Ionospheric Alfvén Resonator (IAR)

Abstract

The topic of this chapter is the ionospheric Alfvén resonator (IAR) which has been the subject of a great deal of research during recent years. The IAR resonance cavity occupies a space between the conducting E layer and the topside ionosphere where there occurs the strong gradient of Alfvén velocity. The IAR accumulates the Alfvén wave energy in the ULF/ELF frequency range, typically between 0.5 and 7 Hz. In this chapter, the structure, models, and possible physical mechanisms for the IAR excitation are studied. Dispersion relation and the IAR resonance spectra at night and daytime conditions are calculated.

Appendices

Appendix C: Vector and Scalar Potentials of Electromagnetic Field

1.1 General Description

In this section we introduce the standard vector and scalar potentials of the electromagnetic field in a conducting medium immersed in the external magnetic field B ₀. To treat the electric and magnetic fields, we need Maxwell’s equations, which, in their full form, are given by Eqs. (1.1)–(1.4). If we seek for the solution of these equations in the form $\mathbf{B} = \mathbf{B}_{0} +\delta \mathbf{B}$, where δ B is the small variation of B ₀, the electromagnetic field can be represented through the vector potential, A, and scalar potential, $\Phi$, as follows (Jackson 2001)

$$\displaystyle{ \delta \mathbf{B} = \nabla \times \mathbf{A,} }$$

(5.73)

$$\displaystyle{ \mathbf{E} = -\nabla \Phi - \partial _{t}\mathbf{A}. }$$

(5.74)

Considering two important cases when the external field B ₀ is a constant value or when B ₀ denotes the Earth’s magnetic field in the dipole approximation given by Eq. (1.32), we have the condition ∇×B ₀ = 0. Taking the notice of this condition and substituting the field presentation given by Eqs. (5.73) and (5.74) into Maxwell equations (1.2) and (1.3) converts these equations into identities.

Let z axis be positive parallel to the external/unperturbed magnetic field B ₀ and $\mathbf{\hat{z}} = \mathbf{B}_{0}/B_{0}$ be a unit vector parallel to B ₀. In this notation the total vector potential can be written as $\mathbf{A =}A\hat{\mathbf{z}} + \mathbf{A}_{\perp }$, where the second term represents the perpendicular component of the vector potential. We choose the calibration equation for the vector potential in the form

$$\displaystyle{ \nabla _{\perp }\cdot \mathbf{A}_{\perp } = 0, }$$

(5.75)

where ∇_⊥ denotes the perpendicular component of the gradient, that is $\nabla _{\perp } = \left (\partial _{x},\partial _{y}\right )$, where the symbols $\partial _{x} = \partial /\partial x$ and $\partial _{y} = \partial /\partial y$ denote the partial derivatives with respect to x and y, respectively. It follows from Eq. (5.75) that the vector A _⊥ can be written in the form

$$\displaystyle{ \mathbf{A}_{\perp } = \nabla _{\perp }\times \left (\Psi \hat{\mathbf{z}}\right )\!, }$$

(5.76)

where $\Psi$ is the second scalar potential. Indeed, substituting Eq. (5.76) for A _⊥ into Eq. (5.75) gives an identity. Hence we get

$$\displaystyle{ \mathbf{A} = A\hat{\mathbf{z}} + \nabla _{\perp }\times \left (\Psi \hat{\mathbf{z}}\right )\!. }$$

(5.77)

Subsisting Eq. (5.76) for A _⊥ into Eqs. (5.73) and (5.74) and rearranging yields

$$\displaystyle{ \delta \mathbf{B} = \left (\nabla _{\perp }A\right ) \times \hat{\mathbf{z}} + \nabla _{\perp }\partial _{z}\Psi -\hat{\mathbf{z}}\nabla _{\perp }^{2}\Psi, }$$

(5.78)

and

$$\displaystyle{ \mathbf{E} = -\nabla _{\perp }\Phi -\nabla _{\perp }\times \left (\hat{\mathbf{z}}\partial _{t}\Psi \right ) -\hat{\mathbf{z}}\left (\partial _{z}\Phi + \partial _{t}A\right )\!. }$$

(5.79)

1.2 Potentials of Shear Alfvén and Compressional Waves in Plasma

The representation of the electromagnetic field via potentials is of frequent use in plasma waves physics. In specific cases the general wave equations can be split into two independent sets of equations in such a way that the scalar potentials $\Phi$ and A describe the shear Alfvén mode while the potential $\Psi$ corresponds to the compressional mode.

As the plasma is immersed in the external magnetic field, the plasma conductivity exhibits anisotropy, which can be described by the tensor of the plasma conductivity (2.5) or by the tensor of dielectric permittivity (2.18). As the field-aligned plasma permittivity $\varepsilon _{\Vert }$, that is, the tensor component parallel to the magnetic field B ₀ tends to infinity, the parallel electric field becomes infinitesimal, that is E _z  = 0. This implies that $\partial _{z}\Phi + \partial _{t}A = 0$, so that the component A can be expressed through $\Phi$. The same is true if the field-aligned plasma conductivity $\sigma _{\Vert } \rightarrow \infty$. In particular, if all perturbed quantities are considered to vary as exp(−iωt), then

$$\displaystyle{ i\omega A = \partial _{z}\Phi. }$$

(5.80)

In fact this means that the shear Alfvén and compressional modes can be described through two scalar potentials, say $\Phi$ and $\Psi$, instead of three potentials. For example, the shear Alfvén mode can be represented via only the potential $\Phi$

$$\displaystyle{ \delta \mathbf{B}_{A} = \frac{\left (\nabla _{\perp }\partial _{z}\Phi \right ) \times \hat{\mathbf{z}}} {i\omega },\ \ \ \mathbf{E}_{A} = -\nabla _{\perp }\Phi. }$$

(5.81)

As can be seen from Eq. (5.81), the magnetic and electric fields of the shear Alfvén mode are both perpendicular to the external magnetic field B ₀. This conclusion is consistent with the analysis made in Chap. 3 and is illustrated in Fig. 1.15. Nevertheless, the total field-aligned Alfvén current, $j_{z_{A}}$, including the conduction and displacement currents, is nonzero. Substituting Eq. (5.81) for δ B _A into Eq. (1.1), yields

$$\displaystyle{ j_{z_{A}} = \frac{i\nabla _{\perp }^{2}\partial _{z}\Phi } {\mu _{0}\omega }. }$$

(5.82)

The FMS/compressional mode can be expressed by the potential $\Psi$ as follows:

$$\displaystyle{ \delta \mathbf{B}_{F} = \nabla _{\perp }\partial _{z}\Psi -\hat{\mathbf{z}}\nabla _{\perp }^{2}\Psi,\ \ \ \mathbf{E}_{ F} = i\omega \nabla _{\perp }\times \left (\hat{\mathbf{z}}\Psi \right )\!. }$$

(5.83)

It follows from Eq. (5.83) that the electrical field of the compressional mode is perpendicular to the external magnetic field as shown in Fig. 1.16, while the parallel current density $j_{z_{C}} = 0$.

1.3 Fourier Transform over Space

As before, we assume that a local coordinate system has the z axis positive parallel to the magnetic field B ₀. The direct and inverse Fourier transforms of the electromagnetic perturbations over the coordinates x and y perpendicular to B ₀ are given by Eqs. (5.3) and (5.4). Applying the same Fourier transform to Eqs. (5.78) and (5.79) gives the relationships (5.7) and (5.8) between the components, b and e, of electromagnetic field and potential functions, A,$\Phi$, and $\Psi$ in the $\left (\omega,\mathbf{k}_{\perp }\right )$ space, where ω is the frequency and $\mathbf{k}_{\perp }\mathbf{=}\left (k_{x}\mathbf{,}k_{y}\right )$ stands for the perpendicular wave vector. In the magnetosphere and ionosphere the potentials A and $\Phi$ are related through Eq. (5.80). Combining this equation and Eqs. (5.7) and (5.8), we come to the two potential field representation

$$\displaystyle{ \mathbf{b} = i\mathbf{k}_{\perp }\partial _{z}\Psi + \frac{\left (\mathbf{k}_{\perp }\times \hat{\mathbf{z}}\right )} {\omega } \partial _{z}\Phi + k_{\perp }^{2}\Psi \hat{\mathbf{z}}, }$$

(5.84)

and

$$\displaystyle{ \mathbf{e} = -i\mathbf{k}_{\perp }\Phi -\left (\mathbf{k}_{\perp }\times \hat{\mathbf{z}}\right )\omega \Psi. }$$

(5.85)

Here the potential $\Phi$ describes the shear Alfvén while the potential $\Psi$ corresponds to the FMS mode.

In a similar fashion we may obtain the Fourier transform of the parallel electric current density produced by the shear Alfvén wave

$$\displaystyle{ j_{z_{A}} = -\frac{ik_{\perp }^{2}\partial _{z}\Phi } {\mu _{0}\omega }. }$$

(5.86)

As we have noted above, the field representation through the vector and scalar potentials satisfies the Faraday law given by Eq. (1.2). It is useful to demonstrate, additionally, that Eqs. (5.84) and (5.85) satisfy a Fourier transform of the Faraday equation given by Eq. (5.6). In other words, we now show that substituting of Eqs. (5.7) and (5.8) for b and e into Eq. (5.6) gives an identity. To verify this statement one should take into account that

$$\displaystyle{ \mathbf{k}_{\perp }\times \left (\hat{\mathbf{z}} \times \mathbf{k}_{\perp }\right ) = k_{\perp }^{2}\hat{\mathbf{z}}, }$$

(5.87)

and

$$\displaystyle{ \hat{\mathbf{z}}\times \left (\mathbf{k}_{\perp }\times \hat{\mathbf{z}}\right ) = \mathbf{k}_{\perp }. }$$

(5.88)

In this notation the first term on the right-hand side of Eq. (5.6) is reduced to

$$\displaystyle{ i\left (\mathbf{k}_{\perp }\times \mathbf{e}\right ) = i\omega k_{\perp }^{2}\Psi \hat{\mathbf{z}}+i\left (i\omega A - \partial _{ z}\Phi \right )\left (\mathbf{k}_{\perp }\times \hat{\mathbf{z}}\right )\!. }$$

(5.89)

The second term of Eq. (5.6) can be converted to

$$\displaystyle{ \hat{\mathbf{z}} \times \partial _{z}\mathbf{e =}i\partial _{z}\Phi \left (\mathbf{k}_{\perp }\times \hat{\mathbf{z}}\right ) -\omega \partial _{z}\Psi \mathbf{k}_{\perp }. }$$

(5.90)

Combining Eqs. (5.89) and (5.90) and rearranging we come to the following equation

$$\displaystyle{ i\left (\mathbf{k}_{\perp }\times \mathbf{e}\right ) +\hat{ \mathbf{z}} \times \partial _{z}\mathbf{e = -}\omega A\left (\mathbf{k}_{\perp }\times \hat{\mathbf{z}}\right ) -\omega \partial _{z}\Psi \mathbf{k}_{\perp } + i\omega k_{\perp }^{2}\Psi \hat{\mathbf{z}} =i\omega \mathbf{b,} }$$

(5.91)

that coincides with Eq. (5.6), which is required to be proved.

1.4 Cylindrical Coordinates

In the course of the main text, some of the phenomena are considered in the cylindrical coordinates $r,\varphi$, and z. On account of the representation of the perpendicular divergence operator in the cylindrical coordinates, the calibration equation (5.75) reduces to

$$\displaystyle{ \frac{1} {r}\partial _{r}\left (rA_{r}\right ) + \frac{1} {r}\partial _{\varphi }A_{\varphi } = 0. }$$

(5.92)

This equation holds true if

$$\displaystyle{ A_{r} = \frac{1} {r}\partial _{\varphi }\Psi,\quad \mbox{ and}\quad A_{\varphi } = -\partial _{r}\Psi. }$$

(5.93)

Finally we arrive at the following representation

$$\displaystyle{ \mathbf{A} =\hat{ \frac{\mathbf{r}} {r}}\partial _{\varphi }\Psi -\hat{\boldsymbol{\varphi }} \partial _{r}\Psi +\hat{ \mathbf{z}}A, }$$

(5.94)

where $\hat{\mathbf{r}}$, $\hat{\boldsymbol{\varphi }}$, and $\hat{\mathbf{z}}$ stand for the unit vectors. Substituting Eq. (5.94) for A into Eq. (5.73) yields

$$\displaystyle\begin{array}{rcl} \delta B_{r} = \frac{1} {r}\partial _{\varphi }A_{z} - \partial _{z}A_{\varphi } = \frac{1} {r}\partial _{\varphi }A + \partial _{rz}^{2}\Psi,& &{}\end{array}$$

(5.95)

$$\displaystyle\begin{array}{rcl} \delta B_{\varphi } = \partial _{z}A_{r} - \partial _{r}A_{z} = \frac{1} {r}\partial _{z\varphi }^{2}\Psi - \partial _{ r}A,& &{}\end{array}$$

(5.96)

$$\displaystyle\begin{array}{rcl} \delta B_{z} = \frac{1} {r}\partial _{r}\left (rA_{\varphi }\right ) -\frac{1} {r}\partial _{\varphi }A_{r} = -\frac{1} {r}\partial _{r}\left (r\partial _{r}\Psi \right ) - \frac{1} {r^{2}}\partial _{\varphi \varphi }^{2}\Psi.& &{}\end{array}$$

(5.97)

Similarly, substituting Eq. (5.94) for A into Eq. (5.74) yields

$$\displaystyle\begin{array}{rcl} E_{r} = -\partial _{r}\Phi + \frac{i\omega } {r}\partial _{\varphi }\Psi,& &{}\end{array}$$

(5.98)

$$\displaystyle\begin{array}{rcl} E_{\varphi } = -\frac{1} {r}\partial _{\varphi }\Phi - i\omega \partial _{r}\Psi,& &{}\end{array}$$

(5.99)

$$\displaystyle\begin{array}{rcl} E_{z} = -\partial _{z}\Phi + i\omega A.& &{}\end{array}$$

(5.100)

Here, as we have noted above, the terms depending on the potentials $\Phi$ and A describe the shear Alfvén mode, whereas the terms depending on the potential $\Psi$ correspond to the compressional mode.

Appendix D: Solutions of the Boundary Problems

2.1 Solution of the Problem Associated with IAR

In the magnetosphere $\left (z > L\right )$ the solution of wave equations for the potentials $\Phi$ and $\Psi$ is given by Eqs. (5.15) and (5.16). Inside the IAR region $\left (0 < z < L\right )$ the solution of Eq. (5.12) describing Alfvén waves can be written as

$$\displaystyle{ \Phi = \Phi \left (0\right )\cos \frac{\omega z} {V _{AI}} + C_{3}\sin \frac{\omega z} {V _{AI}}, }$$

(5.101)

where C ₃ and $\Phi \left (0\right )$ are undetermined coefficients. In order to match the solutions (5.15) and (5.101) at the boundary z = L, one should take into account a requirement of continuity of the potential $\Phi$ and its derivative $\partial _{z}\Phi$. Whence we get

$$\displaystyle{ C_{3}\sin x_{0} + \Phi \left (0\right )\cos x_{0} = C_{1}\exp \left (ix_{0}\right )\!, }$$

(5.102)

$$\displaystyle{ C_{3}\cos x_{0} - \Phi \left (0\right )\sin x_{0} = i\epsilon C_{1}\exp \left (ix_{0}\right )\!. }$$

(5.103)

where $x_{0} =\omega L/V _{AI}$ denotes the dimensionless frequency and $\epsilon = V _{AI}/V _{AM}$. The set of Eqs. (5.102)–(5.103) can be solved for C ₃ to yield

$$\displaystyle{ C_{3} = i\Phi \left (0\right )\left [\frac{1 +\epsilon -\left (1-\epsilon \right )\exp \left (2ix_{0}\right )} {1 +\epsilon +\left (1-\epsilon \right )\exp \left (2ix_{0}\right )}\right ]. }$$

(5.104)

Substituting Eq. (5.104) for C ₃ into Eq. (5.101), we come to Eq. (5.18), which describes the potential $\Phi$ inside the IAR region.

Similarly, the solution of Eq. (5.13) describing FMS waves in the region 0 < z < L can be written as

$$\displaystyle{ \Psi = \Psi \left (0\right )\cosh \frac{\lambda _{I}z} {L} + C_{4}\sinh \frac{\lambda _{I}z} {L}, }$$

(5.105)

where the function λ _I is given by Eq. (5.20). As before C ₄ and $\Psi \left (0\right )$ denote undetermined coefficients. On account of the continuity of the potential $\Psi$ and its derivative $\partial _{z}\Psi$ at the boundary z = L we get

$$\displaystyle{ \Psi \left (0\right )\cosh \lambda _{I} + C_{4}\sinh \lambda _{I} = C_{2}\exp \lambda _{M}, }$$

(5.106)

$$\displaystyle{ \Psi \left (0\right )\lambda _{I}\sinh \lambda _{I} + C_{4}\lambda _{I}\cosh \lambda _{I} = C_{2}\lambda _{M}\exp \lambda _{M}, }$$

(5.107)

where the function λ _M is given by Eq. (5.17). The set of Eqs. (5.106)–(5.107) can be solved for C ₄ to yield

$$\displaystyle{ C_{4} = \Psi \left (0\right )\left [\frac{\lambda _{I} +\lambda _{M} -\left (\lambda _{I} -\lambda _{M}\right )\exp \left (2\lambda _{I}\right )} {\lambda _{I} +\lambda _{M} + \left (\lambda _{I} -\lambda _{M}\right )\exp \left (2\lambda _{I}\right )}\right ]. }$$

(5.108)

Substituting Eq. (5.108) for C ₄ into Eq. (5.105), we come to Eq. (5.19), which describes the potential $\Psi$ inside the IAR.

2.2 Magnetic Field Perturbations in the Atmosphere and in the Solid Earth

Since there are no sources in the neutral atmosphere $\left (-d < z < 0\right )$, the ULF electromagnetic perturbations excited by the ionospheric current in the atmosphere are described by Laplace equation (5.27). A spatial Fourier transform of this equation is given by

$$\displaystyle{ \partial _{zz}^{2}\mathbf{b-}k_{ \perp }^{2}\mathbf{b} = 0. }$$

(5.109)

A vertical electric current j _z flowing from the conducting ionosphere must be zero at the boundary z = 0 and everywhere in the layer − d < z < 0 because the atmosphere is an insulator. Taking Ampere’s law, applying a Fourier transform to this equation, using Eq. (5.87) and the representation (5.7) of the magnetic field via the potentials, we obtain

$$\displaystyle{ \mu _{0}j_{z} = \left (\nabla \times \mathbf{b}\right )_{z} = \left (\mathbf{k}_{\perp }\times \mathbf{b}_{\perp }\right )_{z} = k_{\perp }^{2}A = 0, }$$

(5.110)

whence it follows that A = 0 in the atmosphere including the upper boundary z = 0. Thus the magnetic field in the atmosphere is derivable from only the potential $\Psi$

$$\displaystyle{ \mathbf{b} = \left (k_{\perp }^{2}\hat{\mathbf{z}} + i\mathbf{k}_{ \perp }\partial _{z}\right )\Psi. }$$

(5.111)

Substituting Eq. (5.111) for b into Eq. (5.109) yields

$$\displaystyle{ \partial _{zz}^{2}\Psi - k_{ \perp }^{2}\Psi = 0. }$$

(5.112)

The solid Earth (z < −d) is supposed to be a uniform conductor with a constant conductivity σ _g . The low frequency electromagnetic field in the solid Earth is described by the quasisteady Maxwell equation (5.28). Applying a Fourier transform to this equation, using Eq. (5.7), and rearranging, we obtain

$$\displaystyle{ \partial _{zz}^{2}\Psi -\xi ^{2}\Psi = 0, }$$

(5.113)

where $\xi ^{2} = k_{\perp }^{2} - i\mu _{0}\sigma _{g}\omega$ is the squared “wave” number/propagation factor in the ground.

Now we need to solve Eq. (5.112) and (5.113) for the atmosphere and for the solid Earth, respectively, and then match the solutions at the boundary $z = -d$. The solution of Eq. (5.113) decays at infinity $\left (z \rightarrow -\infty \right )$ and is

$$\displaystyle{ \Psi = \Psi \left (-d\right )\exp \left [\xi \left (z + d\right )\right ],\quad \left (\text{Re}\xi > 0\right )\!. }$$

(5.114)

The solution of Eq. (5.112) can be written as

$$\displaystyle{ \Psi = C_{+}\exp \left (-k_{\perp }z\right ) + C_{-}\exp \left (k_{\perp }z\right )\!, }$$

(5.115)

where the constants C ₊ and C ₋ can be expressed through the constant $\Psi \left (-d\right )$ making allowance for the continuity of $\Psi$ and $\partial _{z}\Psi$ at the boundary $z = -d$. This yields

$$\displaystyle{ C_{\pm } = \frac{1} {2}\Psi \left (-d\right )\left (1 \pm \frac{\xi } {k_{\perp }}\right )\!. }$$

(5.116)

Substituting Eq. (5.116) for C _± into Eq. (5.115) gives the solution of the problem in the region − d < z < 0.

2.3 Boundary Conditions at the E-Layer of the Ionosphere

In Sect. 5.1 we have derived the boundary condition (5.26) at z = 0 in the approximation of an infinitely thin conducting E-layer. Now we shall express this boundary condition through scalar potentials of the electromagnetic field. In deriving this condition we shall take into account the following properties of triple vector and scalar products

$$\displaystyle{ \hat{\mathbf{z}}\times \left (\mathbf{v\times }\hat{\mathbf{z}}\right ) = \mathbf{v}_{\perp }, }$$

(5.117)

and

$$\displaystyle{ \mathbf{k}_{\perp }\cdot \left (\mathbf{v\times }\hat{\mathbf{z}}\right ) = \hat{\mathbf{z}}\cdot \left (\mathbf{k}_{\perp }\times \mathbf{v}\right ) = \left (\mathbf{k}_{\perp }\times \mathbf{v}\right )_{z}. }$$

(5.118)

Eq. (5.26) contains the jump of perpendicular magnetic field across the conducting E-layer. Taking the magnetic field representation (5.7) through the scalar potentials $(A,\Psi )$ we obtain that $\left [\mathbf{b}_{\perp }\right ] = i\mathbf{k}_{\perp }\left [\partial _{z}\Psi \right ] + i\left (\mathbf{k}_{\perp }\times \hat{\mathbf{z}}\right )\left [A\right ]$, where the square brackets denote the jump of the functions across the E layer, for example, $\left [A\right ] = A\left (0+\right ) - A\left (0-\right )$. Substituting $\left [\mathbf{b}_{\perp }\right ]$ into the boundary condition (5.26), taking into account the continuity of e _⊥ at z = 0, using the potentials $(A,\Phi,\Psi )$ according to Eqs. (5.7)–(5.8)), and combining these equations with Eqs. (5.87)–(5.88) and (5.117) we find that

$$\displaystyle\begin{array}{rcl} & & iV _{AI}\left \{\left [A\right ]\mathbf{k}_{\perp } + \left [\partial _{z}\Psi \right ]\left (\hat{\mathbf{z}} \times \mathbf{k}_{\perp }\right )\right \} \\ & & \quad = -\mathbf{k}_{\perp }\left \{\alpha _{H}\omega \Psi + i\alpha _{P}\Phi \right \} + \left (\hat{\mathbf{z}} \times \mathbf{k}_{\perp }\right )\left \{\alpha _{P}\omega \Psi - i\alpha _{H}\Phi \right \} \\ & & \quad + B_{0}\left \{\alpha _{H}\mathbf{v}_{\perp } +\alpha _{P}\left (\mathbf{v\times }\hat{\mathbf{z}}\right )\right \}. {}\end{array}$$

(5.119)

Here we have just used the identity $\mathbf{v \times B}_{0}=B_{0}\left (\mathbf{v\times }\hat{\mathbf{z}}\right )$.

Taking the scalar and cross product of Eq. (5.119) with k _⊥, taking into account Eq. (5.118) and rearranging, we get

$$\displaystyle{ \left [A\left (0\right )\right ] = \frac{i\alpha _{H}x_{0}} {L} \Psi \left (0\right ) - \frac{\alpha _{P}} {V _{AI}}\Phi \left (0\right ) - f_{1}, }$$

(5.120)

$$\displaystyle{ \left [\partial _{z}\Psi \left (0\right )\right ] = -\frac{i\alpha _{P}x_{0}} {L} \Psi \left (0\right ) - \frac{\alpha _{H}} {V _{AI}}\Phi \left (0\right ) + f_{2}, }$$

(5.121)

where x ₀ is a dimensionless frequency defined in Eq. (5.17). Here we made use of the following abbreviations

$$\displaystyle{ f_{1} = \frac{iB} {V _{AI}k_{\perp }^{2}}\left \{\alpha _{P}\left (\mathbf{k}_{\perp }\times \mathbf{v}\right )_{z} +\alpha _{H}\left (\mathbf{k}_{\perp }\cdot \mathbf{v}\right )\right \}, }$$

(5.122)

$$\displaystyle{ f_{2} = \frac{iB} {V _{AI}k_{\perp }^{2}}\left \{\alpha _{P}\left (\mathbf{k}_{\perp }\cdot \mathbf{v}\right ) -\alpha _{H}\left (\mathbf{k}_{\perp }\times \mathbf{v}\right )_{z}\right \} }$$

(5.123)

Depending on the neutral wind velocity, v, the functions f ₁ and f ₂ play a role of forcing functions/sources for the IAR excitation.

Now we use Eq. (5.29) for $\Psi$ in the atmosphere to connect the potential $\Psi$ and its derivative at the interface z = 0 between the atmosphere and the ionosphere

$$\displaystyle{ \partial _{z}\Psi \left (0-\right ) = k_{\perp }\Psi \left (0\right ) \frac{\xi +k_{\perp }\tanh \left (k_{\perp }d\right )} {k_{\perp } +\xi \tanh \left (k_{\perp }d\right )}, }$$

(5.124)

where ξ is given by Eq. (5.30). Here minus in the argument of the function $\Psi$ in Eq. (5.124) denotes that the derivative should be taken just below the E-layer.

As is seen from Eqs. (5.120) and (5.121), the boundary conditions at z = 0 relate the jump of values of $\partial _{z}\Psi$ and A just above and below the E layer of the ionosphere. It follows from Eq. (5.19) that just above E-layer the function $\partial _{z}\Psi$ is

$$\displaystyle{ \partial _{z}\Psi \left (0+\right ) = \Psi \left (0\right )\frac{\beta _{2}\lambda _{I}} {L}, }$$

(5.125)

where the function β ₂ is given in Eq. (5.22). Notice that the values of $\Psi \left (0\right )$ are the same in both Eq. (5.124) and Eq. (5.125) because the function $\Psi$ must be continuous at z = 0. Subtracting Eq. (5.125) from Eq. (5.124) brings about the jump of derivative $\partial _{z}\Psi$ across the E layer

$$\displaystyle{ \left [\partial _{z}\Psi \left (0\right )\right ] = \Psi \left (0\right )\left (\frac{\lambda _{I}\beta _{2}} {L} - k_{\perp } \frac{\xi +k_{\perp }\tanh \left (k_{\perp }d\right )} {k_{\perp } +\xi \tanh \left (k_{\perp }d\right )}\right )\!. }$$

(5.126)

As we have noted above, the potential A = 0 in the atmosphere, so that the jump of function A across the E layer equals the value of A just above E-layer, that is $\left [A\left (0\right )\right ] = A\left (0+\right )$. According to Eqs. (5.80) and (5.18), the jump of A is given by

$$\displaystyle{ \left [A\left (0\right )\right ] = \frac{\left [\partial _{z}\Phi \left (0+\right )\right ]} {i\omega } = \frac{\beta _{1}\Phi (0)} {V _{AI}}, }$$

(5.127)

Substituting Eqs. (5.126) and (5.127) for the jump of functions A and $\partial _{z}\Psi$ into boundary conditions (5.120) and (5.121), we are thus left with the set

$$\displaystyle{ \frac{ix_{0}\alpha _{H}} {L} \Psi \left (0\right ) -\frac{\left (\beta _{1} +\alpha _{P}\right )} {V _{AI}} \Phi \left (0\right ) = f_{1}, }$$

(5.128)

$$\displaystyle{ \frac{\alpha _{H}} {V _{AI}}\Phi \left (0\right ) + \frac{\left (ix_{0}\alpha _{P} - s\right )} {L} \Psi \left (0\right ) = f_{2}, }$$

(5.129)

where the dimensionless frequency x ₀ is again defined in Eq. (5.17), and the functions f ₁ and f ₂ are given by Eqs. (5.122) and (5.123). Equations (5.128) and (5.129) can be solved for $\Psi \left (0\right )$.

Appendix E: Solutions of the Axially Symmetrical Problem

3.1 TM Mode in the Neutral Atmosphere and in the Ground

In Sect. 5.3 we study the electromagnetic field excited by the vertical CG lightning discharge which is located on the vertical z axis in the neutral atmosphere. The problem is axially symmetrical since the geomagnetic field B ₀ is assumed to be directed vertically upward. The components of the electromagnetic perturbations can be expressed through potential functions A, $\Phi$ and $\Psi$ in cylindrical coordinates $z,r,\varphi$ via Eqs. (5.95)–(5.100). For the axially symmetrical problem these equations are simplified to

$$\displaystyle{ \delta B_{\varphi } = -\partial _{r}A,\ \ \ E_{r} = -\partial _{r}\Phi,\ \ \ E_{z} = -\partial _{z}\Phi + i\omega A, }$$

(5.130)

and

$$\displaystyle{ E_{\varphi } = -i\omega \partial _{r}\Psi,\ \ \ \delta B_{r} = \partial _{rz}^{2}\Psi,\ \ \ \delta B_{ z} = -\frac{1} {r}\partial _{r}\left (r\partial _{r}\Psi \right )\!. }$$

(5.131)

To treat the TM mode generated by the vertical CG discharge in the atmosphere, Maxwell’s equations are required, which are given by the set of Eqs. (5.48)–(5.50). As is seen from Eq. (5.130) the TM mode components $\delta B_{\varphi }$, E _r , and E _z are represented by the potentials $\Phi$ and A and do not depend on $\Psi$. Substituting these components into Eqs. (5.48)–(5.50) and rearranging, we obtain that Eq. (5.50) is reduced to identity, while Eqs. (5.48) and (5.49) take the forms

$$\displaystyle{ \partial _{z}A = \frac{i\omega } {c^{2}}\Phi, }$$

(5.132)

$$\displaystyle{ \partial _{zz}^{2}A + \frac{1} {r}\partial _{r}\left (r\partial _{r}A\right ) + \frac{\omega ^{2}} {c^{2}}A = -\frac{\mu _{0}m\left (\omega \right )} {2\pi r} \delta \left (z + d - h\right )\delta \left (r\right )\!, }$$

(5.133)

where $m\left (\omega \right )$ stands for Fourier transform of the lightning current moment and − d < z < 0.

We shall seek for the solution of Eqs. (5.132) and (5.133) in the form of Bessel transform, for example,

$$\displaystyle{ A\left (k,z,\omega \right ) =\int \limits _{ 0}^{\infty }A\left (r,z,\omega \right )J_{ 0}\left (kr\right )rdr, }$$

(5.134)

where $J_{0}\left (kr\right )$ is the Bessel function of the first kind of zero order and the parameter k of the Bessel transform plays a role of the perpendicular wave number. For brevity, here we made use of the same designation for the potential $A\left (r,z,\omega \right )$ and its Bessel transform $A\left (k,z,\omega \right )$. The same representation is true for the potentials $\Phi$ and $\Psi$.

Applying the Bessel transform to Eq. (5.133) we obtain

$$\displaystyle{ \partial _{z}^{2}A - k_{ a}^{2}A = -\frac{\mu _{0}m\left (\omega \right )} {2\pi } \delta \left (z + d - h\right )\!, }$$

(5.135)

where $k_{a}^{2} = k^{2} -\omega ^{2}/c^{2}$, and $A = A\left (k,z,\omega \right )$. We shall restrict our study to the low-frequency limit when k _a  ≈ k. Integrating Eq. (5.135) from $z = h - d-\varepsilon$ to $z = h - d+\varepsilon$ and then formally taking $\varepsilon \rightarrow 0$, we come to the following condition at $z = h - d$

$$\displaystyle{ \left [\partial _{z}A\right ] = -\frac{\mu _{0}m\left (\omega \right )} {2\pi }, }$$

(5.136)

where the square brackets denote the jump of z-derivative of A at $z = h - d$. The solution of Eq. (5.135) in both regions, z < h − d and z > h − d, should be matched via the condition (5.136). The solution of Eq. (5.135) under the requirement that A is continuous at that boundary has the form

$$\displaystyle{ A = C_{1}\exp \left (kz\right ) + C_{2}\exp \left (-kz\right ) -\frac{\mu _{0}m\left (\omega \right )\eta \left (z^{{\prime}}\right )} {2\pi k} \sinh \left (kz^{{\prime}}\right ) }$$

(5.137)

where C ₁ and C ₂ are arbitrary constants, $z^{{\prime}} = z + d - h$, and $\eta \left (z^{{\prime}}\right )$ is the step-function; i.e., η = 1 if z ^′ > 0 and η = 0 if z ^′ < 0.

Now we shall treat electromagnetic fields in the ground, which is considered as a uniform conducting half-space $\left (z < -d\right )$ with constant conductivity σ _g . For the axially symmetrical problem Maxwell’s equations for the ground are given by Eqs. (5.51) and (5.52). Substituting Eq. (5.130) and for the TM mode into those equations, yields

$$\displaystyle{ \partial _{z}A = -\mu _{0}\sigma _{g}\Phi, }$$

(5.138)

$$\displaystyle{ \partial _{zz}^{2}A + \frac{1} {r}\partial _{r}\left (r\partial _{r}A\right ) + i\omega \mu _{0}\sigma _{g}A = 0. }$$

(5.139)

Taking the Bessel transform of Eq. (5.139) we get

$$\displaystyle{ \partial _{z}^{2}A -\xi ^{2}A = 0, }$$

(5.140)

where $A = A\left (k,z,\omega \right )$ and as before $\xi ^{2} = k^{2} - i\mu _{0}\omega \sigma _{g}$ stands for the propagation factor in the ground. We mention in passing that Eq. (5.140) is analogous to Eq. (5.113) for the scalar potential $\Psi$ in the ground. Taking into account that A should tend to zero when z → −∞ we chose the solution of Eq. (5.140) in the form

$$\displaystyle{ A = A\left (-d\right )\exp \left [\xi \left (z + d\right )\right ], }$$

(5.141)

where Reξ > 0 and $A\left (-d\right )$ denotes the value of potential A on the ground surface.

The components $\delta B_{\varphi }$ and E _r must be continuous at $z = -d$ whence it follows that A and $\Phi$ must be continuous at $z = -d$. Taking into account Eqs. (5.132) and (5.138) for A and $\Phi$, the boundary condition on the ground surface takes the following form

$$\displaystyle{ \partial _{z}A\left (-d + 0\right ) = -\frac{i\omega \varepsilon _{0}} {\sigma _{g}} \partial _{z}A\left (-d - 0\right )\!. }$$

(5.142)

Taking into account the boundary conditions at $z = -d$ and combining Eqs. (5.137) and (5.141) gives a set of algebraic equations for undefined constants. These equations can be solved for C ₁ and C ₂ to yield

$$\displaystyle{ C_{1} = \frac{A\left (-d\right )} {2} \left (1 - \frac{i\omega \varepsilon _{0}\xi } {k\sigma _{g}}\right )\exp \left (kd\right )\!, }$$

(5.143)

$$\displaystyle{ C_{2} = \frac{A\left (-d\right )} {2} \left (1 + \frac{i\omega \varepsilon _{0}\xi } {k\sigma _{g}}\right )\exp \left (-kd\right )\!. }$$

(5.144)

Substituting Eqs. (5.143) and (5.144) into Eq. (5.137) yields the solution of the problem in the neutral atmosphere

$$\displaystyle\begin{array}{rcl} A& =& A\left (-d\right )\left \{\cosh \left [k\left (d + z\right )\right ] -\chi \sinh \left [k\left (d + z\right )\right ]\right \} \\ & & -\,\frac{\mu _{0}m\left (\omega \right )\eta \left (z^{{\prime}}\right )} {2\pi k} \sinh \left (kz^{{\prime}}\right )\!. {}\end{array}$$

(5.145)

Here we introduce the dimensionless parameter $\chi = i\omega \varepsilon _{0}\xi /\left (\sigma _{g}k\right )$. Substituting Eq. (5.145) for A into Eq. (5.132), we obtain

$$\displaystyle\begin{array}{rcl} \Phi & =& \frac{kc^{2}} {i\omega } \left [A\left (-d\right )\left \{\sinh \left [k\left (d + z\right )\right ] -\chi \cosh \left [k\left (d + z\right )\right ]\right \}\right. \\ & & \left.-\,\frac{\mu _{0}m\left (\omega \right )\eta \left (z^{{\prime}}\right )} {2\pi k} \cosh \left (kz^{{\prime}}\right )\right ]. {}\end{array}$$

(5.146)

Choosing the typical parameters $\sigma _{g} = 10^{-3}$ S/m, ω = 2π Hz, $k = 10^{-2} \textendash 10^{-4}$ km⁻¹, one can find that $\chi \approx 10^{-5} \textendash 10^{-7}$, i.e., this value can be neglected. This means that the TM mode in the atmosphere is practically independent of the ground conductivity, but that is not the case for the TE mode treated below in any details.

As is seen from Eq. (5.146), there is a discontinuity in the potential $\Phi$ at $z = h - d$ due to the presence of a step function $\eta \left (z^{{\prime}}\right )$ in the last term. This leads to an interesting question of how this discontinuity may influence the electric and magnetic fields in Eqs. (5.130) and (5.131). In other words, this gives rise to the question of whether the spatial field representation will save the continuity despite the presence of the discontinuity in the potential $\Phi$. Some insight into this problem can be achieved by taking into account that this discontinuity results from the use of the point dipole approach for the lightning discharge. In order to clarify the problem, consider for simplicity a point electric dipole immersed in an infinite space. In such a case the solution of the problem can be written as

$$\displaystyle{ b_{\varphi } = \frac{m\left (\omega \right )} {4\pi } \exp \left (-k\left \vert z^{{\prime}}\right \vert \right )\!, }$$

(5.147)

and

$$\displaystyle{ e_{r} = \pm \frac{ikm\left (\omega \right )} {4\pi \varepsilon _{0}\omega } \exp \left (-k\left \vert z^{{\prime}}\right \vert \right )\!, }$$

(5.148)

where the sign plus in Eq. (5.148) corresponds to z′ > 0 whereas the sign minus corresponds to z′ < 0. Applying the inverse Bessel transform to Eq. (5.148) we get

$$\displaystyle{ e_{r} = \pm \frac{im\left (\omega \right )} {4\pi \varepsilon _{0}\omega } \int \limits _{0}^{\infty }k^{2}\exp \left (-k\left \vert z^{{\prime}}\right \vert \right )J_{ 1}\left (kr\right )dk = \frac{im\left (\omega \right )rz^{{\prime}}} {4\pi \varepsilon _{0}\omega \left (z^{{\prime}2} + r^{2}\right )^{5/2}}. }$$

(5.149)

As is seen from Eq. (5.149), we have obtained the function, which is continuous everywhere except for the point $z{\prime} = r = 0$. This means that despite the discontinuity in Eq. (5.148) at z′ = 0 the inverse Bessel transform gives the continuous radial electric field. It follows from this example that the Bessel transform of the field of the point source may be represented by the discontinuous function. So one may expect that the inverse Bessel transform of Eq. (5.146) will result in a continuous function describing a spatial field representation.

3.2 TE Mode in the Neutral Atmosphere and in the Ground

As we have noted in Chap. 5, the TE mode in the atmosphere may occur due to the excitation of secondary sources such as the Hall current in the ionosphere. Since there are no sources of the TE mode in the atmosphere the Maxwell equations (4.1) and (4.2) can be written as

$$\displaystyle\begin{array}{rcl} \partial _{r}\delta B_{z} - \partial _{z}\delta B_{r} = \frac{i\omega } {c^{2}}E_{\varphi },& &{}\end{array}$$

(5.150)

$$\displaystyle\begin{array}{rcl} \partial _{z}E_{\varphi } = -i\omega \delta B_{r},& &{}\end{array}$$

(5.151)

$$\displaystyle\begin{array}{rcl} \partial _{r}E_{\varphi } = i\omega \delta B_{z}.& &{}\end{array}$$

(5.152)

Substituting Eqs. (5.130) and (5.131) into Eqs. (5.150)–(5.152), one can express the TE mode components through the potential $\Psi$. As a result we come to the single wave equation for the potential $\Psi$

$$\displaystyle{ r^{-1}\partial _{ r}\left (r\partial _{r}\Psi \right ) + \partial _{zz}^{2}\Psi = -\frac{\omega ^{2}} {c^{2}}\Psi. }$$

(5.153)

Applying Bessel transform to this equation we get

$$\displaystyle{ \partial _{zz}^{2}\Psi - k_{ a}^{2}\Psi = 0,\qquad \left (-d < z < 0\right )\!, }$$

(5.154)

where $k_{a}^{2} = k^{2} -\omega ^{2}/c^{2} \approx k^{2}$. It should be noted that if one changes the parameter k by the perpendicular “wave number” k _⊥, Eq. (5.154) coincides with Eq. (5.112) for the case of “plane” atmosphere. Similarly, one can derive an equation for the ground that is completely coincident with Eq. (5.113). This means that the solution of the plane problem given by Eqs. (5.115) and (5.116) holds true in the axially symmetrical case.

3.3 The Ionosphere and Magnetosphere

In the model the space z > 0 consists of a solely cold collisionless plasma, which is described by Maxwell equations (4.2) and (5.2) and the plasma dielectric permittivity tensor (2.18). When the cylindrical coordinates are applied, the Maxwell equations are split into two independent sets of equations. The first set includes the components of the shear Alfvén waves, i.e., E _r , E _z and $\delta B_{\varphi }$

$$\displaystyle\begin{array}{rcl} \partial _{z}\delta B_{\varphi } = \frac{i\omega } {V _{A}^{2}}E_{r},& &{}\end{array}$$

(5.155)

$$\displaystyle\begin{array}{rcl} \frac{1} {r}\partial _{r}\left (r\delta B_{\varphi }\right ) = -\frac{i\omega \varepsilon _{\Vert }} {c^{2}}E_{z},& &{}\end{array}$$

(5.156)

$$\displaystyle\begin{array}{rcl} \partial _{z}E_{r} - \partial _{r}E_{z} = i\omega \delta B_{\varphi }.& &{}\end{array}$$

(5.157)

The second one is for the three components of the FMS wave, i.e., $E_{\varphi }$, δ B _r , and δ B _z

$$\displaystyle\begin{array}{rcl} \partial _{r}\delta B_{z} - \partial _{z}\delta B_{r} = \frac{i\omega } {V _{A}^{2}}E_{\varphi },& &{}\end{array}$$

(5.158)

$$\displaystyle\begin{array}{rcl} \partial _{z}E_{\varphi } = -i\omega \delta B_{r},& &{}\end{array}$$

(5.159)

$$\displaystyle\begin{array}{rcl} \partial _{r}E_{\varphi } = i\omega \delta B_{z}.& &{}\end{array}$$

(5.160)

Since $\varepsilon _{\Vert }$ has been assumed to be infinite, the parallel electric field in the magnetosphere becomes infinitesimal, i.e., E _z  → 0. As before, the electromagnetic field is derivable by the scalar potentials A, $\Phi$, and $\Psi$ through Eqs. (5.130), (5.131), and (5.80). Substituting these equations into the set of Eqs. (5.155)–(5.160), and rearranging under the requirement that $i\omega A = \partial _{z}\Phi$, yields

$$\displaystyle{ \partial _{z}^{2}\Phi + \frac{\omega ^{2}} {V _{A}^{2}}\Phi = 0, }$$

(5.161)

$$\displaystyle{ \partial _{zz}^{2}\Psi + \frac{1} {r}\partial _{r}\left (r\partial _{r}\Psi \right ) = - \frac{\omega ^{2}} {V _{A}^{2}}\Psi. }$$

(5.162)

Applying Bessel transforms to Eqs. (5.161) and (5.162), we come to the equations that are completely similar to Eqs. (5.12) and (5.13) for the shear Alfvén and compressional waves in the plane problem.

3.4 E Layer of the Ionosphere

In order to obtain the boundary conditions at the bottom of the ionosphere we now consider the conductive E layer of the ionosphere. In the framework of the axially symmetrical problem the neutral wind velocity is assumed to be independent of azimuthal angle $\varphi$ in the ionosphere. Substituting Eq. (5.24) for the current density into Ampere’s equation (1.5), taking the notice of $\mathbf{B =}\delta \mathbf{B + B}_{0}$, and using cylindrical coordinates, we obtain

$$\displaystyle{ \mu _{0}^{-1}\partial _{ z}\delta B_{\varphi } =\sigma _{H}\left (E_{\varphi } - V _{r}B_{0}\right ) -\sigma _{P}\left (E_{r} + V _{\varphi }B_{0}\right )\!, }$$

(5.163)

$$\displaystyle{ \mu _{0}^{-1}\left (\partial _{ z}\delta B_{r} - \partial _{r}\delta B_{z}\right ) =\sigma _{P}\left (E_{\varphi } - V _{r}B_{0}\right ) +\sigma _{H}\left (E_{r} + V _{\varphi }B_{0}\right )\!, }$$

(5.164)

where σ _P and σ _H are the Pedersen and Hall conductivities, and V _r and $V _{\varphi }$ are the components of the neutral flow velocity. In what follows we use a thin E layer approximation, which is valid if a typical thickness of the E layer, l, is much smaller than the skin-depth in the ionosphere $l_{s} \sim \left (\mu _{0}\sigma _{P}\omega \right )^{-1/2}$. Integrating Eqs. (5.163) and (5.164) with respect to z across the E layer from z = 0 to z = l and making formally l → 0, gives the boundary conditions at z = 0

$$\displaystyle{ \mu _{0}^{-1}\left [\delta B_{\varphi }\right ] = \Sigma _{ H}E_{\varphi } - \Sigma _{P}E_{r} - B_{0}\left (\Sigma _{H}V _{r} + \Sigma _{P}V _{\varphi }\right )\!, }$$

(5.165)

$$\displaystyle{ \mu _{0}^{-1}\left [\delta B_{ r}\right ] = \Sigma _{P}E_{\varphi } + \Sigma _{H}E_{r} + B_{0}\left (\Sigma _{H}V _{\varphi } - \Sigma _{P}V _{r}\right )\!, }$$

(5.166)

where the square brackets denote the jump of magnetic field across the E-layer, and the height-integrated Pedersen and Hall conductivities, $\Sigma _{P}$ and $\Sigma _{H}$, are given by Eq. (5.25). As before the wind velocity V is assumed to be independent of the z-coordinate.

Substituting Eqs. (5.130) and (5.131) for the potentials into Eqs. (5.165) and (5.166) and applying a Bessel transform to these equations, we get

$$\displaystyle{ \left [A\left (0\right )\right ] = \frac{i\alpha _{H}x_{0}} {L} \Psi \left (0\right ) - \frac{\alpha _{P}} {V _{AI}}\Phi \left (0\right ) - F_{1}, }$$

(5.167)

$$\displaystyle{ \left [\partial _{z}\Psi \left (0\right )\right ] = -\frac{i\alpha _{P}x_{0}} {L} \Psi \left (0\right ) - \frac{\alpha _{H}} {V _{AI}}\Phi \left (0\right ) + F_{2}, }$$

(5.168)

where the Bessel transform of the potentials A, $\Phi$ and $\Psi$ are given by Eq. (5.134). Here as before $\alpha _{P} = \Sigma _{P}/\Sigma _{w}$ and $\alpha _{H} = \Sigma _{H}/\Sigma _{w}$ are the ratios of the height-integrated Pedersen and Hall conductivities to the wave conductivity $\Sigma _{w} = \left (\mu _{0}V _{AI}\right )^{-1}$, and the dimensionless frequency x ₀ is again defined in Eq. (5.16). Additionally we made use of the following abbreviations:

$$\displaystyle{ F_{1} = \frac{B_{0}} {kV _{AI}}\left (\alpha _{H}v_{r} +\alpha _{P}v_{\varphi }\right )\!, }$$

(5.169)

$$\displaystyle{ F_{2} = \frac{B_{0}} {kV _{AI}}\left (\alpha _{P}v_{r} -\alpha _{H}v_{\varphi }\right )\!, }$$

(5.170)

where v _r and $v_{\varphi }$ are the Bessel transform of the radial and azimuthal components of the wind velocity

$$\displaystyle{ v_{r,\varphi }\left (\omega,k\right ) =\int \limits _{ 0}^{\infty }V _{ r,\varphi }\left (\omega,r\right )rJ_{1}\left (kr\right )dr. }$$

(5.171)

It is interesting to note that Eqs. (5.167)–(5.170) are identical to Eqs. (5.120)–(5.123) if the parameter k _⊥ is replaced by the factor $\mathbf{k} = ik\hat{\mathbf{r}}$ where $\hat{\mathbf{r}} = \mathbf{r}/r$ is a unite vector directed along the vector r.

3.5 Electromagnetic Perturbations at the Ground Surface

We start with calculation of the jump of the potential A across the E layer. Taking into account that the value $A\left (0+\right )$ can be found from Eq. (5.127) and combining this equation with Eq. (5.53) we obtain that

$$\displaystyle{ \left [A\left (0\right )\right ] = \frac{\beta _{1}} {V _{AI}}\Phi \left (0\right ) - A\left (-d\right )\cosh \left (kd\right ) + \frac{\mu _{0}m\left (\omega \right )} {2\pi k} \sinh \left \{k\left (d - h\right )\right \}. }$$

(5.172)

Substituting Eq. (5.172) into Eq. (5.167) and rearranging leads to

$$\displaystyle{ \frac{ix_{0}\alpha _{H}} {L} \Psi \left (0\right ) -\frac{\left (\beta _{1} +\alpha _{P}\right )} {V _{AI}} \Phi \left (0\right ) = f, }$$

(5.173)

where

$$\displaystyle{ f = \frac{B_{0}} {kV _{AI}}\left (\alpha _{H}v_{r} +\alpha _{P}v_{\varphi }\right ) - A\left (-d\right )\cosh \left (kd\right ) + \frac{\mu _{0}m\left (\omega \right )} {2\pi k} \sinh \left \{k\left (d - h\right )\right \}. }$$

(5.174)

Here one can see an analogy between Eqs. (5.173) and (5.128), which was derived for the plane problem. These two equations differ only in the source functions which stay on the right-hand sides of these equations.

As has already been intimated, considering an analogy between the plane and cylindrical problems, Eq. (5.168) can be reduced to the equation analogous to Eq. (5.129), i.e.

$$\displaystyle{ \frac{\alpha _{H}} {V _{AI}}\Phi \left (0\right ) + \frac{\left (ix_{0}\alpha _{P} - s\right )} {L} \Psi \left (0\right ) = \frac{B_{0}} {kV _{AI}}\left (\alpha _{P}v_{r} -\alpha _{H}v_{\varphi }\right )\!, }$$

(5.175)

Finally, one should take into account the continuity of the potential $\Phi$ at z = 0. As it follows from Eq. (5.132)

$$\displaystyle{ \Phi \left (0\right ) = \frac{kc^{2}} {i\omega } \left \{A\left (-d\right )\sinh \left (kd\right ) -\frac{\mu _{0}m\left (\omega \right )} {2\pi k} \cosh \left [k\left (d - h\right )\right ]\right \}. }$$

(5.176)

The set of Eqs. (5.173), (5.175), and (5.176) can be solved for $A\left (-d\right )$, $\Phi \left (0\right )$ and $\Psi \left (0\right )$ to yield

$$\displaystyle{ A\left (-d\right ) \approx \frac{\mu _{0}m\left (\omega \right )\cosh \left [k\left (d - h\right )\right ]} {2\pi k\sinh \left (kd\right )}, }$$

(5.177)

$$\displaystyle\begin{array}{rcl} \Psi \left (0\right )& \approx & \frac{iL} {kq}\left \{\frac{\mu _{0}\alpha _{H}m\left (\omega \right )\cosh \left (kh\right )} {2\pi \sinh \left (kd\right )} \right. \\ & & \left.+\, \frac{B_{0}} {V _{AI}}\left [\left (\alpha _{H}^{2} +\alpha _{ P}^{2} +\beta _{ 1}\alpha _{P}\right )v_{r} -\beta _{1}\alpha _{H}v_{\varphi }\right ]\right \},{}\end{array}$$

(5.178)

where β ₁ and q are given by Eqs. (5.21) and (5.34). In deriving Eqs. (5.177) and (5.178) we have neglected the terms which contain the factor $\left (V _{AI}/c\right )^{2} \approx 10^{-6} \ll 1$. The potential $\Phi$ is derivable from Eq. (5.177) through

$$\displaystyle{ \Phi \left (-d\right ) = -\frac{i\xi A\left (-d\right )} {\mu _{0}\sigma _{g}}. }$$

(5.179)

Combining Eqs. (5.29) and (5.178) we obtain

$$\displaystyle\begin{array}{rcl} \Psi \left (-d\right )& =& \frac{iL} {kq\beta _{3}}\left \{\frac{\mu _{0}\alpha _{H}m\left (\omega \right )\cosh \left (kh\right )} {2\pi \sinh \left (kd\right )} \right. \\ & & \left.+\, \frac{B_{0}} {V _{AI}}\left [\left (\alpha _{H}^{2} +\alpha _{ P}^{2} +\beta _{ 1}\alpha _{P}\right )v_{r} -\beta _{1}\alpha _{H}v_{\varphi }\right ]\right \},{}\end{array}$$

(5.180)

where the function β ₃ is given by Eq. (5.31).

Applying a Bessel transform to Eqs. (5.130) and (5.131) one can express the components of electromagnetic field through the potentials. The derivatives $\partial _{z}\Psi$ and $\partial _{z}\Phi$ in Eqs. (5.130) and (5.131) are derivable from Eqs. (5.114) and (5.146) via $\partial _{z}\Psi \left (-d\right ) =\xi \Psi \left (-d\right )$ and $\partial _{z}\Phi \left (-d\right ) = -ik^{2}c^{2}A\left (-d\right )/\omega$. On account of these expressions we get

$$\displaystyle{ b_{r}\left (-d\right ) = -k\xi \Psi \left (-d\right )\!,\ \ b_{\varphi }\left (-d\right ) = kA\left (-d\right )\!,\ \ b_{z}\left (-d\right ) = k^{2}\Psi \left (-d\right )\!, }$$

(5.181)

$$\displaystyle{ e_{r}\left (-d\right ) = k\Phi \left (-d\right )\!,\ \ e_{\varphi }\left (-d\right ) = i\omega k\Psi \left (-d\right )\!,\ \ e_{z}\left (-d\right ) \approx ik^{2}c^{2}A\left (-d\right )/\omega. }$$

(5.182)