- \frac{\gy{\mu}}{q_s} \dgav{\evalperturbedgcadjoint{\curl (\gyfone \vc{B}_0)}}\biggr) \diff^3 \gy{u}
\end{equation}

where we have substituted $(B_{s,\shortparallel}^\effective \dot{\gy{\vc{R}}})_0 = B_0 \gy{u}_\shortparallel \hb$ as follows from <ref>.

We need the Fourier component of the adjoint of the gyro-average for evaluating <ref>.
The gyro-average adjoint on $\Omega = \mathbb{R}^3$, with constant $\vc{B}_0$, is given by

\begin{equation}
\int Q \cp(\gy{\vc{r}} + \rho(\gy{\mu}) \hrho(\gy{\theta})) \diff^6 \gy{z}
\int Q(\gy{\vc{r}} - \rho(\gy{\mu}) \hrho(\gy{\theta})) \cp \diff^6 \gy{z}
\quad \implies \quad
\gav{\evalperturbedgcadjoint{Q}} = \gav{Q(\gy{\vc{r}} - \vc{\rho})},
\end{equation}

where we have made use of the coordinate transformation

\begin{equation}
\gy{\vc{r}} \mapsto \gy{\vc{r}} - \rho(\gy{\mu}) \hrho(\gy{\theta}),
\end{equation}

which has unit Jacobian.
It follows that the corresponding Fourier component is given by

\begin{equation}\label{eq:fourier_gavadjoint}
\fourier{\gav{\evalperturbedgcadjoint{Q}}} = J_0(-\kperp \rho) \fourier{Q} = J_0 \fourier{Q}
\end{equation}

by the symmetry of the zeroth-order Bessel function of the first kind.
A similar result holds for the adjoint of the disc average

\begin{equation}
\dgav{\evalperturbedgcadjoint{Q}} = \dgav{Q(\gy{\vc{r}} - \burby \vc{\rho})}
\quad \implies \quad
\fourier{\dgav{\evalperturbedgcadjoint{Q}}} = \frac{2 J_1}{\kperp \rho} \fourier{Q}
\end{equation}

It follows that <ref>, in terms of Fourier components, can be expressed as

\begin{align}
\fourier{\evalperturbedgcadjoint{\vc{\mathcal{J}}}}{}^\free_{1,s} = {} &
\frac{q_s}{m_s} \int \biggl[
\gyfzero \roundpar{
J_0 \fourier{(B_{s,\shortparallel}^\effective \dot{\gy{\vc{R}}})_1}
+ \ii \frac{\gy{\mu}}{q_s} \frac{2 J_1}{\kperp \rho} \fourier{B_{1,s,\shortparallel}^\effective} \hb \times \vc{k}
} \nonumber \\ &
+ \fourier{\gyfone} \roundpar{
J_0 \vc{B}_0 \gy{u}_\shortparallel
+ \ii \frac{\gy{\mu}}{q_s} \frac{2 J_1}{\kperp \rho} \vc{B}_0 \times \vc{k}
\biggr] \diff^3 \gy{u}\label{eq:djdt_tmp_fourier}
\end{align}

for which find that <ref> can be written in terms of Fourier components as follows

\begin{equation}
\fourier{(B_{s,\shortparallel}^\effective \dot{\gy{\vc{R}}})_1} =
\frac{J_0 \gy{u}_\shortparallel}{\omega} (\hb \bcdot \vc{k} \times \fourier{\vc{E}}_1) \hb
+ \frac{1}{\omega} \squarepar{
\gy{u}_\shortparallel J_0 \hb \times (\vc{k} \times \fourier{\vc{E}}_1)
+ \omega J_0 \fourier{\vc{E}}_1
- \ii \frac{\mu}{q_s} \frac{2 J_1}{\kperp \rho} (\vc{k} \times \fourier{\vc{E}}_1 \bcdot \hb) \vc{k}
} \times \hb.
\end{equation}

\begin{equation}\label{eq:gy_eomR_linpart_fourier}
\fourier{(B_{s,\shortparallel}^\effective \dot{\gy{\vc{R}}})_1} = \Rdotmat \fourier{\vc{E}}_1
, \quad
\Rdotmat \defeq
J_0 \roundpar{
- \frac{\gy{u}_\shortparallel}{\omega} \mat{K}
+ \frac{1}{B_0} \mat{B}_0
+ \ii \frac{2 \gy{\mu} J_1}{\omega q_s \kperp \rho} (\vc{k} \times \hb) (\vc{k} \times \hb)^\transpose
\end{equation}

where the matrix $\mat{K}$ is such that <ref> holds.
Moreover, the Fourier transform of $B_{1,s,\shortparallel}^\effective$ is given by

\begin{equation}\label{eq:B1sparallel_fourier}
\fourier{B_{1,s,\shortparallel}^\effective}
= J_0 \fourier{B}_{1,\shortparallel}
= \frac{J_0}{\omega} (\hb \times \vc{k}) \bcdot \fourier{\vc{E}}_1
\end{equation}

by making use of <ref>.

When substituting <ref> into <ref> we find

\begin{equation}\label{eq:fourier_current}
\fourier{\evalperturbedgcadjoint{\vc{\mathcal{J}}}}{}^\free_{1,s} = \pertJmat \fourier{\vc{E}}_1,
\end{equation}

\begin{align}
\pertJmat \defeq {} &  \frac{q_s}{m_s} \int \biggl[\gyfzero \roundpar{
J_0 \Rdotmat
+ \ii \frac{\gy{\mu}}{q_s} \frac{2 J_1}{\kperp \rho} \frac{J_0}{\omega} (\hb \times \vc{k}) (\hb \times \vc{k})^\transpose}
\nonumber \\ &
+ \roundpar{
J_0 \vc{B}_0 \gy{u}_\shortparallel
+ \ii \frac{\gy{\mu}}{q_s} \frac{2 J_1}{\kperp \rho} \vc{B}_0 \times \vc{k}
} \pertfvec^\transpose
\biggr] \diff^3 \gy{u}
\end{align}

§.§ The Fourier component of the polarization and magnetization

Computing the Fourier component of the polarization (<ref>), while substituting Faraday's law (<ref>), results in

\begin{equation}
\omega \fourier{\vc{\mathcal{P}}}_{1,s} = \int \gyfzero \frac{m_s}{B_0^2} \squarepar{\omega \fourier{\vc{E}}_{1,\perp} + \gy{u}_\shortparallel \hb \times (\vc{k} \times \fourier{\vc{E}}_1)} \jacobiangcslab \diff^3 \gy{u},
\end{equation}

\begin{equation}\label{eq:fourier_polarizationmat}
\fourier{\vc{\mathcal{P}}}_{1,s} = \polarizationmat \fourier{\vc{E}}_{1}
, \quad
\polarizationmat =
\int \frac{\gyfzero}{B_0} \squarepar{
\mat{I}_\perp
+ \frac{\gy{u}_\shortparallel}{\omega} (\vc{k} \hb^\transpose - \mat{I}_3 \kpa)
} \diff^3 \gy{u},
\end{equation}

where we have defined the perpendicular projection matrix $\mat{I}_\perp$

\begin{equation}
\mat{I}_\perp
\mat{I}_3 - \hb \hb^\transpose
\quad \implies \quad
\mat{I}_\perp \vc{S}
\vc{S}_\perp
\end{equation}

For the magnetization (<ref>) we find

\begin{multline}
\omega \fourier{\vc{\mathcal{M}}}_{1,s} =
-\int \gyfzero \roundpar{
\frac{m_s \gy{u}_\shortparallel}{B_0^2} \hb \times \squarepar{\omega \fourier{\vc{E}}_{1,\perp} + \gy{u}_\shortparallel \hb \times (\vc{k} \times \fourier{\vc{E}}_1)}
+ \frac{\gy{\mu}}{B_0} (\vc{k} \times \fourier{\vc{E}}_1)_\perp
} \jacobiangcslab \diff^3 \gy{u}\\
- \omega \int \frac{2 J_1}{\kperp \rho} (\gyfzero \fourier{B}_{1,\shortparallel}^\effective + \fourier{\gyfone} B_0) \hb \frac{\gy{\mu}}{m_s} \diff^3 \gy{u}
\end{multline}

\begin{equation}\label{eq:fourier_magnetizationmat}
\fourier{\vc{\mathcal{M}}}_{1,s} =
\magnetizationmat \fourier{\vc{E}}_1
, \quad
\magnetizationmat =
-\int \frac{\gyfzero}{m_s} \squarepar{
\frac{m_s \gy{u}_\shortparallel}{B_0} \roundpar{-\frac{1}{B_0} \mat{B}_0 \mat{I}_\perp + \frac{\gy{u}_\shortparallel}{\omega} \mat{I}_\perp \mat{K}}
- \frac{\gy{\mu}}{\omega} \mat{I}_\perp \mat{K}
} \diff^3 \gy{u}
% \\
% & - \int \frac{2 \gy{\mu} J_1}{m_s \kperp \rho} \hb \roundpar{\gyfzero \frac{J_0}{\omega} \hb \times \vc{k} + B_0 \pertfvec}^\transpose \diff^3 \gy{u}
\end{equation}

§.§ The gyrokinetic susceptibility tensor

We now consider the Fourier component of the linearization of <ref>

\begin{equation}
-\omega^2 \roundpar{\epsilon_0 \fourier{\vc{E}}_1 + \sum_s \fourier{\vc{\mathcal{P}}}_{1,s}}
\frac{1}{\mu_0} \vc{k} \times (\vc{k} \times \fourier{\vc{E}}_1)
- \omega \sum_s \vc{k} \times \fourier{\vc{\mathcal{M}}}_{1,s}
+ \ii\omega \sum_s \fourier{\evalperturbedgcadjoint{\vc{\mathcal{J}}}}_{1,s}
\end{equation}

and therefore, when comparing to <ref>, we find that the gyrokinetic susceptibility tensor $\gyFIXED{\mat{X}}$ is given by

\begin{equation}\label{eq:gy_susceptibility}
\epsilon_0 \gyFIXED{\mat{X}} =
\sum_s \polarizationmat
+ \frac{1}{ \omega} \mat{K} \sum_s \magnetizationmat
+ \ii \frac{1}{\omega} \sum_s \pertJmat,
\end{equation}

where the polarization, magnetization and current matrices are given by Eqs. (<ref>), (<ref>) and (<ref>), respectively.

To facilitate the comparison with the results from [Hasegawa, 1975, Zonta et al., 2021] we choose the following parallel and perpendicular directions

\begin{equation}\label{eq:par_perp_dirs}
\hb = \vc{\hat{e}}_z
, \quad
\vc{k} = \kperp \vc{\hat{e}}_x + \kpa \vc{\hat{e}}_z.
\end{equation}

This results in the following polarization matrix

\begin{equation}\label{eq:fourier_polarizationmat_xz}
\polarizationmat =
\int \frac{\gyfzero}{\omega B_0}
\begin{pmatrix}
\omega - \kpa \gy{u}_\shortparallel & 0 & \kperp \gy{u}_\shortparallel\\
0 & \omega - \kpa \gy{u}_\shortparallel & 0\\
0 & 0 & 0\\
\end{pmatrix}
\diff^3 \gy{u}.
\end{equation}

The vector representation of the perturbed distribution function becomes

\begin{equation}\label{eq:lin_vlasov_fourier_xz}
\pertfvec
\frac{
\omega_{\mathrm{c},s} \pard{\gyfzero}{\gy{u}_\shortparallel}
{\ii B_0 (\omega - \kpa \gy{u}_\shortparallel)}
\begin{pmatrix}
- \ii \frac{2 \gy{\mu} \kpa J_1}{\omega q_s \rho}\\
\end{pmatrix}
\end{equation}

such that the magnetization matrix is given by

\begin{equation}\label{eq:fourier_magnetizationmat_xz}
\begin{aligned}
\magnetizationmat = {} &
-\int \frac{\gyfzero}{\omega B_0} \begin{pmatrix}
0 & (\gy{u}_\shortparallel^2 - \frac{\gy{\mu} B_0}{m_s}) \kpa - \omega \gy{u}_\shortparallel & 0\\
\omega \gy{u}_\shortparallel - (\gy{u}_\shortparallel^2 - \frac{\gy{\mu} B_0}{m_s}) \kpa & 0 & (\gy{u}_\shortparallel^2 - \frac{\gy{\mu} B_0}{m_s}) \kperp\\
0 & \frac{2 B_0 \gy{\mu} J_0 J_1}{m_s \rho} & 0
\end{pmatrix}
\diff^3 \gy{u}
\end{aligned}
\end{equation}

by making use of

\begin{equation}
\mat{B} = \begin{pmatrix}
0 & B_z & -B_y\\
-B_z & 0 & B_x\\
B_y & -B_x & 0
\end{pmatrix}
, \quad
\hb \times \vc{k} = \kperp \vc{\hat{e}}_z \times \vc{\hat{e}}_x = \kperp \vc{\hat{e}}_y.
\end{equation}

The matrix representation of the perturbed gyrocentre EOM can be written as

\begin{equation}\label{eq:gy_eomR_linpart_fourier_xz}
\Rdotmat =
\frac{J_0}{\omega}
\begin{pmatrix}
0 & \omega - \kpa \gy{u}_\shortparallel & 0\\
\kpa \gy{u}_\shortparallel - \omega & \ii \frac{2 \gy{\mu} \kperp J_1}{ q_s \rho J_0} & -\kperp \gy{u}_\shortparallel\\
0 & \kperp \gy{u}_\shortparallel & 0
\end{pmatrix}
\end{equation}

such that the perturbed gyrocentre free current matrix can be written as loop current

\begin{equation}\label{eq:fourier_currentmat_xz}
\begin{aligned}
\pertJmat
= {} &
\omega_{\mathrm{c},s} \int
\frac{\gyfzero J_0^2}{\omega B_0}
\begin{pmatrix}
0 & \omega - \kpa \gy{u}_\shortparallel & 0\\
\kpa \gy{u}_\shortparallel - \omega & \ii \frac{2 \gy{\mu} \kperp J_1}{ q_s \rho J_0} & -\kperp \gy{u}_\shortparallel\\
0 & \kperp \gy{u}_\shortparallel & 0
\end{pmatrix} \diff^3 \gy{u} \\
& + \frac{q_s}{m_s} \int
\gy{u}_\shortparallel J_0 \frac{
\omega_{\mathrm{c},s} \pard{\gyfzero}{\gy{u}_\shortparallel}
{\ii (\omega - \kpa \gy{u}_\shortparallel)}
\begin{pmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
0 & - \ii \frac{2 \gy{\mu} \kpa J_1}{\omega q_s \rho} & J_0
\end{pmatrix}
\diff^3 \gy{u}.
\end{aligned}
\end{equation}

Before we can combine the results, we must compute the curl of the magnetization, which in matrix form is given by

\begin{equation}\label{eq:fourier_curl_magnetizationmat_xz}
\mat{K} \magnetizationmat =
-\int \frac{\gyfzero}{\omega B_0}
\begin{pmatrix}
\kpa \omega \gy{u}_\shortparallel & 0 & 0\\
0 & \kpa \omega \gy{u}_\shortparallel
% + \frac{2 B_0 \gy{\mu} \kperp J_0 J_1}{m_s \rho}
& 0\\
-\kperp \omega \gy{u}_\shortparallel  & 0 & 0
\end{pmatrix}
\diff^3 \gy{u}
- \int \frac{\gyfzero}{\omega B_0} \roundpar{\gy{u}_\shortparallel^2 - \frac{\gy{\mu} B_0}{m_s}}
\begin{pmatrix}
-  \kpa^2 & 0 &  \kperp \kpa\\
0 & - \kpa^2 & 0\\
\kpa \kperp & 0 & - \kperp^2
\end{pmatrix}
\diff^3 \gy{u}
% & - \int \frac{2 \gy{\mu} J_1}{m_s \rho} \frac{
%   \omega_{\mathrm{c},s} \pard{\gyfzero}{\gy{u}_\shortparallel}
% }
% {\ii (\omega - \kpa \gy{u}_\shortparallel)}
% \begin{pmatrix}
%   0 & 0 & 0\\
%   0 & - \ii \frac{2 \gy{\mu} \kpa J_1}{\omega q_s \rho} & J_0\\
%   0 & 0 & 0
% \end{pmatrix} \diff^3 \gy{u}
\end{equation}

Substitution of <ref> into <ref> then
When substituting this into <ref> we find

\begin{equation}\label{eq:gy_susceptibility_sub}
\begin{aligned}
\gyFIXED{\mat{X}} = {} &
\sum_s \frac{\omega_{\mathrm{p},s}^2}{\omega \omega_{\mathrm{c},s}^2} \int \frac{\gyfzero}{\gy{\mathcal{N}}_{00,s}}
\begin{pmatrix}
\omega - 2\kpa \gy{u}_\shortparallel & 0 & \kperp \gy{u}_\shortparallel\\
0 & \omega - 2\kpa \gy{u}_\shortparallel & 0\\
\kperp \gy{u}_\shortparallel & 0 & 0\\
\end{pmatrix}
\jacobiangcslab \diff^3 \gy{u}\\
& - \sum_s \frac{\omega_{\mathrm{p},s}^2}{\omega^2 \omega_{\mathrm{c},s}^2} \int \frac{\gyfzero}{\gy{\mathcal{N}}_{00,s}}
\roundpar{\gy{u}_\shortparallel^2 - \frac{\gy{u}_\tau^2}{2}}
\jacobiangcslab \diff^3 \gy{u} \begin{pmatrix}
- \kpa^2 & 0 & \kperp \kpa\\
0 &- \kpa^2 & 0\\
\kpa \kperp & 0 & -\kperp^2
\end{pmatrix}\\
& + \ii \sum_s \frac{\omega_{\mathrm{p},s}^2}{\omega^2\omega_{\mathrm{c},s}} \int
\frac{\gyfzero}{\gy{\mathcal{N}}_{00,s}} J_0^2
\begin{pmatrix}
0 & \omega - \kpa \gy{u}_\shortparallel & 0\\
\kpa \gy{u}_\shortparallel - \omega & \ii \frac{2\kperp \gy{u}_\tau J_1}{J_0} & -\kperp \gy{u}_\shortparallel\\
0 & \kperp \gy{u}_\shortparallel & 0
\end{pmatrix} \jacobiangcslab \diff^3 \gy{u} \\
& - \sum_s \frac{\omega_{\mathrm{p},s}^2}{\omega^2} \int \frac{
\kpa^2 \gyfzero
{\gy{\mathcal{N}}_{00,s} (\kpa \gy{u}_\shortparallel - \omega)^2}
\begin{pmatrix}
0 & 0 & 0\\
0 & \gy{u}_\tau^2 J_1^2 & \ii \frac{\omega}{\kpa} \gy{u}_\tau J_0 J_1\\
0 & -\ii \frac{\omega}{\kpa} \gy{u}_\tau  J_0 J_1 & \frac{\omega^2}{\kpa^2} J_0^2
\end{pmatrix} \jacobiangcslab \diff^3 \gy{u},
\end{aligned}
\end{equation}

where $\jacobiangcslab = B_0 / m_s$, and the plasma frequency is given by

\begin{equation}\label{eq:plasma_frequency}
\omega_{\mathrm{p},s} = q_s \sqrt{\frac{\gy{\mathcal{N}}_{00,s}}{\epsilon_0 m_s}}.
\end{equation}

In deriving <ref> we have made use of partial integration w.r.t. $\gy{u}_\shortparallel$ and assume that $\gyfzero$ vanishes at $\gy{u}_\shortparallel = \pm \infty$.

§.§ Drift kinetic susceptibility tensor

In order to compare with the results from [Zonta et al., 2021], we now ignore FLR effects by assuming $\kperp \rho \ll 1$ such that we may approximate the Bessel functions as follows

\begin{equation}
J_0 \approx 1
, \quad
J_1 \approx \frac{\kperp \rho}{2}.
\end{equation}

This results in the following ZLR limit of <ref>

\begin{equation}\label{eq:gy_susceptibility_zlr}
\begin{aligned}
\gyFIXED{\mat{X}}^\zlr = {} &
\sum_s \frac{\omega_{\mathrm{p},s}^2}{\omega \omega_{\mathrm{c},s}^2} \int \frac{\gyfzero}{\gy{\mathcal{N}}_{00,s}}
\begin{pmatrix}
\omega - 2\kpa \gy{u}_\shortparallel & 0 & \kperp \gy{u}_\shortparallel\\
0 & \omega - 2\kpa \gy{u}_\shortparallel & 0\\
\kperp \gy{u}_\shortparallel & 0 & 0\\
\end{pmatrix}
\jacobiangcslab \diff^3 \gy{u}\\
& - \sum_s \frac{\omega_{\mathrm{p},s}^2}{\omega^2 \omega_{\mathrm{c},s}^2} \int \frac{\gyfzero}{\gy{\mathcal{N}}_{00,s}}
\roundpar{\gy{u}_\shortparallel^2 - \frac{\gy{u}_\tau^2}{2}}
\jacobiangcslab \diff^3 \gy{u} \begin{pmatrix}
- \kpa^2 & 0 & \kpa \kperp\\
0 &- \kpa^2 & 0\\
\kpa \kperp & 0 & -\kperp^2
\end{pmatrix}\\
& + \ii \sum_s \frac{\omega_{\mathrm{p},s}^2}{\omega^2 \omega_{\mathrm{c},s}^2} \int
\frac{\gyfzero \jacobiangcslab}{\gy{\mathcal{N}}_{00,s}}
\begin{pmatrix}
0 & \omega_{\mathrm{c},s} (\omega - \kpa \gy{u}_\shortparallel) & 0\\
-\omega_{\mathrm{c},s} (\omega - \kpa \gy{u}_\shortparallel) & \ii \kperp^2 \gy{u}_\tau^2 & - \omega_{\mathrm{c},s} \kperp \gy{u}_\shortparallel\\
0 & \omega_{\mathrm{c},s} \kperp \gy{u}_\shortparallel & 0
\end{pmatrix} \diff^3 \gy{u} \\
& - \sum_s \frac{\omega_{\mathrm{p},s}^2}{\omega^2} \int \frac{
\kpa^2 \gyfzero
{\gy{\mathcal{N}}_{00,s} (\kpa \gy{u}_\shortparallel - \omega)^2}
\begin{pmatrix}
0 & 0 & 0\\
0 & \gy{u}_\tau^4 \kperp^2 \frac{1}{4 \omega_{\mathrm{c},s}^2} & \ii \frac{\kperp}{\kpa} \gy{u}_\tau^2 \frac{\omega}{2 \omega_{\mathrm{c},s}}\\
0 & -\ii \gy{u}_\tau^2 \frac{\kperp}{\kpa} \frac{\omega}{2 \omega_{\mathrm{c},s}} & \frac{\omega^2}{\kpa^2}
\end{pmatrix} \jacobiangcslab \diff^3 \gy{u}.
\end{aligned}
\end{equation}

It can be verified that this agrees with <cit.>.

Component wise this yields

\begin{align}
\gyFIXED{\matcmp{X}}_{xx}^\zlr &=
\sum_s \frac{\omega_{\mathrm{p},s}^2}{\omega_{\mathrm{c},s}^2} \int \frac{\gyfzero}{\gy{\mathcal{N}}_{00,s}}
\squarepar{\roundpar{1 - \frac{\kpa \gy{u}_\shortparallel}{\omega}}^2 - \frac{\kpa^2 \gy{u}_\tau^2}{2 \omega^2}}
\jacobiangcslab \diff^3 \gy{u}\\
\gyFIXED{\matcmp{X}}_{xy}^\zlr &=
\ii \sum_s \frac{\omega_{\mathrm{p},s}^2}{\omega \omega_{\mathrm{c},s}} \int
\frac{\gyfzero}{\gy{\mathcal{N}}_{00,s}} \roundpar{1 - \frac{\kpa \gy{u}_\shortparallel}{\omega}} \jacobiangcslab \diff^3 \gy{u}\\
\gyFIXED{\matcmp{X}}_{xz}^\zlr &=
\sum_s \frac{\omega_{\mathrm{p},s}^2}{\omega_{\mathrm{c},s}^2} \int \frac{\gyfzero}{\gy{\mathcal{N}}_{00,s}} \frac{k_\perp}{\kpa}
\squarepar{
\frac{\kpa \gy{u}_\shortparallel}{\omega} \roundpar{1 - \frac{\kpa \gy{u}_\shortparallel}{\omega}}
+ \frac{\gy{u}_\tau^2}{2} \frac{\kpa^2}{\omega^2}
\jacobiangcslab \diff^3 \gy{u}\\
\gyFIXED{\matcmp{X}}_{yx}^\zlr &= - \gyFIXED{\matcmp{X}}_{xy}^\zlr\\
\gyFIXED{\matcmp{X}}_{yy}^\zlr &=
\gyFIXED{\matcmp{X}}_{xx}^\zlr
- \sum_s \frac{\omega_{\mathrm{p},s}^2}{\omega_{\mathrm{c},s}^2} \int \frac{\gyfzero}{\gy{\mathcal{N}}_{00,s}} \roundpar{
\frac{\kperp^2 \gy{u}_\tau^2}{\omega^2}
+ \frac{1}
{(\gy{u}_\shortparallel - \omega / \kpa)^2}
\frac{\kperp^2 \gy{u}_\tau^4}{4 \omega^2}} \jacobiangcslab \diff^3 \gy{u}\\
\gyFIXED{\matcmp{X}}_{yz}^\zlr &=
-\ii \sum_s \frac{\omega_{\mathrm{p},s}^2}{\omega \omega_{\mathrm{c},s}} \int
\frac{\gyfzero}{\gy{\mathcal{N}}_{00,s}} \roundpar{
\frac{\kperp \gy{u}_\shortparallel}{\omega}
+ \frac{\kperp \gy{u}_\tau^2}{2 \kpa (\gy{u}_\shortparallel - \omega / \kpa)^2}
} \jacobiangcslab \diff^3 \gy{u} \\
\gyFIXED{\matcmp{X}}_{zx}^\zlr &= \gyFIXED{\matcmp{X}}_{xz}^\zlr\\
\gyFIXED{\matcmp{X}}_{zy}^\zlr &= - \gyFIXED{\matcmp{X}}_{yz}^\zlr\\
\gyFIXED{\matcmp{X}}_{zz}^\zlr &=
\sum_s \frac{\omega_{\mathrm{p},s}^2}{\omega^2 \omega_{\mathrm{c},s}^2} \int \frac{\gyfzero}{\gy{\mathcal{N}}_{00,s}}
\roundpar{\gy{u}_\shortparallel^2 - \frac{\gy{u}_\tau^2}{2}}
\kperp^2 \jacobiangcslab \diff^3 \gy{u}
- \sum_s \int \frac{\frac{\gyfzero}{\gy{\mathcal{N}}_{00,s}}}
{(\gy{u}_\shortparallel - \omega / \kpa)^2} \frac{\omega_{\mathrm{p},s}^2}{\kpa^2} \jacobiangcslab \diff^3 \gy{u}
\end{align}

§.§ Gyrokinetic Darwin susceptibility tensor

This is very similar to the gauge-invariant gyrokinetic model, except that we must express the additional term in the Darwin polarization (<ref>) in terms of the electric field $\fourier{\vc{E}}_1$.
That is, we must express the perpendicular part of the vector potential in terms of the electromagnetic fields.

From the definition of the electric field (<ref>) we find that

\begin{equation}
\fourier{\vc{E}}_1 = -\ii \vc{k} \fourier{\phi}_1 + \ii \omega \fourier{\vc{A}}_1
\quad \implies \quad
\vc{k}_\perp \bcdot \fourier{\vc{E}}_1 = -\ii \kperp^2 \fourier{\phi}_1,
\end{equation}

after having substituted the Fourier ansatz as well as the constraint (<ref>), $\vc{k}_\perp \bcdot \fourier{\vc{A}}_1 = 0$.
It follows that the perpendicular part of the vector potential is given by

\begin{equation}
\fourier{\vc{A}}_{1,\perp} = \frac{\fourier{\vc{E}}_{1,\perp} \kperp^2 - \vc{k}_\perp \vc{k}_\perp \bcdot \fourier{\vc{E}}_1}{\ii \omega \kperp^2}.
\end{equation}

Using the specific parallel and perpendicular directions given by <ref> we find

\begin{equation}\label{eq:darwin_A_as_E_xz}
\fourier{\vc{A}}_{1,\perp}
\mat{A}_\perp \fourier{\vc{E}}_1
, \quad
\mat{A}_\perp =
\begin{pmatrix}
0 & 0 & 0\\
0 & -\frac{\ii}{ \omega} & 0\\
0 & 0 & 0
\end{pmatrix},
\end{equation}

such that when considering <ref> we find that the gyrokinetic Darwin susceptibility tensor is given by

\begin{equation}
\gyFIXED{\mat{X}}^\darwin = \gyFIXED{\mat{X}} - \ii \omega \sum_s \frac{\omega_{\mathrm{p},s}^2}{\omega_{\mathrm{c},s}^2} \mat{A}_\perp,
\end{equation}

where we have moreover substituted the definition of the plasma frequency (<ref>).

We do not discuss the removal of the $\hb \bcdot (\vc{E}_1 \times \vc{B}_1)$ term from $\gy{H}_{2,s}$ as this does not affect the susceptibility tensor in case of a centred Maxwellian background distribution.

natexlab#1#1#1#1 #1#1 #1#1#1#1#1#1 #1#1#1#1 #1#1 #1#1 #1#1#1#1#1#1#1#1

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