% Id: notes.tex,v 1.6 2021/03/22 07:43:54 brandenb Exp $
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\newcommand{\csisoth}{c_{\rm s}^{\rm(isoth)}}
\newcommand{\GeV}{\,{\rm GeV}}
\def\OmK{{\Omega}_{\rm K}}
\def\OmM{{\Omega}_{\rm M}}
\def\OmGW{{\Omega}_{\rm GW}}
\def\kNy{k_{\rm Ny}}
\def\kGW{k_{\rm GW}}
\def\hX{h_\times}
\def\hT{h_+}
\def\hX{h_\times}
\def\thT{\tilde{h}_+}
\def\thX{\tilde{h}_\times}
\newcommand{\tildeh}{\tilde{h}}
\newcommand{\tildeT}{\tilde{T}}
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\input gafd_journ

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\begin{document}

\section{Exponential evolution of the source in time}

It is convenient to restrict our attention to the case of a purely
monochromatic exponential growth of $\BB$ at the rate $\gamma_0$.
The magnetic energy increases then at the rate $2 \gamma_0$.
In the following, the notation simplifies when letting the argument
$t$ start at $t=0$, so that the actual time is simply $t+1$.
Let us now assume that $\tilde{T}(\kk,t)$ is given by
\EQ
\tilde{T}(\kk,t+1)=\theta(t)\,\tilde{T}_0(k)\,e^{2\gamma t},
\EN
where $\theta(t)=1$ for $t>0$ and $0$ otherwise is the Heaviside
step function, and $\tilde{T}_0(k)$ is the Fourier transform of one of the
two polarization modes of the stress, which is assumed to depend just
on $k=|\kk|$.

During the early growth phase, the expansion of the universe can be
neglected and we can solve the GW equation in closed form as
\begin{equation}
\tilde{h}(k,t+1)=\frac{6\tilde{T}_0(k)}{k}\int_0^t \sin k(t-t') \, e^{2\gamma t'} \, \dd t',
\end{equation}
where $\tilde{h}$ stands for both $\tilde{h}_+$ and $\tilde{h}_\times$.
Using $\sin\phi=(e^{\ii\phi}-e^{-\ii\phi})/2\ii$ for any variable $\phi$,
we have
\begin{equation}
h=\frac{3T_0}{\ii k}\!\int_0^t \left[
e^{(2\gamma-\ii k)t'+\ii kt} -e^{(2\gamma+\ii k)t'-\ii kt} 
\right] \dd t'.\;
\end{equation}
Integrating between the two boundaries yields
\begin{equation}
h(k,t+1)=\frac{3T_0(k)}{\ii k} \left[
 \frac{e^{2\gamma t}-e^{ \ii kt}}{2\gamma-\ii k}
-\frac{e^{2\gamma t}-e^{-\ii kt}}{2\gamma+\ii k} \right].
\label{ht1}
\end{equation}
\Eq{ht1} can be written as
\begin{equation}
\tilde{h}(k,t+1)=\frac{6\tilde{T}_0(k)}{4\gamma^2+k^2} \left(
e^{2\gamma t} -\cos kt -\frac{2\gamma}{k}\sin kt \right),
\end{equation}
or, replacing $2\gamma\to k_0$,
\begin{equation}
\tilde{h}(k,t+1)=\frac{6\tilde{T}_0(k)}{k_0^2+k^2} \left(
e^{k_0 t} -\cos kt -\frac{k_0}{k}\sin kt \right)
\label{hexp}
\end{equation}
and its time derivative
\begin{equation}
k^{-1}\dot{\tilde{h}}(k,t+1)=\frac{6\tilde{T}_0(k)}{k_0^2+k^2} \left(
\frac{k_0}{k} e^{k_0t} + \sin kt -\frac{k_0}{k}\cos kt \right).
\label{dhexp}
\end{equation}
In practice, we are always interested in the case where the
exponential term dominates over the cosine and sine terms.
When $k<2\gamma$, $\Sp(h)$ and $\Sp(\dot{h})$ are proportional to 
$\Sp(T_0)$.
When the source has a white noise spectrum, we have
$\Sp(h)\propto k^2[\tilde{T}_0(k)]^2\propto k^2$.
However, when $k>2\gamma$, we find
$\Sp(h)\propto\Sp(\dot{h})\propto k^2[\tilde{T}_0(k)/k^2]^2\propto k^{-2}$,
with the breakpoint being at $ck=2\gamma_0$,
where we have restored the speed of light for dimensional clarity.

\subsection{Solution for subsequent evolution}

Let us assume that the souce is turned off abruptly at $t=t_0$.
The GW field then continunues in a purely oscillatory fashion with
\begin{equation}
\pmatrix{\tilde{h} \cr \dot{\tilde{h}}/k}(k,t+1)
=\pmatrix{\cos kt_{\rm m} & \sin kt_{\rm m}\cr -\sin kt_{\rm m} & \cos kt_{\rm m}}
\pmatrix{A\cr B},
\end{equation}
where the coefficients $A$ and $B$ have to be determined by matching
$\tilde{h}$ and $\dot{\tilde{h}}$ of this solution to \Eqs{hexp}{dhexp}
at $t=t_{\rm e}$, which we write as
\begin{equation}
\pmatrix{\tilde{h}_{\rm m} \cr \dot{\tilde{h}}_{\rm m}/k}\equiv
\frac{6\tilde{T}_0(k)}{k_0^2+k^2}
\pmatrix{
e^{k_0 t_{\rm m}} -\cos kt_{\rm m} -\frac{k_0}{k}\sin kt_{\rm m} \cr
\frac{k_0}{k} e^{k_0t_{\rm m}} + \sin kt_{\rm m} -\frac{k_0}{k}\cos kt_{\rm m} }.
\end{equation}
Thus, we require
\begin{equation}
\pmatrix{\tilde{h}_{\rm m} \cr \dot{\tilde{h}}_{\rm m}/k}
=\pmatrix{\cos kt_{\rm m} & \sin kt_{\rm m}\cr -\sin kt_{\rm m} & \cos kt_{\rm m}}
\pmatrix{A\cr B},
\end{equation}
with the solution
\begin{equation}
\pmatrix{A\cr B}
=\pmatrix{\cos kt_{\rm m} & \sin kt_{\rm m}\cr -\sin kt_{\rm m} & \cos kt_{\rm m}}
\pmatrix{\tilde{h}_{\rm m} \cr \dot{\tilde{h}}_{\rm m}/k}
\end{equation}
Thus,
\begin{equation}
\pmatrix{\tilde{h} \cr \dot{\tilde{h}}/k}(k,t+1)
=\pmatrix{\cos kt_{\rm m} & \sin kt_{\rm m}\cr -\sin kt_{\rm m} & \cos kt_{\rm m}}
\pmatrix{\cos kt_{\rm m} & \sin kt_{\rm m}\cr -\sin kt_{\rm m} & \cos kt_{\rm m}}
\pmatrix{\tilde{h}_{\rm m} \cr \dot{\tilde{h}}_{\rm m}/k}
\end{equation}
or, using the addition theorems of trigonometric functions,
\begin{equation}
\pmatrix{\tilde{h} \cr \dot{\tilde{h}}/k}(k,t+1)
=\pmatrix{\cos k(t-t_{\rm m}) & \sin k(t-t_{\rm m}) \cr -\sin k(t-t_{\rm m}) & \cos k(t-t_{\rm m}) }
\pmatrix{\tilde{h}_{\rm m} \cr \dot{\tilde{h}}_{\rm m}/k}
\end{equation}
We thus have
\begin{equation}
\tilde{h}(k,t+1)=\frac{6\tilde{T}_0(k)}{k_0^2+k^2}\left[
\cos k(t-t_{\rm m}) \left( e^{k_0 t_{\rm m}} -\cos kt_{\rm m} -\frac{k_0}{k}\sin kt_{\rm m} \right)
+\sin k(t-t_{\rm m}) \left( \frac{k_0}{k} e^{k_0t_{\rm m}} + \sin kt_{\rm m} -\frac{k_0}{k}\cos kt_{\rm m} \right)
\right]
\end{equation}
or
\begin{equation}
\tilde{h}(k,t+1)=\frac{6\tilde{T}_0(k)}{k_0^2+k^2}\left[
\left(\cos k(t-t_{\rm m}) +\frac{k_0}{k} \sin k(t-t_{\rm m})\right) \left(e^{k_0t_{\rm m}} -\cos kt_{\rm m} \right)
-\left( \frac{k_0}{k} \cos k(t-t_{\rm m}) +\sin k(t-t_{\rm m})\right) \sin kt_{\rm m} 
\right]
\end{equation}

\end{document}

