2 Kinetic equation for the duct plasma

The basic equations are the linearized Vlasov equation and the Poisson equation for electrons. Ions are assumed to be stationary and only form the neutralizing background,

(2.1)\begin{equation} \frac{\partial}{\partial t} f({\boldsymbol r}, {\boldsymbol v},t)+{\boldsymbol v}\boldsymbol{\cdot}\frac{\partial f}{\partial {\boldsymbol r}}- \frac{e}{mc}{\boldsymbol v}\times{\boldsymbol B_0}\boldsymbol{\cdot} \frac{\partial f}{\partial{\boldsymbol v}}-\frac{e}{m}{\boldsymbol E}\boldsymbol{\cdot}\frac{\partial f_0}{\partial{\boldsymbol v}}=0,\end{equation}

with $B_0=\hat {{\boldsymbol z}}B_0$ and

(2.2)\begin{gather} \boldsymbol{\nabla}\boldsymbol{\cdot}{\boldsymbol E}={-}4{\rm \pi} e\int f \, {\rm d}^3v+S(x,y,z), \end{gather}

(2.3)\begin{gather}\boldsymbol{\nabla}\times{\boldsymbol E}=0. \end{gather}

In (2.2), $S(x,y,z)$ represents the fictitious charge sheets as defined in (1.1). Additionally, $f({\boldsymbol r}, {\boldsymbol v},t)$ and $f_0({\boldsymbol v})$ are the perturbation and the equilibrium distribution functions, respectively, and the rest of the symbols are standard.

We Fourier transform the above equations by performing $\int _{-\infty }^\infty d^3r~\, {\rm e}^{{\rm i}{\boldsymbol k}\boldsymbol {\cdot }{\boldsymbol r}}(\cdots )$ and by assuming $\partial /\partial t=-{\rm i}\omega$. Then, the wave has a phasor ${\rm e}^{{\rm i}k_zz-{\rm i}\omega t}$. The $(k_z,\omega )$ dependency in the Fourier amplitudes will be suppressed. Using (2.2) and (2.3), we can write

(2.4)\begin{equation} {\boldsymbol E}({\boldsymbol k},\omega)={\rm i}\frac{{\boldsymbol k}}{k^2}\left[ 4{\rm \pi} e \int f({\boldsymbol k},\omega,{\boldsymbol v})\, {\rm d}^3v+S ({\boldsymbol k})\right],\end{equation}

where $S({\boldsymbol k})$ is the Fourier transform of (1.1):

(2.5)\begin{align} S({\boldsymbol k})& =\delta(k_y)[A_1\varSigma_0\, {\rm e}^{{\pm} {\rm i}2nak_x}+A_2\varSigma_1\, {\rm e}^{{\pm} {\rm i}(2n-1)ak_x}]\nonumber\\ & \quad +\delta(k_x)[B_1\varSigma_0\, {\rm e}^{{\pm} {\rm i}2nbk_y}+B_2\varSigma_1\, {\rm e}^{{\pm} {\rm i}(2n-1)bk_y}], \end{align}

where $A$ and $B$ terms may be functions of $k_z, \omega$, the double signs are summed over, and the notation $\varSigma _0$ and $\varSigma _1$ are the summations in (1.1). Introducing cylindrical coordinates in the velocity space such that $v_x=v_\perp \cos \varphi$, $v_y=v_\perp \sin \varphi$, (2.1) is written in the form

(2.6)\begin{align} f({\boldsymbol k}, {\boldsymbol v},\omega)=\frac{e}{m\omega_c}\int_{-\infty}^\varphi \,{\rm d}\varphi' \exp[\varPhi(\varphi)-\varPhi(\varphi')] \left[E_z\frac{\partial f_0}{\partial v_z}+(E_x cos\varphi'+E_y \sin\varphi')\frac{\partial f_0}{\partial v_\perp}\right],\end{align}

where $\omega _c=eB_0/mc$, the gyrofrequency, and

(2.7)\begin{equation} \varPhi(\varphi)=\frac{{\rm i}}{\omega_c}[(\omega-k_zv_z)\varphi-v_\perp k_x \sin\varphi+v_\perp k_y \cos\varphi]. \end{equation}

The $\int {\rm d}\varphi '$ integral can be carried out by introducing a Bessel function series and we can write (2.4) in the form

(2.8)\begin{equation} E_z({\boldsymbol k},\omega)=\frac{{\rm i}k_z}{k^2\varepsilon_L} S({\boldsymbol k}),\end{equation}

where we used $E_x=({k_x}/{k_z})E_z, E_y=({k_y}/{k_z})E_z$, and the dielectric permittivity $\varepsilon _L$ is

(2.9)\begin{align} \varepsilon_L& =1-\frac{i}{k^2}\frac{\omega_p^2}{\omega_c} \int_0^\infty v_\perp \, {\rm d} v_\perp\int_{-\infty}^\infty {\rm d} v_z \int_0^{2{\rm \pi}} {\rm d}\varphi \int_{-\infty}^\varphi \exp \frac{{\rm i}}{\omega_c}[\varPhi(\varphi)-\varPhi(\varphi')]\nonumber\\ & \quad \times \left[k_z \frac{\partial f_0}{\partial v_z}+ \frac{\partial f_0}{\partial v_\perp}(k_x\cos \varphi'+k_y\sin\varphi')\right]. \end{align}

We shall evaluate the velocity integral in the above equation for a cold plasma distribution function

(2.10)\begin{equation} f_0(v_\perp,v_z)=\frac{\delta(v_\perp)}{2{\rm \pi} v_\perp}\delta(v_z). \end{equation}

The angular integrals should be performed first to remove the singularity $1/v_\perp$ in the velocity integral. Following the well-known steps involving Bessel function series expansion, $\varepsilon _L$ becomes after $\int {\rm d}\varphi \int {\rm d}\varphi '$ integration,

(2.11)\begin{equation} \varepsilon_L=1+\frac{\omega_p^2}{k^2} \int_0^\infty v_\perp \, {\rm d} v_\perp\int_{-\infty}^\infty {\rm d} v_z \sum_n\frac{J_n^2(a_e)}{\omega-k_zv_z-n\omega_c} \left(k_z\frac{\partial f_0}{\partial v_z} +\frac{n\omega_c}{v_\perp}\frac{\partial f_0}{\partial v_\perp}\right),\end{equation}

where $\omega _p$ is the plasma frequency and $a_e=k_\perp v_\perp /\omega _c$, $k_\perp =\sqrt {k_x^2+k_y^2}$.

In evaluating the $\partial f_0/\partial v_z$ term in (2.11), $\int dv_\perp$ can be straightforwardly carried out by taking the limit $a_e \rightarrow 0$, which gives $J_n(0)\rightarrow \delta _{n,0}$. Also, by integrating by parts with respect to $\partial /\partial v_z$, one obtains

(2.12)\begin{equation} -\frac{\omega_p^2}{k^2} \frac{k_z^2}{\omega^2}. \end{equation}

For the term $\partial f_0/\partial v_\perp$, integrating by parts and using

(2.13)\begin{equation} \frac{\partial}{\partial v_\perp}J_n^2(a_e)=2 \frac{k_\perp}{\omega_c}J_n(a_e)J_n'(a_e) \end{equation}

and the asymptotic relation

(2.14)\begin{equation} \lim_{a\rightarrow0}\frac{2n}{a}J_n(a)J_n'(a)\rightarrow \frac{1}{2}(\delta_{n,1}-\delta_{n.-1}) \end{equation}

yield

(2.15)\begin{equation} -\frac{\omega_p^2}{k^2}\frac{k_\perp^2}{\omega^2-\omega_c^2}. \end{equation}

Collecting the above results, the cold plasma dielectric permittivity is obtained as

(2.16)\begin{equation} \varepsilon_L=1-\frac{\omega_p^2}{k^2} \left(\frac{k_z^2}{\omega^2}+\frac{k_\perp^2}{\omega^2-\omega_c^2}\right). \end{equation}

Next, we calculate the electric displacement

(2.17)\begin{equation} {\boldsymbol D}({\boldsymbol k},\omega)={\boldsymbol E}({\boldsymbol k},\omega)+\frac{4{\rm \pi} i}{\omega} {\boldsymbol J}({\boldsymbol k},\omega),\end{equation}

where

(2.18)\begin{equation} {\boldsymbol J}({\boldsymbol k},\omega)={-}{\rm e}\int {\rm d}^3v {\boldsymbol v} f({\boldsymbol k},\omega,{\boldsymbol v})\end{equation}

is the current. Using (2.6) and (2.18), we write the $x$ component in (2.17) as an integral

(2.19)\begin{align} D_x& =E_x-\frac{4{\rm \pi} e^2}{m} \frac{i}{\omega_c\omega}\int_0^\infty v_\perp^2 \,{\rm d} v_\perp\int_{-\infty}^\infty dv_z\int_0^{2{\rm \pi}}{\rm d} \varphi \cos\varphi\nonumber\\ & \quad \times\int_{-\infty}^\varphi {\rm d} \varphi' \exp \frac{i}{\omega_c}[(\omega-k_zv_z)(\varphi-\varphi')-v_\perp k_x(\sin\varphi-\sin\varphi') \nonumber\\ & \quad +v_\perp k_y(\cos\varphi-\cos\varphi')]\times \left[E_z \frac{\partial f_0}{\partial v_z}+ \frac{\partial f_0}{\partial v_\perp} (E_x \cos\varphi'+E_y \sin\varphi')\right]. \end{align}

In the above integral, the $\partial f_0/\partial v_z$ term vanishes due to the $\delta$-function, $\delta (v_\perp )$. Integrating by parts with respect to $\partial /\partial v_\perp$, (2.19) becomes

(2.20)\begin{align} D_x=E_x-\frac{4{\rm \pi} e^2}{m} \frac{2i}{\omega_c\omega} \int_0^{2{\rm \pi}}{\rm d}\varphi \cos\varphi\, {\rm e}^{{\rm i}({\omega}/{\omega_c})\varphi}\int_{-\infty}^\varphi {\rm d} \varphi'\, {\rm e}^{-{\rm i}\varphi'({\omega}/{\omega_c})} (E_x\cos\varphi'+E_y\sin\varphi'). \end{align}

In the above equation, integrals are elementary and can be easily carried out. Entirely analogous algebra yields the expression for $D_y$. Summarizing the results, we write

(2.21)\begin{gather} D_x=\varepsilon_xE_x={\rm i} \frac{k_x}{k^2} \frac{\varepsilon_x}{\varepsilon_L}S({\boldsymbol k}), \end{gather}

(2.22)\begin{gather}D_y=\varepsilon_yE_y={\rm i} \frac{k_y}{k^2} \frac{\varepsilon_y}{\varepsilon_L}S({\boldsymbol k}), \end{gather}

where we used (2.8), and the dielectric coefficients $\varepsilon _{x,y}$ are

(2.23)\begin{gather} \varepsilon_x=1-\frac{\omega_p^2}{\omega^2-\omega_c^2}+ \frac{{\rm i}k_y}{k_x} \frac{\omega_c}{\omega} \frac{\omega_p^2}{\omega^2-\omega_c^2}, \end{gather}

(2.24)\begin{gather}\varepsilon_y=1-\frac{\omega_p^2}{\omega^2-\omega_c^2}- \frac{{\rm i}k_x}{k_y}\frac{\omega_c}{\omega} \frac{\omega_p^2}{\omega^2-\omega_c^2}. \end{gather}

We have shown that, for a cold plasma, $\varepsilon _{x,y}$ can be obtained without going through the Bessel function series expansion. Equations (2.8), (2.21) and (2.22) will be connected to the corresponding vacuum side field components to derive the boundary equations.

3 Boundary equations for the duct plasma waves

To apply the boundary conditions on the interface, the electric field and the displacement vector components expressed in the Fourier ${\boldsymbol k}$ space should be inverted to the fields in the ordinary ${\boldsymbol r}$ space by performing $\int _{-\infty }^\infty {\rm d} k_x\int _{-\infty }^\infty {\rm d} k_y\, {\rm e}^{{\rm i}k_xx+{\rm i}k_yy}(\cdots )$. The integrals involve infinite series through the surface charge $S({\boldsymbol k})$ (see (2.5)), but the infinite series are nicely summed at the particular positions corresponding to $x=0, a$ and $y=0, b$ (Lee & Cho Reference Lee and Cho2022; Lee & Lim Reference Lee and Lim2022). Thus, we apply the boundary conditions along the two infinite lines: $(x,y,z)=(0,0,z)$ and $(a,b,z)$ with $-\infty < z<\infty$. The two lines correspond to the two seams of the duct which are diagonally opposite. In the following, we present only the wide steps of the development; the details are referred to the earlier reports (Lee & Cho Reference Lee and Cho2022; Lee & Lim Reference Lee and Lim2022):

(3.1)\begin{align} E_z(0,0,z)& =\lim_{x=0,y=0}\int_{-\infty}^\infty {\rm d} k_x \, {\rm e}^{{\rm i}k_xx}\int_{-\infty}^\infty {\rm d} k_y \, {\rm e}^{{\rm i}k_yy}\frac{{\rm i}k_z}{k^2\varepsilon_L}\nonumber\\ & \quad \times [\delta(k_y)(A_1\varSigma_0\, {\rm e}^{{\pm} {\rm i}2nak_x}+A_2\varSigma_1\, {\rm e}^{{\pm} {\rm i}(2n-1)ak_x})\nonumber\\ & \quad +\delta(k_x) (B_1\varSigma_0\, {\rm e}^{{\pm} {\rm i}2nbk_y}+B_2\varSigma_1\, {\rm e}^{{\pm} {\rm i}(2n-1)bk_y})]\nonumber\\ & =\int_{-\infty}^\infty {\rm d} k_x\frac{{\rm i}k_z}{(k_x^2+k_z^2) \varepsilon_L(k_x)}(A_1\varSigma_0\, {\rm e}^{{\pm} {\rm i}2nak_x}+ A_2\varSigma_1\, {\rm e}^{{\pm} {\rm i}(2n-1)ak_x})\nonumber\\ & \quad +\int_{-\infty}^\infty {\rm d} k_y\frac{{\rm i}k_z}{(k_y^2 +k_z^2)\varepsilon_L(k_y)}(B_1\varSigma_0\, {\rm e}^{{\pm} {\rm i} 2nbk_y}+B_2\varSigma_1\, {\rm e}^{{\pm} {\rm i}(2n-1)bk_y})\nonumber\\ & =\int_{-\infty}^\infty {\rm d} k_x\frac{2{\rm i}k_z}{(k_x^2+k_z^2) \varepsilon_L(k_x)}(A_1S_1(ak_x)+A_2S_2(ak_x))\nonumber\\ & \quad +\int_{-\infty}^\infty {\rm d} k_y\frac{2{\rm i}k_z}{(k_y^2 +k_z^2)\varepsilon_L(k_y)}(B_1S_1(bk_y)+B_2S_2(bk_y)), \end{align}

where

(3.2a)\begin{align} \varepsilon_L(k_y)& =\varepsilon_L(k_x,k_y)\vert_{k_x=0}, \quad \varepsilon_L(k_x)=\varepsilon_L(k_x,k_y)\vert_{k_y=0}, \end{align}

(3.3a)\begin{align}S_1(\chi)& =\tfrac{1}{2}+\, {\rm e}^{2{\rm i} \chi}+\, {\rm e}^{4{\rm i}\chi}+\cdots \quad S_2(\chi)=\, {\rm e}^{{\rm i}\chi}+\, {\rm e}^{3{\rm i}\chi}+\cdots, \end{align}

(3.4)\begin{align} E_z(a,b,z)& =\int_{-\infty}^\infty {\rm d} k_x\frac{2{\rm i} k_z}{(k_x^2+k_z^2)\varepsilon_L(k_x)}(A_1S_2(ak_x)+A_2S_1(ak_x))\nonumber\\ & \quad +\int_{-\infty}^\infty {\rm d} k_y\frac{2{\rm i}k_z}{(k_y^2 +k_z^2)\varepsilon_L(k_y)}(B_1S_2(bk_y)+B_2S_1(bk_y)), \end{align}

(3.5)\begin{align} D_x(0,0,z)& =\lim_{x=0,y=0}\int_{-\infty}^\infty {\rm d} k_x \, {\rm e}^{{\rm i}k_xx}\int_{-\infty}^\infty {\rm d} k_y \, {\rm e}^{{\rm i}k_yy} \frac{{\rm i}k_x}{k^2} \frac{\varepsilon_x(k_x,k_y)}{\varepsilon_L}\nonumber\\ & \quad \times [\delta(k_y)(A_1\varSigma_0\, {\rm e}^{{\pm} {\rm i}2nak_x}+ A_2\varSigma_1\, {\rm e}^{{\pm} {\rm i}(2n-1)ak_x})\nonumber\\ & \quad +\delta(k_x) (B_1\varSigma_0\, {\rm e}^{{\pm} {\rm i}2nbk_y}+ B_2\varSigma_1\, {\rm e}^{{\pm} {\rm i}(2n-1)bk_y})]\nonumber\\ & =\int_{-\infty}^\infty {\rm d} k_x \frac{{\rm i}k_x}{k_x^2+k_z^2} \frac{\nu}{\varepsilon_L(k_x)} (A_1\varSigma_0\, {\rm e}^{{\pm} {\rm i}2nak_x}+A_2\varSigma_1\, {\rm e}^{{\pm} {\rm i}(2n-1)ak_x})\nonumber\\ & ={\rm i}A_1\int_{-\infty}^\infty {\rm d} k_x \frac{k_x}{k_x^2+k_z^2} \frac{\nu}{\varepsilon_L(k_x)}, \end{align}

where

(3.6)\begin{equation} \nu=1-\frac{\omega_p^2}{\omega^2-\omega_c^2}.\end{equation}

Analogously, we obtain

(3.7)\begin{gather} D_x(a,b,z)={-}{\rm i}A_2\int_{-\infty}^\infty {\rm d} k_x \frac{k_x}{k_x^2+k_z^2} \frac{\nu}{\varepsilon_L(k_x)}, \end{gather}

(3.8)\begin{gather}D_y(0,0,z)={\rm i}B_1\int_{-\infty}^\infty {\rm d} k_y \frac{k_y}{k_x^2+k_z^2} \frac{\nu}{\varepsilon_L(k_y)}, \end{gather}

(3.9)\begin{gather}D_y(a,b,z)={-}{\rm i}B_2\int_{-\infty}^\infty {\rm d} k_y \frac{k_y}{k_x^2+k_z^2} \frac{\nu}{\varepsilon_L(k_y)}. \end{gather}

Vacuum solution

We have $\nabla ^2{\boldsymbol E}=0$ in the vacuum region, which is solved by

(3.10)\begin{equation} {\boldsymbol E}\sim \exp {\rm i}k_z\times \exp({\pm} k_xx) \times \exp({\pm} k_yy), \end{equation}

with the important constraint $k_x^2+k_y^2=k_z^2$. We can construct the vacuum solutions as follows:

vacuum region (i). $x<0, y<0$, where

(3.11)\begin{gather} E_z(i)=H \, {\rm e}^{{\rm i}k_zz}\, {\rm e}^{k_xx}\, {\rm e}^{k_yy}, \end{gather}

(3.12)\begin{gather}E_x(i)={-}{\rm i} H \frac{k_x}{k_z}\, {\rm e}^{{\rm i}k_zz} \, {\rm e}^{k_xx}\, {\rm e}^{k_yy}, \end{gather}

(3.13)\begin{gather}E_y(i)={-}{\rm i} H \frac{k_y}{k_z}\, {\rm e}^{{\rm i}k_zz} \, {\rm e}^{k_xx}\, {\rm e}^{k_yy}; \end{gather}

vacuum region (ii). $x>a, y>b$, where

(3.14)\begin{gather} E_z(ii)=G \, {\rm e}^{{\rm i}k_zz}\, {\rm e}^{{-}k_xx}\, {\rm e}^{{-}k_yy}, \end{gather}

(3.15)\begin{gather}E_x(ii)={\rm i} G \frac{k_x}{k_z}\, {\rm e}^{{\rm i}k_zz}\, {\rm e}^{{-}k_xx}\, {\rm e}^{{-}k_yy}, \end{gather}

(3.16)\begin{gather}E_y(ii)={\rm i} G \frac{k_y}{k_z}\, {\rm e}^{{\rm i}k_zz}\, {\rm e}^{{-}k_xx}\, {\rm e}^{{-}k_yy}. \end{gather}

Putting $(x,y)=(0,0)$ or $(a,b)$ in the above equations gives the vacuum side boundary values of the relevant quantities.

4 Dispersion relation for the duct plasma surface wave

We enforce the following boundary conditions to connect the plasma and the vacuum fields: $[E_z]=0, [D_x]=0, [D_y]=0$.

Along lines $(0,0,z)$ and $(a,b,z)$, $[E_z]=0$ respectively gives

(4.1)\begin{gather} A_1I_1+A_2I_2+B_1J_1+B_2J_2=H, \end{gather}

(4.2)\begin{gather}A_1I_2+A_2I_1+B_1J_2+B_2J_1=G\, {\rm e}^{{-}k_xa}\, {\rm e}^{{-}k_yb}, \end{gather}

where

(4.3)\begin{gather} I_i=\int_{-\infty}^\infty {\rm d} k_x\frac{2{\rm i}k_z}{(k_x^2+k_z^2)\varepsilon_L(k_x)}S_i(ak_x)\quad (i=1, 2), \end{gather}

(4.4)\begin{gather}J_i=\int_{-\infty}^\infty {\rm d} k_y\frac{2{\rm i}k_z}{(k_y^2+k_z^2)\varepsilon_L(k_y)}S_i(bk_y)\quad (i=1, 2), \end{gather}

$[D_x]=0$ gives

(4.5)\begin{gather} {\rm i}A_1\int_{-\infty}^\infty {\rm d} k_x \frac{k_x}{k_x^2+k_z^2} \frac{\nu}{\varepsilon_L(k_x)} ={-}{\rm i}H \frac{k_x}{k_z}, \end{gather}

(4.6)\begin{gather}-{\rm i}A_2\int_{-\infty}^\infty {\rm d} k_x \frac{k_x}{k_x^2+k_z^2} \frac{\nu}{\varepsilon_L(k_x)} ={\rm i}G \frac{k_x}{k_z}\, {\rm e}^{{-}k_xa}\, {\rm e}^{{-}k_yb}, \end{gather}

$[D_y]=0$ gives

(4.7)\begin{gather} {\rm i}B_1\int_{-\infty}^\infty {\rm d} k_y \frac{k_y}{k_x^2+k_z^2} \frac{\nu}{\varepsilon_L(k_y)} ={-}{\rm i}H\frac{k_y}{k_z}, \end{gather}

(4.8)\begin{gather}-{\rm i}B_2\int_{-\infty}^\infty {\rm d} k_y \frac{k_y}{k_x^2+k_z^2} \frac{\nu}{\varepsilon_L(k_y)} ={\rm i}G \frac{k_y}{k_z}\, {\rm e}^{{-}k_xa}\, {\rm e}^{{-}k_yb}. \end{gather}

Let us rewrite (4.3) and (4.4) in slightly different forms:

(4.9)\begin{gather} I_i=2{\rm i}k_z\mu \int_{-\infty}^\infty \, {\rm d} \kappa\frac{S_i(a\kappa)}{\kappa^2+\xi^2 k_z^2} \quad (i=1, 2), \end{gather}

(4.10)\begin{gather}J_i=2{\rm i}k_z\mu \int_{-\infty}^\infty \, {\rm d} \kappa\frac{S_i(b\kappa)}{\kappa^2+\xi^2 k_z^2} \quad (i=1, 2), \end{gather}

where

(4.11)\begin{gather} \mu=\frac{\omega^2-\omega_p^2}{\omega^2-\omega_P^2-\omega_c^2}, \end{gather}

(4.12)\begin{gather}\xi=\sqrt{\frac{(\omega^2-\omega_c^2)(\omega^2- \omega_p^2)}{\omega^2(\omega^2-\omega_P^2-\omega_c^2)}} =\sqrt{\left(1-\frac{\omega_p^2}{\omega^2}\right) \left/\left(1-\frac{\omega_p^2}{\omega^2-\omega_c^2}\right)\right.}. \end{gather}

Integrals in (4.5)–(4.8) can be contour-integrated, giving the value ${\rm i}{\rm \pi}$. Therefore, they become

(4.13)\begin{gather} {\rm \pi}\eta A_1={\rm i}H\frac{k_x}{k_z}, \end{gather}

(4.14)\begin{gather}{\rm \pi} \eta A_2={\rm i}G\frac{k_x}{k_z}\, {\rm e}^{{-}k_xa}\, {\rm e}^{{-}k_yb}, \end{gather}

(4.15)\begin{gather}{\rm \pi} \eta B_1={\rm i}H\frac{k_y}{k_z}, \end{gather}

(4.16)\begin{gather}{\rm \pi} \eta B_2={\rm i}G \frac{k_y}{k_z}\, {\rm e}^{{-}k_xa}\, {\rm e}^{{-}k_yb}, \end{gather}

where

(4.17)\begin{equation} \eta=\frac{\omega^2-\omega_p^2}{\omega^2-\omega_c^2}. \end{equation}

We have six unknowns, $A_1,A_2,B_1,B_2,H,G$ in six equations (4.1), (4.2), (4.13)–(4.16). Eliminating $H, G, A_1, A_2$ gives

(4.18)\begin{gather} B_1(k_xI_1+k_yJ_1+{\rm i}{\rm \pi} \eta~k_z)+B_2(J_2k_y+I_2 k_x)=0, \end{gather}

(4.19)\begin{gather}B_1(I_2k_x+J_2k_y)+B_2(I_1k_x+J_ik_y+{\rm i}{\rm \pi} \eta k_z)=0, \end{gather}

which give the dispersion relation in the form

(4.20)\begin{equation} k_x(I_1\pm I_2)+k_y(J_1\pm J_2)={-}{\rm i}{\rm \pi} \eta k_z.\end{equation}

The $+$ ($-$) sign corresponds to symmetric (anti-symmetric) mode. Since we have the constraint $k_x^2+k_y^2=k_z^2$, let us put

(4.21a)\begin{equation} k_x=\frac{bk_z}{\sqrt{a^2+b^2}}, \quad k_y=\frac{ak_z}{\sqrt{a^2+b^2}}, \end{equation}

so that the $y$-direction becomes ignorable when $b\rightarrow \infty$. Using the formula (Lee & Cho Reference Lee and Cho2022; Lee & Lim Reference Lee and Lim2022)

(4.22)\begin{equation} \int_{-\infty}^\infty {\rm d} \kappa\frac{S_i(a\kappa)\pm S_2(a\kappa)}{\kappa^2 +\xi^2 k_z^2} =\frac{1}{2}\int_{-\infty}^\infty \frac{{\rm d} \kappa}{\kappa^2+\xi^2k_z^2} \frac{1\pm \, {\rm e}^{{\rm i}a\kappa}}{1\mp \, {\rm e}^{{\rm i}a\kappa}}. \end{equation}

Equation (4.20) takes the form

(4.23)\begin{equation} \frac{b}{\sqrt{a^2+b^2}}\int_{-\infty}^\infty \frac{{\rm d} \kappa}{\kappa^2+\xi^2k_z^2} \frac{1\pm \, {\rm e}^{{\rm i}a\kappa}}{1\mp \, {\rm e}^{{\rm i}a\kappa}} +\frac{a}{\sqrt{a^2+b^2}}\int_{-\infty}^\infty \frac{{\rm d} \kappa}{\kappa^2+\xi^2k_z^2} \frac{1\pm \, {\rm e}^{{\rm i}b\kappa}}{1\mp \, {\rm e}^{{\rm i}b\kappa}} ={-}\frac{{\rm \pi}\eta}{k_z\mu}.\end{equation}

Picking up the simple pole at $\kappa =i\xi k_z$, (4.23) becomes

(4.24)\begin{equation} \frac{b}{\sqrt{a^2+b^2}}\tanh \frac{a}{2}\xi k_z+\frac{a}{\sqrt{a^2+b^2}}\tanh \frac{b}{2}\xi k_z={-}\frac{\eta\xi}{\mu}={-}\nu\xi,\end{equation}

for the lower sign (symmetric mode), where $\nu$ is defined by (3.6). For the upper sign (anti-symmetric mode), $\tanh$ is replaced by $\coth$. Taking the limit $b\rightarrow \infty$ gives the slab dispersion relation

(4.25)\begin{equation} \tanh \frac{a}{2}\xi k_z={-}\nu\xi. \end{equation}

Taking the limit $a\rightarrow \infty$ in the above equation gives the dispersion relation for semi-infinite plasma, which is $\nu \xi =-1$:

(4.26)\begin{equation} 1-\frac{\omega_p^2}{\omega^2-\omega_c^2}+\sqrt{ \left(1-\frac{\omega_p^2}{\omega^2-\omega_c^2}\right)\left/ \left(1-\frac{\omega_p^2}{\omega^2}\right)\right. }=0,\end{equation}

which agrees with the electrostatic dispersion relation obtained from the cold fluid equation (Lee Reference Lee1995). The slab dispersion relations obtained by taking $b\rightarrow \infty$ that contain the symmetric and anti-symmetric modes agree with the earlier results which were obtained by fluid equation (Gradov & Stenflo Reference Gradov and Stenflo1983).

5 Cylindrical plasma

In this section, we derive the surface wave dispersion relations in a cold magnetized plasma contained in infinite-length cylinder of radius ‘$a$’, interfaced with vacuum. We use the following fluid equations written in cylindrical coordinates:

(5.1)\begin{gather} -{\rm i}\omega{\boldsymbol v}(r,\theta,z)=\frac{e}{m}\boldsymbol{\nabla}\phi-\omega_c{\boldsymbol v}\times\hat{{\boldsymbol z}}, \end{gather}

(5.2)\begin{gather}-{\rm i}\omega n(r,\theta,z)+N(r)\boldsymbol{\nabla}\boldsymbol{\cdot}{\boldsymbol v} +v_r\frac{\partial N}{\partial r}=0, \end{gather}

(5.3)\begin{gather}\nabla^2\phi=4{\rm \pi} en. \end{gather}

Here, we assume that the equilibrium density $N(r)$ has density gradients in the radial direction. At the interface $r=a$, $N$ changes abruptly with infinite gradient, otherwise is homogeneous. This artificiality is designed solely to derive the boundary conditions at the interface by ‘infinitesimal integration’ across the interface. Equation (5.1) yields

(5.4)\begin{gather} v_r=\frac{e}{m} \frac{\omega}{\omega^2-\omega_c^2} \left({\rm i}\frac{\partial\phi}{\partial r}+\frac{\omega_c}{\omega} \frac{1}{r} \frac{\partial\phi}{\partial\theta}\right), \end{gather}

(5.5)\begin{gather}v_\theta=\frac{e}{m} \frac{\omega}{\omega^2-\omega_c^2} \left(-\frac{\omega_c}{\omega} \frac{\partial\phi}{\partial r}+{\rm i}\frac{1}{r} \frac{\partial\phi}{\partial\theta}\right), \end{gather}

(5.6)\begin{gather}v_z=\frac{{\rm i}}{\omega} \frac{e}{m} \frac{\partial\phi}{\partial z}. \end{gather}

Equations (5.2)–(5.6) give the wave equation written in terms of $\phi$:

(5.7)\begin{align} & \frac{\partial}{\partial r}\left[\left(1-\frac{\omega_p^2}{\omega^2 -\omega_c^2}\right)\frac{\partial\phi}{\partial r}\right] +\left(1-\frac{\omega_p^2}{\omega^2-\omega_c^2}\right) \frac{1}{r} \frac{\partial\phi}{\partial r}+ \frac{\partial}{\partial z}\left[\left(1-\frac{\omega_p^2}{\omega^2} \right)\frac{\partial\phi}{\partial z}\right]\nonumber\\ & \quad +\frac{{\rm i}\omega_c}{\omega} \frac{1}{r} \frac{\partial}{\partial r}\left(\frac{\omega_p^2}{\omega^2- \omega_c^2}\right) \frac{\partial\phi}{\partial\theta} +\left(1-\frac{\omega_p^2}{\omega^2-\omega_c^2}\right) \frac{1}{r^2} \frac{\partial^2\phi}{\partial\theta^2}=0, \end{align}

where $\omega _p^2=4{\rm \pi} e^2N(r)/m$. If we take the infinitesimal integration across the interface $r=a$ in the manner $\int _{-\epsilon }^\epsilon (\cdots )\, {\rm d} r$, we can extract useful boundary conditions. In this operation, we pick up only the perfect differentials that survive the limit $\epsilon \rightarrow 0$. Equation (5.4) yields

(5.8)\begin{equation} [\phi]_{r=a}=0.\end{equation}

Integrating (5.7) yields

(5.9)\begin{equation} \left[\left(1-\frac{\omega_p^2}{\omega^2-\omega_c^2}\right) \frac{\partial\phi}{\partial r}\right]_{r=a}=0,\end{equation}

where the square bracket denotes the jump across the interface. Putting $\omega _p=$ const. or zero in (5.7) gives the wave equation in the plasma or in the vacuum. We also restrict to an azimuthally symmetric field $(\partial /\partial \theta =0$). We have

(5.10)\begin{gather} \frac{\partial^2\phi}{\partial r^2}+\frac{1}{r} \frac{\partial\phi}{\partial r}+ \frac{\partial^2\phi}{\partial z^2}=0, \quad \text{in vacuum}, \end{gather}

(5.11)\begin{gather}\frac{\partial^2\phi}{\partial r^2}+\frac{1}{r} \frac{\partial\phi}{\partial r}+\xi^2 \frac{\partial^2\phi}{\partial z^2}=0, \quad \text{in plasma}, \end{gather}

where $\xi$ is defined in (4.12).

Now let us solve the above equations for an infinite cylindrical plasma. Since the $z$-direction is infinite, we can assume $\phi (z)\sim \, {\rm e}^{{\rm i}k_zz}$ and the plasma equation in (5.11) becomes

(5.12)\begin{equation} \frac{\partial^2\phi(r)}{\partial r^2}+\frac{1}{r} \frac{\partial\phi}{\partial r}-\xi^2k_z^2\phi(r)=0,\end{equation}

where we suppressed the $z$-dependent phase. Equation (5.12) is solved by the functions $I_0$ and $K_0$, the modified Bessel functions of first and second kind of order zero, respectively. Here, $K_0$ is rejected since it blows up at $r=0$, so the plasma solution takes the form

(5.13)\begin{equation} \phi_p(r)=A I_0(k_z\xi r)\quad 0< r< a. \end{equation}

In vacuum, the function $I_0$ should be rejected since it blows up at $r=\infty$. We have

(5.14)\begin{equation} \phi_v(r)=BK_0(k_zr)\quad r>a, \end{equation}

where $A$, $B$ are constants to be determined by boundary conditions. Equations (5.8) and (5.9) give

(5.15)\begin{gather} AI_0(\xi k_za)=BK_0(k_za), \end{gather}

(5.16)\begin{gather}\nu\xi A I_0'(\xi k_za)=BK_0'(k_za), \end{gather}

where prime denotes the derivative with respect to the argument and $\nu$ is defined in (3.6). The above two equations yield the dispersion relation

(5.17)\begin{equation} \nu\xi \frac{I_0'(\xi k_za)}{I_0(\xi k_za)}=\frac{K_0'(k_za)}{K_0(k_za)}.\end{equation}