%$Id: notes.tex,v 1.102 2022/04/05 19:25:44 brandenb Exp $
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%\title{Turbulent magnetic energy survival after reheating}
%\title{Effect of magnetic and axion fields after reheating}
%\title{Effect of magnetic fields and the inflaton after reheating}
\title{Electromagnetic energy conversion into kinetic and thermal energies}
\author{}
\date{\today,~ $ $Revision: 1.102 $ $}
\begin{document}
\maketitle

\section{Introduction}

In hydrodynamic turbulence, dissipation of energy is in principle
straightforward: everything that gets in does get out eventually.
But when magnetic fields are involved, this becomes more complicated,
because there are now two exit channels, and it is {\em a priori} not
clear, which of the two takes the lion's share in specific situations.
A similar problem may also occur when the electric energy reservoir
is involved.
Before getting to that, let us first recall the different situations
in hydrodynamic and hydromagnetic turbulence, which are illustrated in
\Fig{psketch}(a).

\Fig{psketch}(a) presumes that kinetic energy can be tapped by the dynamo
action and thereby converted into magnetic energy \citep{Mof78}.
This is a generic process that we now know works in virtually all types
of turbulent systems provided the electric conductivity is large enough
\citep{Tobias21}.
And here comes already the first problem.
Large conductivity means small magnetic diffusivity and therefore
also less dissipation \citep{Bra14}.
Looking at \Fig{psketch}(a) however, this cannot be quite true---at least
not in the steady state when the dynamo term must be just as large as
the dissipation term.

\begin{figure}[b!]\begin{center}
\includegraphics[width=.49\textwidth]{psketch}
\includegraphics[width=.49\textwidth]{psketch2}
\end{center}\caption[]{
Energy flow diagrams for (a) a standard dynamo
and (b) a dynamo at large scales (LS) with a reversed dynamo at small scales (SS).
The dashed arrows indicate relatively weak flows of energy.
Adapted from \cite{BR19}.
}\label{psketch}\end{figure}

The puzzle of efficient dynamo action, but inefficient dissipation
was solved by realizing that at large conductivities (and especially
large magnetic Prandtl number, which is the ratio $\Pm\equiv\nu/\eta$ of
kinematic viscosity $\nu$ to magnetic diffusivity $\eta$, is much larger
than unity), a second conversion occurs at smaller length scales where
magnetic energy can be converted back into kinetic energy; see \Fig{psketch}(b).
This process was termed a reversed dynamo \citep{BR19}, and it happens
at small scales when $\Pm\gg1$.

Large magnetic Prandtl numbers occur in all systems were the density is
very low, notably, in stellar coronae, in galaxies, and also in early
universe plasmas.
This is true in collisional plasmas, when viscosity and magnetic diffusivity
can be described by a Spitzer-like formulae, which leads to
%\begin{equation}
%\Pm\approx1.1 \times10^{-4}
%\left({T\over10^6\K}\right)^4
%\left({\rho\over0.1\g\cm^{-3}}\right)^{-1}
%\left({\ln\Lambda\over20}\right)^{-2}.
%\label{Pm}
%\end{equation}
$\Pm\approx T_6^4/\rho_{-5}$, where $T_6$ is the temperature in units of
$10^6\K$ and $\rho_{-5}$ is the density in units of $10^{-5}\g\cm^{-3}$
\citep{BS05}.
Note the high power in which temperature enters here.
Nevertheless, the density factor is often more important.
Compare, for example, the solar surface, where the temperature is
$10^4\K$, with the solar corona where $T\approx10^6\K$.
The density drops much faster, from $10^{-6}\g\cm^{-3}$ to
$10^{-14}\g\cm^{-3}$, and therefore, although the temperature and density
factors work in the same direction, the density factor is here at least
as important as the temperature factor.
In galaxies, the situation is even more impressive because there the
density can be as low at $10^{-24}\g\cm^{-3}$.

%\section{Electromagnetic energy conversion after reheating}

While the energy conservation between magnetic and kinetic energies
is reasonably well understood, not much is known about the conversion
from electromagnetic energy into magnetic energy when the electric
conductivity gradually increases.
Such a process is important at the end of cosmological inflation.
A stochastic, electromagnetic field may have been produced during
inflation and reheating \citep{Sharma+17}.
At the end of reheating, the electric conductivity of the universe
increased.
As discussed in \cite{BS21}, significant magnetic field losses can occur
if the increase of conductivity takes very long, especially when the
magnetic diffusivity is at an intermediate level for a long time.
In the two extreme cases of very large diffusivity (corresponding to a
vacuum with undamped electromagnetic waves), and very small diffusivity
(corresponding to perfect conductivity), no significant losses are
expected.
It is only during the period when the magnetic diffusivity is at an
intermediate level that significant resistive losses can occur.

Once the conductivity has reached large values, i.e., when the magnetic,
diffusivity is small, strong turbulent flows will be driven, causing the
magnetic energy to undergo {\em turbulent} decay with inverse cascading,
which has been studied intensively since the mid 1990s \citep{BEO96,
CHB01, BJ04, Kahn+13, BKT15, BK17, HS22}.
At some point around the time of recombination, the photon mean free path
becomes very large and a process called Silk damping becomes important
\citep{Silk68}.
It results from the interactions between photons and the gas and damps out
all inhomogeneities in the photon--baryon plasma \citep{Kolb+Turner90}.

In \cite{BEO97}, this was modeled as a strongly increased viscosity,
making the magnetic Prandtl number even larger.
However, a more physical approach is to add a friction term of the form
$-\uu/\tau$ on the right hand side of the momentum equation \citep{BJ04}.
It is generally taken for granted that the magnetic field just survives
Silk damping without much additional loss, and that the magnetic field
is just frozen into the plasma.
However, the details of this process have not yet been modeled.

The goal here is to understand more quantitatively how much magnetic
energy survives during the conversions from electromagnetic fields to
MHD turbulence and finally, from MHD turbulence to a frozen-in state.

\section{1-D Maxwell equations with imposed field}
We begin by studying here a simple one-dimensional model, where an
electromagnetic wave with equal amounts of electric and magnetic energies
gets converted into magnetic and kinetic energies as the conductivity
increases and the electric field short-circuits.

The evolution of the electric and magnetic fields, $\EE$ and $\BB$,
respectively, is given by the Maxwell equations, here written in SI units:
\begin{equation}
\frac{1}{c^2} \frac{\partial\EE}{\partial t}=\nab\times\BB-\mu_0\JJ
\label{dEdt}
\end{equation}
\begin{equation}
\frac{\partial\BB}{\partial t}=-\nab\times\EE,
\label{dBdt}
\end{equation}
To close the equations, we use Ohm's law,
\begin{equation}
\JJ=\sigma\,(\EE+\uu\times\BB).
\label{Ohm0}
\end{equation}

The emergence of electric conductivity leads to a certain electromagnetic
loss given by $\JJ\cdot\EE$.
Using Ohm's law in the form $\EE=\JJ/\sigma-\uu\times\BB$,
\begin{equation}
\EE=\JJ/\sigma-\uu\times\BB,
\label{Ohm3}
\end{equation}
we see that
$\bra{\JJ\cdot\EE}=\bra{\JJ^2}/\sigma-\bra{\JJ\cdot(\uu\times\BB)}$,
or, using $-\JJ\cdot(\uu\times\BB)=\uu\cdot(\JJ\times\BB)$.
Thus, we have
\begin{equation}
\bra{\JJ\cdot\EE}=\bra{\JJ^2}/\sigma-\bra{\uu\cdot(\JJ\times\BB)},
\label{Ohm4}
\end{equation}
so part of the electromagnetic energy turns into Joule heating and another
part is converted into kinetic energy through work done by the Lorentz
force, or against the Lorentz force if $\bra{\uu\cdot(\JJ\times\BB)}$
is negative, as in a dynamo.

\begin{figure}[b!]\begin{center}
\includegraphics[width=\columnwidth]{pboxes_BEU}
\end{center}\caption[]{
Energy conversion from magnetic to kinetic energies via the
electric energy reservoir.
%\url{pboxes_BEU}
}\label{pboxes_BEU}\end{figure}

To illustrate the basic change from electromagnetic fields to MHD,
let us write the basic equations in the following form:
\begin{equation}
\frac{\partial\AAA}{\partial t}=-\EE,
\label{Induction}
\end{equation}
\begin{equation}
-\left(1+\underbrace{(\eta/c^2)\,\partial/\partial t}_{\mbox{non-MHD}}\right)\EE=
\uu\times\BB+\eta\left(\nabla^2\AAA-\nab\nab\cdot\AAA\right),
\label{AmpereMaxwell}
\end{equation}
\begin{equation}
\mu_0\JJ=-(\nabla^2\AAA-\nab\nab\cdot\AAA)
\underbrace{+\dot{\EE}/c^2}_{\mbox{non-MHD}},
\label{Ohm}
\end{equation}
\begin{equation}
\rho\frac{\DD\uu}{\DD t}=-\nab p+\JJ\times\BB+\nab\cdot(2\rho\nu\SSSS),
\label{Momentum}
\end{equation}
\begin{equation}
\rho\frac{\DD\ln\rho}{\DD t}=-\nab\cdot\uu,
\label{Continuity}
\end{equation}
where $\mu_0$ is the vacuum permeability, $\eta$ is the magnetic
diffusivity, and the magnetic field $\BB=\nab\times\AAA$ has been
expressed in terms of its vector potential $\AAA$.
The dot in $\dot{\EE}$ indicates just ordinary time differentiation,
i.e., $\dot{\EE}\equiv\partial\EE/\partial t$.
In MHD, the Faraday displacement current, $\dot{\EE}/c^2$, here written
in blue, is omitted.

This equation shows us that, if we were to neglect the $\dot{\EE}$ term,
\Eq{AmpereMaxwell} becomes just the right-hand side of \Eq{Induction},
a standard diffusion equation.
Likewise, instead of using Ohm's law in the form
$\JJ=\sigma\,(\EE+\uu\times\BB)$, it is advantageous to compute $\mu_0\JJ$
as $\nab\times\BB+\dot{\EE}/c^2$, which shows that the current is a
consequence of the magnetic field having a finite curl, rather than
being proportional to $\sigma$ times the comoving electric field.

We emphasize here that the difference between the two versions of the
Ampere--Maxwell law in \Eqs{AmpereMaxwell}{Ohm} is that in the former,
Ohm's law,
\begin{equation}
\mu_0\JJ=(\EE+\uu\times\BB)/\eta.
\label{Ohm2}
\end{equation}
has been included in \Eq{AmpereMaxwell}.
This allows us to see more readily that, when the Faraday displacement
current is omitted, the right-hand side (rhs) of the induction equation \eq{Induction}
is just $\uu\times\BB$ along with a diffusion term for $\AAA$.
Neglecting the Faraday displacement current is well justified when $\eta$
is small, or the conductivity $\sigma=1/\mu_0\eta$ is large.
Also from a numerical point of view, it is important to write the
governing equations in the form \Eqss{Induction}{Momentum}, because
they allow us to treat \Eq{AmpereMaxwell} either as an evolution
equation for $\EE$ when $\eta$ is large, or as an implicit equation
for $\EE$ when $\eta$ is small.
In the latter, in practice, we can omit the $\dot{\EE}$ terms.
In some of our turbulence simulations, we do this when $\eta$ is below
a certain critical value, $\eta_{\rm crit}$.

The volume average of the electric, magnetic,
kinetic, and thermal energy densities can be expressed as
$\bra{\frac{1}{2}\epsilon_0\EE^2}$,$\bra{\frac{1}{2\mu_0}\BB^2}$,$\bra{\frac{1}{2}\rho\uu^2}$
and $\bra{\rho e}$ respectively.

We can get the evolution equation from here.
\begin{equation}
\frac{\partial \EE}{\partial t}=\EE\cdot\nab\times\BB-\mu_0\JJ\
\end{equation}
Taking the dot product with $\EE$ we get,
\begin{equation}
\EE\cdot\frac{\partial \EE}{\partial t}=\BB\cdot\nab\times\EE-\nab(\EE\times\BB)-\mu_0 \JJ\cdot\EE
\end{equation}
which can also take the form,
\begin{equation}
\frac{\partial}{\partial t}\frac{E^2}{2}=(\uu\times\BB)\JJ-\frac{J^2}{\sigma}
\end{equation}
Ohm's law was used here.
Similarly, for magnetic field we get,
\begin{equation}
\begin{split}
\frac{\partial}{\partial t}\frac{B^2}{2}&=\EE\cdot\nab\times\BB-\nab(\EE\times\BB)\\
&=-\uu(\JJ\times\BB)-\frac{J^2}{\sigma}-\nab(\EE\times\BB)
\end{split}
\end{equation}
Here, $\nab(\EE\times\BB)$ is the Poynting vector.

Now we can get the time derivatives of the volume averages of the energies.
\begin{eqnarray}
\frac{d}{dt}\bra{\frac{1}{2\mu_0}\BB^2}=-\frac{1}{\mu_0}\bra{\BB\cdot\nab\times\EE}\\
\frac{d}{dt}\bra{\frac{1}{2}\epsilon_0\EE^2}=\frac{1}{\mu_0}\bra{\BB\cdot\nab\times\EE}-\bra{\JJ\cdot\EE} \\
\frac{d}{dt}\bra{\frac{1}{2}\rho\uu^2}=\bra{\uu(\JJ\times\BB)}-\bra{2\rho\nu\SSSS^2}\\
\frac{d}{dt}\bra{\rho e}=\bra{\frac{J^2}{\sigma}}+\bra{2\rho\nu\SSSS^2}
\end{eqnarray}

From these we can get the following equations for kinetic energy $\mathcal{E}_K$ and magnetic energy $\mathcal{E}_M$.
\begin{eqnarray}
\bra{\uu\cdot\JJ\times\BB}=\mathcal{E}_K+\frac{d}{dt}\bra{\frac{1}{2}\rho\uu^2}\\
\bra{\BB\cdot\nab\times\EE}=\mathcal{E}_M+\frac{d}{dt}\bra{\frac{\epsilon_0}{2}E^2}+\bra{\uu\cdot(\JJ\times\BB)}
\end{eqnarray}
Later the term $\bra{\uu\cdot\JJ\times\BB}$ has been denoted by $W_L$(work done against Lorentz force).

In \Fig{presults_Tdep}, we show the evolution of various energy reservoirs.
We see that the magnetic energy decays and gives off energy to the
electric energy reservoir through the term $\bra{\BB\cdot\nab\times\EE}$.
Note that the electric energy was initially equal to the magnetic one,
but as the conductivity increases, there is a rapid loss of electric energy ($\dot{\EEEl}<0$),
and most of it gets dissipated thermally, and only a small fraction
($<10\%$ for $\Delta t=10$) gets transferred into kinetic energy.

%At first, electric energy gets transferred to, which was initially
%equal to the magnetic energy, but then quickly loses electric energy
%($\dot{\mathcal{E}_E}<0$) and most of it gets dissipated thermally, and
%only a small fraction ($\leq 10\%$) gets transformed to kinetic energy
%($T>10$), and magnetic energy dissipation is almost linear.

%\begin{enumerate}\itemsep-0.2mm
%\item
%\end{enumerate}

\subsection{Experiments with different temporal conductivity variations}

In \Fig{pcomp_switch}, we show the evolution of $\Brms$, $\urms$,
and $B_y$ at one point $x=x_*$ in the domain for a run fully with the
displacement current included (black), compared with runs with different
values of $\eta_*$.

We now introduce two quantities with the same dimension as time t, $\tau_c$ and $\tau_k$, which can be expresses as 
\begin{eqnarray}
\frac{1}{\tau_c}=\frac{c^2}{\eta}\\
\frac{1}{\tau_k}=\eta k^2
\end{eqnarray}
where k is the wave number. The combined value of $\tau$ can be expresses as
\begin{equation}
\frac{1}{\tau_{comb}}=min\left(\frac{1}{\tau_c},\frac{1}{\tau_k}\right)   
\label{combined_tau}
\end{equation}
\Eq{combined_tau} can be simplified into
\begin{equation}
\frac{1}{\tau_{\rm comb}}=\frac{1}{\tau_c+\tau_k}.
\end{equation}
The total (integrated) effect on dissipation is therefore
\begin{equation}
\int\frac{dt}{\tau_{\rm comb}}=\int\frac{dt}{\tau_c+\tau_k}.
\end{equation}
For a magnetic field that that behaves as a wave
\begin{equation}
B(t,z) \sim B_0(t)\cos kz
\end{equation}
where the amplitude(?) decays as
\begin{equation}
B_0(t) \sim b_0 \exp (-t/\tau_{comb})
\end{equation}
If $\tau_{comb}$ is large, it decays slowly. Now we introduce a decay (?) factor that describes the decay behavior,
\begin{equation}
D=\int{\frac{dt}{\tau_{comb}}}
\end{equation}
which is large for slow transitions and small for fast transitions.


Assume $\BB_0=(B_0,0,0)$, so
\begin{equation}
\frac{\partial A_z}{\partial t}=-E_z,
\end{equation}
\begin{equation}
\frac{\partial E_z}{\partial t}=-\frac{\partial^2 A_z}{\partial x^2}+J_z,
\end{equation}
Ohm's law,
\begin{equation}
J_z=\sigma(E_z-u_y B_0\blue{+u_x B_y})
\end{equation}
where $B_y=-\partial_x A_z$,
and the momentum equation,
%\begin{eqnarray}
%\frac{\partial u_x}{\partial t}&=&-\cs^2\frac{\partial\ln\rho}{\partial x}\blue{-\frac{1}{\rho}J_zB_y}\cr
%&&+\fourthird\nu\left(\frac{\partial^2 u_y}{\partial x^2}\blue{+\frac{\partial u_y}{\partial x}\frac{\partial\ln\rho}{\partial x}}\right),
%\end{eqnarray}
\begin{eqnarray}
\frac{\partial u_x}{\partial t}=-\cs^2\frac{\partial\ln\rho}{\partial x}\blue{-\frac{1}{\rho}J_zB_y}
+\fourthird\nu\left(\frac{\partial^2 u_y}{\partial x^2}\blue{+\frac{\partial u_y}{\partial x}\frac{\partial\ln\rho}{\partial x}}\right),
\end{eqnarray}
\begin{eqnarray}
\frac{\partial u_y}{\partial t}=\frac{1}{\rho}J_zB_0
+\nu\left(\frac{\partial^2 u_y}{\partial x^2}\blue{+\frac{\partial u_y}{\partial x}\frac{\partial\ln\rho}{\partial x}}\right),
\end{eqnarray}
along with the continuity equation
\begin{eqnarray}
\frac{\partial\ln\rho}{\partial t}=-\blue{u_x\frac{\partial\ln\rho}{\partial x}}-\frac{\partial u_x}{\partial x}
\end{eqnarray}
In Fourier space,
\begin{eqnarray}
\dot{\tilde{B}}= -ik \times \tilde{E}\\
\left(\mu_0\sigma+\frac{1}{c^2}\frac{d}{dt}\right)\tilde{\dot{E}}=-ik \times \dot{\tilde{B}}\\
\left(\mu_0\sigma+\frac{1}{c^2}\frac{d}{dt}\right)\frac{dE}{dt}= -k^2 E\\
\frac{1}{c^2}\ddot{E}+\mu_0\sigma\dot{E}+k^2 E=0
\end{eqnarray}
In the code, we use the initial condition {\tt initaa='Alfven-x'}, which means
\begin{equation}
A_z=-\frac{\cal A}{k}\cos k(x-\vA t)
\end{equation}
\begin{equation}
E_z=-{\cal A}\vA\sin k(x-\vA t)
\end{equation}
\begin{equation}
u_y={\cal A}\sin k(x-\vA t)
\end{equation}
We use
\begin{equation}
\ln\eta(t)=\ln\eta_0+(\ln\eta_{\max}-\ln\eta_0)\,\left[1-\Theta(t)\right],
\end{equation}
with
\begin{equation}
\Theta=\max\left[\min\left(\frac{t-t_0}{\tau_0},\,1\right),\,0\right]
\end{equation}
being a piecewise linear function that goes from 0 (for $t\leq t_0$)
to 1 (for $t\geq t_0+\tau_0$).

As in \cite{BS21}, we also studied a linear increase of $\sigma$ of the form
\begin{equation}
\sigma(t)=\sigma_{\min}+(\sigma_0-\sigma_{\min})\,\Theta(t),
\end{equation}
where $\sigma_0=1/\eta_0$ and $\sigma_{\min}=1/\eta_{\max}$.

As we see from \Figss{pcomp_diff_eta_prof_03}{pcomp_diff_eta_prof_01},
the time spent while $\eta k^2/\omega_{\rm em}=O(1)$ is longest for
the exponential decay profile (typically in the range
$20\leq\omega_{\rm em}t\leq34$).

%cvs co protiti/AlfvenWave_Maxwell/D128e
\begin{figure}[h!]\begin{center}
\includegraphics[width=\columnwidth]{pvid_last_D128e}
\end{center}\caption[]{
$\nu=\eta_0=2\times10^{-3}$, $\eta_{\max}=200$, 
$t_0=10$, $\tau_0=1$, $dt=5e-3$.
%\url{pvid_last_D128b}
}\label{pvid_last_D128e}\end{figure}

\begin{figure}[h!]\begin{center}
\includegraphics[width=\columnwidth]{pvid_last_D128j_T50}
\end{center}\caption[]{
$T=50$.
}\label{pvid_last_D128j_T50}\end{figure}

\begin{figure}[h!]\begin{center}
\includegraphics[width=\columnwidth]{pvid_last_t_D128j}
\end{center}\caption[]{
$T=1$.
}\label{pvid_last_t_D128j}\end{figure}

\begin{figure}[h!]\begin{center}
\includegraphics[width=\columnwidth]{pvid_last_t_D128j_T10}
\end{center}\caption[]{
$T=10$.
}\label{pvid_last_t_D128j_T10}\end{figure}

\begin{figure}[h!]\begin{center}
\includegraphics[width=\columnwidth]{pvid_last_t_D128j_T50}
\end{center}\caption[]{
$T=50$.
}\label{pvid_last_t_D128j_T50}\end{figure}

\begin{figure}[h!]\begin{center}
\includegraphics[width=\columnwidth]{pcomp_diff_eta_prof_03}
\end{center}\caption[]{
Evolution of $B_y(x_*,t)$ for the log-switch-on profile (black)
and the linear-sigma profile (red).
$T=50$, $B_0=0.3$.
The run directories are D128j\_T50b (for log-switch-on) and
D128j\_T50c (for linear-sigma) in {\tt protiti/AlfvenWave\_Maxwell}.
% D128j_T50b  and  D128j_T50c  in  protiti/AlfvenWave_Maxwell
% dir1='D128j_T50b_b01' & dir2='D128j_T50c_b01'
% pcomp_diff_eta_prof_b03.pro
The idl procedure is {\tt pcomp\_diff\_eta\_prof\_b03.pro}
in {\tt protiti/AlfvenWave\_Maxwell/idl}
}\label{pcomp_diff_eta_prof_03}\end{figure}

\begin{figure}[h!]\begin{center}
\includegraphics[width=\columnwidth]{pcomp_diff_eta_prof_01}
\end{center}\caption[]{
$T=50$, $B_0=0.1$.
}\label{pcomp_diff_eta_prof_01}\end{figure}

\begin{table}[htb]\caption{
Time step constraint.
}\vspace{12pt}\centerline{\begin{tabular}{lccccccc}
$\eta_0$ & $\sigma$ & $\delta t$ \\
\hline
$2\times10^{-3}$ &  500 & $5\times10^{-3}$ \\ % D128g
$1\times10^{-3}$ & 1000 & $2\times10^{-3}$ \\ % D128h
$5\times10^{-4}$ & 2000 & $1\times10^{-3}$ \\ % D128i
\label{Ttimescale}\end{tabular}}\end{table}

\begin{table}[htb]\caption{
Time step constraint.
}\vspace{12pt}\centerline{\begin{tabular}{lccccccc}
$T$ & $\EEM$ \\
\hline
 1 & 0.30798491
10 & 0.30285558
50 & 0.30285558
\label{Ttimescale}\end{tabular}}\end{table}

\begin{table}[htb]\caption{
Decay results
}\vspace{12pt}\centerline{\begin{tabular}{lccccccc}
$\omega_{\rm A}$ & log-switch-on & linear-sigma \\
\hline
0.1 & $(0.0014,-5.6\times10^{-6})$ & $(0.0588817-4.2\times10^{-4})$ \\ %D128j_T50b_b01  D128j_T50c_b01
0.3 & $(0.0025,-2.8\times10^{-6})$ & $(0.0520724-2.6\times10^{-4})$ \\ %D128j_T50b      D128j_T50c
\label{Tdecay}\end{tabular}}\end{table}

\begin{figure}[h!]\begin{center}
\includegraphics[width=\columnwidth]{pcomp_switch}
\end{center}\caption[]{
Evolution of $\Brms$, $\urms$, and $B_y$ at one point $x=x_*$ in the
domain for a run fully with the displacement current included (black),
compared with runs with different values of $\eta_*$:
0.1 (red), 0.5 (blue), and 2 (orange).
}\label{pcomp_switch}\end{figure}

\begin{figure}[h!]\begin{center}
\includegraphics[width=\columnwidth]{pcomp_switch_b}
\end{center}\caption[]{
Turbulence run, linear sigma profile.
}\label{pcomp_switch_b}\end{figure}

\begin{figure}[h!]\begin{center}
\includegraphics[width=\columnwidth]{pcomp_switch_c}
\end{center}\caption[]{
Turbulence run, log profile,
black: variable time step,
red: $\delta t=5\times10^{-4}$, and
blue: $\delta t=2\times10^{-4}$.
}\label{pcomp_switch_c}\end{figure}

\begin{figure}[h!]\begin{center}
\includegraphics[width=\columnwidth]{pcomp_switch_d}
\end{center}\caption[]{
Turbulence run, $128^3$, log-switch-on profile,
with $\nu=\eta=10^{-2}$, and
$\eta_{\rm switch}=5\times10^{-2}$ (black),
0.1 (red), and 0.2 (blue).
The orange line is the case with 
$\eta_{\rm switch}=0.1$ and $\JJ=\nab\times\BB/\mu_0$
at all times.
}\label{pcomp_switch_d}\end{figure}

\begin{table}[htb]\caption{Decay factor}
\vspace{12pt}\centerline{\begin{tabular}{lccccccc}
$k$ & D(log-switch-on) & D(linear-sigma) \\
\hline
0.05 & ? & $4.84$ \\
0.1 & ? & $1.62$ \\
0.2 & ? & $1.57$
\label{Tdecay}\end{tabular}}\end{table}

\begin{figure}[h!]\begin{center}
\includegraphics[width=\columnwidth]{pdec_D128j_T1a_etaE0}
\end{center}\caption[]{
Initially, all the energy is in $\bra{\epsilon_0\EE^2/2}+\bra{\BB^2/2\mu_0}$.
The red line gives the integrated Ohmic energy gain, $\int\epsM\,\dd t$.
At the end, this energy gets distributed to equal amounts to kinetic
energy, $\bra{\rho\uu^2/2}$ and magnetic energy $\bra{\BB^2/2\mu_0}$.
\url{pdec_D128j_T1a_etaE0}
}\label{pdec_D128j_T1a_etaE0}\end{figure}

\begin{figure}[h!]\begin{center}
\includegraphics[width=\columnwidth]{pdec_D128j_T10a_etaE0}
\end{center}\caption[]{
\url{pdec_D128j_T10a_etaE0}
}\label{pdec_D128j_T10a_etaE0}\end{figure}

\begin{figure}[h!]\begin{center}
\includegraphics[width=\columnwidth]{pdec_D128j_T20a_etaE0}
\end{center}\caption[]{
\url{pdec_D128j_T20a_etaE0}
The orange line shows $\bra{\BB^2}/2+\bra{\rho\uu^2}/2$.
Here, $\nu=5\times10^{-4}$, and also $\eta_{\rm fin}=5\times10^{-4}$.
The yellow and blue lines give just $\bra{\BB^2}/2$ and $\bra{\rho\uu^2}/2$ separately.
The red line is $\int\epsM\,\dd t$.
}\label{pdec_D128j_T20a_etaE0}\end{figure}

\begin{figure}[h!]\begin{center}
\includegraphics[width=\columnwidth]{pdec_D128j_T50a_etaE0}
\end{center}\caption[]{
\url{pdec_D128j_T50a_etaE0}
}\label{pdec_D128j_T50a_etaE0}\end{figure}

\begin{figure}[h!]\begin{center}
\includegraphics[width=\columnwidth]{pdec_D128j_T20a_etaE0_nu1em1}
\end{center}\caption[]{
\url{pdec_D128j_T20a_etaE0_nu1em1} with $\nu=0.1$.
The dashed red line is $\int(\epsM+\epsK)\,\dd t$.
}\label{pdec_D128j_T20a_etaE0_nu1em1}\end{figure}

\begin{figure}[h!]\begin{center}
\includegraphics[width=\columnwidth]{pdec_D128j_T20a_etaE0_nu5em1}
\end{center}\caption[]{
\url{pdec_D128j_T20a_etaE0_nu5em1} with $\nu=0.5$.
}\label{pdec_D128j_T20a_etaE0_nu5em1}\end{figure}

\begin{figure}[h!]\begin{center}
\includegraphics[width=\columnwidth]{pdec_D128j_T10a_etaE0_nu1em1}
\end{center}\caption[]{
\url{pdec_D128j_T10a_etaE0_nu1em1}
The dashed red line is $\int(\epsM+\epsK)\,\dd t$.
}\label{pdec_D128j_T10a_etaE0_nu1em1}\end{figure}

\begin{figure}[h!]\begin{center}
\includegraphics[width=\columnwidth]{pdec_D128j_T10a_etaE0_nu5em1}
\end{center}\caption[]{
\url{pdec_D128j_T10a_etaE0_nu5em1}
}\label{pdec_D128j_T10a_etaE0_nu5em1}\end{figure}

\begin{figure}[h!]\begin{center}
\includegraphics[width=\columnwidth]{pdec_D128j_T10a_etaE0_nu1em2}
\end{center}\caption[]{
\url{pdec_D128j_T10a_etaE0_nu1em2} with $\nu=0.01$.
It turns out that the mean magnetic and electric energy densities
decay like $\exp(-\nu k^2 t)$ (without a factor of 2).
Furthermore, the ratio is here $\EEEl/\EEM\approx10$ for $\Pm=20$.
}\label{pdec_D128j_T10a_etaE0_nu1em2}\end{figure}

\begin{figure}[h!]\begin{center}
%\includegraphics[width=\columnwidth]{panalyze}
\includegraphics[width=\columnwidth]{presults_Tdep}
\end{center}\caption[]{
Evolution of energy fluxes for the model with
the 'log-switch-on' conductivity profile.
$\nu=\eta=5\times10^{-4}$.
In all cases, the initial diffusivity is $\eta_{\rm ini}=10^4$.
The only difference to the run with a larger viscosity
is that $\epsK$ is larger.
\Fig{panalyze_nu5em4}
%\url{panalyze}
}\label{presults_Tdep}\end{figure}

\begin{figure}[h!]\begin{center}
\includegraphics[width=\columnwidth]{panalyze_nu5em4}
\end{center}\caption[]{
\url{panalyze_nu5em4}
}\label{panalyze_nu5em4}\end{figure}

\begin{figure}[h!]\begin{center}
\includegraphics[width=\columnwidth]{presults_Tdep_energies}
\end{center}\caption[]{
\url{presults_Tdep_energies}
}\label{presults_Tdep_energies}\end{figure}

\section{}

\begin{table}[htb]\caption{
Regimes of energy transfer.
}\vspace{12pt}\centerline{\begin{tabular}{lccccccc}
$t_{\rm dur}<10$ & $t_{\rm dur}>10$ \\
\hline
$W_{\rm L}/\epsM\sim0.1$ & $\propto t_{\rm dur}^{-1.5}$ \\
\label{Tregimes}\end{tabular}}\end{table}

\section{Friedmann equation and conformal time}
The second Friedmann equation can be written in the form
\begin{equation}
\frac{\ddot{a}}{a}=-\frac{H^2}{2}(1+3w)
\end{equation}
where w is the equation of state and H is the Hubble parameter, which can be expressed as 
\begin{equation}
H \equiv \frac{\dot{a}}{a}
\end{equation}
Now let us introduce b such as $H=\frac{b}{a}$. And
\begin{eqnarray}
b=\dot{a}=Ha\\
\begin{split}
\dot{b}=\ddot{a}&=-\frac{a}{2}H^2(1+3w)\\
&=-\frac{1}{2}\frac{b^2}{a}(1+3w)
\end{split}
\end{eqnarray}
Now, to get the conformal time $\eta$, we need to solve the Friedmann equation.
\begin{equation}
d\eta=\frac{dt}{a}
\end{equation}
which gives
\begin{equation}
a=\frac{dt}{d\eta}
\end{equation}
Now, taking the first and second derivatives,
\begin{eqnarray}
\dot{a}=\frac{da}{dt}=\frac{1}{a}\frac{da}{d\eta}\\
\begin{split}
\ddot{a}&=-\frac{1}{a^2}\dot{a}\frac{da}{d\eta}+\frac{1}{a}\frac{d}{dt}\frac{da}{d\eta}\\
&=-\frac{1}{a^3}\frac{da}{d\eta}+\frac{1}{a^2}\frac{d^2a}{d\eta^2}
\end{split}
\end{eqnarray}

\section{Conformal time}

\subsection{$\tilde{\AAA}$}

\begin{equation}
\AAA=a^{-1}\tilde{\AAA}
\end{equation}
\begin{equation}
\AAA'=-a^{-2}a'\tilde{\AAA}+a^{-1}\tilde{\AAA}'
\end{equation}
\begin{equation}
\AAA''=2a^{-3}(a')^2\tilde{\AAA}-a^{-2}a''\tilde{\AAA}-a^{-2}a'\tilde{\AAA}'
-a^{-2}a'\tilde{\AAA}'+a^{-1}\tilde{\AAA}''
\end{equation}
or
\begin{equation}
\AAA''=\left[2a^{-3}(a')^2-a^{-2}a''\right]\tilde{\AAA}-2a^{-2}a'\tilde{\AAA}'
+a^{-1}\tilde{\AAA}''
\end{equation}
or
\begin{equation}
\AAA''=\left[2(a'/a)^2-(a''/a)\right]\tilde{\AAA}/a
-2(a'/a)\tilde{\AAA}'/a
+\tilde{\AAA}''/a
\end{equation}

\subsection{$\AAAA$}

Define $\AAAA=f\AAA$, so $\AAA=f^{-1}\AAAA$, and therefore
\begin{equation}
\AAA'=-f^{-2}f'\AAAA+f^{-1}\AAAA'
\end{equation}
\begin{equation}
\AAA''=2f^{-3}(f')^2\AAAA-f^{-2}f''\AAAA-f^{-2}f'\AAAA'
-f^{-2}f'\AAAA'+f^{-1}\AAAA''
\end{equation}
or
\begin{equation}
\AAA''=\left[2f^{-3}(f')^2-f^{-2}f''\right]\AAAA-2f^{-2}f'\AAAA'
+f^{-1}\AAAA''
\end{equation}
or
\begin{equation}
\AAA''=\left[2(f'/f)^2-(f''/f)\right]\AAAA/f
-2(f'/f)\AAAA'/f
+\AAAA''/f
\end{equation}
So
\begin{equation}
\AAA''+2(f'/f)\AAA'
=\left[2(f'/f)^2-(f''/f)\right]\AAAA/f
-2(f'/f)\AAAA'/f
+\AAAA''/f
+2(f'/f)\AAA'
\end{equation}
or
\begin{equation}
\AAA''+2(f'/f)\AAA'
=\left[2(f'/f)^2-(f''/f)\right]\AAAA/f
-2(f'/f)\AAAA'/f
+\AAAA''/f
+2(f'/f)\left[-(f'/f)\AAAA+\AAAA'\right]/f
\end{equation}
or
\begin{equation}
\AAA''+2(f'/f)\AAA'
=\left[2(f'/f)^2-(f''/f)\right]\AAAA/f
-2(f'/f)\AAAA'/f
+\AAAA''/f
-2(f'/f)^2\AAAA/f+2(f'/f)\AAAA'/f
\end{equation}
The $2(f'/f)^2\AAAA/f$ and $2(f'/f)\AAAA'/f$ terms cancel, so
\begin{equation}
\AAA''+2(f'/f)\AAA'
=-(f''/f)\AAAA/f+\AAAA''/f
\end{equation}

\subsection{Power laws}

\begin{equation}
f=t^\alpha,\quad
f'=\alpha t^{\alpha-1},\quad
f''=\alpha(\alpha-1) t^{\alpha-2},
\end{equation}
so
\begin{equation}
f'/f=\alpha t^{-1},\quad
f''/f=\alpha(\alpha-1) t^{-2},
\end{equation}
${\cal A}$ equation:
\begin{equation}
{\cal A}''=(f''/f){\cal A}
\end{equation}
Solution: ${\cal A}=t^n$, ${\cal A}'=nt^{n-1}$, and ${\cal A}''=n(n-1)t^{n-2}$, so
\begin{equation}
n(n-1)=\alpha(\alpha-1)
\end{equation}
so one solution is $n=\alpha$, so $n$ has the {\em same sign} as $\alpha$.
But this is only {\em one} out of two solutions.
The other (more relevant) one is 
$n=1-\alpha$, so $n$ has the {\em opposite sign} as $\alpha$.
Note also that $A={\cal A}/f=t^{n-\alpha}$.

Directly $A$ equation:
\begin{equation}
A''+2(f'/f)A'=0
\end{equation}
or, substituting $A'=-E$,
\begin{equation}
E'+2(f'/f)E=0
\end{equation}
Solution: $E=t^m$, $E'=mt^{m-1}$, so
\begin{equation}
mt^{m-1}+2\alpha t^{m-1}=0,
\end{equation}
so
\begin{equation}
m=-2\alpha
\end{equation}
{\em opposite sign}.
Now, if $A=t^p$, then 
$E=-A'=-pt^{p-1}=t^m$, so $m=p-1$.
Question: $p=n-\alpha$? This is only compatible with the
solution above for $\alpha=1/2$.
The relevant branch has $p=1-2\alpha$.

Thus, we have $A=t^{-2\alpha}$ and $E=t^{1-2\alpha}$.
Interestingly, this can imply growth for $A$, but decay for $E$,
when $0\leq\alpha\leq1/2$.

\bibliographystyle{jpp}
\bibliography{ref}
\end{document}
