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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{Scaled versions of $\AAA$}

In the following, we denote by $t$ conformal time,
and derivatives with respect to $t$ are denoted by primes.

\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'
=\Big[\AAAA''-(f''/f)\AAAA\Big]/f,
\end{equation}
which relates the harmonic and damped oscillator equations to each other.

\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 $A={\cal A}/f=t^{n-\alpha}$.

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

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 $0\leq\alpha\leq1/2$.

\section{Electric energy equation}

Use
\begin{equation}
\EEEl\equiv\bra{\epsilon_0(f\EE)^2/2},
\end{equation}
so that
\begin{equation}
\dot\EEEl=f'\EEE+f\dot{\EEE}.
\end{equation}
or
\begin{equation}
\frac{1}{f}\dot\EEEl=\dot{\EEE}+\frac{f'}{f}\EEE
\end{equation}
Using
\begin{equation}
\left[\frac{1}{c^2}\left(\frac{\partial^2}{\partial t^2}
+2\frac{f'}{f}\frac{\partial}{\partial t}\right)
-\nabla^2\right]\AAA=0.
\label{nocurlyA}
\end{equation}
we have
\begin{equation}
\dot\EEEl=-2\frac{f'}{f}\EEEl+f^2\bra{\EE\cdot\nab\times\BB}+f^2\bra{\JJ\cdot\EE}
\end{equation}
so the growth term is by a factor 2 smaller
and the energy transfer terms attain $f^2$ factors.


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