \lambda_{\sigma}^R \label{eq:BC2}, \Seqnsplit
-k\varphi_{\mu}^{-1}\left(-\lambda_r^I\dfrac{I_0(k)}{I_1(k)}+\lambda_r^K\dfrac{K_0(k)}{K_1(k)} \right) &amp;=&amp;
k_{\sigma}\lambda_{\sigma}^R \dfrac{I_0(k_{\sigma})}{I_1(k_{\sigma})}+ k_{\mu}\lambda_{\mu}^R\dfrac{I_0(k_{\mu})}{I_1(k_{\mu})}. \label{eq:BC3} \Seqnsplit
\lambda_{r}^I \dfrac{I_1(k(1+e))}{I_1(k)} + \lambda_{r}^K \dfrac{K_1(k(1+e))}{K_1(k)} &amp;=&amp; \lambda_{\sigma}^S + \lambda_{\mu}^S \label{eq:BC1b}, \Seqnsplit
\varphi_{\mu}^{-1}\left(\lambda_{\theta}^I\dfrac{I_1(k(1+e))}{I_1(k)}+\lambda_{\theta}^K\dfrac{K_1(k(1+e))}{K_1(k)}\right) &amp;=&amp; \lambda_{\sigma}^S \label{eq:BC2b}, \Seqnsplit
k \varphi_{\mu}^{-1}\left(-\lambda_r^I\dfrac{I_0(k(1+e))}{I_1(k)}+\lambda_r^K\dfrac{K_0(k(1+e))}{K_1(k)} \right) &amp;=&amp; 
k_{\sigma}\lambda_{\sigma}^S \dfrac{K_0(k_{\sigma}(1+e))}{K_1(k_{\sigma}(1+e))} + k_{\mu}\lambda_{\mu}^S \dfrac{K_0(k_{\mu}(1+e))}{K_1(k_{\mu}(1+e))}. 
\label{eq:BC3b}
\end{eqnarray}}\unskip

Finally, to apply the two last boundary conditions (\ref{eq:Ezcont1})
and (\ref{eq:Ezcont2}), we need to calculate the $z$-component of the
current density, that is given by
{\begin{equation}
J_z = \tilde{\varphi}_{\mu}^{-1}\dfrac{1}{r}\dfrac{\partial}{\partial
r}(r\tilde{\phi}), \quad \mbox{with } \tilde{\phi}=\tilde{\mu} cs B_r +
(1+\tilde{\mu} s^2)B_{\theta} \label{Jz}.
\end{equation}}\unskip
Replacing $B_r$ and $B_{\theta}$ given by (\ref{eq:Brlambda}) and
(\ref{eq:Btetalambda}), in (\ref{Jz}b) leads to
{\begin{eqnarray}\label{eq:phi}
\tilde{\phi} = \begin{cases}
r&lt; 1,     &amp;  c \lambda_{\sigma}^R\dfrac{I_1(k_{\sigma} r)}{I_1(k_{\sigma})} \\[.75pc]
1&lt;r&lt; 1+e, &amp;  c \left(\lambda_{\theta}^I\dfrac{I_1(k r)}{I_1(k)}+  \lambda_{\theta}^K\dfrac{K_1(k r)}{K_1(k)}\right) \\[.75pc] 
r&gt; 1+e,   &amp;  c \lambda_{\sigma}^S\dfrac{K_1(k_{\sigma} r)}{K_1(k_{\sigma}(1+e))},
\end{cases}
\end{eqnarray}}\unskip
which is nothing else than $B_{\theta}$ for $\mu=0$, implying that
$\bJ_z$ is not affected by the anisotropy of the magnetic permeability.
Applying the relations $\partial_r(rI_1(\nu r))=\nu r I_0(\nu r)$ and
$\partial_r(rK_1(\nu r))=-\nu r K_0(\nu r)$ to (\ref{Jz}a),
leads to
{\begin{eqnarray}\label{eq:jz}
J_z = \begin{cases}
r&lt; 1,    &amp; c k_{\sigma} \lambda_{\sigma}^R\dfrac{I_0(k_{\sigma} r)}{I_1(k_{\sigma})} \\[.75pc]
1&lt;r&lt;1+e, &amp; c k\varphi_{\mu}^{-1}\left( \lambda_{\theta}^I\dfrac{I_0(k r)}{I_1(k)}- \lambda_{\theta}^K\dfrac{K_0(k r)}{K_1(k)}\right) \\[.75pc]
r&gt; 1+e,  &amp;-c k_{\sigma} \lambda_{\sigma}^S\dfrac{K_0(k_{\sigma} r)}{K_1(k_{\sigma}(1+e))}.
\end{cases}
\end{eqnarray}}\unskip

Replacing (\ref{eq:Brlambda}) and (\ref{eq:jz}) in (\ref{eq:Ezcont1})
and (\ref{eq:Ezcont2}), leads to
{\begin{eqnarray} \label{eq:BC4}
\varphi_{\sigma}^{-1}\varphi_{\mu}^{-1}ck 
\left(\lambda_{\theta}^I\dfrac{I_0(k)}{I_1(k)}- \lambda_{\theta}^K\dfrac{K_0(k)}{K_1(k)}\right) - 
\omega s (\lambda_r^I+\lambda_r^K)
= c k_{\sigma} \lambda_{\sigma}^R \dfrac{I_0(k_{\sigma})}{I_1(k_{\sigma})}
-\Omega s (\lambda_{\sigma}^R+\lambda_{\mu}^R)
\end{eqnarray}}\unskip
%{\vspace*{-1pc}}
{\begin{eqnarray}
&amp;&amp; \varphi_{\sigma}^{-1}\varphi_{\mu}^{-1}ck
\left(\lambda_{\theta}^I\dfrac{I_0(k(1+e))}{I_1(k)}- \lambda_{\theta}^K\dfrac{K_0(k(1+e))}{K_1(k)}\right) - 
\omega s (1+e) 
\left(\lambda_r^I\dfrac{I_1(k(1+e))}{I_1(k)}+\lambda_r^K\dfrac{K_1(k(1+e))}{K_1(k)}\right)
\nonumber \\
&amp;&amp; \quad 
= -c k_{\sigma} \lambda_{\sigma}^S \dfrac{K_0(k_{\sigma}(1+e))}{K_1(k_{\sigma}(1+e))}.
\label{eq:BC4b}
 \end{eqnarray}}\unskip

A dispersion relation can be derived, writing that the system composed
of the eight equations (\ref{eq:BC1})--(\ref{eq:BC3b}), (\ref{eq:BC4}) and
(\ref{eq:BC4b}), with the eight unknowns $\lambda_{\sigma}^R,
\lambda_{\mu}^R, \lambda_{r}^I, \lambda_{r}^K,
\lambda_{\theta}^I,\lambda_{\theta}^K,\lambda_{\sigma}^S$ and
$\lambda_{\mu}^S$, is singular (zero determinant). It leads to the
dynamo threshold $\Omega^c$ such that
{\begin{equation}\label{eq:dynamo threshold}
\overline{\Omega^c} = 
\dfrac{F_{\mu} F_{\sigma} -\overline{\omega} [ - M_{\mu \sigma} E_{\mu}
E_{\sigma} + L_{\mu \sigma}  ( D B A' - B C B' ) ]}
{ M_{\mu \sigma} (D B A' - B C B' ) - L_{\mu \sigma} ( B' + K_{\mu}  D )
( B' + K_{\sigma}  D )   - M_{\mu \sigma} L_{\mu \sigma} D^2
\overline{\omega}},
\end{equation}}\unskip
with
{\begin{eqnarray}
\overline{\Omega^c} &amp;=&amp; \varphi _{\mu} \varphi _{\sigma} (\Omega^c -
\omega ) \dfrac{s}{c}, \quad \overline{\omega} = \varphi _{\mu}  \varphi
_{\sigma} \omega \dfrac{s}{c} (1+e),\quad A = \dfrac{I_0 (k)}{I_1 (k)},
\quad  B = k \left( \dfrac{I_0 (k)}{I_1 (k)} + \dfrac{K_0 (k)}{K_1 (k)}
\right), \Seqnsplit
C &amp;=&amp; \dfrac{I_1 (k')}{I_1 (k)}, \quad D = \dfrac{I_1 (k')}{I_1 (k)} -
\dfrac{K_1 (k')}{K_1 (k)} , \quad  A' = k \dfrac{I_0 (k')}{I_1 (k)},
\quad  B' = k \left( \dfrac{I_0 (k')}{I_1 (k)} + \dfrac{K_0 (k')}{K_1
(k)} \right), \Seqnsplit
I_{\mu} &amp;=&amp; k_{\mu} \varphi _{\mu} \dfrac{I_0 (k_{\mu})}{I_1 (k_{\mu})},
\quad I_{\sigma} = k_{\sigma} \varphi _{\sigma} \dfrac{I_0
(k_{\sigma})}{I_1 (k_{\sigma})}, \quad	K_{\mu} = k_{\mu}
\varphi _{\mu} \dfrac{K_0 (k'_{\mu})}{K_1 (k'_{\mu})}, \quad K_{\sigma}
= k_{\sigma} \varphi _{\sigma} \dfrac{K_0 (k'_{\sigma})}{K_1
(k'_{\sigma})}, \Seqnsplit
L_{\mu \sigma} &amp;=&amp;  \varphi _{\sigma}^{-1} I_{\sigma} - \varphi
_{\mu}^{-1} I_{\mu} , \quad M_{\mu \sigma}=  \varphi_{\sigma}^{-1}
K_{\sigma} - \varphi_{\mu}^{-1} K_{\mu} , \quad E_{\mu} = A D - B C -
D I_{\mu}, \quad  E_{\sigma} = A D - B C - D I_{\sigma}, \nonumber \\
\Seqnsplit
F_{\mu} &amp;=&amp; B ( A' + K_{\mu} C ) - (B' + K_{\mu} D )  (A - I_{\mu} ),
\quad  F_{\sigma} = B ( A' + K_{\sigma} C ) - (B' + K_{\sigma} D )  (A
- I_{\sigma} ) , \Seqnsplit
k' &amp;=&amp; (1+e)k, \quad k'_{\mu}=(1+e)k_{\mu}, \quad, 
k'_{\sigma}=(1+e)k_{\sigma}.
\end{eqnarray}}\unskip
The derivation of (\ref{eq:dynamo threshold}) is given in 
Appendix~\ref{Appendix: threshold}. At
this stage we can make two checks, one for $e=0$, the other for
$e\rightarrow \infty$.

For $e=0$, we have $k'=k$, $k'_{\mu}=k_{\mu}$,
$k'_{\sigma}=k_{\sigma}$, $A'=A$, $B'=B$, $C=1$, $D=0$,
$E_{\mu}=E_{\sigma}=-B$, $F_{\mu}=B(K_{\mu}+I_{\mu})$,
$F_{\sigma}=B(K_{\sigma}+I_{\sigma})$. After some algebra and using the
Wronskian relation
$x(({I_0(x)})/({I_1(x)}) + ({K_0(x)})/({K_1(x)})) = [I_1(x)K_1(x)]^{-1}$, 
the following dynamo threshold is found
{\begin{equation}\label{eq:Omegac_e=0}
\Omega^c = \dfrac{c}{s} (I_1(k_{\sigma}) K_1(k_{\sigma})-I_1(k_{\mu})K_1(k_{\mu}))^{-1},
\end{equation}}\unskip
which is nothing else than the expression given in~\cite{Plunian2021}. 

In the limit $e\rightarrow \infty$, we have $k' \gg 1$,  $k'_{\mu} \gg
1$, $k'_{\sigma} \gg 1$. Knowing that for $x\rightarrow \infty$,
$I_0(x)\equiv I_1(x) \approx e^x/\sqrt{2\pi x} \rightarrow \infty$ and
$K_0(x)\equiv K_1(x) \approx e^{-x}\sqrt{\pi /2x} \rightarrow 0$, we
then have $B'\equiv A'$, $D\equiv C$, $K_{\mu}\equiv k_{\mu}
\varphi_{\mu}$, $K_{\sigma}\equiv k_{\sigma} \varphi_{\sigma}$, $M_{\mu
\sigma}\equiv k_{\sigma}-k_{\mu}$. In (\ref{eq:dynamo threshold}), the
factors multiplied by $\overline{\omega}$ are of the same order of
magnitude as the terms that are not proportional to
$\overline{\omega}$. As $\overline{\omega}\equiv e \gg 1$,
Equation (\ref{eq:dynamo threshold}) then simplifies into
{\begin{equation}\label{eq:dynamo threshold e infini}
\overline{\Omega^c} = 
\dfrac{ - M_{\mu \sigma} E_{\mu} E_{\sigma} + L_{\mu \sigma}  ( D B A' - B C B' )}
{M_{\mu \sigma} L_{\mu \sigma} D^2} \equiv 
\dfrac{ - E_{\mu} E_{\sigma}}{ L_{\mu \sigma} D^2}.
\end{equation}}\unskip

Then we can show that 
{\begin{equation}\label{eq:Omegac_einfty}
\Omega^c - \omega = - \dfrac{c}{s}
\dfrac{(\Phi(k_{\sigma})+\varphi_{\sigma}^{-1}\Psi(k))(\Phi(k_{\mu})+\varphi_{\mu}^{-1}\Psi(k))}
{\Phi(k_{\sigma})-\Phi(k_{\mu})},
\end{equation}}\unskip
where 
\changeEqNo{57a,b}
{\begin{subequations}\label{eq:PhiPsi}
\begin{equation}
\Phi(x) = x I_0(x)/I_1(x), \quad 
\Psi(x) = x K_0(x)/K_1(x),
\end{equation}
\end{subequations}}\unskip
\resetEqNo{57}\unskip
\noindent
which is identical to the dynamo threshold expression given
in~\cite{Plunian2023} for a rotor immersed in an isotropic medium, both
rotating at angular velocities $\Omega$ and $\omega$ respectively.

Finally, we note that in (\ref{eq:dynamo threshold}), exchanging
$\sigma$ and $\mu$ corresponds to changing the angular velocities
$\overline{\Omega^c}$ and $\overline{\omega}$ into their opposite
$-\overline{\Omega^c}$ and $-\overline{\omega}$, as already shown
by~\cite{Favier2013,Marcotte2021} for the induction equation with
isotropic electromagnetic conductivity and permeability. 

In our model, $\omega$ is assumed to be constant with $r$. Therefore,
taking $\omega=\beta \Omega$ with $\beta \in [0,1]$, and replacing in 
(\ref{eq:dynamo threshold}) leads to
{\begin{equation}\label{eq:dynamo threshold*}
\Omega^c = \dfrac{c}{s}\varphi_{\sigma}^{-1}\varphi_{\mu}^{-1}
\dfrac{Q(1-\beta)+P(1+e)\beta \pm \sqrt{\Delta}}{2(1+e)\beta(1-\beta)M_{\mu \sigma}L_{\mu \sigma}D^2},
\end{equation}}\unskip
with
{\begin{eqnarray}
P &amp;=&amp; [ - M_{\mu \sigma} E_{\mu} E_{\sigma} + L_{\mu \sigma} (D B A' - B C B') ], \Seqnsplit
Q &amp;=&amp; M_{\mu \sigma} (D B A' - B C B' ) - L_{\mu \sigma} ( B' + K_{\mu}  D ) ( B' + K_{\sigma}  D ) , \Seqnsplit
\Delta &amp;=&amp; [Q(1-\beta)+P(1+e)\beta]^2-4F_{\mu} F_{\sigma} M_{\mu \sigma}L_{\mu \sigma} D^2(1+e)\beta(1-\beta).
\end{eqnarray}}\unskip
In the limit $\beta(1-\beta)\approx 0$, we have $|\Omega^c| \approx
|({c}/{s})\varphi_{\sigma}^{-1}\varphi_{\mu}^{-1}F_{\mu}F_{\sigma}[Q(1-\beta)+P(1+e)\beta]^{-1}|$, leading to 
{\begin{equation}\label{eq:beta 0 or 1}
|\Omega^c| = \begin{cases}
\left|\dfrac{c}{s}\varphi_{\sigma}^{-1}\varphi_{\mu}^{-1} F_{\mu}F_{\sigma}Q^{-1} \right| \\
\left|\dfrac{c}{s}\varphi_{\sigma}^{-1}\varphi_{\mu}^{-1} F_{\mu}F_{\sigma}P^{-1}(1+e)^{-1}\right|,
\end{cases} \quad \mbox{for }
\beta = \begin{cases}
0 \\
1.
\end{cases} 
\end{equation}}\unskip

\section{Results}\label{sec:Results}

As in~\cite{Plunian2023}, and with experimental applications in mind,
we consider three cases characterized by the materials that compose
both the rotor and the stator. The first case denoted
copper$^{(\perp)}$/kapton$^{(\parallel)}$, corresponds to the choice
made for the Fury dynamo experiment, in which the rotor and stator are
composed of copper as the main conducting material and kapton as an
electrical insulator. In Fury, vertical grooves have been cut following
logarithmic spirals as in Figure~\ref{fig:anisotopric cylinders}. These
grooves have been filled with kapton sheets to force the electrical
currents to follow curved trajectories, in a way reproducing
anisotropic electrical conductivity~\cite{Alboussiere2022}. In the
other two cases studied here, copper is replaced by iron or
$\mu$-metal, both of which have a relative magnetic permeability well
in excess of unity. Kapton is replaced by tin, whose relative magnetic
permeability is equal to unity. The resulting arrangements 
iron$^{(\perp)}$/tin$^{(\parallel)}$ and
$\mu$-metal$^{(\perp)}$/tin$^{(\parallel)}$ mimic materials with an
anisotropic magnetic permeability. In all three cases the gap is filled
with galinstan, an alloy of gallium, indium and tin, which is liquid at
room temperature and essential for ensuring good electrical contact
between the rotor and the stator. The values of the electrical
conductivities and magnetic permeabilities corresponding to these three
cases are collected in Table~\ref{table:materials}, together with the
corresponding values of $\sigma$, $\mu$, $\varphi_{\sigma}$ and
$\varphi_{\mu}$.

%tab1
\begin{table}
\caption{\label{table:materials}Electromagnetic properties of the
materials composing the rotor, the gap and the stator}
\tabcolsep=4pt
\begin{tabular}{ccccccccccc}
\thead
&amp; \multicolumn{6}{c}{Rotor/Stator} &amp; \multicolumn{4}{c}{Gap (Galinstan)} \\
\cline{2-7}\cline{8-11}
Materials &amp; $\sigma^{\perp}$ &amp; $\mu^{\perp}$ &amp; $\sigma^{\parallel}$ &amp;
$\mu^{\parallel}$ &amp; $\sigma$ &amp; $\mu$ &amp; $\sigma_g$ &amp; $\mu_g$ &amp;
$\varphi_{\sigma}$ &amp; $\varphi_{\mu}$\\
\endthead
Cu$^{(\perp)}$ Ka$^{(\parallel)}$ &amp; 63 &amp; 1 &amp; $5\times10^{-22}$ &amp; 1 &amp;
$13\times10^{22}$ &amp; 0 &amp; 3.86 &amp; 1 &amp; $0.06$ &amp; 1 \\  
Fe$^{(\perp)}$ Sn$^{(\parallel)}$ &amp; 10.2 &amp; $5\times10^3$ &amp; 9.9 &amp; 1 &amp;
0.03 &amp; $5\times10^3$ &amp; 3.86 &amp; 1 &amp; 0.38 &amp; $2\times10^{-4}$\vspace*{4pt} \\ 
$\mu$m$^{(\perp)}$ Sn$^{(\parallel)}$ &amp; 2.1 &amp; $1.5\times10^5$ &amp; 9.9 &amp; 1
&amp;${-}$0.79 &amp; $1.5\times10^5$ &amp; 3.86 &amp; 1 &amp; 1.85 &amp; $6.7\times10^{-6}$
\botline
\end{tabular}
\tabnote{The electrical conductivities
$\sigma^{\perp}$, $\sigma^{\parallel}$ and $\sigma_g$ are given in unit
of $10^6~\Omega^{-1}{\cdot}\mathrm{m}^{-1}$. The magnetic
permeabilities $\mu^{\perp}$, $\mu^{\parallel}$ and $\mu_g$ are given
in unit of $4\pi\times 10^{-7}~\mathrm{H}{\cdot}\mathrm{m}^{-1}$.
The parameters $\sigma$,
$\mu$, $\varphi_{\sigma}$ and $\varphi_{\mu}$ are dimensionless. The
numerical values of the electrical conductivity and magnetic
permeability of the materials are taken from appropriate
references~\cite{Jiles1991,Smithells,Wu2008,Dupont2022}.\vspace*{6pt}}
\end{table}

\subsection{Dimensionless results}

In Figure~\ref{fig:RGS}, the dynamo threshold $\Omega^c$ is plotted
either (left) versus the magnetic wave number $k$ for a pitch angle
$\alpha/\pi=0.15 \pi$ or (right) versus $\alpha$ for $k=1.5$.  The
three rows of subfigures correspond, from top to bottom, to the
copper$^{(\perp)}$/kapton$^{(\parallel)}$,
iron$^{(\perp)}$/tin$^{(\parallel)}$ and
$\mu$-metal$^{(\perp)}$/tin$^{(\parallel)}$ cases. In each subfigure
six gap thicknesses $e$ including the case without gap ($e=0$), and
three values of the angular velocities ratio $\beta$ are considered.
The value $e=1\%$ corresponds to that of Fury. The other non-zero
values of $e$ correspond to larger gaps, up to $71.4\%$ of the rotor
radius. Table~\ref{table:values} summarises the geometric
characteristics of Fury and those of a slightly larger model with
several values of galinstan gap. 

\begin{figure}[t!]
{\vspace*{-4pt}}
\includegraphics{fig03}
{\vspace*{-4pt}}
\caption{\label{fig:RGS}The dynamo threshold $|\Omega^c|$ is plotted,
either (left) versus  the wave number $k$ for a pitch angle
$\alpha=0.15 \pi$, or (right) versus $\alpha$ for $k=1.5$. The top
row corresponds to the copper$^{(\perp)}$/kapton$^{(\parallel)}$ case,
the middle row to the iron$^{(\perp)}$/tin$^{(\parallel)}$ case, the
bottom row to the $\mu$-metal$^{(\perp)}$/tin$^{(\parallel)}$ case.
Each curve corresponds to a couple $(e,\beta)$, with $e\in\{0;10^{-2};
2.9\times 10^{-2}; 2.57\times 10^{-1};4.86\times 10^{-1};7.14\times
10^{-1};\}$ and $\beta\in\{0;0.5;1\}$.}
{\vspace*{-4pt}}
\end{figure}

%tab2
\begin{table}
\caption{\label{table:values}Geometrical characteristics of the Fury
dynamo experiment~\cite{Alboussiere2022} and another experimental
project with larger galinstan gaps}
\begin{tabular}{ccccccc}
\thead
&amp; \multicolumn{4}{c}{Dimensional} &amp; \multicolumn{2}{c}{Dimensionless} \\ 
\cline{2-5}\cline{6-7}
Parameters &amp; $R$ &amp; $e$ &amp; $h$ &amp; $\alpha/\pi$ &amp; $e$ (\%) &amp; $k=2\pi R /h$ \\ 
Units &amp; cm &amp; mm &amp; cm &amp; rad &amp; -- &amp; -- \\
\endthead
Fury &amp; 5 &amp; 0.5 &amp; 20.9 &amp; 0.15 &amp; 1 &amp; 1.5\vspace*{4pt} \\
\morerows{3}{Larger gap} &amp; \morerows{3}{7} &amp; 2 &amp; \morerows{3}{30} &amp; \morerows{3}{0.15} &amp; \02.9 &amp; \morerows{3}{1.5} \\
&amp; &amp; 18 &amp; &amp; &amp; 25.7 &amp;\\
&amp; &amp; 34 &amp; &amp; &amp; 48.6 &amp;\\
&amp; &amp; 50 &amp; &amp; &amp; 71.4 &amp;
\botline
\end{tabular}
\tabnote{The rotor's radius $R$, the gap thickness $e$, the height $h$
of the experiment, are dimensional quantities. The pitch angle $\alpha$
of anisotropy is given in radian. The same notation $e$ is used for the
dimensional and dimensionless gap thickness, the relation between both
being $e\;(\%)=10^2(e/R)\;$. The value of $k$ corresponds to a complete
period of the magnetic field along the height of the experiment.}
\end{table}

From the top subfigures of Figure~\ref{fig:RGS} we see that, for
$e=2.9\%$, taking $\alpha=0.15$ and $k=0.15$ is not far from the
minimum dynamo threshold. This is less true when $e$ is increased. We
could think that increasing the galinstan gap thickness is necessarily
detrimental to the dynamo  because the shear between the rotor and the
stator will be weaker. This is mainly the case, except for example in
the top left sub-figure, which shows curve crossings for $\beta=0.5$,
$e=48.6\%$ and $e=71.4\%$. Taking $\beta=0$ leads to the largest dynamo
threshold unless $k$ is sufficiently large as can be seen e.g.\ at the
top left subfigure, with curve crossings  for $e=71.4\%$, $\beta=0.5$
and $\beta=0$. We note that increasing $e$ shifts the minimum threshold
to smaller values of $k$ and $\alpha$. 

Finally there is a stricking difference between the
copper$^{(\perp)}$/kapton$^{(\parallel)}$ case and the two others
iron$^{(\perp)}$/tin$^{(\parallel)}$ and
$\mu$-metal$^{(\perp)}$/tin$^{(\parallel)}$. Indeed, whereas at $e=0$
the minimum dynamo thresholds of the three cases are comparable,
$\min_k\left(\Omega^c(k)\right)\approx \{14.6, 14.8,8.3\}$, increasing
$e$ leads to much higher dynamo thresholds for the two last cases, 
$\min_k\left(\Omega^c(k)\right) \approx \{15.1, 84, 710\}$ for $e=1\%$,
$\min_k\left(\Omega^c(k)\right) \approx \{16, 165, 1600\}$ for
$e=2.9\%$. This is presumably because in these cases the galinstan gap
corresponds to an abrupt radial jump of the magnetic permeability, of
at least four order of magnitude as given in
Table~\ref{table:materials} by the small values of $\varphi_{\mu}$. In
comparison,  in the copper$^{(\perp)}$/kapton$^{(\parallel)}$ case, 
the radial jump of electrical conductivity of around two orders of
magnitude as given by $\varphi_{\sigma}$ in
Table~\ref{table:materials}, is much less detrimental to the dynamo.
This significant increase in the dynamo threshold for the cases of
iron$^{(\perp)}$/tin$^{(\parallel)}$ and
$\mu$-metal$^{(\perp)}$/tin$^{(\parallel)}$ must however be revised
when dimensional quantities are considered, as explained in
Section~\ref{sec:Dimensional results}.

In Figure~\ref{fig:lignesdechamp} the electric current density and
magnetic field lines are plotted at the threshold, in the three cases
copper$^{(\perp)}$/kapton$^{(\parallel)}$,
iron$^{(\perp)}$/tin$^{(\parallel)}$ and
$\mu$-metal$^{(\perp)}$/tin$^{(\parallel)}$, for $k=1.5, \;
\alpha/\pi=0.15, \; \beta=0$ and several values of $e$. The magnetic
field lines are plotted from the expressions of $B_r$ and $B_{\theta}$
given in (\ref{eq:Brlambda}) and (\ref{eq:Btetalambda}). The electric
current density can be calculated from (\ref{eq:Ampere}), leading to
the dimensionless following expression
{\begin{eqnarray}
J_r &amp;=&amp;  -\mathrm{i}k\tilde{\varphi}_{\mu}^{-1}\tilde{\phi}, \quad \mbox{with } 
\tilde{\phi} = \tilde{\mu} cs B_r + (1+\tilde{\mu} s^2)B_{\theta}, \label{Jr} \Seqnsplit
J_{\theta} &amp;=&amp; \mathrm{i}k^{-1}\tilde{\varphi}_{\mu}^{-1}
[D_k(B_r)+\tilde{\mu}c^2k^2B_r+\tilde{\mu}csk^2B_{\theta}]. \label{Jteta}
\end{eqnarray}}\unskip
At the dynamo threshold, $\gamma=0$, Equation (\ref{eq:gammaBr})
implies that
{\begin{equation}
D_k(B_r) = (1+\tilde{\sigma} s^2)^{-1} [ cs(\tilde{\sigma}
-\tilde{\mu}) k^2 B_{\theta} -\tilde{\mu} c^2 k^2 B_r ],
\end{equation}}\unskip
leading to 
{\begin{equation}\label{Jteta2}
J_{\theta} = \mathrm{i}k
\dfrac{cs\tilde{\sigma}}{1+s^2\tilde{\sigma}}\tilde{\varphi}_{\mu}^{-1}\tilde{\phi}.
\end{equation}}\unskip
Then (\ref{Jr}) and (\ref{Jteta2}) lead to the following relation
{\begin{equation}\label{eq:JrJteta}
(1+s^2\tilde{\sigma})J_{\theta}+cs\tilde{\sigma}J_r =0,
\end{equation}}\unskip
corresponding to logarithmic spiral lines for $\tilde{\sigma}\ne 0$ and
radial lines for $\tilde{\sigma}=0$ as plotted in
Figure~\ref{fig:lignesdechamp}. 

\begin{figure}
{\vspace*{-2pt}}
\includegraphics{fig04}
{\vspace*{-2pt}}
\caption{\label{fig:lignesdechamp}Plots of the magnetic field lines
(full-blue) and the electric current lines (dashed-red) in the
horizontal plane $(x, y)$, for the three cases
Cu$^{(\perp)}$Ka$^{(\parallel)}$ (left column), Fe$^{(\perp)}$
Sn$^{(\parallel)}$ (middle column), $\mu$m$^{(\perp)}$
Sn$^{(\parallel)}$ (right column), for $\alpha/\pi = 0.15$,
$\beta=0$, $k=1.5$ and, from top to bottom, $e=1\%, 25.7\%, 71.4\%$.}
{\vspace*{-2pt}}
\end{figure}

In Figure~\ref{fig:JzBz}, $B_z(r)$ and $J_z(r)$ are plotted, again in
the three cases copper$^{(\perp)}$/kapton$^{(\parallel)}$,
iron$^{(\perp)}$/tin$^{(\parallel)}$ and
$\mu$-metal$^{(\perp)}$/tin$^{(\parallel)}$, and for the same
parameters as in Figure~\ref{fig:lignesdechamp}. The expression of
$J_z(r)$ is given in (\ref{eq:jz}). The expression of $B_z(r)$ can be
derived from (\ref{eq:divB3}). Applying the relations
$\partial_r(rI_1(\nu r))=\nu r I_0(\nu r)$ and $\partial_r(rK_1(\nu
r))=-\nu r K_0(\nu r)$ to (\ref{eq:Brlambda}), leads to
{\begin{eqnarray}\label{eq:Bzexpression}
B_z = \mathrm{i}k^{-1} \begin{cases}
r&lt; 1,    &amp; -s\left(\lambda_{\sigma}^R k_{\sigma} \dfrac{I_0(k_{\sigma} r)}{I_1(k_{\sigma})}+ \lambda_{\mu}^R k_{\mu} \dfrac{I_0(k_{\mu} r)}{I_1(k_{\mu})}\right) \\[.75pc]
1&lt;r&lt;1+e  &amp; -sk\left(\lambda_{r}^I \dfrac{I_0(k r)}{I_1(k)}- \lambda_{r}^K \dfrac{K_0(k r)}{K_1(k)}\right) \\[.75pc]
r&gt; 1+e,  &amp; s\left(\lambda_{\sigma}^S k_{\sigma} \dfrac{K_0(k_{\sigma} r)}{K_1(k_{\sigma}(1+e))}+ \lambda_{\mu}^S k_{\mu} \dfrac{K_0(k_{\mu} r)}{K_1(k_{\mu}(1+e))}\right).
\end{cases}
\end{eqnarray}}\unskip
For $e=0$, we can show from (\ref{eq:BC3}) and (\ref{eq:BC3b}) that
$B_z$ given by (\ref{eq:Bzexpression}) is continuous at $r=1$, as
plotted in Figure~\ref{fig:JzBz}.

\begin{figure}
{\vspace*{-2pt}}
\includegraphics{fig05}
\caption{\label{fig:JzBz}Plots of $B_z(r)/B_{\max}$ (top) and
$J_z(r)/J_{\max}$ (bottom), for the three cases
Cu$^{(\perp)}$Ka$^{(\parallel)}$ (left column), Fe$^{(\perp)}$
Sn$^{(\parallel)}$ (middle column), $\mu$m$^{(\perp)}$
Sn$^{(\parallel)}$ (right column), for $\alpha/\pi = 0.15$,
$\beta=0$, $k=1.5$, and $e= 0\%, 1\%, 2.9\%, 25.7\%, 48.6\%, 71.4\%$.}
\end{figure}

\subsection{Dimensional results}\label{sec:Dimensional results}

In Figure~\ref{fig:RGSdimensional}, the critical frequency $f^c$ is
plotted versus the rotor's radius $R$ for several values of the pitch
angle $\alpha$ and of the gap thickness $e$ (which after being
multiplied by $R$ is now dimensional). The magnetic wave number is
fixed at $k=1.5$, corresponding to an experiment with an aspect ratio
$h/R=2\pi/k$ of about 4.2, after assuming that the wavelength of the
magnetic field is equal to the vertical size of the experiment. This
aspect ratio of 4.2, which is the one used in Fury, is a good
compromise between theoretical predictions and the robustness of the
device. The angular velocities ratio is fixed at $\beta=0$,
corresponding to a gap at rest. The frequency is obtained from
{\begin{equation}
f^c = \dfrac{|\Omega^c|}{2\pi R^2 \sigma^{\perp} \mu^{\perp}}.
\end{equation}}\unskip

\begin{figure}
\includegraphics{fig06}
\caption{\label{fig:RGSdimensional}The critical frequency $f^c$ (in Hz)
is plotted versus the rotor's radius $R\in[0, 7~\mathrm{cm}]$ for (left)
the copper$^{\perp}$/kapton$^{\parallel}$, (middle) the
iron$^{\perp}$/tin$^{\parallel}$ and (right) the
$\mu$-metal$^{\perp}$/tin$^{\parallel}$ cases, with six or seven gap
thicknesses $e\in[0,  50~\mathrm{mm}]$, and three values of $\alpha$. The
solid, dahed and dotted curves correspond to $\alpha/\pi=0.15, 0.1,
0.05$. The wave number and angular velocities ratio are fixed at
$k=1.5$ and $\beta=0$.}
{\vspace*{-2pt}}
\end{figure}

Taking $R=7$ cm and $\alpha=0.15\pi$ (solid lines), and for a rather
large gap of galinstan $e=50$ mm leads to a dynamo threshold $f^c=35,
28$ and 14 Hz,  for respectively the
copper$^{(\perp)}$/kapton$^{(\parallel)}$, 
iron$^{\perp}$/tin$^{\parallel}$ and 
$\mu$-metal$^{\perp}$/tin$^{\parallel}$ cases. These values are
entirely achievable experimentally, and we note that the use of a
galinstan gap with a rotor and stator of anisotropic magnetic
permeability is not as detrimental as the dimensionless results
suggested. It is even a much better system if a small rotor's radius is
considered together with a small gap of galinstan. Indeed, for $R=3$ cm
and $e=0.5$ mm, the dynamo threshold is $f^c=1.3$ and 3 Hz for
respectively the iron$^{\perp}$/tin$^{\parallel}$ and 
$\mu$-metal$^{\perp}$/tin$^{\parallel}$ cases, whereas it is about 38 Hz
for the copper$^{(\perp)}$/kapton$^{(\parallel)}$ case.  

In the left subfigure of Figure~\ref{fig:RGSdimspetite}, $f^c$ is
plotted versus $e$, for values of $e$ and $R$ smaller than the one used
in Fury ($e\le 0.5$ mm and $R\le 3$ cm), $\beta=0$ and
$\alpha=0.15\pi$. The results suggest the possibility of building a
miniature dynamo with a rotor's radius $R$ of 1--2 cm and a gap of
galinstan $e$ of 0.3 to 0.5 mm, using iron$^{\perp}$/tin$^{\parallel}$. 

\begin{figure}
{\vspace*{-2pt}}
\includegraphics{fig07}
{\vspace*{-2pt}}
\caption{\label{fig:RGSdimspetite}The critical frequency $f^c$ (in Hz)
is plotted: (left) versus the galinstan gap thickness $e$ for the
iron$^{\perp}$/tin$^{\parallel}$ and the
$\mu$-metal$^{\perp}$/tin$^{\parallel}$ cases, for three values of
rotor's radius $R\in\{1, 2, 3\}$ cm, for $k=1.5$ and
$\alpha/\pi=0.15$; (middle) versus the magnetic vertical wavelength
$h$ for  the iron$^{\perp}$/tin$^{\parallel}$ case, for four values of
gap's thickness $e\in\{0,0.3, 0.4, 0.5\}$ mm, for $R=1.5$ cm and
$\alpha/\pi=0.15$; (right) versus the magnetic vertical wavelength
$h$ for  the iron$^{\perp}$/tin$^{\parallel}$ case, for four values of 
$\alpha/\pi \in\{0.01,0.05, 0.1, 0.15\}$, for $R=1.5$ cm and 
$e=0.3$~mm. In all cases, $\beta=0$. In the middle and right subfigures two
values of $\mu_g$ are considered $\mu_g \in\{4\pi\times 10^{-7},
20\pi\times 10^{-7}\}$.} 
{\vspace*{-2pt}}
\end{figure}

In the middle and right subfigures of Figure~\ref{fig:RGSdimspetite},
$f^c$ is plotted  versus $h$ in the iron$^{\perp}$/tin$^{\parallel}$
case, for $R=1.5$ cm and $\beta=0$. In the middle subfigure, $e$ is
varied and $\alpha$ is fixed to $0.15\pi$, while in the right
subfigure $e$ is fixed to 0.3 mm and $\alpha$ is varied. In addition,
two values of $\mu_g$ are considered, either 
$\mu_g = 4  \pi\times 10^{-7}~\mathrm{H}{\cdot}\mathrm{m}^{-1}$ 
for a gap filled with galinstan or 
$\mu_g = 20 \pi\times 10^{-7}~\mathrm{H}{\cdot}\mathrm{m}^{-1}$ 
for a gap filled with a magnetic liquid metal (MLM) made of iron
particules in liquid eGaIn. The electric conductivity of such MLM is
equal to $3.9\times 10^6~\Omega^{-1}{\cdot}\mathrm{m}^{-1}$, so
identical to the one of galinstan, but with a magnetic permeability
five times larger~\cite{Carle2017}.

From the middle subfigure of Figure~\ref{fig:RGSdimspetite} we see that
for $e=0.3$ mm and $h=10$ cm the dynamo threshold is $f^c=3.4$ Hz for a
gap filled with galinstan, and $f^c=0.8$ Hz for gap filled with MLM.
Assuming that the stator's radius is twice larger than the rotor's
radius $R=1.5$ cm, the ensemble constituted of the rotor gap and stator
will have a mass of about 2.2 kg (for an iron density equal to 7.86
g${\cdot}$cm$^3$). These results correspond to $\alpha/\pi=0.15$. 

From the right subfigure of Figure~\ref{fig:RGSdimspetite}, wee see
that taking smaller value of $\alpha$ down to $\alpha/\pi=0.05$ leads
to a smaller dynamo threshold. Therefore, for $R=1.5$ cm, $e=0.3$ mm
and $\alpha/\pi=0.05$, taking $h=10$ cm leads to a dynamo threshold 
$f^c=2.2$ Hz for a gap filled with galinstan, and $f^c=0.7$ Hz for gap
filled with MLM. This corresponds again to a device mass of about 2.2
kg. For a gap filled with MLM and device height $h=5$ cm the dynamo
threshold is about 1.5 Hz and the mass about 1.1 kg. The results are
summarized in Table~\ref{table:results}.

%tab3
\begin{table}
\caption{\label{table:results}Dynamo threshold frequency for the
iron$^{\perp}$/tin$^{\parallel}$ combination, $R=1.5$ cm, $e=0.3$~mm,
two values of the pitch angle $\alpha$, two values of the magnetic wave
length $h$ and two types of liquid metal filling the gap, galinstan or
magnetic liquid metal{\vspace*{-2pt}}}
\begin{tabular}{ccccc}
\thead
&amp; \multicolumn{2}{c}{$h=10$ cm} &amp; \multicolumn{2}{c}{$h=5$ cm} \\ 
\cline{2-3}\cline{4-5}
&amp; $f^c_{\mathrm{Ga}}$ (Hz) &amp; $f^c_{\mathrm{MLM}}$ (Hz) &amp; $f^c_{\mathrm{Ga}}$ (Hz) &amp; $f^c_{\mathrm{MLM}}$ (Hz)\vspace*{2pt}\\
\endthead
$\alpha/\pi=0.15$ &amp; 3.4 &amp; 0.8 &amp; 8.5 &amp; 1.9 \\
$\alpha/\pi=0.05$ &amp; 2.2 &amp; 0.7 &amp; 6 &amp; 1.5
\botline
\end{tabular}
{\vspace*{-2pt}}
\end{table}

\section{Conclusion}

In this paper, we studied the presence of a gap between the rotor and
stator of an anisotropic dynamo in cylindrical geometry. The rotor and
the stator have a logarithmic anisotropy given by the same pitch angle
$\alpha$. The gap between the rotor and the stator is filled with an
isotropic material, such as liquid metal. The dynamo threshold
corresponding to the minimum angular velocity of the rotor, above which
a magnetic field is generated, was calculated for different values of
the gap thickness. We considered three cases depending on the materials
composing both the rotor and stator, 
copper$^{\perp}$/kapton$^{\parallel}$, iron$^{\perp}$/tin$^{\parallel}$
and $\mu$-metal$^{\perp}$/tin$^{\parallel}$, the first case
corresponding to an electrical conductivity anisotropy, the two others
to a magnetic permeability anisotropy. For the material filling the gap
we mainly considered galinstan which is liquid at room\break temperature. 

When the magnetic wavelength is not too large compared with the rotor
radius, three general trends were found: (i) the presence of a gap is
always detrimental to the dynamo, and (ii) the most detrimental case
corresponds to a gap at rest, which is therefore the case selected for
dimensioning a new experimental application, (iii) increasing the gap
thickness increases the dynamo threshold.

The study focused more specifically on two objectives. The first one
concerns the dimensioning of a new experiment with a gap larger than
that of Fury so that, once the dynamo is established, it can react
dynamically with the flow within the gap by means of Lorentz forces.
But of course, to do this, we first need to be able to start a dynamo
with a large gap. Although the copper$^{\perp}$/kapton$^{\parallel}$
combination appears as slightly less interesting than the other two, it
still offers reasonable values of the dynamo threshold, with rotor's
frequencies $f^c$ between 10Hz and 35 Hz for a rotor's radius $R=7$ cm
and a gap thickness $e$ between 0.2 cm and 5 cm. By way of comparison,
in the case of Fury, the dynamo threshold frequency was $f^c=$24 Hz
(for $R=5$ cm, $h=20.9$ cm and $e=0.5$ mm). Assuming a rotor radius
$R=7$ cm and a height $h=$30 cm (corresponding to $k=1.5$ as in Fury), a
gap thickness $e=5$ cm and a stator thickness of 4 cm, with copper and
galinstan densities of 9 g${\cdot}$cm$^{-3}$ and 6.44 g${\cdot}$cm$^{-3}$
respectively, we obtain a mass of about 200 kg, almost five times
heavier than Fury but nevertheless achievable on a laboratory scale.
The advantage of such copper$^{\perp}$/kapton$^{\parallel}$ combination
is that it precisely does not contain any magnetic material garanteeing
a dynamo starting without any help of a remanent magnetic field
contrary to previous experiments~\cite{Miralles2013}. In addition, Fury
has demonstrated experimentally that such a combination works like a
dynamo, at least for a thin galinstan gap. At the moment no anisotropic
dynamo experiment based on iron$^{\perp}$/tin$^{\parallel}$ or
$\mu$-metal$^{\perp}$/tin$^{\parallel}$ has yet been built.  This
brings us to our second objective, which is to build a miniature dynamo
experiment with a mass of the order of one kilogram.  

To design a miniature dynamo experiment, we need to reduce the rotor
radius, its height and the gap thickness. Of the three combinations
studied, the iron$^{\perp}$/tin$^{\parallel}$ combination appears to
be the most appropriate with the lowest dynamo threshold. Assuming a
rotor radius $R=1.5$ cm, a height $h=10$ cm (corresponding to
$k=0.94$), and a galinstan gap thickness $e=0.3$ mm, the dynamo
threshold corresponds to a rotor frequency $f^c=2.2$ Hz (resp.\ 3.4 Hz)
for an anisotropic pitch angle $\alpha/\pi=0.05$ (resp.\ 0.15). For a
stator thickness of 1.5 cm and an iron density of 7.86
g${\cdot}$cm$^{-1}$, we obtain a mass of about 2.2 kg. If, instead of
galinstan, the gap is filled with a magnetic liquid metal made up of
iron particles in liquid eGaIn, the height and mass of the experiment
can even be divided by a factor of 2, corresponding to a mass around
1.1 kg, and a dynamo threshold rotor frequency $f^c=1.5$ Hz for
$\alpha/\pi=0.05$.  

\section*{Declaration of interests}

The authors do not work for, advise, own shares in, or receive funds
from any organization that  could benefit from this article, and have
declared no affiliations other than their research organizations.

\section*{Funding}

FP acknowledges the French Academy of Sciences and Electricit\'e de
France for granting the Amp\`ere prize to the ISTerre Geodynamo team in
2021. This work was supported by the PEPR Origins.

\back{}

\appendix{}
\def\appendixlabel{Appendix \Alph{section}}

\section{}\label{Appendix: threshold}

The matrix obtained from Equations (\ref{eq:BC1})--(\ref{eq:BC3b}), 
(\ref{eq:BC4}) and (\ref{eq:BC4b}) is written
below, with the following order of the variables $\lambda _{\sigma}^R$,
$\lambda _{\mu}^R$, $\lambda _r^I$, $\lambda _r^K$, $\lambda
_{\theta}^I$, $\lambda _{\theta}^K$, $\lambda _{\sigma}^S$ and $\lambda
_{\mu}^S$, using the following notations
{\begin{eqnarray*}
A &amp;=&amp; k \dfrac{I_0 (k)}{I_1 (k)}, \quad
B=k \left( \dfrac{I_0 (k)}{I_1 (k)} + \dfrac{K_0 (k)}{K_1 (k)} \right), \\
A' &amp;=&amp; k \dfrac{I_0 (k')}{I_1 (k)}, \quad 
B' = k \left( \dfrac{I_0 (k')}{I_1 (k)} + \dfrac{K_0 (k')}{K_1 (k)} \right), \\
C &amp;=&amp; \dfrac{I_1 (k')}{I_1 (k)}, \quad 
D = \dfrac{I_1 (k')}{I_1 (k)} - \dfrac{K_1 (k')}{K_1 (k)}, \\
I_{\mu} &amp;=&amp; k_{\mu} \varphi _{\mu} \dfrac{I_0 (k_{\mu})}{I_1 (k_{\mu})}, \quad 
I_{\sigma} = k_{\sigma} \varphi _{\sigma} \dfrac{I_0 (k_{\sigma})}{I_1 (k_{\sigma})}, \\
K_{\mu} &amp;=&amp; k_{\mu} \varphi _{\mu} \dfrac{K_0 (k'_{\mu})}{K_1 (k'_{\mu})}, \quad 
K_{\sigma} = k_{\sigma} \varphi _{\sigma} \dfrac{K_0 (k'_{\sigma})}{K_1 (k'_{\sigma})}.
\end{eqnarray*}}\unskip

The system that must be singular (zero determinant) is thus represented
by the matrix

{\fontsize{9}{10.5}\selectfont
\begin{equation*}
\begin{matrix}
-1 &amp; -1 &amp; 1 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
-1 &amp; 0 &amp; 0 &amp; 0 &amp; \varphi_{\mu}^{-1} &amp; \varphi_{\mu}^{-1} &amp; 0 &amp; 0 \\
- \varphi _{\sigma}^{-1} I_{\sigma} &amp; - \varphi _{\mu}^{-1} I_{\mu} &amp; \varphi_{\mu}^{-1} A   &amp; - \varphi_{\mu}^{-1} (B-A) &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; C &amp; C-D &amp; 0 &amp; 0 &amp; -1 &amp; -1 \\
0 &amp; 0 &amp; 0 &amp; 0 &amp; \varphi_{\mu}^{-1} C &amp; \varphi_{\mu}^{-1} (C-D) &amp; -1 &amp; 0 \\
0 &amp; 0 &amp; - \varphi_{\mu}^{-1} A' &amp;  \varphi_{\mu}^{-1} (B' - A') &amp; 0 &amp; 0 &amp; -\varphi_{\sigma}^{-1} K_{\sigma} &amp; - \varphi_{\mu}^{-1} K_{\mu}  \\
-c \varphi _{\sigma}^{-1} I_{\sigma} + \Omega s &amp; \Omega s &amp; -\omega s &amp; -\omega s &amp; \varphi_{\sigma}^{-1} \varphi_{\mu}^{-1} c A &amp; - \varphi_{\sigma}^{-1} \varphi_{\mu}^{-1} c (B-A)  &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; -\omega s (1+e) C &amp; -\omega s (1+e) (C-D) &amp; \varphi_{\sigma}^{-1} \varphi_{\mu}^{-1} c A'  &amp; -\varphi_{\sigma}^{-1} \varphi_{\mu}^{-1} c (B' - A') &amp; c  \varphi _{\sigma}^{-1} K_{\sigma} &amp; 0.
\end{matrix}
\end{equation*}}\unskip

Let's modify the matrix in the following ways: exchange column 1 and 2
(the determinant is opposite), replace column 2 by 1-2 (the determinant
is opposite), replace column 4 by 4-3 (no change in the determinant),
multiply columns 3, 4, 5 and 6 by $\varphi _{\mu}$ (the determinant is
multiplied by $\varphi _{\mu}^4$), replace column 6 by 6-5 (no change
in the determinant), replace column 7 by 8-7 (the determinant is
opposite), add $\varphi _{\mu}$ times column 1 to column 3 (no change
in the determinant), divide lines 7 and 8 by $c$ (the determinant is
divided by $c^2$), replace column 5 by 5-2 (no change in the
determinant) and multiply lines 7 and 8 by $\varphi _{\sigma}$ (the
determinant is multiplied by $\varphi _{\sigma}^2$). We now have the
matrix

{\fontsize{8.8}{10.5}\selectfont
\begin{equation*}
\begin{matrix}
-1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; 1 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\
- \varphi _{\mu}^{-1} I_{\mu} &amp; \varphi _{\sigma}^{-1} I_{\sigma} - \varphi _{\mu}^{-1} I_{\mu} &amp; A - I_{\mu}    &amp; - B &amp; - \varphi _{\sigma}^{-1} I_{\sigma} + \varphi _{\mu}^{-1} I_{\mu} &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; \varphi _{\mu} C &amp; - \varphi _{\mu} D &amp; 0 &amp; 0 &amp; 0 &amp; -1 \\
0 &amp; 0 &amp; 0 &amp; 0 &amp; C &amp; -D &amp; 1 &amp; 0 \\
0 &amp; 0 &amp; - A' &amp;   B' &amp; 0 &amp; 0 &amp; \varphi_{\sigma}^{-1} K_{\sigma} - \varphi_{\mu}^{-1} K_{\mu} &amp; - \varphi_{\mu}^{-1} K_{\mu}  \\
\varphi _{\sigma} \Omega \dfrac{s}{c} &amp; I_{\sigma} &amp; \varphi _{\mu} \varphi _{\sigma} ( \Omega -\omega ) \dfrac{s}{c} &amp; 0 &amp; A - I_{\sigma}  &amp; - B  &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; - \varphi _{\mu} \varphi _{\sigma} \omega \dfrac{s}{c} (1+e) C &amp;  \varphi _{\mu} \varphi _{\sigma} \omega \dfrac{s}{c} (1+e) D  &amp; A'  &amp; -B' &amp; - K_{\sigma}  &amp; 0.
\end{matrix}
\end{equation*}}\unskip

Its determinant is $- c^{-2} \varphi _{\mu}^4 \varphi _{\sigma}^2$
times the original determinant. With an opposite sign, it is also the
determinant of the $6\times 6$ bottom-right sub-matrix
{\begin{equation*}
\begin{matrix}
A - I_{\mu}    &amp; - B &amp; - \varphi _{\sigma}^{-1} I_{\sigma} + \varphi _{\mu}^{-1} I_{\mu} &amp; 0 &amp; 0 &amp; 0 \\
\varphi _{\mu} C &amp; - \varphi _{\mu} D &amp; 0 &amp; 0 &amp; 0 &amp; -1 \\
0 &amp; 0 &amp; C &amp; -D &amp; 1 &amp; 0 \\
- A' &amp;  B' &amp; 0 &amp; 0 &amp; \varphi_{\sigma}^{-1} K_{\sigma} - \varphi_{\mu}^{-1} K_{\mu} &amp; - \varphi_{\mu}^{-1} K_{\mu}  \\
\varphi _{\mu} \varphi _{\sigma} ( \Omega -\omega ) \dfrac{s}{c} &amp; 0 &amp; A - I_{\sigma}  &amp; - B  &amp; 0 &amp; 0 \\
- \varphi _{\mu}  \varphi _{\sigma} \omega \dfrac{s}{c} (1+e) C &amp;  \varphi _{\mu} \varphi _{\sigma} \omega \dfrac{s}{c} (1+e) D  &amp; A'  &amp; -B' &amp; - K_{\sigma} &amp; 0.
\end{matrix}
\end{equation*}}\unskip

We introduce some more notations
{\begin{eqnarray*}
L_{\mu \sigma}    &amp;=&amp; k_{\sigma} \dfrac{I_0 (k_{\sigma})}{I_1 (k_{\sigma})} - k_{\mu} \dfrac{I_0 (k_{\mu})}{I_1 (k_{\mu})} = \varphi _{\sigma}^{-1} I_{\sigma} - \varphi _{\mu}^{-1} I_{\mu}, \\
M_{\mu \sigma}    &amp;=&amp; k_{\sigma} \dfrac{K_0 (k'_{\sigma})}{K_1 (k'_{\sigma})} - k_{\mu} \dfrac{K_0 (k'_{\mu})}{K_1 (k'_{\mu})} = \varphi_{\sigma}^{-1} K_{\sigma} - \varphi_{\mu}^{-1} K_{\mu}, \\
\overline{\Omega} &amp;=&amp; \varphi _{\mu}  \varphi _{\sigma} (\Omega - \omega ) \dfrac{s}{c}, \nonumber \\
\overline{\omega} &amp;=&amp; \varphi _{\mu}  \varphi _{\sigma} \omega \dfrac{s}{c} (1+e),
\end{eqnarray*}}\unskip
and the matrix can be written 
{\begin{equation*}
\begin{matrix}
 A - I_{\mu}    &amp; - B &amp; - L_{\mu \sigma} &amp; 0 &amp; 0 &amp; 0 \\
\varphi _{\mu} C &amp; - \varphi _{\mu} D &amp; 0 &amp; 0 &amp; 0 &amp; -1 \\
0 &amp; 0 &amp; C &amp; -D &amp; 1 &amp; 0 \\
- A' &amp;  B' &amp; 0 &amp; 0 &amp; M_{\mu \sigma} &amp; - \varphi_{\mu}^{-1} K_{\mu}  \\
\overline{\Omega} &amp; 0 &amp; A - I_{\sigma}  &amp; - B  &amp; 0 &amp; 0 \\
- \overline{\omega} C &amp;  \overline{\omega} D  &amp; A'  &amp; -B' &amp; - K_{\sigma}  &amp; 0.
\end{matrix}
\end{equation*}}\unskip

Removing $\varphi _{\mu}^{-1} K_{\mu}$ times line 2 to line 4, leads to
the matrix 
{\begin{equation*}
\begin{matrix}
A - I_{\mu}    &amp; - B &amp; - L_{\mu \sigma} &amp; 0 &amp; 0 &amp; 0 \\
\varphi _{\mu} C &amp; - \varphi _{\mu} D &amp; 0 &amp; 0 &amp; 0 &amp; -1 \\
0 &amp; 0 &amp; C &amp; -D &amp; 1 &amp; 0 \\
- A' - K_{\mu} C &amp;  B' + K_{\mu} D  &amp; 0 &amp; 0 &amp; M_{\mu \sigma} &amp; 0  \\
\overline{\Omega}  &amp; 0 &amp; A - I_{\sigma}  &amp; - B  &amp; 0 &amp; 0 \\
- \overline{\omega} C &amp;  \overline{\omega} D  &amp; A'  &amp; -B' &amp; - K_{\sigma}  &amp; 0.
\end{matrix}
\end{equation*}}\unskip
Its determinant is just opposite to the determinant of the $5\times 5$
sub-matrix obtained when developing the last column
{\begin{equation*}
\begin{matrix}
A - I_{\mu}    &amp; - B &amp; - L_{\mu \sigma} &amp; 0 &amp; 0  \\
0 &amp; 0 &amp; C &amp; -D &amp; 1  \\
- A' - K_{\mu} C &amp;  B' + K_{\mu} D  &amp; 0 &amp; 0 &amp; M_{\mu \sigma}   \\
\overline{\Omega}  &amp; 0 &amp; A - I_{\sigma}  &amp; - B  &amp; 0  \\
- \overline{\omega} C &amp;  \overline{\omega} D  &amp; A'  &amp; -B' &amp; - K_{\sigma}.
\end{matrix}
\end{equation*}}\unskip
Removing $M_{\mu \sigma}$ times the second line to the third and $-
K_{\sigma} $ times the second line to the fifth, the determinant is
equal to that of the $4\times 4$ submatrix
{\begin{equation*}
\begin{matrix}
A - I_{\mu}    &amp; - B &amp; - L_{\mu \sigma} &amp; 0   \\
- A' - K_{\mu} C &amp;  B' + K_{\mu} D  &amp; - M_{\mu \sigma} C &amp; M_{\mu \sigma} D   \\
\overline{\Omega} &amp; 0 &amp; A - I_{\sigma}  &amp; - B   \\
- \overline{\omega} C &amp;  \overline{\omega} D  &amp; A' + K_{\sigma} C  &amp; -B' - K_{\sigma} D.
\end{matrix}
\end{equation*}}\unskip
Adding $M_{\mu \sigma} D B^{-1}$ times the third line to the second
line, then adding $B^{-1} (- B' - K_{\sigma} D  ) $ times the third
line to the fourth line, leads to the determinant of a $3\times 3$ submatrix
{\begin{equation*}
\begin{matrix}
A - I_{\mu}    &amp; - B &amp; - L_{\mu \sigma}    \\
- A' - K_{\mu} C + M_{\mu \sigma} D B^{-1} \overline{\Omega} &amp;  
B' + K_{\mu} D  &amp; - M_{\mu \sigma} ( C - (A - I_{\sigma} ) D B^{-1} ) \\
- \overline{\omega} C - (B' + K_{\sigma} D  ) B^{-1} \overline{\Omega} &amp; 
\overline{\omega} D  &amp; A' + K_{\sigma} C -  (B' + K_{\sigma} D  ) B^{-1} (A - I_{\sigma}).
\end{matrix}
\end{equation*}}\unskip
Adding $(A - I_{\mu})B^{-1}$ times the second column to the first
column and subtracting $L_{\mu \sigma} B^{-1}$ times the second column
to the third column, leads to the $2\times 2$ submatrix
{\begin{equation*}
\begin{matrix}
- A' - K_{\mu} C + M_{\mu \sigma} D B^{-1} \overline{\Omega} + (A - I_{\mu})B^{-1} ( B' + K_{\mu} D ) &amp; 
- M_{\mu \sigma} ( C - (A - I_{\sigma} ) D B^{-1} ) - L_{\mu \sigma} B^{-1} ( B' + K_{\mu} D )  \\
- \overline{\omega} C - (B' + K_{\sigma} D  ) B^{-1} \overline{\Omega}  + (A - I_{\mu})B^{-1} \overline{\omega} D &amp; 
A' + K_{\sigma} C -  (B' + K_{\sigma} D  ) B^{-1} (A - I_{\sigma} ) - L_{\mu \sigma} B^{-1} \overline{\omega}  D.
\end{matrix}
\end{equation*}}\unskip
Multiplying all coefficients by $B$ and rearranging leads to
{\begin{equation*}
\begin{matrix}
- A' B - K_{\mu} B C + M_{\mu \sigma} D \overline{\Omega} + (A - I_{\mu}) ( B' + K_{\mu} D ) &amp; 
M_{\mu \sigma} ( - B C + A D  - I_{\sigma}  D) - L_{\mu \sigma} ( B' + K_{\mu} D ) \\
- (B' + K_{\sigma} D  ) \overline{\Omega}  + (A D - I_{\mu} D -B C) \overline{\omega} &amp; 
B (A' + K_{\sigma} C ) -  (B' + K_{\sigma} D  ) (A - I_{\sigma} ) - L_{\mu \sigma} D  \overline{\omega}.
\end{matrix}
\end{equation*}}\unskip
We introduce the following last notations
{\begin{eqnarray*}
E_{\mu} &amp;=&amp; A D - B C - D I_{\mu}, \quad 
E_{\sigma} = A D - B C - D I_{\sigma}, \\
F_{\mu} &amp;=&amp; B ( A' + K_{\mu} C ) - (B' + K_{\mu} D )  (A - I_{\mu}), \quad 
F_{\sigma} = B ( A' + K_{\sigma} C ) - (B' + K_{\sigma} D )  (A - I_{\sigma} ),
\end{eqnarray*}}\unskip
so that the $2\times 2$ matrix can be written
{\begin{equation*}
\begin{matrix}
-F_{\mu} + M_{\mu \sigma} D \overline{\Omega} &amp; 
E_{\sigma} M_{\mu \sigma} - L_{\mu \sigma} ( B' + K_{\mu} D )  \\
-(B' + K_{\sigma} D  ) \overline{\Omega} + E_{\mu}  \overline{\omega} &amp; 
F_{\sigma} - L_{\mu \sigma} D  \overline{\omega}.
\end{matrix}
\end{equation*}}\unskip
Its determinant can be readily calculated
{\begin{eqnarray*}
\mathrm{det} &amp;=&amp; - F_{\mu} F_{\sigma} + \overline{\Omega} 
[ M_{\mu \sigma} (D F_{\sigma} + B' E_{\sigma}  + K_{\sigma} D E_{\sigma} ) - L_{\mu \sigma} ( B' + K_{\mu} D ) ( B' + K_{\sigma} D ) ] \\ 
&amp;&amp; +\, \overline{\omega} [ - M_{\mu \sigma} E_{\mu} E_{\sigma} + L_{\mu \sigma}  (D F_{\mu} + B' E_{\mu}  + K_{\mu} D E_{\mu})] - 
M_{\mu \sigma} L_{\mu \sigma} D^2 \overline{\Omega} \overline{\omega}.
\end{eqnarray*}}\unskip
Now, we remark that the coefficients we have introduced are not
independent and we have
{\begin{eqnarray*}
D F_{\sigma} + B' E_{\sigma} + K_{\sigma} D E_{\sigma} &amp;=&amp; D B A' - B C B', \\
D F_{\mu} + B' E_{\mu} + K_{\mu} D E_{\mu} &amp;=&amp; D B A' - B C B',
\end{eqnarray*}}\unskip
making these two combinations equal and independent of $\sigma$ or
$\mu$. So the determinant is now
{\begin{eqnarray*}
\mathrm{det} &amp;=&amp; - F_{\mu} F_{\sigma} + \overline{\Omega} [M_{\mu
\sigma} (D B A' - B C B' ) - L_{\mu \sigma} ( B' + K_{\mu} D ) ( B' +
K_{\sigma} D)] \\ 
&amp;&amp; +\, \overline{\omega} [ - M_{\mu \sigma} E_{\mu} E_{\sigma} + L_{\mu
\sigma}  ( D B A' - B C B')] -  M_{\mu \sigma} L_{\mu \sigma} D^2
\overline{\Omega} \overline{\omega}.
\end{eqnarray*}}\unskip
Because $M_{\mu \sigma}$ and $L_{\mu \sigma}$ are both antisymmetrical
(they change sign when $\sigma$ and $\mu$ are switched), we can see
that the determinant obeys the symmetry
{\begin{equation*}
\mathrm{det} (\mu, \sigma, \overline{\Omega}, \overline{\omega}) = 
\mathrm{det} (\sigma, \mu, -\overline{\Omega}, -\overline{\omega}).
\end{equation*}}\unskip

\printbibliography
\refinput{crphys20241029-reference.tex}

\end{document}
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