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\newcommand{\ygafd}[3]{, Geophys. Astrophys. Fluid Dynam. {\bf #2}, #3 (#1).}

\title{Analytic calculations on magnetic Beltrami field with partial helicity.}

%\date{Received 5 July 2016; published 3 February 2017,~ $ $Revision: 1.8 $ $}
%\preprint{NORDITA-2016-82 --- Phys. Rev. Lett. 118, 055102 (2017)}
\date{\today,~ $ $Revision: 1.8 $ $}
%\preprint{NORDITA-2018-xx}

\begin{document}
\maketitle

\section{Single--mode 1D Beltrami field model}

\subsection{Magnetic field}

We start assuming a one--dimensional
single--mode Beltrami field,
with characteristic wave number $k_0 \in \mathbb{R}$,
partial helicity $\sigma \in [0, 1]$, and
amplitude $B_0 \in [0, \sqrt{2}]$.

\begin{equation}
\boxed{\BB(x, t) = \sqrt{\frac{2}{1 + \sigma^2}}
B_0 \Theta(t)
\Vector{0}{\sigma \sin k_0 x}{\cos k_0 x}.}
\label{B0}
\end{equation}

The Heaviside function $\Theta(t)$ is included to indicate that
this field abruptly appears at $t = 0$, which refers
to the time of generation.
This model considers the magnetic field as an initial
condition.
The prefactor is included, so that
the total (normalized) magnetic energy density
is $\OmM = \half \Bra{\BB^2} = \half B_0^2 \in [0, 1]$,
where angle brackets denote averaging in physical
space.
\begin{equation}
\OmM (t) = \half \Bra{\BB^2} = \frac{1}{1 + \sigma^2} B_0^2
\Theta(t) \BBra{\sigma^2 \sin^2 k_0 x + \cos^2 k_0x} =
\half B_0^2 \Theta(t),
\label{OmM}
\end{equation}
where we have used $\Bra{\cos^2 k_0 x} = \Bra{\sin^2 k_0 x}
= \half$.
Hence, the normalized amplitude of the magnetic fields is
$B_0 \in [0, \sqrt{2}]$.

In Fourier space, the magnetic field is
\begin{equation}
\tBB(k_x, t) =\half \sqrt{\frac{2}{1 + \sigma^2}}
B_0 \Theta(t)
\Vector{0}{i \sigma \Para{\deltp - \deltn}}
{\deltp + \deltn},
\end{equation}
where $\delt$ is the Dirac's
delta function, and
we have used the convention
\begin{equation}
\tBB(\kk, t) = \int \BB(\xx, t) e^{-i \kk \cdot \xx} \dd \xx, \quad
\BB(\xx, t) = \frac{1}{(2 \pi)^3} \int \tBB (\kk, t)
e^{-i \kk \cdot \xx} \dd \kk.
\end{equation}

The magnetic spectrum is obtained by shell--integration of $\half
\BB(\kk, t) \conj{\BB}(\kk, t)$.
In $D$-dimensions, this is multiplied by a
factor $k^{D - 1}$, and integrated over the solid angle
$\Omega_{\rm D}$.
The resulting spectrum is only function of the
wave number, $k = \sqrt{\kk^2}$.
\begin{equation}
\boxed{\EM (k, t) = \half \int_{\Omega_1} \tBB (k_x, t)
\cdot \conj{\BB} (k_x, t) \dd \Omega_1 = 
\half B_0^2 \Theta(t) \deln,}
\end{equation}
where $\Omega_1 = 2$ is the one-dimensional solid angle.
The total magnetic energy density is
\begin{equation}
\boxed{\OmM (t) = \int_0^{\infty} \EM (k, t) \dd k = 
\half B_0^2 \Theta(t),}
\end{equation}
which is consistent with the initial definition of
the amplitude $\OmM = \half B_0^2$, shown
in \Eq{OmM}.

The magnetic spectral peak is at wavenumber
\begin{equation}
\boxed{\kM = \frac{ \OmM (t)}
{\int_0^{\infty} k^{-1} \EM (k, t)\dd k} = |k_0|.}
\end{equation}

\subsection{Magnetic helicity}

The magnetic field is solenoidal, i.e., $\nab \cdot \BB = 0$.
Hence, we can define a magnetic vector potential $\AAA$ such
that $\nab \times \AAA = \BB$
\begin{equation}
\nab \times \AAA (x) = \sqrt{\frac{2}{1 + \sigma^2}} B_0
\Theta(t)
\nab \times
\Vector{f_1(x)}{f_2(x)}{f_3 (x)} =
\BB (x) \Rightarrow 
\left\{ \begin{array}{c c}
\partial_x f_3 = &  - \sigma \sin k_0 x \\
\partial_x f_2 =  & \cos k_0 x
\end{array}, \right.
\end{equation}
which leads to the solutions $f_2(x) = \sin k_0 x/k_0 + A_2$,
$f_3(x) = \sigma \cos k_0 x/k_0 + A_3$, for some constants
$A_2, A_3 \in \mathbb{R}$.
Hence,
\begin{equation}
\boxed{\AAA = \sqrt{\frac{2}{1 + \sigma^2}} \frac{B_0}
{k_0} \Theta(t) \left[ \Vector{0}{\sin k_0 x}{\sigma \cos k_0 x} +
\AAA_{\rm g} \right],}
\end{equation}
where $\AAA_{\rm g} = (f_1(x), A_2, A_3)$ is the
gauge--dependent part of the vector potential.
However, $\AAA_{\rm g}$ does not contribute to the
magnetic helicity
\begin{equation}
\Bra{\BB \cdot \AAA_{\rm g}} = \sqrt{\frac{2}{1 + \sigma^2}}
B_0 \Theta(t) \Bra{A_2 \sigma \sin k_0x
+ A_3 \cos k_0 x} = 0,
\end{equation}
where $\Bra{\cos k_0 x} = \Bra{\sin k_0 x} = 0$.
Hence, we don't need to assume a specific gauge.

The normalized magnetic helicity density is
\begin{equation}
\boxed{\XiM(t) = \Bra{\BB \cdot \AAA} = \frac{2 \sigma}{1 + \sigma^2}
\frac{B_0^2}{k_0} \Theta(t) = \frac{2 \sigma}{1 + \sigma^2}
\frac{2 \OmM (t)}{k_0}.}
\label{XiM}
\end{equation}

In Fourier space, the magnetic vector potential is
\begin{equation}
\tAA (k_x, t) = \half \sqrt{\frac{2}{1 + \sigma^2}} \frac{B_0}
{k_0} \Theta(t) \left[\Vector{0}{i \Para{\deltp 
- \deltn}}{\sigma \Para{\deltp + \deltn}} + 
\Vector{2 \tilde{f}_1 (k_x)}{2 A_2 \delt}{2 A_3 \delt} \right].
\end{equation}

The magnetic helical spectrum is
\begin{align}
\HM (k, t) & = \int_{\Omega_1} \tBB(k_x, t) \cdot \conj{\AAA} (k_x, t)
\dd \Omega_1 = \frac{2}{1 + \sigma^2} \frac{B_0^2}{k_0}
\Theta(t) \quarter 2 \sigma \Para{\deltp + \deltn} \nonumber \\
 & = \boxed{
\frac{2 \sigma}{1 + \sigma^2}\frac{B_0^2}{k_0}
\Theta(t) \deln .}
\end{align}

The total magnetic helicity is computed from $\Xi(t)$, and
should be consistent with \Eq{XiM}, due to Parseval
theorem.
\begin{equation}
\boxed{\XiM(t) = \int_0^\infty \HM(k, t) \dd k = 
\frac{2 \sigma}{1 + \sigma^2} \frac{2 \OmM (t)}{k_0}.}
\end{equation}

The fractional magnetic helicity is
\begin{equation}
\boxed{\eM = \frac{\kM\BBra{\BB(x, t) \cdot \AAA(x, t)}}
{\BBra{\BB^2 (x, t)}} = 
\frac{\kM\XiM(t)}{2 \OmM(t)} = 
\frac{2 \sigma}{1 + \sigma^2},}
\end{equation}
which reduces to $+1, 0,$ and $-1$,
for $\sigma = +1, 0,$ and $-1$, respectively.

We can see that for the case $\sigma = \pm 1$, if we take
the gauge $\AAA_{\rm g} = \nullvector$, we get
$\AAA = \pm \BB/k_0$.
This can be seen as an eigenvalue problem, where
\begin{equation}
\AAA = \left(\nab \times\right)^{-1} \BB = \lambda \BB
\Rightarrow \left[ \left(\nab \times\right)^{-1} - \lambda \II
\right] \BB = \nullvector,
\end{equation}
which has solution $\lambda = \pm k_0^{-1}$, only when
$\sigma = \pm 1$.
This fact shows that in the case of fully positive or
negative helical, the Beltrami field is an eigenfunction
of the inverse rotational operator $(\nab \times)^{-1}$.

It can be interested to compute the parameter
$\sigma$ in terms of the partial helicity $\eM$.
This leads to
\begin{equation}
\sigma = \frac{1}{\eM} \left(1 - \sqrt{1 - \eM^2}\right),
\end{equation}
which is not defined for $\eM = 0$.
When $\eM = 0$, we can take the limit
\begin{equation}
\lim_{\eM \rightarrow 0} \sigma = 
\lim_{\eM \rightarrow 0}
\frac{1}{\eM} \frac{\eM^2}{2} = 0.
\end{equation}
Hence, $\sigma (\eM = 0) = 0$, as expected.

A different proposal considered to define the
fractional magnetic helicity is
\begin{equation}
\eM = \kM \left\langle \frac{\BB(x, t) \cdot \AAA(x, t)}{\BB^2(x, t)}
\right\rangle.
\end{equation}
Let's show that for the considered field, this is
not a relevant way to define the fractional helicity,
since it is constant and equal to $1$ for all
values of $\sigma$.
\red{
In addition, two problems arise: 
\begin{itemize}
\item In first place,
it is not clear if for a general periodic magnetic
field this definition of the fractional magnetic
helicity is gauge--independent.
Based on this model, I suggest this should be the case,
so maybe this would not be an issue.
\item In second place, it is not clear how
to define the spectrum of such a quantity, since
it is not obtained from the contraction of two vectors,
as for the energy density or the magnetic helicity.
Hence, computing a spectrum for this quantity,
and then compute the integral over all wave
numbers does not seem a plausible option.
\end{itemize} 
}

We have
\begin{equation}
\eM = \sigma \left\langle \frac{1 + A_1 \sin k_0 x
+ A_2 \cos k_0 x}{\sigma^2 \sin^2 k_0 x + \cos^2 k_0 x}
\right\rangle = \sigma \Bra{\bigl(\sigma^2 \sin^2 k_0 x
+ \cos^2 k_0 x\bigr)^{-1}},
\end{equation}
where the gauge--dependent terms from $A_1$ and $A_2$
are zero, because the periodicity of these functions
is $2\pi/k_0$.
In the range $[-\pi/k_0, \pi/k_0]$, the functions
$\bigl(\sin k_0 x, \cos k_0 x\bigr)/\bigl(\sigma^2 \sin^2 k_0 x + 
\cos^2 k_0 x\bigr)$ are odd, because the denominator
is an even function, and the numerator is odd in
both cases.
Hence, the integral over one period is the integral
of an odd function and hence, its value is zero.

Finally, the remaining term is a function of $\sin^2
k_0 x$ and $\cos^2 k_0 x$ and, hence, its period is
$\pi/k_0$.
Therefore, its mean value over a period is
\begin{equation}
\eM = \sigma \Bra{\bigl(\sigma^2 \sin^2 k_0 x
+ \cos^2 k_0 x\bigr)^{-1}} = \frac{\sigma}{\pi}
\int_0^\pi \frac{\dd x}{\sigma^2 \sin^2 x +
\cos^2 x}.
\end{equation}
The indefinite integral is given by the
function $\arctan\bigl(\sigma \tan (x) \bigr)$,
which presents discontinuities at $(2n + 1)\pi/2$,
for $n \in \mathbb{Z}$.
Hence, the analytic calculation requires
some additional method to solve it.
It can easily be computed numerically, and it leads to
\begin{equation}
\eM = \frac{\sigma}{\pi} \int_0^\pi \frac{\dd x}{\sigma^2
\sin^2 x + \cos^2 x} = 1,
\end{equation}
which is independent of the value of $\sigma$.
This shows that this definition is not physically
meaningful, since we start with a magnetic
field with partial helicity, and we end up
with a ``fractional helicity'' always equal to
$1$ using this definition.

Note that numerically, one needs to use $10^{N}$
points of numerical integration to get a convergent
solution of the integral down to values of
$|\sigma| \sim 10^{-N}$, which in a practical sense,
it is enough, since we can show easily that for values of
$|\sigma| > 10^6$, $\eM$ obtained is independent of
$\sigma$ and always equal to 1.
However, to rigurously show that this result is independent of
$\sigma$ for values of $\sigma$ arbitrarily small,
we need to compute the integral analytically.

For this purpose, we want to compute
\begin{equation}
\frac{\sigma}{\pi} \int_0^\pi \frac{\dd x}{\sigma^2+
\sin^2 x + \cos^2 x} = \frac{2\sigma}{\pi}
\int_0^\pi f^{-1}(x) \dd x,
\end{equation}
where $f(x) = 1 + \sigma^2 + (1 - \sigma^2)
\cos 2 x$.
In first place, let's extend the real function $f(x)$
to the complex function $f(z)$, where $z \in \mathbb{C}$.
Now, we can look for the singularities (poles in this case) in the
complex plane of the function $f(z)^{-1}$, which
correspond to the zeros of $f(z)$.
\begin{gather}
f(z) = 1 + \sigma^2 + (1 - \sigma^2) \cos 2 z =
1 + \sigma^2 + \half \bigl(1 - \sigma^2\bigr)
\bigl(e^{2iz} + e^{-2iz} \bigr) = 0, \nonumber \\
\half (1 - \sigma^2) e^{4iz} + (1 + \sigma^2) e^{2iz}
+ \half (1 - \sigma^2) = 0, \nonumber \\
e^{2 iz} = \frac{-(1 + \sigma^2) \pm \sqrt{(1 + \sigma^2)^2
+ (1 - \sigma^2)^2}}{1 - \sigma^2} = - \frac{1 + \sigma^2}
{1 - \sigma^2} \pm \frac{2|\sigma|}{1 - \sigma^2} = - 
\frac{1 \pm |\sigma|}{1 \mp |\sigma|}, \nonumber \\ 
2 i z = {\rm Ln}\left(- \frac{1 \pm |\sigma|}
{1 \mp |\sigma|} \right) = \ln \left(\frac{1 \pm |\sigma|}
{1 \mp |\sigma|}\right) + \pi(1 + 2 k) i,
\end{gather}
where Ln denotes the complex logarithmic function, and
$k \in \mathbb{Z}$, such that we have infinite solutions
$z_s$ along the real axis, with periodicity $2\pi$ and
starting at $\pi/2$ in absolute value, as expected
from the form of the indefinite integral.
We note that both solutions, corresponding to $+$
and $-$ sign within the real logarithmic
function are always positive and bounded for all values of
$\sigma \in (-1, 1)$, where we exclude the
value $|\sigma| = 1$ that can be easily calculated as a
separate case. 
Hence, the poles of $f^{-1} (z)$ are
\begin{equation}
z_p = -\half i \ln \left(\frac{1 \pm |\sigma|}
{1 \mp |\sigma|}\right) + \pi\Bigl(\half + k\Bigr).
\end{equation}
Therefore, we can define a contour in the complex plane,
such that it goes around the singularity counterclockwise,
as shown in the following scheme.
\begin{figure}
\begin{center}
\includegraphics[width=.5\textwidth]{residue}
\end{center}
\caption{Scheme of the contour integration
taken in the complex plane, and the position of
the singularity within the connected domain.}
\end{figure}

In this case, we take the contour from 0 to $\pi$,
in the real axis, such that we only have 
one singularity within the contour, which corresponds to
$z_{p0} = \half \pi + \half i \ln \left(\frac{1 + |\sigma|}
{1 - |\sigma|}\right).$
The value of $f^{-1}(z)$ in the contour away from the
real axis can be estimated as the limit of $f(z)$, 
when $z$ is purely imaginary, i.e., $z = ai$, for
some $a \in \mathbb{R}$, and $a \rightarrow \infty$.
In this limit, $f(z) \propto \cos^{-1} (z)$,
and we have
\begin{equation}
\lim_{a \rightarrow \infty} f(ia) =
\lim_{a \rightarrow \infty} \cos^{-1} (ia) = \lim_{a \rightarrow
\infty}\left(e^{-a} + e^{a}\right)^{-1} = 0.
\end{equation}
Therefore, the value of the integral along the total
contour vanishes when we are in the part of the path
that does not correspond to the real interval
$[0, \pi]$. Hence,
\begin{equation}
\oint_C f^{-1}(z)  \dd z = \int_0^{\pi} f^{-1}(x) \dd x.
\end{equation}
Finally, we can use the residue Cauchy theorem, since the
interior of the contour can be continuously stretched to the
surrounding of the singularity, i.e., the interior of the
integral path is connected, and contains one point
singularity.
In this case (for counterclockwise contour, and
one point singularity), the integral can be computed
as
\begin{equation}
\oint_C f^{-1}(z) \dd z = 2 \pi i {\rm Res} (f, z_{p0}).
\end{equation}
We can compute the residue of $f(z)$ at $z_{p0}$
as
\begin{align}
{\rm Res}^{-1} (f, z_{p0}) = f'(z_{p0}) & = i(1 - \sigma^2)
\Bigl(e^{2iz_{p0}} - e^{-2iz_{p0}}\Bigr)  \nonumber \\  & 
= -i (1 - \sigma^2) \left( \frac{1 - |\sigma|}
{1 + |\sigma|} - \frac{1 + |\sigma|}{1 - |\sigma|} \right)
= 4i |\sigma|.
\end{align}
Hence, we finally get that
\begin{equation}
\boxed{\eM = \frac{2\sigma}{\pi} \int_0^\pi f^{-1} (x) = 
\frac{2 \sigma}{\pi} \bigl(2 \pi i\bigr)/
\bigl(4i|\sigma|\bigr) = {\rm sgn} (\sigma)},
\end{equation}
from which we observe that $\eM$ has always value $\pm 1$
where the sign is the same as the sign of $\sigma$.
For $\sigma = \pm 1$, we have the simple integral
\begin{equation}
\eM \bigl(\sigma = \pm 1\bigr) = \frac{\pm 1}{\pi} \int_0^\pi
\dd x = \pm 1,
\end{equation}
which can also be described by the general expression
previously obtained.

The conclusion is that the proposed definition for $\eM$
is less complete than the first one, since for this
simple 1D Beltrami field with a single mode, it only
describes the sign of the helicity, but not the
intensity with respect to the total magnetic
energy density.

\subsection{Current helicity}

We can also define a helicity based uniquely on the magnetic field
and its derivatives.
In general, helicity is defined for some vector field $\JJ$,
such that
\begin{equation}
\Xi_{\rm J} (t) = \Bra{\JJ \cdot (\nab \times)^{-1} \JJ},
\end{equation}
where if $\JJ$ denotes the current density, then
$\Xi_{\rm J}$ is the normalized current helicity.
This can be computed from the magnetic field since,
using MHD equations (and neglecting the electric
displacement), we have $\JJ = \nab \times \BB$.
Hence,
\begin{equation}
\boxed{\Xi_{\rm J} (t) = \Bra{\BB \cdot \nab \times \BB} = 
\frac{2 \sigma}{1 + \sigma^2} |k_0| B_0^2 \Theta(t) = 
\frac{2 \sigma}{1 + \sigma^2} 2 \kM \OmM (t).}
\end{equation}
We see that for $\sigma = \pm 1$, since $\BB$
is an eigenfunction of the operator
$(\nab \times)^{-1} = -|k_0|^{-1}$, from where
we can see that $\nab \times = -|k_0|$ and hence,
$\nab \times \BB = |k_0| \BB$, as it's obtained. 

The current helicity spectrum is
\begin{equation}
\boxed{H_{\rm J} (k, t) = \int_{\Omega_1} \tBB(k_x, t) \cdot (- i \kk
\times \tBB (k_x, t))^\ast \dd \Omega_1 = 
\frac{2 \sigma}{1 + \sigma^2} |k_0| B_0^2  \Theta(t) \deln,}
\end{equation}
where $\kk = (k_x, 0, 0)$.
The total (normalized) current helicity density is
\begin{equation}
\boxed{\Xi_{\rm J} (t) = \int_0^{\infty} \HM (k, t) \dd k =
\frac{2 \sigma}{1 + \sigma^2} |k_0|B_0^2 \Theta(t).}
\end{equation}

The corresponding energy density can be computed for
$JJ = \nab \times \BB$ as
\begin{align}
\Omega_{\rm J}(t) = & \half \Bra{\JJ^2} = \half \Bra{(\nab \times \BB)^2} = 
\frac{2}{1 + \sigma^2} B_0^2 k_0^2
\Theta(t) \Bra{\sin^2 k_0 x + 
\sigma^2 \cos^2 k_0 x} \nonumber \\
= & \OmM(t) k_0^2.
\end{align}

We can also define a fractional helicity for the current
density as
\begin{equation}
\boxed{\varepsilon_{\rm J} = \frac{\kM \Bra{\BB \cdot
(\nab \times \BB)}}{2 \Bra{\JJ^2}} = \frac{2 \sigma}{1 + \sigma^2}.}
\end{equation}
This is the same relation as the one obtained for the
magnetic partial helicity.

We can observe that the magnetic helicity has been defined
in terms of the magnetic vector potential $\AAA$, and
a kinetic helicity can be defined in terms of the
vorticity $\omega = \nab \times \uu$, where $\uu$
is the velocity field.

\subsection{Stress tensor}

The stress tensor for magnetic fields is
\begin{equation}
\Tij (x, t) = - B_i (x, t) B_j (x, t) +
\half \delta_{ij} \BB^2 (x, t),
\end{equation}
%In first place, we can observe that the total energy density
%associated to the stress tensor is
%\begin{equation}
%\Omega_{\rm T} = \half \Bra{T_{ij}(x, t)
%T_{ij} (x, T)} = \half
%\end{equation}
where
\begin{equation}
\BB^2 (x, t) = \frac{2 \OmM}{1 + \sigma^2}
\Theta(t) \left[(1 + \sigma^2) + (1 - \sigma^2)
\cos 2k_0 x \right],
\end{equation}
and
\begin{equation}
-B_i B_j (\xx, t) = - \frac{2 \OmM}{1 + \sigma^2}
\Theta(t)
\Matrixsym{0}{0}{0}{\sigma^2 (1 - \cos 2 k_0 x)}
{\sigma \sin 2 k_0 x}{1 + \cos 2 k_0 x}.
\end{equation}

Combining both we obtain the stress tensor
\begin{equation}
\Tij (\xx, t) =  \frac{\OmM}{1 + \sigma^2} \Theta(t) 
\Matrixsym{T_{11} (x)}{0}{0}{T_{22}(x)}{T_{23}(x)}{-T_{22} (x)},
\label{Tij}
\end{equation}
where
\begin{gather}
T_{11} (x) = (1 + \sigma^2) + (1 - \sigma^2) \cos 2k_0 x, \\
T_{22} (x) = (1 - \sigma^2) + (1 + \sigma^2) \cos 2k_0 x, \\
T_{23} (x) = - 2 \sigma \sin 2k_0 x.
\end{gather}

\subsection{TT projection}

Since GWs are produced by linear perturbations over the metric tensor,
and the stress tensor components are also perturbations over background
fields, constant modes ($\kk = \nullvector$) do not produce GWs.
For this reason, we can neglect the terms that are homogeneous in space.
This can be done by computing $\tTk_{ij} (\kk, t)
= \tTij (\kk, t) [1 - \delta(\kk)]$.
The stress obtained is the same as in \Eq{Tij}, replacing $T_{11} (\xx)$
and $T_{22} (\xx)$ by  
\begin{gather}
T_{11}^{\kk} (\xx) = (1 - \sigma^2) \cos 2k_0 x, \\
T_{22}^{\kk} (\xx) = (1 + \sigma^2) \cos 2k_0 x. \\
\end{gather}

In Fourier space, the resulting stress tensor is
\begin{equation}
\tTk_{ij} (\kk, t) = \frac{\OmM}{1 + \sigma^2} \Theta(t)
\left( \begin{array}{c c c}
\tTk_{11} (\kk) & 0 & 0 \\
0 & \tTk_{22} (\kk) & \tTk_{12} (\kk) \\
0 & \tTk_{12} (\kk) & - \tTk_{22} (\kk)
\end{array} \right),
\end{equation}
where
\begin{gather}
\tTk_{11} (\kk) = \half (1 - \sigma^2)
\bigl[\deltp2 + \deltn2 \bigr], \\
\tTk_{22} (\kk) = \half (1 + \sigma^2)
\bigl[\deltp2 + \deltn2 \bigr], \\
\tT_{12} (\kk) = -i \sigma  
\bigl[ \deltp2 - \deltn2 \bigr].
\end{gather}
%
%If we neglect the terms proportional to $\delta_{ij}$
%
%\begin{equation}
%T_{ij}^{\rm TT} (\xx, t) = - \Omega_{\rm M} \Theta(t) \left(\begin{array}{c c c}
%0 & 0 & 0 \\
%0 & - \frac{1}{2} (1 + \sigma^2) \cos 2 k_0 x & 2 \sigma \sin 2k_0 x \\
%0 & 2\sigma \sin 2k_0 x & \frac{1}{2} (1 + \sigma^2) \cos 2k_0 x
%\end{array} \right).
%\end{equation}

The transverse and traceless projection of the stress tensor is
\begin{equation}
\tTijTT (\kk, t) = \Proj \tTk_{lm} (\kk, t),
\end{equation}
where $P_{ij} = \delta_{ij} - \hatk_i \hatk_j$ is the
projection operator being $\hatkk = \kk/\sqrt{\kk^2}$.

The only wave vectors that leads to non-zero stress in
Fourier space are $\kk = \pm 2 \kk_0 = (\pm 2 k_0, 0, 0)$.
The projection operator for these wave vectors is
$P_{ij} (\pm 2\kk_0) = \delta_{ij} - \delta_{i1} \delta_{j1}$.
This leads to
\begin{gather}
P_{il} (\pm 2 \kk_0) P_{jm} (\pm 2 \kk_0) \tTk_{lm}
(\pm 2 \kk_0, t) = \frac{\OmM}{1 + \sigma^2} \Theta(t) \left( \begin{array}{c c c}
0 & 0 & 0 \\ 0 & \tTk_{22} (\pm 2 \kk_0) & 
\tT_{12} (\pm 2 \kk_0) \\ 0 & \tT_{12} (\pm 2 \kk_0) & 
- \tTk_{22} (\pm 2 \kk_0)
\end{array} \right), \\
P_{ij} (\pm 2 \kk_0) P_{lm} (\pm 2 \kk_0) \tTk_{lm} (\pm 2 \kk_0) = 
0_{ij}.
\end{gather}

Therefore, the transverse and traceless projected stress tensor is
\begin{gather} 
\tTijkTT (\kk, t) = \frac{\OmM}{1 + \sigma^2} \Theta(t)
\left[\Matrixsym{0}{0}{0}{\half (1 + \sigma^2)}{i\sigma}
{- \half (1 + \sigma^2)}  \deltp2 + \dots \right. \nonumber \\
\left. \dots + \Matrixsym{0}{0}{0}{\half (1 + \sigma^2)}
{-i\sigma}{-\half (1 + \sigma^2)}
\deltn2  \right].
\end{gather}

Back to physical space we have
\begin{equation}
\TijkTT (\xx, t) = \frac{\OmM}{1 + \sigma^2} \Theta(t)
\left( \begin{array}{c c c} 0 & 0 & 0 \\
0 & (1 + \sigma^2) \cos 2k_0 x & 2 \sigma \sin 2 k_0 x \\
0 & 2 \sigma \sin 2 k_0 x  & - (1 + \sigma^2) \cos 2k_0 x
\end{array} \right).
\label{Tij2}
\end{equation}

\subsection{Polarization modes}

We can define $+$ and $\times$ polarization modes, such that the
stress tensor is decomposed as $\tTij (\kk, t) =
\tTT (\kk, t) \eT (\kk) + \tTX
(\kk, t) \eX (\kk)$, where $\eTX$
are the linear polarization basis tensors, defined as
\begin{equation}
\eT (\kk) = e_i^1 e_j^1 - e_i^2 e_j^2, \quad 
\eX (\kk) = e_i^1 e_j^2 + e_i^2 e_j^1,
\end{equation}
where $\ee^1$, $\ee^2$, and $\ee^3 = \hatkk$
are orthonormal unit vectors.

If these vectors are orthonormal we can see that
the linear polarization basis tensors have the following
property
\begin{equation}
\eA (\kk) \eAp (\kk) = 2 \delta_{A A'},
\end{equation}
where $A, A' = +, \times$, and $\delta_{A A'}$ is the
Kronecker delta.
Using this property we can derive the polarization modes
of the stress tensor.
\begin{equation}
\tT_{A} (\kk, t)  = \frac{1}{2} \eA (\kk)
\tTij (\kk, t).
\end{equation}

These vectors are defined such that for half of all
the wave vectors $\kk$, $\ee^1, \ee^2, \ee^3$ form a
right-handed basis, and for the corresponding $-\kk$,
they form a left-handed basis.
This is required to obtain unit vectors $\ee^1$, and
$\ee^2$ that are even operators under the transformation
$\kk \rightarrow -\kk$.
Hence, $\eTX (\kk)$ are both even
operators, as explained in \cite{RPBKKM20}.

Let's show this by {\em reductio ad absurdum}.
We choose $\ee^1, \ee^2, \ee^3$ to form a right-handed
basis for all wave vectors $\kk$.
In our case, the only non-zero contributions to the
Fourier stress tensor are for $\kk = \pm 2 \kk_0$.
We have unit vector $\ee^3 = (\pm 1, 0, 0)$, and we 
can choose orthonormal $\ee^1$ and $\ee^2$.
For simplicity we take $\ee^1 = (0, 1, 0)$, and
$\ee^2 = (0, 0, \pm 1)$.
Note that this choice is not unique.

We construct polarization basis tensors $\eTX$
for $\pm 2 \kk_0$.
\begin{equation}
\eT (2 \kk_0) = \eT (- 2 \kk_0) = \left(\begin{array}{c c c}
0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array} \right), \quad 
\eX (2 \kk_0) = - \eX (- 2 \kk_0) = \left(\begin{array}{c c c}
0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right),
\end{equation}
where we observe, as expected, that $e_{ij}^+ (\kk)$ is an even operator,
and $\eX (\kk)$ is an odd operator under transformation
$\kk \rightarrow -\kk$.

The components of the stress tensor are
\begin{gather}
\tTT (\kk, t) = \quarter \OmM \Theta(t) (1 + \sigma^2)
\bigl[\delttwop + \delttwon \bigr],
\label{Tpw} \\ 
\tTX (\kk, t) = \half \OmM \Theta(t) i \sigma
[\delttwop + \delttwon ].
\label{Tcw}
\end{gather}
We can observe from \Eqs{Tpw}{Tcw} that one is proportional to the other by a
factor $2 i \sigma/(1 + \sigma^2)$.
Therefore, the functions obtained for $\tTTX (\kk, t)$ cannot 
be two linearly independent modes of the tensor $\tTij (\kk, t)$ as
they are supposed to be, leading to an absurd.

We can observe that this is due to the fact that $\eX$ is an odd 
operator under transformation
$\kk \rightarrow -\kk$ if we choose $\ee^1, \ee^2, \ee^3$ to form a right-handed
basis for all wave vectors.

Therefore, if we choose $\ee^1$ and $\ee^2$ to be the same for $\pm 2 \kk_0$,
we solve this issue.
For simplicity let's take $\ee^1 = (0, 1, 0)$, and
$\ee^2 = (0, 0, 1)$, which forms a right-handed basis for $\ee^3 = (1, 0, 0)$,
and a left-handed basis for $\ee^3 = (-1, 0, 0)$.
The procedure to choose $\ee^1$ and $\ee^2$ for general wave vectors $\kk$
is described in \cite{RPBKKM20}.

In this case
\begin{equation}
\eT (2 \kk_0) = \eT (- 2 \kk_0) = \left(\begin{array}{c c c}
0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array} \right), \quad 
\eX (2 \kk_0) = \eT (- 2 \kk_0) = \left(\begin{array}{c c c}
0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right),
\end{equation}
\begin{gather}
\tTT (\kk, t) = \quarter \OmM \Theta(t) (1 + \sigma^2)
\bigl[\delttwop + \delttwon \bigr],
\label{Tpr} \\ 
\tTX (\kk, t) = \half \OmM \Theta(t) i \sigma
\bigl[\delttwop - \delttwon \bigr],
\label{Tcr}
\end{gather}
which are now linearly independent modes, and hence, have physical meaning,
to describe the two polarization modes of the stress tensor.

Back to physical space, this is
\begin{gather}
\TT (\xx, t) = \half \OmM \Theta(t) (1 + \sigma^2)
\cos 2 k_0 x, \\ 
\TX (\xx, t) = \OmM \Theta(t) \sigma
\sin 2 k_0 x.
\end{gather}

Note that if we transform $\ee^2 \rightarrow - \ee^2$, we just switch
$\ee^3 = (1, 0, 0)$ to form a left-handed basis, while $\ee^3 = (-1, 0, 0)$
to form a right-handed basis, which relies on the arbitrary decision
of which half of the wave vectors are assumed positive.
This leads to
\begin{equation}
\TX (\xx, t) = - \OmM \Theta(t) \sigma
\sin 2 k_0 x.
\end{equation}
Therefore, since this is as valid as previous assumption, the sign of
$\TX (\xx, t)$ is not well-defined and both signs are correct.

On the other hand, we can also choose the basis vectors $\ee^1$ and
$\ee^2$ to be odd operators under the transformation
$\kk \rightarrow - \kk$.
In this case, the resulting polarization modes $\tTTX
(\kk, t)$ are unchanged.

\subsection{Gravitational waves}

The tensor-mode perturbations over a flat universe, neglecting
the effect of expansion, can be computed from the GW equation
\begin{equation}
\Bigl(\partial_t^2 - c^2 \nab^2\Bigr) h_{ij} (\xx, t) = 
6 H_\ast^2 \TijTT (\xx, t).
\end{equation}

The general solution to the GW equation for a general
one-dimensional sourcing term is
\begin{equation}
\hij (x, t) = \frac{3 H_\ast^2}{c} \int_0^t
\int_{x - ct + cs}^{x + ct - c s} \TijTT (z, s)
\dd z \dd s. 
\end{equation} 

Hence, we can compute the two linear polarization modes of
the GW tensor-mode perturbations $h_{+, \times} (\xx, t)$ directly
from $\TTX (\xx, t)$ as
\begin{equation}
\hT (x, t) = \frac{3 H_\ast^2}{2 c} \OmM (1 + \sigma^2)
\int_0^t \int_{x - ct + cs}^{x + ct - c s}
\Theta(s) \cos 2k_0 z \dd z \dd s.
\end{equation}
Let's compute in first place the integral over $z$:
\begin{align}
\int_{x - ct + cs}^{x + ct - c s} \cos 2k_0 z \dd z&  = 
\frac{1}{2 k_0} \Bigl[ \sin 2k_0 (x + ct - cs) - \sin 2k_0
(x - ct + cs) \Bigr] \nonumber \\
& = \frac{1}{k_0} \cos 2 k_0 x \sin 2 c k_0 (t - s).
\end{align}
The integral over $s$ is
\begin{align}
\frac{1}{k_0} \cos 2 k_0 x \int_0^t \Theta(s) \sin 2 c k_0 (t - s) ds & = 
\frac{1}{2 c k_0^2} \cos 2k_0 x \Theta(t) (1 - \cos 2 ck_0 t ) \\
& = \frac{1}{c k_0^2} \cos 2k_0 x \Theta(t) \sin^2 c k_0 t. 
\end{align}

This finally leads to
\begin{equation}
\hT (x, t) = \frac{3 H_\ast^2}{2c^2 k_0^2} \OmM (1 + \sigma^2)
\Theta(t) \cos 2k_0x \sin^2 c k_0 t.
\end{equation}

To compute $\hX (x, t)$, we can use previous results.
The integral over $z$  is
\begin{align}
\int_{x - ct + cs}^{x + ct - c s} \sin 2k_0 z \dd z&  = 
\frac{1}{2 k_0} \Big[ \cos 2k_0 (x + ct - cs) - \cos 2k_0
(x - ct + cs) \Big] \nonumber \\
& = - \frac{1}{k_0} \sin 2 k_0 x \sin 2 c k_0 (t - s).
\end{align}
The integral over $s$ is
\begin{equation}
- \frac{1}{k_0} \sin 2 k_0 x \int_0^t
\Theta(s) \sin 2 c k_0 (t - s) ds = -
\frac{1}{c k_0^2} \sin 2k_0 x \Theta(t) \sin^2 c k_0 t. 
\end{equation}

This finally leads to
\begin{equation}
\hX(x, t) = - \frac{3 H_\ast^2}{c^2 k_0^2} \OmM \sigma
\Theta(t) \sin 2k_0x \sin^2 c k_0 t.
\end{equation}
Note that we can switch the sign from negative to positive, since
the sign of $\TX(x, t)$ is not well-defined.
Hence, we'll take its positive value, and we will see that this
does not affect the results.

To summarize, we have obtained
\begin{align}
\hT (x, t) = \frac{3 H_\ast^2}{2c^2 k_0^2} & \OmM (1 + \sigma^2)
\Theta(t) \cos 2k_0x \sin^2 c k_0 t, \\
\hX (x, t) = \frac{3 H_\ast^2}{c^2 k_0^2} & \OmM \sigma
\Theta(t) \sin 2k_0x \sin^2 c k_0 t.
\end{align}
If we denote $A(t) = (3 H_\ast^2)/(c^2 k_0^2) \OmM
\Theta (t) \sin^2 ck_0 t$, we can express
\begin{align}
\hT (x, t) & =  \half A(t) (1 + \sigma^2) \cos 2k_0x, \\
\hX (x, t) & = A(t) \sigma \sin 2k_0 x.
\end{align}

In Fourier space, the tensor-mode perturbations are
\begin{align}
\thT (\kk, t) & = \quarter A(t) (1 + \sigma^2)
[\delttwop + \delttwon ], \\
\thX (\kk, t) & = \half A(t) i \sigma
[\delttwop - \delttwon].
\end{align}

The GW tensor is
\begin{equation}
\thij (\kk, t) = \eT \thT (\kk, t) + \eX
\thX (\kk, t) = \left(\begin{array}{c c c}
0 & 0 & 0 \\
0 & \thT (\kk, t) & \thX (\kk, t) \\
0 & \thX (\kk, t) & - \thT (\kk, t)
\end{array} \right).
\end{equation}

Back to physical space, this is
\begin{equation}
\hij (\xx, t) = \frac{3 H_\ast^2}{c^2 k_0^2} \OmM \Theta(t)
\sin^2 c k_0 t \left(\begin{array}{c c c} 0 & 0 & 0 \\
0 & \half (1 + \sigma^2)\cos 2 k_0 x & \sigma \sin 2k_0 x \\
0 & \sigma \sin 2k_0 x & - \half (1 + \sigma^2)\cos 2 k_0 x
\end{array} \right).
\end{equation}

\subsection{GW spectra}

We can compute now the symmetric spectral function $\Sh (k, t)$
from the strains $\thTX (\kk, t)$.

\begin{align}
\Sh (k, t) = & \int_{\Omega_1} \Bigl( \thT (\kk, t)
\thTc (\kk, t) + \thX (\kk, t)
\thXc (\kk, t)\Bigr) \dd \Omega_1 \\ 
= & A^2 (t) \Bigl[\sigma^2 + \quarter (1 + \sigma^2)^2
\Bigr] \deltwon, \\
\Sh (k, t) = & \Biggl(\frac{3 H_\ast^2}{ c^2 k_0^2}
\OmM \sin^2 c k_0 t \Biggr)^2 \Theta(t) \Biggl[\sigma^2 + \quarter
(1 + \sigma^2)^2 \Biggr] \deltwon.
\end{align}
In the case of a Beltrami field, $\sigma = \pm 1$, we recover Eq.(39)
of Ref.~\cite{RPBKKM20}.
\begin{equation}
\Sh (k, t) = 2\Biggl(\frac{3 H_\ast^2}{ c^2 k_0^2}
\OmM \sin^2 c k_0 t \Biggr)^2 \Theta(t) \deltwon.
\end{equation}
On the other hand, for a non-helical field, $\sigma = 0$, we have
\begin{equation}
\Sh (k, t) = \quarter \Biggl(\frac{3 H_\ast^2}{ c^2 k_0^2}
\OmM \sin^2 c k_0 t \Biggr)^2 \Theta(t) \deltwon.
\end{equation}
We can observe a difference of 8 in the spectral function for the cases
of fully helical Beltrami field vs non-helical field.

The spectrum of the characteristic amplitude of GWs $\hc(k, t)$, per unit of 
lorgarithmic wave number, is
\begin{equation}
\hc(k, t) = \sqrt{k \Sh (k, t)} = 
\frac{6 H_\ast^2}{ c^2 |k_0|}
\OmM \sin^2 c k_0 t \Theta(t) f(\sigma)
\deltwon,
\end{equation}
where $f(\sigma) = \Bigl(\sigma^2 + \quarter(1 +
\sigma^2)^2 \Bigr)^{1/2}$ represents the dependence with
$\sigma$.
We can average over time oscillations to get
\begin{equation}
\hc(k) = 
\frac{3 H_\ast^2}{ c^2 |k_0|}
\OmM \sqrt{\sigma^2 + \quarter
(1 + \sigma^2)^2} \deltwon.
\end{equation}

For non-dimensional wave number, we have
\begin{equation}
\hc(k) = k \frac{3 \OmM}{2 \bar k_0^2} \sqrt{\sigma^2 +
\quarter (1 + \sigma^2)^2} \deltwon.
\end{equation}

We can integrate over all wave numbers to get
\begin{equation}
\hc = \int_0^\infty \hc(k)/k \dd k = 
\frac{3 \OmM}{2 \bar k_0^2} \sqrt{\sigma^2 +
\quarter (1 + \sigma^2)^2}.
\end{equation}

We can use this value to obtain estimates at the present time.
The characteristic strain would shift as $\hc \propto a^{-1}$.
Therefore, the value at the present time would be
\begin{equation}
\hc = \frac{3 \OmM}{2 \bar k_0^2} a_0^{-1} \sqrt{\sigma^2 +
\quarter (1 + \sigma^2)^2}.
\end{equation}
Using values $\OmM = 0.1$, $N = \bar k_0/2\pi = 100$,
$a_0 = 1.254 \cdot 10^{15}$, we get
\begin{equation}
\hc = 3.0288 \cdot 10^{-22} (\OmM/0.1) (N/100)^{-1}
(T_\ast/100 \GeV)^{-1}(\gS/100)^{-1/3} f(\sigma),
\end{equation}
where $f(\sigma) = \Bigl(\sigma^2 + \quarter(1 +  \sigma^2)^2 \Bigr)^{1/2}$ is the
dependence on the partial helicity.
The function $f(\sigma)$ is even with respect to $\sigma$ so positive
and negative helical magnetic fields will affect equally to the
characteristic amplitude.
The minimum $f(\sigma)$ is obtained when $\sigma = 0$, and
$f(0) = 1/2$, while the maximum is obtained when $\sigma = \pm 1$
with $f(\pm 1) = \sqrt{2}$
\begin{figure}\begin{center}
\includegraphics[width=.45\textwidth]{hc_sigma_OmegaN}
\includegraphics[width=.45\textwidth]{hc_f_sigma}
\end{center} \end{figure}
\begin{figure}\begin{center}
\includegraphics[width=.32\textwidth]{fsigma}
\includegraphics[width=.32\textwidth]{f_epsilon}
\includegraphics[width=.32\textwidth]{f_epsilon_sigma}
\end{center} \end{figure}

The antisymmetric spectrum is zero in this case, for all wave numbers.
\begin{equation}
\Ah (k, t) = 2 \int_{\Omega_1} \Imag (\thT (\kk, t)
\thXc (\kk, t))\dd \Omega_1 = 0. 
\end{equation}

\subsection{GW energy density}

The time derivatives of the strains are
\begin{align}
\dhT (\xx, t) & = \half C(t) (1 + \sigma^2) \cos 2 k_0 x, \nonumber \\
\dhX (\xx, t) & = C(t) \sigma \sin 2 k_0 x, \nonumber \\
\tdhT (\kk, t) & = \quarter C(t) (1 + \sigma^2)
\bigl[ \delttwop + \delttwon \bigr], \nonumber \\
\tdhX (\kk, t) & = \half C(t) i \sigma
\bigl[ \delttwop - \delttwon \bigr],
\end{align}
where we have now defined $C(t) = (3 H_\ast^2)/(ck_0) \OmM
\Theta(t) \sin 2 c k_0 t$.

The GW energy density spectrum is
\begin{equation}
\Sdh (k, t) = \int_{\Omega_1} 
\end{equation}


\begin{thebibliography}{}

\bibitem{RPBKKM20}
Roper Pol, A., Brandenburg, A., Kahniashvili, T.,
Kosowsky, A., Mandal, S.\ygafd{2020}{114}{130}

\end{thebibliography}

\end{document}
