Generation of geodesic acoustic mode by nonlinear coupling of magnetic island and island-driven beta-induced Alfvén eigenmode

The geodesic acoustic mode (GAM) is a high frequency (few tens kHz) zonal flow induced by the component of the field line curvature tangential to the magnetic surface [1]. The GAM has attracted much attention during last decades due to the ability to regulate anomalous transport via turbulence suppression [2]. It can be excited parametrically by plasma turbulence [3] as well as by energetic ions [4] (in the latter case, there exists also a specific GAM branch, so called EGAM, induced by energetic ions itself [5]). It is possible to excite the GAM by an external antenna [6].

Recently, axis-symmetric (i.e. with zero toroidal mode number, n = 0) modes identified as GAM, which accompanied standing beta-induced Alfvén eigenmodes (BAE) [11, 12] driven by magnetic islands, have been observed on HL-2A tokamak [9]. This observation motivated present communication. Below we demonstrate that this GAM has frequency (in the island frame) and mode structure identical to those of BAE, and calculate the GAM/BAE amplitude ratio.

We proceed introducing the toroidal sidebands of the axis-symmetric (A) and helical (B) electron response

$$\begin{eqnarray*}&&{f}_{e}^{(A)}(x,\theta ,t)=[{f}_{e}^{c(A)}(x)\cos \theta \\ &&\quad +{f}_{e}^{s(A)}(x)\sin \theta ]\exp (-{\rm{i}}\omega t),\\ &&\quad {f}_{e}^{(B)}(x,\xi ,\theta ,t)=[{f}_{e}^{c(B)}(x,\xi )\cos \theta \\ &&\quad +{f}_{e}^{s(B)}(x,\xi )\sin \theta ]\exp (-{\rm{i}}\omega t),\end{eqnarray*}$$

where x is the distance from the rational surface, θ is the poloidal angle, $\xi =m\theta -n\zeta -{\omega }^{(I)}t$ is the island phase, and upper index c(s) denotes amplitude of the up-down symmetric (asymmetric) toroidal sideband. Note that the choice of the island phase in the BAE response is natural, since island-driven BAE observed in experiments [7–10] have the same helicity as magnetic island, and form standing wave in the island frame. For this reason, up-down symmetric part of the electron response to BAE generates up-down asymmetric, axis-symmetric perturbation of the electron distribution via magnetic flutter in the dominant parallel streaming term of the kinetic equation

$$\begin{eqnarray}&&-{\rm{i}}\omega {f}_{e}^{c(A)}+\displaystyle \frac{{v}_{\parallel }}{{qR}}{f}_{e}^{s(A)}\\ &&+\,{v}_{\parallel }{\left\langle \displaystyle \frac{\delta {B}_{r}^{(I)}(x,\xi )}{{B}_{0}}\displaystyle \frac{\partial {f}_{e}^{c(B)}(x,\xi )}{\partial x}\right\rangle }_{\xi }=0,\end{eqnarray} \tag{ 1 }$$

where $\delta {B}_{r}^{(I)}$ is the perturbation of the radial magnetic field induced by the island, and

$$\begin{eqnarray*}&&\langle ...{\rangle }_{\xi }\equiv \displaystyle \frac{1}{2\pi }{\int }_{-\pi }^{\pi }(...){\rm{d}}\xi .\end{eqnarray*}$$

Under the condition $\omega \ll {v}_{{Te}}/{qR}$, with v_Te the electron thermal velocity, the solution of equation (1) is given by

$$\begin{eqnarray}&&{f}_{e}^{s(A)}=-{qR}{\left\langle \displaystyle \frac{\delta {B}_{r}^{(I)}}{{B}_{0}}\displaystyle \frac{\partial {f}_{e}^{c(B)}}{\partial x}\right\rangle }_{\xi }.\end{eqnarray} \tag{ 2 }$$

In the previous work [13], we have neglected ${f}_{e}^{c(B)}\propto {k}_{\theta }$ assuming that $\partial /\partial x\gg {k}_{\theta }$ in the inertial layer around the rational surface, a reasonable assumption for the almost 'zonal' BAE with ${k}_{\theta }{\rm{\Delta }}\ll 1$, where Δ is the BAE half-width defined shortly. Below we calculate this sideband, which contributes to equation (2) and to the corresponding flux-surface averaged radial current ${J}_{r}^{e(A)}$.

The sideband ${f}_{e}^{c(B)}$ satisfies the kinetic equation

$$\begin{eqnarray}&&-{\rm{i}}\omega {f}_{e}^{s(B)}-\displaystyle \frac{{v}_{\parallel }}{{qR}}{f}_{e}^{c(B)}=-\displaystyle \frac{{{eE}}_{\parallel }^{s(B)}}{{T}_{e}}{v}_{\parallel }{F}_{{Me}},\end{eqnarray} \tag{ 3 }$$

where $e\gt 0$ is the elementary charge, ${E}_{\parallel }^{s}(B)$ is the sideband of the parallel electric field, and F_Me is the equilibrium (Maxwellian) distribution function. In the limit $\omega \ll {v}_{{Te}}/{qR}$, equation (3) yields

$$\begin{eqnarray}&&{f}_{e}^{c(B)}=\displaystyle \frac{{{eqRE}}_{\parallel }^{s(B)}}{{T}_{e}}{F}_{{Me}}.\end{eqnarray} \tag{ 4 }$$

Note that neglect of the ${f}_{e}^{s(B)}$ in equation (3) is justified provided $1\gg {k}_{\theta }{\rm{\Delta }}\gg \sqrt{{m}_{e}/{m}_{i}}$.

The sideband ${E}_{\parallel }^{s(B)}$ can be found from the neutrality condition for the sideband density perturbations. For the ions we have

$$\begin{eqnarray}&&-{\rm{i}}\omega {f}_{{i}}^{c(B)}+\displaystyle \frac{{v}_{\parallel }}{{qR}}{f}_{{i}}^{s(B)}=-\displaystyle \frac{c}{{B}_{0}R}{{\rm{i}}{k}}_{\theta }{{\rm{\Phi }}}^{(B)}\displaystyle \frac{{m}_{{i}}}{{T}_{{i}}}\left(\displaystyle \frac{{v}_{\perp }^{2}}{2}+{v}_{\parallel }^{2}\right){F}_{{Mi}},\end{eqnarray} \tag{ 5 }$$

where ${{\rm{\Phi }}}^{(B)}$ is the BAE electric potential. In the limit $\omega \gg {v}_{{Ti}}/{qR}$, equation (5) yields

$$\begin{eqnarray}&&{f}_{i}^{c(B)}=\displaystyle \frac{{{cm}}_{i}}{\omega {B}_{0}R}\displaystyle \frac{{F}_{{Mi}}}{{T}_{i}}\left(\displaystyle \frac{{v}_{\perp }^{2}}{2}+{v}_{\parallel }^{2}\right){k}_{\theta }{{\rm{\Phi }}}^{(B)}.\end{eqnarray} \tag{ 6 }$$

Equations (4) and (6), and neutrality condition ${n}_{e}^{c(B)}={n}_{i}^{c(B)}$ yield

$$\begin{eqnarray}&&\displaystyle \frac{{eqR}}{{T}_{e}}{E}_{\parallel }^{s(B)}=\displaystyle \frac{2c}{\omega {B}_{0}R}{k}_{\theta }{{\rm{\Phi }}}^{(B)},\end{eqnarray} \tag{ 7 }$$

$$\begin{eqnarray}&&{f}_{e}^{s(A)}=-\displaystyle \frac{2{qc}}{\omega {B}_{0}}{k}_{\theta }{\left\langle \displaystyle \frac{\delta {B}_{r}^{(I)}}{{B}_{0}}\displaystyle \frac{\partial {{\rm{\Phi }}}^{(B)}}{\partial x}\right\rangle }_{\xi }{F}_{{Me}},\end{eqnarray} \tag{ 8 }$$

$$\begin{eqnarray}{\langle {J}_{r}^{e(A)}\rangle }_{\theta } & \equiv & -\displaystyle \frac{e}{2\pi }{\displaystyle \int }_{0}^{2\pi }{\rm{d}}\theta \displaystyle \int {{\rm{d}}}^{3}{{vf}}_{e}^{s(A)}\sin \theta {V}_{{Dr}}^{e}\\ & = & \displaystyle \frac{2{{qn}}_{0}{T}_{e}}{\omega R}{\left(\displaystyle \frac{c}{{B}_{0}}\right)}^{2}{k}_{\theta }{\left\langle \displaystyle \frac{\delta {B}_{r}^{(I)}}{{B}_{0}}\displaystyle \frac{\partial {{\rm{\Phi }}}^{(B)}}{\partial x}\right\rangle }_{\xi },\end{eqnarray} \tag{ 9 }$$

where ${V}_{{Dr}}^{e}\propto \sin \theta$ is the radial component of the electron magnetic drift.

To keep the charge neutrality, the plasma responds to the current (9) by generation of a geodesic acoustic mode (G). Corresponding expressions for the flux-surface averaged radial currents are well known (e.g. see [14–16]) and given by

$$\begin{eqnarray}&&{\langle {J}_{r}^{e(G)}\rangle }_{\theta }=-2{\rm{i}}\displaystyle \frac{{n}_{0}{T}_{e}}{\omega {R}^{2}}{\left(\displaystyle \frac{c}{{B}_{0}}\right)}^{2}\displaystyle \frac{\partial {{\rm{\Phi }}}^{(G)}}{\partial x},\end{eqnarray} \tag{ 10 }$$

$$\begin{eqnarray}&&{\langle {J}_{r}^{i(G)}\rangle }_{\theta }=-\displaystyle \frac{7}{2}{\rm{i}}\displaystyle \frac{{n}_{0}{T}_{i}}{\omega {R}^{2}}{\left(\displaystyle \frac{c}{{B}_{0}}\right)}^{2}\displaystyle \frac{\partial {{\rm{\Phi }}}^{(G)}}{\partial x},\end{eqnarray} \tag{ 11 }$$

$$\begin{eqnarray}&&{J}_{p}={\rm{i}}\omega \displaystyle \frac{{c}^{2}}{4\pi {V}_{A}^{2}}\displaystyle \frac{\partial {{\rm{\Phi }}}^{(G)}}{\partial x}.\end{eqnarray} \tag{ 12 }$$

Equations (10) and (11) represent radial currents of electrons and ions, respectively, both induced by the geodesic plasma compressibility, whereas equation (12) is the radial polarization current. Balance of the all radial currents, ${J}_{p}+{\langle {J}_{r}^{i(G)}\rangle }_{\theta }\,+{\langle {J}_{r}^{e(G)}\rangle }_{\theta }+{\langle {J}_{r}^{e(A)}\rangle }_{\theta }=0$, yields

$$\begin{eqnarray}&&({\omega }^{2}-{\omega }_{G}^{2})\displaystyle \frac{\partial {\hat{{\rm{\Phi }}}}^{(G)}}{\partial x}={\rm{i}}{q}\displaystyle \frac{{T}_{e}}{{m}_{i}}\displaystyle \frac{{k}_{\theta }}{R}\displaystyle \frac{\delta {\hat{B}}_{r}^{(I)}}{{B}_{0}}\displaystyle \frac{\partial {\hat{{\rm{\Phi }}}}^{(B)}}{\partial x},\end{eqnarray} \tag{ 13 }$$

where hats denote the mode amplitude. In the 'constant Ψ' approximation, $({{\rm{\Delta }}}^{\prime }w,{{\rm{\Delta }}}^{\prime }{\rm{\Delta }})\ll 1$, where ${{\rm{\Delta }}}^{\prime }\sim {k}_{\theta }$ is the tearing mode stability index and w is the magnetic island half-width, we have $\delta {\hat{B}}_{r}^{(I)}(x)\simeq {\rm{const}}$. Then it follows from equation (13) that the generated secondary GAM has a mode structure identical to that of the primary BAE. When the ion and electron Landau damping are neglected, the latter is determined by the BAE eigenmode equation [13]

$$\begin{eqnarray}&&\displaystyle \frac{3}{4}{\rho }_{i}^{2}{\omega }^{2}\displaystyle \frac{{{\rm{d}}}^{4}{\hat{{\rm{\Phi }}}}^{(B)}}{{{\rm{d}}{x}}^{4}}+\displaystyle \frac{{\rm{d}}}{{\rm{d}}{x}}\\ &&\ \ \ \times \left[\left({\omega }^{2}-{\omega }_{G}^{2}-{k}_{\theta }^{2}{V}_{A}^{2}\displaystyle \frac{{x}^{2}}{{L}_{s}^{2}}\right)\displaystyle \frac{{\rm{d}}{\hat{{\rm{\Phi }}}}^{(B)}}{{\rm{d}}{x}}\right]+{k}_{\theta }^{4}{V}_{A}^{2}\displaystyle \frac{{x}^{2}}{{L}_{s}^{2}}{\hat{{\rm{\Phi }}}}^{(B)}=0.\end{eqnarray} \tag{ 14 }$$

The eigenvalues of this kinetic, tearing parity BAE versus the poloidal mode number are shown in figure 1(a), and corresponding mode structures are shown in figure 1(b). Here, $d\equiv {k}_{\theta }{\rm{\Delta }}$ with the BAE half-width Δ given by

$$\begin{eqnarray}&&{\rm{\Delta }}={\left[\displaystyle \frac{3}{4}{\left(\displaystyle \frac{\omega {\rho }_{i}{L}_{s}}{{k}_{\theta }{V}_{A}}\right)}^{2}\right]}^{1/4},\end{eqnarray} \tag{ 15 }$$

where ${\rho }_{i}({L}_{s})$ is the ion gyro-radius (magnetic shear length). Furthermore,

$$\begin{eqnarray}&&{{\rm{\Omega }}}^{2}=\displaystyle \frac{{\omega }^{2}-{\omega }_{G}^{2}}{{k}_{\theta }^{2}{V}_{A}^{2}}\displaystyle \frac{{L}_{s}^{2}}{{{\rm{\Delta }}}^{2}},\end{eqnarray} \tag{ 16 }$$

$$\begin{eqnarray}&&{\omega }_{G}^{2}\equiv \displaystyle \frac{2{v}_{{Ti}}^{2}}{{R}^{2}}\left(\displaystyle \frac{7}{4}+\displaystyle \frac{{T}_{e}}{{T}_{i}}\right).\end{eqnarray} \tag{ 17 }$$

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From figure 1 and equation (16), we have for ${k}_{\theta }{\rm{\Delta }}\ll 1$ and for the lowest frequency mode (shown in blue in figure 1)

$$\begin{eqnarray}&&{\omega }^{2}-{\omega }_{G}^{2}\simeq 3{k}_{\theta }^{2}{V}_{A}^{2}\displaystyle \frac{{{\rm{\Delta }}}^{2}}{{L}_{s}^{2}}.\end{eqnarray} \tag{ 18 }$$

The perturbation of the radial magnetic field induced by the island is given by

$$\begin{eqnarray}&&\delta {\hat{B}}_{r}^{(I)}={({\bf{b}}\times {\rm{\nabla }}\psi )}_{r}={{\rm{i}}{k}}_{\theta }\displaystyle \frac{{B}_{0}{w}^{2}}{4{L}_{s}},\end{eqnarray} \tag{ 19 }$$

where ψ is the helical flux, and we have used the fact that the island and the primary BAE have the same mode number. Combining equations (13)–(19), we arrive at the final expression for the GAM/BAE radial electric field amplitude ratio necessary to compensate the flux-surface averaged radial current (9) induced by the island-BAE coupling

$$\begin{eqnarray}&&\displaystyle \frac{{\hat{E}}_{r}^{(G)}}{{\hat{E}}_{r}^{(B)}}=-\displaystyle \frac{{q}^{2}}{12\hat{s}}{\beta }_{e}{\left(\displaystyle \frac{w}{{\rm{\Delta }}}\right)}^{2},\end{eqnarray} \tag{ 20 }$$

where $\hat{s}$ is the magnetic shear and ${\beta }_{e}$ is the electron beta. As expected, to keep the plasma neutrality, the phase of the secondary GAM is shifted by π with respect to the primary BAE phase. It is difficult to infer about the actual value of $w/{\rm{\Delta }}$ in the experiment [9]. However, the color code in figure 3 of [9] implies that intensity of the secondary GAM is about two orders of magnitude smaller than that of the primary BAE, qualitatively consistent with equation (20).

In summary, coupling of the magnetic island with its 'daughter' BAE generates axis-symmetric, up-down asymmetric electron response. As a result, finite flux-surface averaged radial current is generated. To keep the neutrality, plasma responds to this current by GAM generation. This secondary GAM has the same frequency and mode structure as the primary BAE. The GAM/BAE amplitude ratio given by equation (20) can be comparable to the plasma/magnetic pressure ratio.

Acknowledgments