Coherent Curvature Radio Emission and Polarization from Pulsars

Although almost 53 yr have passed since the pulsar discovery, their radio emission mechanism still remains unknown. This is one of the most challenging problems of modern astrophysics. Curvature radiation seems to be the most natural and practically unavoidable emission process in the pulsar magnetosphere. Pulsars emit radio waves in an intense, narrow beam that sweeps past the observer once per rotation period of the neutron star to produce the observed pulses (Hewish et al. 1968; Manchester & Taylor 1977; Lyne & Smith 2006). Pulsars are believed to produce radio waves in the open field line region of the polar cap. The shapes of average pulses are quite unique for any given pulsar, but the single pulses display erratic behavior on a timescale faster than the spin period. The pulse components' width, shape, and phase location in the pulsar average profiles are expected to reflect the emission region arrangement in the pulsar magnetosphere (Rankin 1983). The pulsar radiation is highly polarized in general, and the circular polarization is generally strongest in the central regions of profiles but is by no means confined to these regions (Han et al. 1998). The polarization angle (PA) swings like the "S"-shaped curve, explained by the rotating vector model (Radhakrishnan & Cooke 1969).

Gangadhara (2010) developed an incoherent curvature radiation model by including viewing geometry in the pulsar dipolar magnetosphere and thereby explained the polarization of pulsar radio emission and also showed that the correlation between PA swing and the sense reversal of circular polarization (Radhakrishnan & Rankin 1990) is a geometric property of the emission process. The incoherent curvature radiation model of Gangadhara (2010) was further developed by Kumar & Gangadhara (2012a, 2012b, 2013) and Wang et al. (2012) to include the perturbations, such as the aberration–retardation (A/R) effects and polar cap currents, and shown that they can change the pulse shapes and polarization.

Pulsars provide a clear case of coherent radio emission among celestial bodies. The very high brightness temperature 10²⁴–10³¹ K (e.g., Cordes 1979) of radio pulsars implies that the emission is coherent. Since the radio emission is extremely bright, it requires a coherent emission mechanism because incoherent emission, in which each of the sources (plasma particles) radiates independently of the others, is inadequate (e.g., Sturrock 1971; Ruderman & Sutherland 1975; Totani 2013). The beamed radio emission from relativistic plasma particles, which are constrained to move along the curved trajectories, occurs in the direction of the velocity (e.g., Jackson 1999). The tangent model, which was developed and analyzed by Gangadhara (2004), is exact in the sense that the line of sight always remains tangent to the field line (Yuen & Melrose 2014).

The mechanism of coherent emission in pulsars is still an outstanding problem. By coherence, it is meant that the emitted radio waves are in phase, and mathematically speaking, coherent radiation can be represented by an almost perfect sine wave at a well-defined frequency. However, Katz (2018) showed that the coherent radiation need not be an almost perfect sine wave. In fact, in both pulsars and fast radio bursts (FRBs), it is usually fairly broadband (more so in pulsars than FRBs, where it may be confined to bands with Δν/ν ∼ 0.03). If there is a spatial clumpy radiating charge distribution, then it can produce the observed spectral structure; in that case, the characteristic curvature radiation frequency could be higher than the observed frequencies.

The collective plasma radiation process acts coherently on a timescale of nanoseconds and below in generating pulsar radio waves. Coherent curvature radiation has been considered as a natural emission mechanism for interpreting pulsar observations (Ruderman & Sutherland 1975). Based on the ideas of Sturrock (1971), they ascribed the radiation to coherence induced by a specific beam-plasma instability in the pulsar magnetosphere. The pulsar radio emission is believed to arise due to the growth of plasma instabilities in the relativistic secondary electron–positron plasma streaming out along the curved open dipolar magnetic field lines. Ruderman & Sutherland (1975) developed a magnetospheric model; however, the beam-plasma instability they envisioned does not grow rapidly enough to produce appreciable coherence (Buschauer & Benford 1976). The simplified expression for the radiation electric field (Equation (14.67) in Jackson 1999, used as Equation (17) by Tridb & Gangadhara 2019) is independent of acceleration; hence, the predictions of polarization are very approximate and cannot predict the correlation between PA swing and sense reversal of circular polarization. The models of coherent curvature radiation that have been developed in the literature consider a single arbitrary curved trajectory having almost no relevance to dipolar field geometry and do not consider the polarization of the emitted radiation. Hence, it is hard to make any polarization predictions that can be tested in pulsar observations.

The mystery of coherent radio emission from pulsars is worth solving, as it resolves one of the outstanding problems in pulsar astronomy. The pulsar magnetosphere consists of relativistic plasma, which can support plasma instabilities. The growth of instabilities due to wave–particle or wave–wave interactions can lead to large-amplitude plasma waves, which can impose correlations among particles. Further, if those particles are accelerated, they can radiate electromagnetic (EM) waves nearly in phase (e.g., Katz 2014; Ghisellini & Locatelli 2018). This condition is easily satisfied when the velocities are equal and aligned with the dipolar magnetic field line tangents. It is necessary to understand the general theoretical framework, including pulsar electrodynamics, the properties of the pulsar plasma that populates the relevant regions of a pulsar magnetosphere, and the properties of the wave modes excited (generated). So far, there has been no theory applied to the concept of coherence to broadband radiation with a wide and continuous range of frequencies and phases. In this regard, the initiation and progress made by Katz (2018) are essential. To date, there are no satisfactory explanations for the enigmatic behavior of pulsar radiation. It is of particular interest from the plasma physics viewpoint to understand how familiar plasma physics concepts and methods need to be modified and adapted to the extreme environment of the pulsar magnetosphere. Cordes (1979) reviewed the pulsar radio observations with a particular emphasis on the properties of radio emission relevant for understanding the emission mechanism.

In this paper, we consider a relativistic plasma moving along a curved trajectory with a radius of curvature ρ much longer than the wavelengths. We consider resonance as a phenomenon in which a small-amplitude driving force (charge density and current perturbations) could produce large-amplitude EM waves. These longitudinal electrostatic waves order the particles to phase up to emitting coherently. In Section 2, we develop the coherent curvature model and present the relevant analytical formulation. In Section 3, we simulate the pulse profiles of a typical pulsar and estimate the brightness temperature by considering different viewing geometries. Finally, we give the discussion and conclusion in Sections 4 and 5, respectively.

The rotationally induced electric field near the polar cap is believed to pull out electrons from the neutron star surface. They are called primary charged particles having a Lorentz factor γ_p ∼ 10⁶ (Ruderman & Sutherland 1975). In the presence of the strong magnetic field, these particles radiate γ-rays via synchrotron and curvature emissions, which in turn decay into secondary pair (e⁺, e⁻) plasma, and the process continues until the secondary plasma Lorentz factor reaches γ ∼ 10²–10³. Some of the secondary electrons are expected to go back to the star surface to neutralize the neutron star from getting positively charged. The interaction of primary plasma with the secondary pair plasma can excite an instability such as the electrostatic-streaming instability in the secondary pair plasma (Buschauer & Benford 1976). The electrostatic instabilities make the (e⁺, e⁻) plasma into a phased array, which behaves like a radiating antenna (e.g., Lu & Kumar 2018) This proves to be much more efficient than the usual local picture of coherence, which extends only over one wavelength. The mathematical details related to the estimation of the radiation electric field due to the accelerated charged particles are given, for example, in Jackson (1999) and Gangadhara (2010). Here we extend them to the collective plasma emission.

2.1. Radiation Electric Field due to an Accelerated Plasma

For certain radiating systems, by choosing geometry involving symmetry, it is possible to find analytic/semianalytic solutions for the radiation fields and energy. The constrained (one-dimensional) motion of relativistic plasma along the superstrong magnetic field can be treated in analogy with the radiating current that flows in a finite conducting medium of simple geometry, such as in thin antennas of arbitrary length. Consider a Cartesian coordinate system XYZ, such that the Z-axis is parallel to the rotation axis $\hat{{\rm{\Omega }}}$ and the XZ-plane defines the fiducial plane. The line of sight can be chosen, without loss of generality, to lie in the XZ-plane,

$$\begin{eqnarray}&&\hat{n}=\{\sin \zeta ,0,\cos \zeta \},\end{eqnarray} \tag{ 1 }$$

where ζ = α + ϱ, and α is the inclination angle of the magnetic axis $\hat{m}$ with respect to $\hat{{\rm{\Omega }}},$ and ϱ is the line-of-sight impact angle with respect to $\hat{m}.$ We shall treat the system XYZ as an inertial frame or a laboratory frame in which the neutron star spins (see Figure 1). Next, let us consider a coordinate system $x^{\prime} y^{\prime} z^{\prime}$ as the rest frame of the plasma and another coordinate system, xyz, which is the fixed frame embedded with the field lines. The axes of the frame xyz are chosen in such a way that the x-axis is parallel to the field line tangent $\hat{b},$ the z-axis is antiparallel to the field line curvature vector $\hat{K}$, and the y-axis is parallel to the binormal ${\hat{b}}_{{\rm{N}}}=\hat{b}\times \hat{K}.$ The primed frame ${x}^{{\rm{{\prime} }}}{y}^{{\rm{{\prime} }}}{z}^{{\rm{{\prime} }}}$ moves with velocity v in the x-direction with respect to the unprimed frame xyz. The reference frames coincide at $t=t^{\prime} =0.$ Consider a plasma that moves relativistically along the curved dipolar magnetic field lines above the pulsar polar cap. Then the charge density and current density perturbations excited due to the instability in the plasma rest frame ${x}^{{\rm{{\prime} }}}{y}^{{\rm{{\prime} }}}{z}^{{\rm{{\prime} }}}$ can be expressed as

$$\begin{eqnarray}\begin{array}{rcl}\sigma ^{\prime} (x^{\prime} ,t^{\prime} ) & = & {\sigma }_{0}^{{\prime} }{e}^{i({k}_{0}^{{\prime} }x^{\prime} -{\omega }_{0}^{{\prime} }t^{\prime} )},\\ {\boldsymbol{J}}^{\prime} (x^{\prime} ,t^{\prime} ) & = & {{\boldsymbol{J}}}_{0}^{{\prime} }{e}^{i({k}_{0}^{{\prime} }x^{\prime} -{\omega }_{0}^{{\prime} }t^{\prime} )}\quad {\rm{for}}\quad | x^{\prime} | \leqslant \displaystyle \frac{{s}_{0}^{{\prime} }}{2},\end{array}\end{eqnarray} \tag{ 2 }$$

where ${\rho }_{0}^{{\prime} }$ and ${{\boldsymbol{J}}}_{0}^{{\prime} }$ are the amplitudes, ${s}_{0}^{{\prime} }$ is the length of the plasma column, and ${\omega }_{0}^{{\prime} }$ and ${k}_{0}^{{\prime} }$ are the frequency and wavenumber of the longitudinal plasma wave, respectively. The phase velocity of the plasma wave is ${v}_{{\rm{p}}}^{{\prime} }={\omega }_{0}^{{\prime} }/{k}_{0}^{{\prime} },$ which is taken to be ∼0, for ${v}_{{\rm{p}}}^{\prime} \ll c.$ Then the Lorentz transformation of the four-current ${J}_{\mu }^{{\prime} }=(c\sigma ^{\prime} ,{\boldsymbol{J}}^{\prime} )$ gives the charge and current densities in the frame xyz,

$$\begin{eqnarray}&&\sigma =\gamma \left(\sigma ^{\prime} +\displaystyle \frac{v}{{c}^{2}}{J}_{{\rm{x}}}^{\prime} \right),\quad {J}_{{\rm{x}}}=\gamma ({J}_{{\rm{x}}}^{\prime} +v\sigma ^{\prime} ),\end{eqnarray} \tag{ 3 }$$

where $\gamma =1/\sqrt{1-{v}^{2}/{c}^{2}}$ and c is the velocity of light. From the equation of continuity, we have ${\omega }_{0}^{{\prime} }{\sigma }_{0}^{{\prime} }={k}_{0}^{{\prime} }{J}_{0}^{{\prime} }.$ Since the phase of the plane wave is Lorentz invariant, the charge and current densities can be expressed as

$$\begin{eqnarray}\begin{array}{rcl}\sigma (x,t) & = & {\sigma }_{0}{e}^{i({k}_{0}x-{\omega }_{0}t)},\\ {\boldsymbol{J}}(x,t) & = & {{\boldsymbol{J}}}_{0}{e}^{i({k}_{0}x-{\omega }_{0}t)}\quad {\rm{for}}\\ {vt}-\displaystyle \frac{{s}_{0}}{2} & \leqslant & x\leqslant {vt}+\displaystyle \frac{{s}_{0}}{2},\end{array}\end{eqnarray} \tag{ 4 }$$

where

$$\begin{eqnarray}&&{\sigma }_{0}=\gamma {\sigma }_{0}^{{\prime} }\left(1+\displaystyle \frac{{{vv}}_{{\rm{p}}}^{{\prime} }}{{c}^{2}}\right),\quad {J}_{0}=\gamma {\sigma }_{0}^{{\prime} }\left(v+{v}_{{\rm{p}}}^{{\prime} }\right).\end{eqnarray} \tag{ 5 }$$

The radiation field due to an accelerated charge is given by (e.g., Jackson 1999)

$$\begin{eqnarray}\begin{array}{rcl}{\boldsymbol{E}}({\boldsymbol{r}},\omega ) & = & \displaystyle \frac{1}{\sqrt{2\pi }}\displaystyle \frac{q}{c}\underset{-\infty }{\overset{+\infty }{\displaystyle \int }}\displaystyle \frac{\hat{n}\times [(\hat{n}-{\boldsymbol{\beta }})\times \dot{{\boldsymbol{\beta }}}]}{R\,{\left(1-{\boldsymbol{\beta }}\cdot \hat{n}\right)}^{2}}\\ & & \times \ {e}^{i\omega \{{t}_{{\rm{r}}}+R({t}_{{\rm{r}}})/c\}}{{dt}}_{{\rm{r}}},\end{array}\end{eqnarray} \tag{ 6 }$$

where t_r = t − R(t_r)/c is the retarded time; t is the observer time; q, β = v /c, and $\dot{{\boldsymbol{\beta }}}=\dot{{\boldsymbol{v}}}/c$ are the charge, velocity, and acceleration of the particle, respectively; and R(t_r) is the distance between the source and the observer. It can be approximated as $R({t}_{{\rm{r}}})\simeq {R}_{0}-\hat{n}\cdot {\boldsymbol{r}}({t}_{{\rm{r}}})$, where r (t_r) is the position vector of the emission point at the retarded time. Now, Equation (6) reduces to

$$\begin{eqnarray}\begin{array}{rcl}{\boldsymbol{E}}({\boldsymbol{r}},\omega ) & = & \displaystyle \frac{1}{\sqrt{2\pi }}\displaystyle \frac{q}{{{cR}}_{0}}{e}^{{{iR}}_{0}\omega /c}\underset{-\infty }{\overset{+\infty }{\displaystyle \int }}\\ & & \times \ \displaystyle \frac{\hat{n}\times [(\hat{n}-{\boldsymbol{\beta }})\times \dot{{\boldsymbol{\beta }}}]}{{\left(1-{\boldsymbol{\beta }}\cdot \hat{n}\right)}^{2}}{e}^{i\omega (t-\hat{n}\cdot {\boldsymbol{r}}/c)}{dt},\end{array}\end{eqnarray} \tag{ 7 }$$

where the suffix "r" on the time variable has been omitted for brevity. To deduce the polarization state of the emitted radiation, we consider the equation of the radiation electric field, which includes acceleration (see Jackson 1999, Equation (14.65)), as it is consistent with the relativistic beaming; i.e., when $\hat{n}\parallel {\boldsymbol{\beta }},$ the intensity and linear polarization received by the observer go to maximum in the case of $\dot{{\boldsymbol{\beta }}}\perp {\boldsymbol{\beta }}.$

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We make the following assumptions:

• 1.  
  The relativistically beamed radio emission is received during the time t_em ∼ 2ρ/(γ κ c), where ρ is the radius of curvature of the local trajectory of the plasma. The parameter κ specifies the speed of the plasma column as a fraction of c, and it is about 1.
• 2.  
  The changes in the size and direction of the plasma packet's motion are minimal during the time t_em.
• 3.  
  The growth time of the instability is longer than t_em.
• 4.  
  The wave packet's phase variation relative to a pure harmonic wave is minimal over the packet's effective extent.

The position vector r of an arbitrary point on an inclined magnetic dipole field line in the Cartesian coordinate system XYZ is given by Gangadhara (2004),

$$\begin{eqnarray}&&{\boldsymbol{r}}={r}_{{\rm{e}}}{\sin }^{2}\theta \,\hat{r},\end{eqnarray} \tag{ 8 }$$

where the unit vector

$$\begin{eqnarray}\begin{array}{rcl}\hat{r} & = & \{\cos \theta \cos {\phi }^{{\rm{{\prime} }}}\sin \alpha +\sin \theta \\ & & \times \,(\cos \alpha \cos \phi \cos {\phi }^{{\rm{{\prime} }}}-\sin \phi \sin {\phi }^{{\rm{{\prime} }}}),\\ & & \cos {\phi }^{{\rm{{\prime} }}}\sin \theta \sin \phi +\sin {\phi }^{{\rm{{\prime} }}}\\ & & \times \,(\cos \theta \sin \alpha +\cos \alpha \cos \phi \sin \theta ),\\ & & \cos \alpha \cos \theta -\cos \phi \sin \alpha \sin \theta \},\end{array}\end{eqnarray} \tag{ 9 }$$

r_e is the field line constant, and $\phi ^{\prime}$ is the rotation phase in the lab frame. The angles θ and ϕ are the magnetic colatitude and azimuth, respectively. The field line tangent is given by b = ∂ r /∂θ. The velocity of the plasma in the frame xyz is given by ${\boldsymbol{u}}=\kappa c\,\hat{b},$ where $\kappa =\sqrt{1-(1/{\gamma }_{{\rm{u}}}^{2})}.$ The unit tangent vector $\hat{b}={\boldsymbol{b}}/| {\boldsymbol{b}}|$, and $| {\boldsymbol{b}}| ={r}_{{\rm{e}}}\sin \theta \sqrt{5+3\cos (2\theta )}/\sqrt{2}.$

We can generalize Equation (7) for the collection of charges accelerated in the plasma column, which is constrained to move along dipolar field lines. In the limit of the continuum distribution of point charges, we can replace

$$\begin{eqnarray}\begin{array}{rcl}{{qe}}^{-i(\omega /c)\hat{n}\cdot {\boldsymbol{r}}(t)} & \to & \displaystyle \int \sigma ({\boldsymbol{r}},t){e}^{-i(\omega /c)\hat{n}\cdot {\boldsymbol{r}}}{dV},\\ q{\boldsymbol{\beta }}{e}^{-i(\omega /c)\hat{n}\cdot {\boldsymbol{r}}(t)} & \to & \displaystyle \frac{1}{c}\displaystyle \int {\boldsymbol{J}}({\boldsymbol{r}},t){e}^{-i(\omega /c)\hat{n}\cdot {\boldsymbol{r}}}{dV}\end{array}\end{eqnarray} \tag{ 10 }$$

in Equation (7), where dV is the volume element. Then we have

$$\begin{eqnarray}\begin{array}{rcl}{\boldsymbol{E}}(\omega ) & = & \displaystyle \frac{1}{\sqrt{2\pi }}\displaystyle \frac{{e}^{{{iR}}_{0}\omega /c}}{{c}^{2}{R}_{0}}\underset{-\infty }{\overset{+\infty }{\iint }}\\ & & \times \ \displaystyle \frac{\hat{n}\times [(c\sigma \hat{n}-{\boldsymbol{J}})\times {\dot{{\boldsymbol{\beta }}}}_{u}]}{{\left(1-{{\boldsymbol{\beta }}}_{u}\cdot \hat{n}\right)}^{2}}{e}^{i\omega (t-\hat{n}\cdot {\boldsymbol{r}}/c)}{dt}\,{dV},\end{array}\end{eqnarray} \tag{ 11 }$$

where β _u = u /c and ${\dot{{\boldsymbol{\beta }}}}_{{\rm{u}}}=\dot{{\boldsymbol{u}}}/c$ are the normalized plasma velocity and acceleration in the lab frame.

Next, by substituting σ and J from Equation (4) into Equation (11), we obtain

$$\begin{eqnarray}\begin{array}{rcl}{\boldsymbol{E}}(\omega ) & = & \displaystyle \frac{1}{\sqrt{2\pi }}\displaystyle \frac{{e}^{{{iR}}_{0}\omega /c}}{{c}^{2}{R}_{0}}\underset{-\infty }{\overset{+\infty }{\iint }}{\sigma }_{0}\\ & & \times \ \displaystyle \frac{\hat{n}\times [(c\hat{n}-{v}_{{\rm{p}}}\hat{b})\times {\dot{{\boldsymbol{\beta }}}}_{u}]}{{\left(1-{{\boldsymbol{\beta }}}_{u}\cdot \hat{n}\right)}^{2}}\\ & & \times \ {e}^{i\{{k}_{0}x-k\hat{n}\cdot {\boldsymbol{r}}-({\omega }_{0}-\omega )t\}}{dt}\,{dV},\end{array}\end{eqnarray} \tag{ 12 }$$

where k = ω/c is the wavenumber of radiation, and ${v}_{{\rm{p}}}=(u+{v}_{{\rm{p}}}^{{\prime} })/(1+{{uv}}_{{\rm{p}}}^{{\prime} }/{c}^{2})\sim u$ is the phase velocity of the plasma wave in the frame xyz. We shall first evaluate the retarded time integral in Equation (12).

2.2. Evaluation of Time Integration

The distance (arc length) covered by relativistic plasma in an incremental time dt is given by ds = κ cdt = ∣ b ∣d θ (Gangadhara 2010). We can integrate ds by choosing the initial condition that t = s = 0 at θ = 0 and obtain

$$\begin{eqnarray}&&t=\displaystyle \frac{1}{\kappa c}s,\end{eqnarray} \tag{ 13 }$$

where s is the arc length of the magnetic field line (for derivation, see Appendix A) given by

$$\begin{eqnarray}&&\begin{array}{l}s={r}_{{\rm{e}}}\left[1-\displaystyle \frac{1}{2\sqrt{2}}\cos \theta \sqrt{5+3\cos (2\theta )}\right.\\ \quad \ +\ \left.\displaystyle \frac{1}{2\sqrt{3}}\mathrm{log}\left(\displaystyle \frac{\sqrt{2}(2+\sqrt{3})}{\sqrt{6}\cos \theta +\sqrt{5+3\cos (2\theta )}}\right)\right].\end{array}\end{eqnarray} \tag{ 14 }$$

If we substitute s into Equation (13), we get the retarded time t as a function of θ for any given r_e. In Figure 2, we have shown the relativistically beamed emission cones centered on the field line tangent $\hat{b}$ at different colatitudes, θ₁, θ₀, and θ₂ on a field line C, which lies in the fiducial plane ($\hat{{\rm{\Omega }}}$-$\hat{n}$ plane) at which the rotation phase $\phi ^{\prime} =0.$ The observer receives radiation over the range θ₁ ≤ θ ≤ θ₂ with a maximum at θ = θ₀ at which $\cos \eta =\hat{b}\cdot \hat{n}=1,$ where η is the angle between $\hat{b}$ and $\hat{n}.$ At θ_1,2, we have η ∼ ∓ 1/γ, and at θ₀, η = 0. For example, using the spin period P = 1 s, Lorentz factor γ = 400, α = 30°, ϱ = 2°, and r_e = 10r_LC, where r_LC = Pc/(2π) is the light cylinder radius, for any θ predicted by Equation (30), we can calculate s using Equation (14) and then t using Equation (13). In Table 1, we have tabulated the values of θ, s, and t at the three points.

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Table 1. Arc Length and Retarded Time at Different Points on a Magnetic Field Line

Point	η (deg)	θ (deg)	s (km)	t (ms)
1	−0.14	1.24	222.81	0.74
2	0.00	1.33	258.51	0.86
3	0.14	1.43	296.85	0.99

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The component of r in the direction of $\hat{n}$ is given by

$$\begin{eqnarray}\begin{array}{rcl}{r}_{\parallel } & = & \hat{n}\cdot {\boldsymbol{r}}\\ & = & \ {r}_{{\rm{e}}}{\sin }^{2}\theta [\cos {\rm{\Gamma }}\cos \theta -\sin \theta \{\sin \zeta (\sin \phi \sin \phi ^{\prime} \\ & & -\ \cos \alpha \cos \phi \cos \phi ^{\prime} )+\sin \alpha \cos \zeta \cos \phi \}],\end{array}\end{eqnarray} \tag{ 15 }$$

where ${\rm{\Gamma }}=\arccos [\cos \alpha \cos \zeta +\sin \alpha \,\sin \zeta \,\cos \phi ^{\prime} ]$ is the half-opening angle of the emission beam, i.e., the angle between $\hat{m}$ and $\hat{n}.$ The coordinates of the emission point (θ = θ₀, ϕ = ϕ₀) at which $\hat{n}$ aligns with $\hat{b},$ are deduced in the tangent model (Gangadhara 2004). The magnetic colatitude θ₀ is given by

$$\begin{eqnarray}&&{\theta }_{0}=\displaystyle \frac{1}{2}\arccos \left[\displaystyle \frac{1}{3}(\cos {\rm{\Gamma }}\,\sqrt{8+{\cos }^{2}{\rm{\Gamma }}}-{\sin }^{2}{\rm{\Gamma }})\right].\end{eqnarray} \tag{ 16 }$$

Since $\sin {\phi }_{0}=-\sin \zeta \,\sin \phi ^{\prime} \csc {\rm{\Gamma }}$ and $\cos {\phi }_{0}=(\cos \alpha \,\sin \zeta \,\cos \phi ^{\prime} \,-\cos \zeta \,\sin \alpha )\csc {\rm{\Gamma }},$ the magnetic azimuth ϕ₀ is given by

$$\begin{eqnarray}&&{\phi }_{0}=\arctan \left(\displaystyle \frac{\sin \zeta \,\sin \phi ^{\prime} }{\cos \zeta \,\sin \alpha -\cos \alpha \sin \zeta \,\cos \phi ^{\prime} }\right).\end{eqnarray} \tag{ 17 }$$

The magnetic colatitude of an arbitrary point on a magnetic field line is given by

$$\begin{eqnarray}&&\theta =\displaystyle \frac{1}{2}\arccos \left[\displaystyle \frac{1}{3}(\cos \tau \sqrt{8+{\cos }^{2}\tau }-{\sin }^{2}\tau )\right],\end{eqnarray} \tag{ 18 }$$

where τ is the angle between $\hat{m}$ and $\hat{b},$ and τ = Γ when $\hat{b}\parallel \hat{n}.$

Since ${{\boldsymbol{k}}}_{0}\parallel \hat{b},$ we have k₀ x ≈ k₀ s over the emission receiving time t_em. We change the variable of integration from t to θ using Equation (13) and keep the limits of integration at ±∞ for mathematical convenience, as the contributions to the integral vanish outside the range θ₁ ≤ θ ≤ θ₂ (see Figure 2). So, we have

$$\begin{eqnarray}\begin{array}{rcl}{\boldsymbol{E}}(\omega ) & = & \displaystyle \frac{1}{\sqrt{2\pi }}\displaystyle \frac{{e}^{{{iR}}_{0}\omega /c}}{{R}_{0}\kappa {c}^{3}}\displaystyle \int {\sigma }_{0}\underset{-\infty }{\overset{+\infty }{\displaystyle \int }}| {\boldsymbol{b}}| \\ & & \times \ \displaystyle \frac{\hat{n}\times [(c\hat{n}-{v}_{{\rm{p}}}\hat{b})\times {\dot{{\boldsymbol{\beta }}}}_{u}]}{{\left(1-{{\boldsymbol{\beta }}}_{u}\cdot \hat{n}\right)}^{2}}\\ & & \times \ {e}^{i\{{k}_{0}s-k{r}_{\parallel }-\tfrac{1}{\kappa c}({\omega }_{0}-\omega )s\}}d\theta \,{dV}.\end{array}\end{eqnarray} \tag{ 19 }$$

Let us define the argument of the integral as

$$\begin{eqnarray}&&{\boldsymbol{ \mathcal E }}(\theta )=| {\boldsymbol{b}}| \displaystyle \frac{\hat{n}\times [(c\hat{n}-{v}_{{\rm{p}}}\hat{b})\times {\dot{{\boldsymbol{\beta }}}}_{u}]}{{\left(1-{{\boldsymbol{\beta }}}_{u}\cdot \hat{n}\right)}^{2}},\end{eqnarray} \tag{ 20 }$$

and power series expanding the components in powers of θ of about θ₀. Then we obtain

$$\begin{eqnarray}\begin{array}{rcl}{{ \mathcal E }}_{{\rm{x}}} & = & {a}_{{\rm{x}}0}+{a}_{{\rm{x}}1}(\theta -{\theta }_{0})+{a}_{{\rm{x}}2}{\left(\theta -{\theta }_{0}\right)}^{2}\\ & & +\ {a}_{{\rm{x}}3}{\left(\theta -{\theta }_{0}\right)}^{3}+O[{\left(\theta -{\theta }_{0}\right)}^{4}],\\ {{ \mathcal E }}_{{\rm{y}}} & = & {a}_{{\rm{y}}0}+{a}_{{\rm{y}}1}(\theta -{\theta }_{0})+{a}_{{\rm{y}}2}{\left(\theta -{\theta }_{0}\right)}^{2}\\ & & +\ {a}_{{\rm{y}}3}{\left(\theta -{\theta }_{0}\right)}^{3}+O[{\left(\theta -{\theta }_{0}\right)}^{4}],\\ {{ \mathcal E }}_{{\rm{z}}} & = & {a}_{{\rm{z}}0}+{a}_{{\rm{z}}1}(\theta -{\theta }_{0})+{a}_{{\rm{z}}2}{\left(\theta -{\theta }_{0}\right)}^{2}\\ & & +\ {a}_{{\rm{z}}3}{\left(\theta -{\theta }_{0}\right)}^{3}+O[{\left(\theta -{\theta }_{0}\right)}^{4}],\end{array}\end{eqnarray} \tag{ 21 }$$

where a_xi , a_yi , and a_zi ; i = 0, 1, 2, and 3 are the series expansion coefficients; and their expressions are given by Equations (B1)–(B12) in Appendix B. By expanding the (θ − θ₀) terms and rewriting in powers of θ, we obtain

$$\begin{eqnarray}\begin{array}{rcl}{{ \mathcal E }}_{{\rm{x}}} & = & {{ \mathcal E }}_{{\rm{x}}0}+{{ \mathcal E }}_{{\rm{x}}1}\theta +{{ \mathcal E }}_{{\rm{x}}2}{\theta }^{2}+{{ \mathcal E }}_{{\rm{x}}3}{\theta }^{3}+O[{\left(\theta -{\theta }_{0}\right)}^{4}],\\ {{ \mathcal E }}_{{\rm{y}}} & = & {{ \mathcal E }}_{{\rm{y}}0}+{{ \mathcal E }}_{{\rm{y}}1}\theta +{{ \mathcal E }}_{{\rm{y}}2}{\theta }^{2}+{{ \mathcal E }}_{{\rm{y}}3}{\theta }^{3}+O[{\left(\theta -{\theta }_{0}\right)}^{4}],\\ {{ \mathcal E }}_{{\rm{z}}} & = & {{ \mathcal E }}_{{\rm{z}}0}+{{ \mathcal E }}_{{\rm{z}}1}\theta +{{ \mathcal E }}_{{\rm{z}}2}{\theta }^{2}+{{ \mathcal E }}_{{\rm{z}}3}{\theta }^{3}+O[{\left(\theta -{\theta }_{0}\right)}^{4}],\end{array}\end{eqnarray} \tag{ 22 }$$

where

$$\begin{eqnarray}\begin{array}{rcl}{{ \mathcal E }}_{{\rm{x}}0} & = & {a}_{{\rm{x}}0}-{a}_{{\rm{x}}1}{\theta }_{0}+{a}_{{\rm{x}}2}{\theta }_{0}^{2}-{a}_{{\rm{x}}3}{\theta }_{0}^{3},\\ {{ \mathcal E }}_{{\rm{x}}1} & = & {a}_{{\rm{x}}1}-2{a}_{{\rm{x}}2}{\theta }_{0}+3{a}_{{\rm{x}}3}{\theta }_{0}^{2},\\ {{ \mathcal E }}_{{\rm{x}}2} & = & {a}_{{\rm{x}}2}-3{a}_{{\rm{x}}3}{\theta }_{0},\quad {{ \mathcal E }}_{{\rm{x}}3}={a}_{{\rm{x}}3},\\ {{ \mathcal E }}_{{\rm{y}}0} & = & {a}_{{\rm{y}}0}-{a}_{{\rm{y}}1}{\theta }_{0}+{a}_{{\rm{y}}2}{\theta }_{0}^{2}-{a}_{{\rm{y}}3}{\theta }_{0}^{3},\\ {{ \mathcal E }}_{{\rm{y}}1} & = & {a}_{{\rm{y}}1}-2{a}_{{\rm{y}}2}{\theta }_{0}+3{a}_{{\rm{y}}3}{\theta }_{0}^{2},\\ {{ \mathcal E }}_{{\rm{y}}2} & = & {a}_{{\rm{y}}2}-3{a}_{{\rm{y}}3}{\theta }_{0},\quad {{ \mathcal E }}_{{\rm{y}}3}={a}_{{\rm{y}}3},\\ {{ \mathcal E }}_{{\rm{z}}0} & = & {a}_{{\rm{z}}0}-{a}_{{\rm{z}}1}{\theta }_{0}+{a}_{{\rm{z}}2}{\theta }_{0}^{2}-{a}_{{\rm{z}}3}{\theta }_{0}^{3},\\ {{ \mathcal E }}_{{\rm{z}}1} & = & {a}_{{\rm{z}}1}-2{a}_{{\rm{z}}2}{\theta }_{0}+3{a}_{{\rm{z}}3}{\theta }_{0}^{2},\\ {{ \mathcal E }}_{{\rm{z}}2} & = & {a}_{{\rm{z}}2}-3{a}_{{\rm{z}}3}{\theta }_{0},\quad \,{{ \mathcal E }}_{{\rm{z}}3}={a}_{{\rm{z}}3}.\end{array}\end{eqnarray} \tag{ 23 }$$

Similarly, the power series expansion of the argument of the exponential is given by

$$\begin{eqnarray}&&\begin{array}{l}{k}_{0}s-k\,{r}_{\parallel }-\displaystyle \frac{1}{\kappa c}({\omega }_{0}-\omega )s={a}_{0}+{a}_{1}(\theta -{\theta }_{0})\\ \quad \ +\ {a}_{2}{\left(\theta -{\theta }_{0}\right)}^{2}+{a}_{3}{\left(\theta -{\theta }_{0}\right)}^{3}+O[{\left(\theta -{\theta }_{0}\right)}^{4}],\end{array}\end{eqnarray} \tag{ 24 }$$

where a_i ; i = 0, 1, 2, and 3 are the series expansion coefficients; and their expressions are given by Equations (B13)–(B16) in Appendix B. By expanding the (θ − θ₀) terms and rewriting in powers of θ, we have

$$\begin{eqnarray}&&\begin{array}{l}{k}_{0}s-k\,{r}_{\parallel }-\displaystyle \frac{1}{\kappa c}({\omega }_{0}-\omega )s\\ \ ={c}_{0}+{c}_{1}\theta +\ {c}_{2}{\theta }^{2}+{c}_{3}{\theta }^{3}+O[{\left(\theta -{\theta }_{0}\right)}^{4}],\end{array}\end{eqnarray} \tag{ 25 }$$

where

$$\begin{eqnarray}\begin{array}{rcl}{c}_{0} & = & {a}_{0}-{a}_{1}{\theta }_{0}+{a}_{2}{\theta }_{0}^{2}-{a}_{3}{\theta }_{0}^{3},\\ {c}_{1} & = & {a}_{1}-2{a}_{2}{\theta }_{0}+3{a}_{3}{\theta }_{0}^{2},\\ {c}_{2} & = & {a}_{2}-3{a}_{3}{\theta }_{0},\quad {c}_{3}={a}_{3}.\end{array}\end{eqnarray} \tag{ 26 }$$

Therefore, the components of E in Equation (19) are

$$\begin{eqnarray}\begin{array}{rcl}{E}_{{\rm{x}}}(\omega ) & = & {E}_{0}\displaystyle \int {\sigma }_{0}\left(\underset{-\infty }{\overset{+\infty }{\displaystyle \int }}({{ \mathcal E }}_{{\rm{x}}0}\right.\\ & & +\ \left.{{ \mathcal E }}_{{\rm{x}}1}\,\theta +{{ \mathcal E }}_{{\rm{x}}2}\,{\theta }^{2}+{{ \mathcal E }}_{{\rm{x}}3}{\theta }^{3}){e}^{i({c}_{1}\theta +{c}_{2}{\theta }^{2}+{c}_{3}{\theta }^{3})}d\theta \right){dV},\\ {E}_{{\rm{y}}}(\omega ) & = & {E}_{0}\displaystyle \int {\sigma }_{0}\left(\underset{-\infty }{\overset{+\infty }{\displaystyle \int }}({{ \mathcal E }}_{{\rm{y}}0}\right.\\ & & +\ \left.{{ \mathcal E }}_{{\rm{y}}1}\,\theta +{{ \mathcal E }}_{{\rm{y}}2}\,{\theta }^{2}+{{ \mathcal E }}_{{\rm{y}}3}\,{\theta }^{3}){e}^{i({c}_{1}\theta +{c}_{2}{\theta }^{2}+{c}_{3}{\theta }^{3})}d\theta \right){dV},\\ {E}_{{\rm{z}}}(\omega ) & = & {E}_{0}\displaystyle \int {\sigma }_{0}\left(\underset{-\infty }{\overset{+\infty }{\displaystyle \int }}({{ \mathcal E }}_{{\rm{z}}0}\right.\\ & & +\ \left.{{ \mathcal E }}_{{\rm{z}}1}\,\theta +{{ \mathcal E }}_{{\rm{z}}2}\,{\theta }^{2}+{{ \mathcal E }}_{{\rm{z}}3}\,{\theta }^{3}){e}^{i({c}_{1}\theta +{c}_{2}{\theta }^{2}+{c}_{3}{\theta }^{3})}d\theta \right){dV},\end{array}\end{eqnarray} \tag{ 27 }$$

where

$$\begin{eqnarray*}&&{E}_{0}=\displaystyle \frac{1}{\sqrt{2\pi }}\displaystyle \frac{{e}^{i\left[\left(\omega {R}_{0}/c\right)+{c}_{0}\right]}}{{R}_{0}\kappa {c}^{3}}.\end{eqnarray*}$$

Now, by substituting the integral solutions S₀ = I₀, S₁ = I₁, S₂ = I₂, and S₃ = I₃, given in Appendix C, into Equation (27), we obtain

$$\begin{eqnarray}\begin{array}{rcl}{E}_{{\rm{x}}}(\omega ) & = & {E}_{0}\displaystyle \int {\sigma }_{0}({{ \mathcal E }}_{{\rm{x}}0}{S}_{0}+{{ \mathcal E }}_{{\rm{x}}1}{S}_{1}+{{ \mathcal E }}_{{\rm{x}}2}{S}_{2}+{{ \mathcal E }}_{{\rm{x}}3}{S}_{3})\,{dV},\\ {E}_{{\rm{y}}}(\omega ) & = & {E}_{0}\displaystyle \int {\sigma }_{0}({{ \mathcal E }}_{{\rm{y}}0}{S}_{0}+{{ \mathcal E }}_{{\rm{y}}1}{S}_{1}+{{ \mathcal E }}_{{\rm{y}}2}{S}_{2}+{{ \mathcal E }}_{{\rm{y}}3}{S}_{3})\,{dV},\\ {E}_{{\rm{z}}}(\omega ) & = & {E}_{0}\displaystyle \int {\sigma }_{0}({{ \mathcal E }}_{{\rm{z}}0}{S}_{0}+{{ \mathcal E }}_{{\rm{z}}1}{S}_{1}+{{ \mathcal E }}_{{\rm{z}}2}{S}_{2}+{{ \mathcal E }}_{{\rm{z}}3}{S}_{3})\,{dV}.\end{array}\end{eqnarray} \tag{ 28 }$$

All of the information about the frequency behavior of E (t) is contained in E (ω). Having solved the time integral, we need to evaluate the volume integral.

2.3. Evaluation of Volume Integration

The field line constant r_e is constant for any given magnetic field line, but it does vary from field line to field line in such a way that it is larger at the magnetic axis and lower at the outer field lines. It is the equatorial distance from the magnetic axis to the field line, i.e., r_e = r at θ = π/2. Hence, its range is 0 ≤ r_e ≤ ∞, where r_e = ∞ for the magnetic axis. Since r_e varies across the field lines in the poloidal (θ) direction, we treat it as a variable while dealing with plasma motion encompassing many field lines. We consider a dipole field line coordinate system (r_e, θ, ϕ), which is defined in terms of the usual spherical polar coordinates (r, θ, ϕ), to solve the volume integral in Equation (28), since the plasma is constrained to move along the field lines. Since r_e gives the location of any arbitrary field line relative to the magnetic axis $\hat{m},$ the angle ${\rm{\Gamma }}=\arccos (\hat{b}\cdot \hat{m})$ becomes the half-opening angle of the emission beam when $\hat{b}\cdot \hat{n}=1.$ Hence, we believe that the field line coordinate system allows for a deeper understanding of the geometry of the emission region. In Appendix D, we deduce the infinitesimal elements of area $d{\boldsymbol{A}}={r}_{{\rm{e}}}{\sin }^{4}\theta \,{{dr}}_{{\rm{e}}}\,d\theta \,\hat{\phi }$ (see Figure 3) and volume ${dV}={r}_{{\rm{e}}}^{2}{\sin }^{7}\theta \,{{dr}}_{{\rm{e}}}\,d\theta \,d\phi \,$ (see Figure 4) for the plasma that streams out along the open dipolar magnetic field lines. So, the vectorial form of Equation (28) is given by

$$\begin{eqnarray}&&{\boldsymbol{E}}(\omega )={E}_{0}{\displaystyle \int }_{{\phi }_{1}}^{{\phi }_{2}}\,{\displaystyle \int }_{{\theta }_{1}}^{{\theta }_{2}}\,{\displaystyle \int }_{{r}_{{\rm{e}}1}}^{{r}_{{\rm{e}}2}}{\sigma }_{0}{\boldsymbol{ \mathcal E }}(\omega )\,{r}_{{\rm{e}}}^{2}{\sin }^{7}\theta \,{{dr}}_{{\rm{e}}}\,d\theta \,d\phi .\end{eqnarray} \tag{ 29 }$$

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The secondary pair plasma cloud, which is constrained to move along the open dipolar magnetic field lines, can have arbitrary shapes and sizes depending upon its creation and evolution mechanism, which may be nonsteady. In the superstrong dipolar magnetic field, the plasma cloud density tends to fall with height as 1/r³ in the diverging field lines. At any given rotation phase, due to relativistic beaming, a distant observer tends to receive radiation from a specific region (beaming region) allowed by the viewing geometry. To find the net coherently radiated electric field of radiation, we need to know the integral limits in Equation (29) allowed by the relativistic beaming conditions.

2.3.1. The Beaming Region

Due to relativistic beaming, at any given rotation phase $\phi ^{\prime} ,$ a distant observer tends to receive the radiation from all of those field lines whose tangents lie within the angle ∼1/γ with respect to the sight line $\hat{n}$ (Gangadhara 2010). Let η be the angle between $\hat{b}$ and $\hat{n},$ then $\cos \eta =\hat{b}\cdot \hat{n},$ and the maximum value of η is ∼1/γ. At any given ϕ₀, we can solve $\cos (1/\gamma )=\hat{b}\cdot \hat{n}$ for τ and find the allowed range (Γ − 1/γ) ≤ τ ≤ (Γ + 1/γ) of τ or −1/γ ≤ η ≤ 1/γ of η, which in turn allows us to find the range of θ with the help of Equation (18). So, at any given rotation phase $\phi ^{\prime}$ for τ = Γ + η and η = ±1/γ, we can find the allowed range of θ using

$$\begin{eqnarray}&&\begin{array}{l}\theta =\displaystyle \frac{1}{2}\arccos \left[\displaystyle \frac{1}{3}(\cos ({\rm{\Gamma }}+\eta )\right.\\ \quad \ \times \ \left.\sqrt{8+{\cos }^{2}({\rm{\Gamma }}+\eta )}-{\sin }^{2}({\rm{\Gamma }}+\eta ))\right].\end{array}\end{eqnarray} \tag{ 30 }$$

The range of η is nearly fixed at ∼±1/γ at all of the rotation phase. So, for mathematical convenience, we prefer to change the variable of integration from θ to η in Equation (29). At any given $\phi ^{\prime} ,$ we have

$$\begin{eqnarray}&&\begin{array}{l}d\theta =\displaystyle \frac{\partial \theta }{\partial \eta }d\eta \ =\displaystyle \frac{\sin ({\rm{\Gamma }}+\eta )\left(\cos ({\rm{\Gamma }}+\eta )\left(\sqrt{{\cos }^{2}({\rm{\Gamma }}+\eta )+8}+\cos ({\rm{\Gamma }}+\eta )\right)+4\right)}{\sqrt{{\cos }^{2}({\rm{\Gamma }}+\eta )+8}\sqrt{{\sin }^{2}({\rm{\Gamma }}+\eta )\left(\sqrt{2}\sqrt{\cos (2({\rm{\Gamma }}+\eta ))+17}\cos ({\rm{\Gamma }}+\eta )+\cos (2({\rm{\Gamma }}+\eta ))+9\right)}}\,d\eta .\end{array}\end{eqnarray} \tag{ 31 }$$

For any given η within its range, we find ϕ by solving $\cos \eta =\hat{b}\cdot \hat{n}.$ This gives (ϕ₀ − δ ϕ) ≤ ϕ ≤ (ϕ₀ + δ ϕ), where

$$\begin{eqnarray}&&\begin{array}{l}\cos (\delta \phi )\\ \ =\ \displaystyle \frac{\sin {\rm{\Gamma }}[\cos (1/\gamma )\csc ({\rm{\Gamma }}+\eta )-\cos {\rm{\Gamma }}\cot ({\rm{\Gamma }}+\eta )]}{{\left(\cos \zeta \sin \alpha -\cos \alpha \cos \phi ^{\prime} {\sin }^{2}\zeta \right)}^{2}+{\sin }^{2}\zeta {\sin }^{2}\phi ^{\prime} }.\end{array}\end{eqnarray} \tag{ 32 }$$

Next, using the characteristic frequency of curvature radiation, ν = 3γ³ c/(4π ρ) (Ruderman & Sutherland 1975), and the radius of curvature of the dipolar magnetic field line,

$$\begin{eqnarray}&&\rho =\displaystyle \frac{{r}_{{\rm{e}}}\sin \theta {\left(5+3\cos (2\theta )\right)}^{3/2}}{3\sqrt{2}(3+\cos (2\theta ))},\end{eqnarray} \tag{ 33 }$$

we can find the range of r_e . Substituting for ρ into the expression of ν and solving for r_e gives

$$\begin{eqnarray}&&{r}_{{\rm{e}}}=\displaystyle \frac{9c{\gamma }^{3}(3+\cos (2\theta ))\csc (\theta )}{2\sqrt{2}\pi {\left(5+3\cos (2\theta )\right)}^{3/2}}\displaystyle \frac{1}{\nu }.\end{eqnarray} \tag{ 34 }$$

If ν is the center frequency of observation with bandwidth δ ν, then we will have a range of frequency between ν₁ = ν + (δ ν/2) and ν₂ = ν − (δ ν/2). At any given rotation phase $\phi ^{\prime}$ and Lorentz factor γ, we can find r_e1 by subsisting ν₁ for ν in Equation (34). Similarly, we can find r_e2 from ν₂.

The density of electron–positron plasma near the pulsar is envisioned to be ${n}_{{}_{\mathrm{NS}}}={10}^{14}$ cm⁻³ (Ruderman & Sutherland 1975; Benford & Buschauer 1977) in the frame xyz, which streams outward after it is created above a magnetospheric gap at the polar cap, and it evolves with altitude as $n={n}_{{}_{\mathrm{NS}}}{({r}_{{}_{\mathrm{NS}}}/{r}_{0})}^{3}$ in the diverging field lines, where r_NS is the radius of the neutron star and ${r}_{0}={r}_{{\rm{e}}}{\sin }^{2}\theta$ is the altitude (emission height). Then the density of particles in the plasma rest frame ${x}^{{\rm{{\prime} }}}{y}^{{\rm{{\prime} }}}{z}^{{\rm{{\prime} }}}$ becomes $n^{\prime} =n/\gamma$, and the charge density ${\sigma }_{0}^{\prime} ={qn}^{\prime} .$ Since ${v}_{{\rm{p}}}^{\prime} \sim 0$, we have the charge density ${\sigma }_{0}=\gamma {\sigma }_{0}^{{\prime} }$ in the frame xyz. The Lorentz transformation of wavenumber ${k}_{0}^{{\prime} }$ is given by ${k}_{0}=\gamma {k}_{0}^{{\prime} }(1+{{uv}}_{{\rm{p}}}^{\prime} /{c}^{2}).$ For ${v}_{{\rm{p}}}^{\prime} \sim 0$, we have ${k}_{0}\approx \gamma {k}_{0}^{{\prime} }$ and frequency ω₀ = v_p k₀, where v_p ∼ u is the phase velocity of the plasma wave (ω₀, k₀).

2.3.2. Clumpy Plasma Distribution

In most cases, the pulsar average profiles show distinct emission components. Ruderman & Sutherland (1975) suggested that the entire polar cap does not radiate; only some selected regions radiate, which may be organized into a central core emission and coaxial conal emissions. This has overwhelming support from observations (e.g., Rankin 1990, 1993). Hence, the radiating region above the polar is believed to have a central column of emission (core) and a few coaxial conal regions of emission cones (e.g., Gil & Krawczyk 1997; Gangadhara & Gupta 2001; Gupta & Gangadhara 2003). The structure of pulsar emission beams are interpreted as nested cones around a core (Rankin 1993) or patchy based on detection of the number of components and their distribution in rotation phase (Lyne & Manchester 1988; Han & Manchester 2001). The emission components in pulsar average profiles are often identified by decomposing the average pulses into individual Gaussians by fitting one with each of the subpulse components (e.g., Kramer et al. 1994). Hence, the clumpy distributions of plasma such as bunches and columns are inevitable in the emission region. When the line of sight crosses the emission region during neutron star spin, it encounters a pattern in intensity. So we assume that in the emission region, the plasma density is Gaussian-modulated in the azimuthal direction. Hence, we define a modulation function f as

$$\begin{eqnarray}&&f(\phi )={f}_{0}\,\exp \left[-{\left(\displaystyle \frac{\phi -{\phi }_{{\rm{p}}}}{{\sigma }_{\phi }}\right)}^{2}\right],\end{eqnarray} \tag{ 35 }$$

where ϕ_p is the peak location of the Gaussian function and f₀ is the amplitude. If w_ϕ is the FWHM, then ${\sigma }_{\phi }={w}_{\phi }/(2\sqrt{\mathrm{ln}2}).$ Therefore, the modulated charge density can be expressed as

$$\begin{eqnarray}&&{\sigma }_{0{\rm{m}}}=f{\sigma }_{0}=\gamma f{\sigma }_{0}^{{\prime} }.\end{eqnarray} \tag{ 36 }$$

2.3.3. Net Radiation Electric Field from Beaming Region

Having defined the limits of volume integration, Equation (29) can be rewritten as

$$\begin{eqnarray}&&{\boldsymbol{E}}(\omega )={{qE}}_{0}\,{n}_{{}_{\mathrm{NS}}}\,{r}_{{}_{\mathrm{NS}}}^{3}{\displaystyle \int }_{{r}_{{\rm{e}}1}}^{{r}_{{\rm{e}}2}}\,{\displaystyle \int }_{{\eta }_{1}}^{{\eta }_{2}}\,{\displaystyle \int }_{{\phi }_{1}}^{{\phi }_{2}}{\boldsymbol{ \mathcal E }}\,\displaystyle \frac{\sin \theta }{{r}_{{\rm{e}}}}\,\displaystyle \frac{\partial \theta }{\partial \eta }d\phi \,d\eta \,{{dr}}_{{\rm{e}}},\end{eqnarray} \tag{ 37 }$$

where the limits are η₁ = −1/γ, η₂ = 1/γ, ϕ₁ = ϕ₀ − δ ϕ, and ϕ₂ = ϕ₀ + δ ϕ, and the r_e limits are mentioned above. The expressions for θ and ∂θ/∂η are given by Equations (30) and (31), respectively. Since the argument of the integrals in Equation (37) is a very lengthy expression, it is challenging to find an analytic solution; therefore, we opt for numerical solutions. Hence, by incorporating all of the abovementioned substitutions, including the charge density σ_0m, and taking into account the assumptions and limits of the beaming region, we numerically solve Equation (37) and estimate the radiation electric field.

To find the PA of the radiation field E (ω), we need to specify two reference directions perpendicular to the sight line $\hat{n}.$ One could be the projected spin axis on the plane of the sky, ${\hat{\epsilon }}_{\parallel }=\{-\cos \zeta ,0,\sin \zeta \},$ and then the other direction is specified by ${\hat{\epsilon }}_{\perp }={\hat{\epsilon }}_{\parallel }\times \hat{n}=\hat{Y},$ where $\hat{Y}$ is a unit vector parallel to the Y-axis. Then the components of E (ω) in the directions ${\hat{\epsilon }}_{\parallel }$ and ${\hat{\epsilon }}_{\perp }$ are given by

$$\begin{eqnarray}\begin{array}{rcl}{E}_{\parallel } & = & {\hat{\epsilon }}_{\parallel }\cdot {\boldsymbol{E}}=-\cos \zeta \,{E}_{{\rm{x}}}+\sin \zeta \,{E}_{{\rm{z}}},\\ {E}_{\perp } & = & {\hat{\epsilon }}_{\perp }\cdot {\boldsymbol{E}}={E}_{{\rm{y}}}.\end{array}\end{eqnarray} \tag{ 38 }$$

The Stokes parameters for the radiation electric field can be defined as

$$\begin{eqnarray}\begin{array}{rcl}I & = & \mu ({E}_{\parallel }{E}_{\parallel }^{* }+{E}_{\perp }{E}_{\perp }^{* }),\quad Q=\mu ({E}_{\parallel }{E}_{\parallel }^{* }-{E}_{\perp }{E}_{\perp }^{* }),\\ U & = & 2\mu \,\mathrm{Re}[{E}_{\parallel }^{* }{E}_{\perp }],\quad V=2\mu \,\mathrm{Im}[{E}_{\parallel }^{* }{E}_{\perp }],\end{array}\end{eqnarray} \tag{ 39 }$$

where μ = c/T is the proportionality factor. It is chosen in such a way that I is precisely the intensity or flux density of the pulse that repeats on an average timescale T.

Let us choose a typical normal pulsar having a spin period P = 1 s, α = 30°, ϱ = 2°, and T ∼ 5.5 ms. Consider a relativistic plasma with Gaussian distribution in charge density σ_0m in the azimuthal direction and Lorentz factor γ = 400. Assume that an inertial observer situated at a distance R₀ = 2 kpc receives the pulses at frequency ν = ω/2π = 600 MHz with a narrow bandwidth δ ν ∼ 2 MHz. In the chosen geometry, the observer receives maximum emission at (θ₀ = 133, ϕ₀ = 0°) due to the accelerated plasma along the dipolar magnetic field lines. Using f₀ = 1, σ_ϕ = 0.25, and ϕ_p = 0°, we have simulated the single component (core) polarization profiles due to coherently emitted radiation. In Figure 5, we have plotted the profiles in the two cases of frequency and wavenumber matching between the EM wave (ω, k) and plasma wave (ω₀, k₀). In Figures 5(a) and (c), we plot the intensity I with the black curve, linear polarization L with the dashed red curve, and circular polarization V with the green curve. The coherently emitted radiation is highly polarized; hence, I² = L² + V², as in the case of 100% elliptically polarized radiation, where L² = Q² + U². Since ∣V∣ < 20% and L > 94% within the pulse window $(-1^\circ \leqslant \phi ^{\prime} \leqslant 1^\circ ),$ the curves of I and L are superposed, as L dominates over V. The PA is plotted in Figures 5(b) and (d) and in close agreement with the rotating vector model (Radhakrishnan & Cooke 1969). The sense reversal of circular polarization is correlated with the swing (S-curve) of PA (Radhakrishnan & Rankin 1990; Gangadhara 2010). We observe that in the resonant case, i.e., when waves are resonantly coupled (ω, k) = (ω₀, κ k₀), an efficient transfer of energy takes place from plasma to EM wave.

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The peak intensity ${I}_{\max }=9.10\times {10}^{-14}$ erg cm⁻² s⁻¹ Hz⁻¹ in the resonant case (see Figure 5(a)) is nearly four times larger than that of the nonresonant case, ${I}_{\max }=2.26\times {10}^{-14}$ erg cm⁻² s⁻¹ Hz⁻¹ (see Figure 5(c)). Hence, by knowing the peak values of emitted radiation flux density ${s}_{\nu }={I}_{\max }$, we can estimate the brightness temperature T_b using the expression deduced by Cordes (1979) in the Rayleigh–Jeans regime,

$$\begin{eqnarray}&&{T}_{{\rm{b}}}=\displaystyle \frac{1}{2{k}_{{\rm{B}}}}\displaystyle \frac{{\lambda }^{2}\,{s}_{\nu }}{{\left(c{\rm{\Delta }}t/R\right)}^{2}},\end{eqnarray} \tag{ 40 }$$

where k_B is the Boltzmann constant, λ is the wavelength of the emitted radiation, and Δt is the timescale of s_ν variation. Using Δt ∼ T = 5.5 ms, λ = 50 cm, and the peak values of I in the two cases of Figures 5(a) and (c), we obtain T_b ≈ 1.15 × 10³³ and 7.16 × 10³¹ K, respectively. They are in the ranges of brightness temperature estimated from the conventional radio pulsar observations.

Similarly, choosing α = 45° and ϱ = 5°, with the remaining parameters the same as in Figure 5, we have plotted the polarization profiles in Figure 6. The peak values of I in Figures 6(a) and (c) are ${I}_{\max }=9.23\times {10}^{-14}$ and $2.31\,\times {10}^{-14}$ erg cm⁻² s⁻¹ Hz⁻¹, respectively, and the respective values of T_b are ≈1.17 × 10³³ and 7.31 × 10³¹ K.

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By choosing α = 60° and ϱ = − 4°, we have reproduced a triple-component profile by considering three Gaussians, one centered at (θ₀ = 266, ϕ_c = 0°) and the other two at (θ₀ = 354, ϕ_l = −40°) and (θ₀ = 354, ϕ_t = 40°) in Figure 7. For this simulation, we considered three Gaussians, one centered on the meridional plane and the other two located symmetrically on either side,

$$\begin{eqnarray}&&\begin{array}{l}f={f}_{0{\rm{l}}}{e}^{-{\left(\phi -{\phi }_{{\rm{l}}}\right)}^{2}/{\sigma }_{\phi l}^{2}}+{f}_{0}{e}^{-{\left(\phi -{\phi }_{{\rm{c}}}\right)}^{2}/{\sigma }_{\phi }^{2}}\\ \quad \ +\ {f}_{0t}{e}^{-{\left(\phi -{\phi }_{{\rm{t}}}\right)}^{2}/{\sigma }_{\phi t}^{2}},\end{array}\end{eqnarray} \tag{ 41 }$$

where chosen for the core (f₀ = 1, σ_ϕ = 0.25), the leading component (f_0l = 0.7, σ_{ϕ l} = 0.2), and the trailing component (f_0t = 0.7, σ_{ϕ t} = 0.2). The peak intensities of the central component (core) in Figures 7(a) and (c) are ${I}_{\max }=9.18\,\times {10}^{-14}$ and $2.30\times {10}^{-14}$ erg cm⁻² s⁻¹ Hz⁻¹, respectively, and the respective values of T_b are ≈1.16 × 10³³ and 7.28 × 10³¹ K. We observe that the circular polarization is more significant in the core emission than in the conal emissions.

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If the beaming region emissions are incoherently emitted, then we have to add intensities, i.e., Stokes parameters, instead of radiation fields. Choosing α = 30° and ϱ = 4°, similar to Figure 7, simulated a triple-component profile due to incoherent curvature radiation; it is plotted in Figure 8. The profile shows a significantly unpolarized emission as I² > L² + V². The profile at the central (core) and conal component peaks is nearly 56% linearly and 15% circularly polarized. The value of I at the core peak is ${I}_{\max }=5.75\,\times {10}^{-29}$ erg cm⁻² s⁻¹ Hz⁻¹, and the corresponding brightness temperature is T_b ≈ 1.76 × 10¹⁷ K, which is far below the observed brightness temperature of pulsars.

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Coherent emission requires a large number of relativistic charges, which can be treated as continuous distribution, and the production of significant fields. Following Equations (4) and (11), it is clear that we treated the radiation mechanism as a hybrid of coherent plasma-curvature radiation, as suggested by Katz (2018). In Figures 5–7, we considered monochromatic waves, which are completely, or 100%, polarized, as the electric vector displays a simple, nonrandom directional behavior in time. The polarization signatures, including PA swing, are in agreement with the observations. The brightness temperature estimated from the model is in the range of ∼10³¹–10³³ K and comparable with the observations. In practice, the radio receiver may not see a single monochromatic component but rather a superposition of many components emitted from different subsections of the beaming region, each with its own polarization state. However, if the amplitude and phase of the waves vary slowly over time, on short timescales of the order 1/ω, they look completely polarized with a definite state of polarization. But over much longer times, Δt ≫ 1/ω, characterizing the times over which amplitudes and phases change significantly, the polarization state can change substantially. Such a wave is no longer monochromatic; the frequency spread Δω can be estimated as Δω > 1/Δt, such that Δω ≪ ω. For this reason, the wave is called quasi-monochromatic. Then the frequency spread Δω becomes the bandwidth of the wave, and the time Δt can be taken as a coherence timescale. For a quasi-monochromatic wave, we have I² ≥ Q² + U² + V², as some portion of the intensity may not be polarized, as shown in Figure 8. An important property of the Stokes parameters is that they are additive for a superposition of independent waves.

Though coherence and resonance are required in the pulsar emission to account for the high brightness temperature, to some extent, coherence fluctuations and nonresonance (quasi-monochromatic) are also needed to explain the observed intensity variations. From single-pulse studies, we infer that magnetospheric pulsar plasma is highly variable and dynamic; hence, coherence fluctuates in time and space. The single pulses, which depict micropulses and subpulses, are, in some sense, incoherent ensembles of coherent emissions. The profiles presented in Figures 5–7 indicate 100% polarization, but in reality, there is some depolarized component as well. This component could arise due to an incoherent emission in some regions where the plasma conditions are not conducive for coherence. For example, the profile in Figure 5(a) is due to resonant and coherent emission where the plasma wave frequency matches that of an EM wave of 600 MHz, whereas in the nonresonant case in Figure 5(c), the plasma wave frequency does not match the EM wave of 600 MHz, and hence the intensity is lower. But it could be resonantly matched with an EM wave of 1200 MHz in some other region.

• 1.  
  We have developed a coherent radio emission model for radio pulsars.
• 2.  
  The collective behavior among the radiating plasma particles and the resonance coupling between the plasma wave and an EM wave are crucial for the enhanced emission.
• 3.  
  The model explains the emission of coherent curvature radiation and its polarization state.
• 4.  
  We confirm that the correlation between the PA swing and the sense reversal of circular polarization is a geometric effect.
• 5.  
  The coherent curvature radiation is nearly 100% polarized in the case of a monochromatic wave.
• 6.  
  The brightness temperatures, ∼10³¹–10³³ K, predicted by the model are closely aligned with the observations.

R.T.G. is grateful to the CAS President's International Fellowship Initiative (PIFI) for supporting this work. J.L.H. is supported by National Natural Science Foundation of China (NSFC) grant No. 11988101. P.F.W. is supported by NSFC grant No. 11873058. We thank the anonymous referee for useful comments.

The incremental arc length of the field line (Gangadhara 2010) is given by

$$\begin{eqnarray}&&{ds}=| {\boldsymbol{b}}| d\theta ,\end{eqnarray} \tag{ A1 }$$

where $| {\boldsymbol{b}}| ={r}_{{\rm{e}}}\sqrt{5+3\cos (2\theta )}\sin \theta /\sqrt{2}$ = ${r}_{{\rm{e}}}\sqrt{1+3{\cos }^{2}\theta }\sin \theta$ is the magnitude of the field line tangent. Next, by substituting for ∣ b ∣ in Equation (A1) and taking the integral on both sides, we have

$$\begin{eqnarray}&&s=\int {ds}={r}_{{\rm{e}}}\int \sqrt{1+3{\cos }^{2}\theta }\sin \theta d\theta .\end{eqnarray} \tag{ A2 }$$

Let $\sqrt{3}\cos \theta =x$; then we have $-\sqrt{3}\sin \theta d\theta ={dx}.$ Therefore, Equation (A2) reduces to

$$\begin{eqnarray}&&s=-\displaystyle \frac{{r}_{{\rm{e}}}}{\sqrt{3}}\int \sqrt{1+{x}^{2}}{dx}.\end{eqnarray} \tag{ A3 }$$

Next, by defining $x=\tan u$ and ${dx}={\sec }^{2}u\,{du},$ we have

$$\begin{eqnarray}&&s=-\displaystyle \frac{{r}_{{\rm{e}}}}{\sqrt{3}}\int {\sec }^{3}u\,{du}.\end{eqnarray} \tag{ A4 }$$

By taking $\sec u$ as the first function and ${\sec }^{2}u$ as the second function, we solve the integration by parts.

Consider

$$\begin{eqnarray}&&\int \sec u\,{\sec }^{2}u\,{du}=\sec u\tan u-\int {\tan }^{2}u\sec u\,{du}.\end{eqnarray} \tag{ A5 }$$

Using ${\tan }^{2}u={\sec }^{2}u-1,$ we can express it as

$$\begin{eqnarray}&&2\int {\sec }^{3}u\,{du}=\tan u\sec u+\int \sec u\,{du},\end{eqnarray} \tag{ A6 }$$

$$\begin{eqnarray}&&\int {\sec }^{3}u\,{du}=\displaystyle \frac{1}{2}\left(\tan u\sec u+\mathrm{log}(\tan u+\sec u)\right)+C,\end{eqnarray} \tag{ A7 }$$

where C is the integration constant. Hence, Equation (A4) can be written as

$$\begin{eqnarray}&&s=-\displaystyle \frac{{r}_{{\rm{e}}}}{2\sqrt{3}}\left(\tan u\sec u+\mathrm{log}(\tan u+\sec u)\right)+C.\end{eqnarray} \tag{ A8 }$$

By substituting for $\tan u=x$ and $\sec u=\sqrt{1+{x}^{2}},$ we get

$$\begin{eqnarray}&&s=-\displaystyle \frac{{r}_{{\rm{e}}}}{2\sqrt{3}}\left(x\sqrt{1+{x}^{2}}+\mathrm{log}(x+\sqrt{1+{x}^{2}})\right)+C.\end{eqnarray} \tag{ A9 }$$

Next, by substituting back $x=\sqrt{3}\cos \theta$ and using the initial condition that s = 0 at the magnetic colatitude θ = 0°, we obtain

$$\begin{eqnarray}&&s=\displaystyle \frac{{r}_{{\rm{e}}}}{12}\left(12+\sqrt{3}\mathrm{log}\left(14+8\sqrt{3}\right)-3\cos \theta \sqrt{10+6\cos (2\theta )}-2\sqrt{3}\mathrm{log}\left(\sqrt{6}\cos \theta +\sqrt{5+3\cos (2\theta )}\right)\right).\end{eqnarray} \tag{ A10 }$$

After combining the log terms and simplifying, we get Equation (14):

$$\begin{eqnarray}&&s={r}_{{\rm{e}}}\left[1-\displaystyle \frac{1}{2\sqrt{2}}\cos \theta \sqrt{5+3\cos (2\theta )}+\displaystyle \frac{1}{2\sqrt{3}}\mathrm{log}\left(\displaystyle \frac{\sqrt{2}(2+\sqrt{3})}{\sqrt{6}\cos \theta +\sqrt{5+3\cos (2\theta )}}\right)\right].\end{eqnarray} \tag{ A11 }$$

The series coefficients appearing in Equation (21) are

$$\begin{eqnarray}&&{a}_{{\rm{x}}0}=\displaystyle \frac{6{c}^{2}{\kappa }^{2}\cos \zeta {e}_{10}{e}_{14}}{{e}_{9}^{2}\left(-2{e}_{9}+\sqrt{2}\kappa {e}_{13}\right){}^{2}},\end{eqnarray} \tag{ B1 }$$

$$\begin{eqnarray}&&{a}_{{\rm{x}}1}=\displaystyle \frac{6{c}^{2}{\kappa }^{2}\cos \zeta \left(2{e}_{15}{e}_{19}+3{e}_{9}{e}_{10}\left(2{e}_{9}-\sqrt{2}\kappa {e}_{13}\right){e}_{21}\right)}{{e}_{9}^{4}\left(2{e}_{9}-\sqrt{2}\kappa {e}_{13}\right){}^{3}},\end{eqnarray} \tag{ B2 }$$

$$\begin{eqnarray}&&{a}_{{\rm{x}}2}=\displaystyle \frac{3{c}^{2}{\kappa }^{2}\cos \zeta \left(-12{e}_{9}{e}_{29}{e}_{30}\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)-4{e}_{15}{e}_{26}+{e}_{31}\right)}{{e}_{9}^{6}\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right){}^{4}},\end{eqnarray} \tag{ B3 }$$

$$\begin{eqnarray}&&{a}_{{\rm{x}}3}=-\displaystyle \frac{{c}^{2}{\kappa }^{2}\cos \zeta \left(3{e}_{9}{e}_{10}{e}_{38}\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right){}^{3}+18{e}_{9}{e}_{40}{e}_{43}\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right){}^{2}+{e}_{44}\right)}{{e}_{9}^{8}\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right){}^{5}},\end{eqnarray} \tag{ B4 }$$

$$\begin{eqnarray}&&{a}_{{\rm{y}}0}=-\displaystyle \frac{3{c}^{2}{\kappa }^{2}{e}_{10}{e}_{47}}{2\sqrt{2}{e}_{9}^{2}\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right){}^{2}},\end{eqnarray} \tag{ B5 }$$

$$\begin{eqnarray}&&{a}_{{\rm{y}}1}=\displaystyle \frac{3{c}^{2}{\kappa }^{2}\left(-12{e}_{9}{e}_{10}{e}_{49}\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right){\cos }^{2}\zeta -6\sqrt{2}{e}_{9}^{2}{e}_{10}{e}_{16}{e}_{51}\kappa +{e}_{53}+{e}_{54}\right)}{\sqrt{2}{e}_{9}^{4}\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right){}^{3}},\end{eqnarray} \tag{ B6 }$$

$$\begin{eqnarray}&&{a}_{{\rm{y}}2}=\displaystyle \frac{3{c}^{2}{\kappa }^{2}\left({e}_{9}{e}_{10}{e}_{61}\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right){}^{2}+{e}_{59}{e}_{65}+{e}_{66}\right)}{2\sqrt{2}{e}_{9}^{6}\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right){}^{4}},\end{eqnarray} \tag{ B7 }$$

$$\begin{eqnarray}&&{a}_{{\rm{y}}3}=-\displaystyle \frac{{c}^{2}{\kappa }^{2}\left(6{e}_{61}{e}_{76}{e}_{9}^{6}\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)+{e}_{67}{e}_{83}+{e}_{86}\right)}{2\sqrt{2}{e}_{9}^{13}\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right){}^{4}},\end{eqnarray} \tag{ B8 }$$

$$\begin{eqnarray}&&{a}_{{\rm{z}}0}=-\displaystyle \frac{6{c}^{2}{e}_{10}{e}_{90}{\kappa }^{2}\sin \zeta }{{e}_{9}^{2}\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right){}^{2}},\end{eqnarray} \tag{ B9 }$$

$$\begin{eqnarray}&&{a}_{{\rm{z}}1}=-\displaystyle \frac{6{c}^{2}{\kappa }^{2}\left(3{e}_{9}{e}_{10}{e}_{94}\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)+2{e}_{91}{e}_{93}\right)\sin \zeta }{{e}_{9}^{4}\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right){}^{3}},\end{eqnarray} \tag{ B10 }$$

$$\begin{eqnarray}&&{a}_{{\rm{z}}2}=-\displaystyle \frac{3{c}^{2}\left(-4{e}_{91}{e}_{98}-12{e}_{9}{e}_{97}{e}_{99}+3{e}_{9}{e}_{10}{e}_{100}\right){\kappa }^{2}\sin \zeta }{{e}_{9}^{6}\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right){}^{4}},\end{eqnarray} \tag{ B11 }$$

$$\begin{eqnarray}&&{a}_{{\rm{z}}3}=-\displaystyle \frac{{c}^{2}{\kappa }^{2}\left(3{e}_{9}{e}_{10}{e}_{103}\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right){}^{3}+36{e}_{9}{e}_{106}{e}_{107}\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)-4{e}_{91}{e}_{110}+{e}_{111}\right)\sin \zeta }{{e}_{9}^{8}\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right){}^{5}},\end{eqnarray} \tag{ B12 }$$

where

$$\begin{eqnarray*}\begin{array}{rcl}{e}_{1} & = & \cos \alpha \cos \phi \sin \phi ^{\prime} +\sin \phi \cos \phi ^{\prime} ,\\ {e}_{2} & = & \sin \alpha \cos \zeta \cos \phi +\sin \zeta \sin \phi \sin \phi ^{\prime} ,\\ {e}_{3} & = & \cos \alpha \cos \phi \cos \phi ^{\prime} -\sin \phi \sin \phi ^{\prime} ,\\ {e}_{4} & = & \sin (2\alpha )\cos \phi \sin \left(2\phi ^{\prime} \right)-4\sin \alpha \sin \phi {\sin }^{2}\phi ^{\prime} ,\\ {e}_{5} & = & \sqrt{2}\kappa \left(\cos \alpha \sin \phi \cos \phi ^{\prime} +\cos \phi \sin \phi ^{\prime} \right),\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{e}_{6} & = & \cos \alpha \cos \phi \cos \phi ^{\prime} ,\\ {e}_{7} & = & \sin \alpha \sin \phi ^{\prime} ,\\ {e}_{8} & = & \sin \alpha \cos \phi ^{\prime} ,\\ {e}_{9} & = & \sqrt{5+3\cos \left(2{\theta }_{0}\right)},\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{e}_{10} & = & 3+\cos \left(2{\theta }_{0}\right),\\ {e}_{11} & = & \cos \left(2{\theta }_{0}\right)\left(6{e}_{8}\kappa -3\sqrt{2}{e}_{9}\sin \zeta \right)-\sqrt{2}{e}_{9}\sin \zeta +10{e}_{8}\kappa ,\\ {e}_{12} & = & -3\sqrt{2}{e}_{9}\sin \zeta \sin \left(2{\theta }_{0}\right)+\kappa \left(9{\sin }^{2}\left(2{\theta }_{0}\right)\left({e}_{6}-\sin \phi \sin \phi ^{\prime} \right)+{e}_{3}\right)+9{e}_{3}\kappa {\cos }^{2}\left(2{\theta }_{0}\right)+6{e}_{3}\kappa \cos \left(2{\theta }_{0}\right),\\ {e}_{13} & = & \cos \alpha \cos \zeta \left(3\cos \left(2{\theta }_{0}\right)+1\right)+\sin \zeta \left(3{e}_{6}\sin \left(2{\theta }_{0}\right)+3{e}_{8}\cos \left(2{\theta }_{0}\right)+{e}_{8}\right)-3{e}_{2}\sin \left(2{\theta }_{0}\right),\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{e}_{14} & = & {e}_{11}\sin \alpha \cos \phi +{e}_{12}\cos \alpha -\sqrt{2}{e}_{9}\cos \zeta \left(-3{e}_{8}\sin \left(2{\theta }_{0}\right)+3{e}_{3}\cos \left(2{\theta }_{0}\right)+{e}_{3}\right),\\ {e}_{15} & = & {e}_{11}\sin \alpha \cos \phi +{e}_{12}\cos \alpha -\sqrt{2}{e}_{9}\cos \zeta \left(-3{e}_{8}\sin \left(2{\theta }_{0}\right)+3{e}_{3}\cos \left(2{\theta }_{0}\right)+{e}_{3}\right),\\ {e}_{16} & = & \cos \alpha \cos \zeta \sin \left(2{\theta }_{0}\right)+{e}_{6}\sin \zeta {\sin }^{2}{\theta }_{0}+{e}_{8}\sin \zeta \sin \left(2{\theta }_{0}\right)+{e}_{6}(-\sin \zeta ){\cos }^{2}{\theta }_{0}+{e}_{2}\cos \left(2{\theta }_{0}\right),\\ {e}_{17} & = & \sin \left(2{\theta }_{0}\right)\left(\sqrt{2}\sin \zeta \left(9\cos \left(2{\theta }_{0}\right)+11\right)-4{e}_{8}{e}_{9}\kappa \right),\\ {e}_{18} & = & \displaystyle \frac{1}{2}\left(22{e}_{3}\sin \left(2{\theta }_{0}\right)+9{e}_{3}\sin \left(4{\theta }_{0}\right)+20{e}_{8}\cos \left(2{\theta }_{0}\right)+9{e}_{8}\cos \left(4{\theta }_{0}\right)+3{e}_{8}\right),\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{c}\begin{array}{c}\begin{array}{rcl}{e}_{19} & = & \left(3{e}_{10}-{e}_{9}^{2}\right)\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)\sin \left(2{\theta }_{0}\right)-6{e}_{9}{e}_{10}\left(\sqrt{2}{e}_{9}{e}_{16}\kappa -\sin \left(2{\theta }_{0}\right)\right),\\ {e}_{20} & = & \sqrt{2}{e}_{9}^{2}\sin \zeta \sin \left(4{\theta }_{0}\right)+6{e}_{9}\kappa \sin \left(4{\theta }_{0}\right)\sin \left(2{\theta }_{0}\right)\left({e}_{3}-{e}_{6}+\sin \phi \sin {\phi }^{{\rm{{\prime} }}}\right)+4{e}_{3}{e}_{9}\kappa {\sin }^{2}\left(2{\theta }_{0}\right)-3\sqrt{2}\sin \zeta {\sin }^{3}\left(2{\theta }_{0}\right),\\ {e}_{21} & = & -\displaystyle \frac{1}{2}{e}_{20}\cos \alpha \csc {\theta }_{0}\sec {\theta }_{0}+{e}_{17}\sin \alpha \cos \phi +\sqrt{2}{e}_{18}\cos \zeta ,\\ {e}_{22} & = & \cos \alpha \cos \zeta \cos \left(2{\theta }_{0}\right)+\sin \left(2{\theta }_{0}\right)\left({e}_{6}\sin \zeta -{e}_{2}\right)+{e}_{8}\sin \zeta \cos \left(2{\theta }_{0}\right),\\ {e}_{23} & = & 12{\cos }^{2}\left(2{\theta }_{0}\right)\left(13\sqrt{2}\sin \zeta -2{e}_{8}{e}_{9}\kappa \right)+10\cos \left(2{\theta }_{0}\right)\left(11\sqrt{2}\sin \zeta -4{e}_{8}{e}_{9}\kappa \right)+\\ & & 54\sqrt{2}\sin \zeta {\cos }^{3}\left(2{\theta }_{0}\right)-3\sqrt{2}\sin \zeta {\sin }^{2}\left(2{\theta }_{0}\right)\left(9\cos \left(2{\theta }_{0}\right)+19\right),\end{array}\end{array}\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{c}\begin{array}{c}\begin{array}{ccc}{e}_{24} & = & \frac{1}{4}({e}_{3}(575\cos (2{\theta }_{0})+426\cos (4{\theta }_{0})+81\cos (6{\theta }_{0})+198)-{e}_{8}(517\sin (2{\theta }_{0})+420\sin (4{\theta }_{0})+81\sin (6{\theta }_{0})))\,,\\ {e}_{25} & = & \displaystyle \frac{\sin \left(2{\theta }_{0}\right)\left(420\cos \left(2{\theta }_{0}\right)+81\cos \left(4{\theta }_{0}\right)+299\right)}{\sqrt{2}},\\ {e}_{26} & = & -54{e}_{9}^{2}{e}_{10}\left(\sin \left(2{\theta }_{0}\right)-\sqrt{2}{e}_{9}{e}_{16}\kappa \right){}^{2}+12{e}_{9}\left({e}_{9}^{2}-3{e}_{10}\right)\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)\sin \left(2{\theta }_{0}\right)\left(\sin \left(2{\theta }_{0}\right)-\sqrt{2}{e}_{9}{e}_{16}\kappa \right)\\ & & +\,\displaystyle \frac{3}{2}{e}_{9}{e}_{10}\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)\left(4\sqrt{2}{e}_{22}{e}_{9}^{3}\kappa -20\cos \left(2{\theta }_{0}\right)-3\cos \left(4{\theta }_{0}\right)-9\right)\\ & & +\,\left({e}_{9}^{2}-3{e}_{10}\right)\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right){}^{2}\left({e}_{9}^{2}\cos \left(2{\theta }_{0}\right)+6{\sin }^{2}\left(2{\theta }_{0}\right)\right),\end{array}\end{array}\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{e}_{27} & = & -{e}_{25}\sin \zeta -24{e}_{9}^{3}\kappa \cos \left(4{\theta }_{0}\right)\left({e}_{6}-\sin \phi \sin \phi ^{\prime} \right)-24{e}_{3}{e}_{9}^{3}\kappa {\sin }^{2}\left(2{\theta }_{0}\right)+24{e}_{3}{e}_{9}^{3}\kappa {\cos }^{2}\left(2{\theta }_{0}\right)+8{e}_{3}{e}_{9}^{3}\kappa \cos \left(2{\theta }_{0}\right),\\ {e}_{28} & = & \sqrt{2}{e}_{9}^{2}\sin \zeta \sin \left(4{\theta }_{0}\right)+6{e}_{9}\kappa \sin \left(4{\theta }_{0}\right)\sin \left(2{\theta }_{0}\right)\left({e}_{3}-{e}_{6}+\sin \phi \sin \phi ^{\prime} \right)+4{e}_{3}{e}_{9}\kappa {\sin }^{2}\left(2{\theta }_{0}\right)-3\sqrt{2}\sin \zeta {\sin }^{3}\left(2{\theta }_{0}\right),\\ {e}_{29} & = & -\displaystyle \frac{1}{2}{e}_{28}\cos \alpha \csc {\theta }_{0}\sec {\theta }_{0}+{e}_{17}\sin \alpha \cos \phi +\sqrt{2}{e}_{18}\cos \zeta ,\\ {e}_{30} & = & 6{e}_{9}{e}_{10}\left(\sqrt{2}{e}_{9}{e}_{16}\kappa -\sin \left(2{\theta }_{0}\right)\right)-\left(3{e}_{10}-{e}_{9}^{2}\right)\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)\sin \left(2{\theta }_{0}\right),\\ {e}_{31} & = & 3{e}_{9}{e}_{10}\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right){}^{2}\left({e}_{23}\sin \alpha \cos \phi -{e}_{27}\cos \alpha +\sqrt{2}{e}_{24}\cos \zeta \right),\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{c}\begin{array}{rcl}{e}_{32} & = & \displaystyle \frac{1}{2}\sin \left(2{\theta }_{0}\right)\left(120\cos \left(2{\theta }_{0}\right)-9\cos \left(4{\theta }_{0}\right)+49\right),\\ {e}_{33} & = & \displaystyle \frac{19160\cos \left(2{\theta }_{0}\right)+17436\cos \left(4{\theta }_{0}\right)+6120\cos \left(6{\theta }_{0}\right)+729\cos \left(8{\theta }_{0}\right)+7755}{4\sqrt{2}},\\ {e}_{34} & = & \displaystyle \frac{1}{4}\sin \left(2{\theta }_{0}\right)\left(576{\cos }^{2}\left(2{\theta }_{0}\right)\left({e}_{8}{e}_{9}\kappa -23\sqrt{2}\sin \zeta \right)+3\cos \left(2{\theta }_{0}\right)\left(640{e}_{8}{e}_{9}\kappa -5267\sqrt{2}\sin \zeta \right)\right.\\ & & +\,\left.1600{e}_{8}{e}_{9}\kappa -3240\sqrt{2}\sin \zeta {\cos }^{3}\left(2{\theta }_{0}\right)+\sqrt{2}\sin \zeta \left(486\cos \left(4{\theta }_{0}\right)+81\cos \left(6{\theta }_{0}\right)-4886\right)\right),\\ {e}_{35} & = & \displaystyle \frac{1}{8}\left(16882{e}_{3}\sin \left(2{\theta }_{0}\right)+17502{e}_{3}\sin \left(4{\theta }_{0}\right)+6138{e}_{3}\sin \left(6{\theta }_{0}\right)+729{e}_{3}\sin \left(8{\theta }_{0}\right)\right.\\ & & +\,\left.19160{e}_{8}\cos \left(2{\theta }_{0}\right)+17436{e}_{8}\cos \left(4{\theta }_{0}\right)+6120{e}_{8}\cos \left(6{\theta }_{0}\right)+729{e}_{8}\cos \left(8{\theta }_{0}\right)+7755{e}_{8}\right),\end{array}\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{e}_{36} & = & 18{e}_{9}\left(\sin \left(2{\theta }_{0}\right)-\sqrt{2}{e}_{9}{e}_{16}\kappa \right){}^{2}-\displaystyle \frac{1}{2}\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)\left(4\sqrt{2}{e}_{22}{e}_{9}^{3}\kappa -20\cos \left(2{\theta }_{0}\right)-3\cos \left(4{\theta }_{0}\right)-9\right),\\ {e}_{37} & = & -\displaystyle \frac{1}{4}\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right){}^{2}\left(-16\sqrt{2}{e}_{16}{e}_{9}^{5}\kappa +301\sin \left(2{\theta }_{0}\right)+60\sin \left(4{\theta }_{0}\right)+9\sin \left(6{\theta }_{0}\right)\right)-24{e}_{9}{e}_{36}\left(\sqrt{2}{e}_{9}{e}_{16}\kappa -\sin \left(2{\theta }_{0}\right)\right)\\ & & +\ 15{e}_{9}\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)\left(\sqrt{2}{e}_{9}{e}_{16}\kappa -\sin \left(2{\theta }_{0}\right)\right)\left(4\sqrt{2}{e}_{22}{e}_{9}^{3}\kappa -20\cos \left(2{\theta }_{0}\right)-3\cos \left(4{\theta }_{0}\right)-9\right),\\ {e}_{38} & = & -\displaystyle \frac{1}{2}\sin (2\alpha )\csc \alpha \left({e}_{33}\sin \zeta +16{e}_{9}^{5}\kappa \left(6\sin \left(4{\theta }_{0}\right)\left({e}_{3}-{e}_{6}+\sin \phi \sin \phi ^{\prime} \right)+{e}_{3}\sin \left(2{\theta }_{0}\right)\right)\right)-{e}_{34}\sin \alpha \cos \phi +\sqrt{2}{e}_{35}\cos \zeta ,\\ {e}_{39} & = & -\displaystyle \frac{1}{2}\sin (2\alpha )\csc \alpha \csc \left(2{\theta }_{0}\right)\left(\sqrt{2}{e}_{9}^{2}\sin \zeta \sin \left(4{\theta }_{0}\right)+6{e}_{9}\kappa \sin \left(4{\theta }_{0}\right)\sin \left(2{\theta }_{0}\right)\left({e}_{3}-{e}_{6}+\sin \phi \sin \phi ^{\prime} \right)\right.\\ & & +\ \left.4{e}_{3}{e}_{9}\kappa {\sin }^{2}\left(2{\theta }_{0}\right)-3\sqrt{2}\sin \zeta {\sin }^{3}\left(2{\theta }_{0}\right)\right)+{e}_{17}\sin \alpha \cos \phi +\sqrt{2}{e}_{18}\cos \zeta ,\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{e}_{40} & = & -\cos \alpha \left(-{e}_{25}\sin \zeta -24{e}_{9}^{3}\kappa \cos \left(4{\theta }_{0}\right)\left({e}_{6}-\sin \phi \sin \phi ^{\prime} \right)-24{e}_{3}{e}_{9}^{3}\kappa {\sin }^{2}\left(2{\theta }_{0}\right)+24{e}_{3}{e}_{9}^{3}\kappa {\cos }^{2}\left(2{\theta }_{0}\right)+8{e}_{3}{e}_{9}^{3}\kappa \cos \left(2{\theta }_{0}\right)\right)\\ & & +\ {e}_{23}\sin \alpha \cos \phi +\sqrt{2}{e}_{24}\cos \zeta ,\\ {e}_{41} & = & 2\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right){}^{3}\left(-54{e}_{9}^{2}{\sin }^{3}\left(2{\theta }_{0}\right)+{e}_{9}^{6}\sin \left(2{\theta }_{0}\right)-9{e}_{9}^{4}\sin \left(4{\theta }_{0}\right)+3{e}_{10}{e}_{32}\right)\\ & & -\ 18{e}_{9}\left({e}_{9}^{2}-3{e}_{10}\right){e}_{36}\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)\sin \left(2{\theta }_{0}\right)\\ & & +\ 36{e}_{9}\left({e}_{9}^{2}-3{e}_{10}\right)\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right){}^{2}\left(\sqrt{2}{e}_{9}{e}_{16}\kappa -\sin \left(2{\theta }_{0}\right)\right)\left({e}_{9}^{2}\cos \left(2{\theta }_{0}\right)+6{\sin }^{2}\left(2{\theta }_{0}\right)\right)+3{e}_{9}{e}_{10}{e}_{37},\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{e}_{42} & = & -54{e}_{9}^{2}{e}_{10}\left(\sin \left(2{\theta }_{0}\right)-\sqrt{2}{e}_{9}{e}_{16}\kappa \right){}^{2}+12{e}_{9}\left({e}_{9}^{2}-3{e}_{10}\right)\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)\sin \left(2{\theta }_{0}\right)\left(\sin \left(2{\theta }_{0}\right)-\sqrt{2}{e}_{9}{e}_{16}\kappa \right)\\ & & +\ \displaystyle \frac{3}{2}{e}_{9}{e}_{10}\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)\left(4\sqrt{2}{e}_{22}{e}_{9}^{3}\kappa -20\cos \left(2{\theta }_{0}\right)-3\cos \left(4{\theta }_{0}\right)-9\right)\\ & & +\ \left({e}_{9}^{2}-3{e}_{10}\right)\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right){}^{2}\left({e}_{9}^{2}\cos \left(2{\theta }_{0}\right)+6{\sin }^{2}\left(2{\theta }_{0}\right)\right),\\ {e}_{43} & = & 6{e}_{9}{e}_{10}\left(\sqrt{2}{e}_{9}{e}_{16}\kappa -\sin \left(2{\theta }_{0}\right)\right)-\left(3{e}_{10}-{e}_{9}^{2}\right)\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)\sin \left(2{\theta }_{0}\right),\\ {e}_{44} & = & 36{e}_{9}{e}_{39}{e}_{42}\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)-4{e}_{15}{e}_{41},\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{e}_{45} & = & 9{e}_{4}{\sin }^{2}\left(2{\theta }_{0}\right)+18{e}_{3}{e}_{7}\cos \left(4{\theta }_{0}\right)+24\left({e}_{3}{e}_{7}-{e}_{1}{e}_{8}\right)\cos \left(2{\theta }_{0}\right)+22{e}_{3}{e}_{7}-40{e}_{1}{e}_{8},\\ {e}_{46} & = & 8{e}_{9}\sin \zeta \left(-3{e}_{7}\sin \left(2{\theta }_{0}\right)+3{e}_{1}\cos \left(2{\theta }_{0}\right)+{e}_{1}\right)+\sqrt{2}{e}_{45}\kappa ,\\ {e}_{47} & = & -8{e}_{5}\cos \zeta \left(3\cos \left(2{\theta }_{0}\right)+5\right)+8{e}_{9}{\cos }^{2}\zeta \left(-3{e}_{7}\sin \left(2{\theta }_{0}\right)+3{e}_{1}\cos \left(2{\theta }_{0}\right)+{e}_{1}\right)+{e}_{46}\sin \zeta ,\\ {e}_{48} & = & 8{e}_{9}\sin \zeta \left(-3{e}_{7}\sin \left(2{\theta }_{0}\right)+3{e}_{1}\cos \left(2{\theta }_{0}\right)+{e}_{1}\right)+\sqrt{2}{e}_{45}\kappa ,\\ {e}_{49} & = & \displaystyle \frac{1}{2}\left(22{e}_{1}\sin \left(2{\theta }_{0}\right)+9{e}_{1}\sin \left(4{\theta }_{0}\right)+20{e}_{7}\cos \left(2{\theta }_{0}\right)+9{e}_{7}\cos \left(4{\theta }_{0}\right)+3{e}_{7}\right),\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{e}_{50} & = & -4\sin \left(2{\theta }_{0}\right)\left(-3\left({e}_{4}-4{e}_{3}{e}_{7}\right)\cos \left(2{\theta }_{0}\right)+4{e}_{3}{e}_{7}-4{e}_{1}{e}_{8}\right),\\ {e}_{51} & = & -8{e}_{5}\cos \zeta \left(3\cos \left(2{\theta }_{0}\right)+5\right)+8{e}_{9}{\cos }^{2}\zeta \left(-3{e}_{7}\sin \left(2{\theta }_{0}\right)+3{e}_{1}\cos \left(2{\theta }_{0}\right)+{e}_{1}\right)+{e}_{48}\sin \zeta ,\\ {e}_{52} & = & 22{e}_{1}\sin \left(2{\theta }_{0}\right)+9{e}_{1}\sin \left(4{\theta }_{0}\right)+20{e}_{7}\cos \left(2{\theta }_{0}\right)+9{e}_{7}\cos \left(4{\theta }_{0}\right)+3{e}_{7},\\ {e}_{53} & = & {e}_{51}\left(\sqrt{2}{e}_{13}{e}_{9}^{2}\kappa -3\sqrt{2}{e}_{10}{e}_{13}\kappa -2{e}_{9}^{3}+12{e}_{10}{e}_{9}\right)\sin \left(2{\theta }_{0}\right),\\ {e}_{54} & = & 3{e}_{9}{e}_{10}\left(8{e}_{5}{e}_{9}\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)\cos \zeta \sin \left(2{\theta }_{0}\right)+{e}_{52}\left(2\sqrt{2}{e}_{13}\kappa -4{e}_{9}\right){\sin }^{2}\zeta +{e}_{9}{e}_{50}\kappa \left(\sqrt{2}{e}_{9}-{e}_{13}\kappa \right)\sin \zeta \right),\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{e}_{55} & = & 12\left({e}_{4}-4{e}_{3}{e}_{7}\right)\cos \left(4{\theta }_{0}\right)-16\left({e}_{3}{e}_{7}-{e}_{1}{e}_{8}\right)\cos \left(2{\theta }_{0}\right),\\ {e}_{56} & = & 8{e}_{9}\sin \zeta \left(-3{e}_{7}\sin \left(2{\theta }_{0}\right)+3{e}_{1}\cos \left(2{\theta }_{0}\right)+{e}_{1}\right)+\sqrt{2}{e}_{45}\kappa ,\\ {e}_{57} & = & \displaystyle \frac{1}{4}\left(-517{e}_{7}\sin \left(2{\theta }_{0}\right)-420{e}_{7}\sin \left(4{\theta }_{0}\right)-81{e}_{7}\sin \left(6{\theta }_{0}\right)+575{e}_{1}\cos \left(2{\theta }_{0}\right)+426{e}_{1}\cos \left(4{\theta }_{0}\right)+81{e}_{1}\cos \left(6{\theta }_{0}\right)+198{e}_{1}\right),\\ {e}_{58} & = & \displaystyle \frac{535}{4}\cos \left(2{\theta }_{0}\right)+105\cos \left(4{\theta }_{0}\right)+\displaystyle \frac{81}{4}\cos \left(6{\theta }_{0}\right)+45,\\ {e}_{59} & = & -8{e}_{5}\cos \zeta \left(3\cos \left(2{\theta }_{0}\right)+5\right)+8{e}_{9}{\cos }^{2}\zeta \left(-3{e}_{7}\sin \left(2{\theta }_{0}\right)+3{e}_{1}\cos \left(2{\theta }_{0}\right)+{e}_{1}\right)+{e}_{56}\sin \zeta ,\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{e}_{60} & = & 3{e}_{7}\sin \zeta \left(517\sin \left(2{\theta }_{0}\right)+420\sin \left(4{\theta }_{0}\right)+81\sin \left(6{\theta }_{0}\right)\right),\\ {e}_{61} & = & -\sin \zeta \left(-6{e}_{1}\sin \zeta \left(2{e}_{58}+20\cos \left(2{\theta }_{0}\right)+3\cos \left(4{\theta }_{0}\right)+9\right)+3\sqrt{2}{e}_{55}{e}_{9}^{3}\kappa +{e}_{60}\right)-48{e}_{5}{e}_{9}^{3}\cos \zeta \cos \left(2{\theta }_{0}\right)+12{e}_{57}{\cos }^{2}\zeta ,\\ {e}_{62} & = & \left(6{e}_{10}-{e}_{9}^{2}\right)\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)\sin \left(2{\theta }_{0}\right)-6\sqrt{2}{e}_{10}\kappa \left({e}_{9}^{2}{e}_{16}-{e}_{13}\sin {\theta }_{0}\cos {\theta }_{0}\right),\\ {e}_{63} & = & 12{e}_{9}^{4}\left(\cos \alpha \cos \zeta \cos \left(2{\theta }_{0}\right)+\sin \left(2{\theta }_{0}\right)\left({e}_{6}\sin \zeta -{e}_{2}\right)+{e}_{8}\sin \zeta \cos \left(2{\theta }_{0}\right)\right)+36{e}_{16}{e}_{9}^{2}\sin \left(2{\theta }_{0}\right)-30{e}_{13}\cos \left(2{\theta }_{0}\right)+\displaystyle \frac{9}{2}{e}_{13}\left(\cos \left(4{\theta }_{0}\right)-5\right),\\ {e}_{64} & = & 48{e}_{7}\sin \zeta {\cos }^{2}\left(2{\theta }_{0}\right)+80{e}_{7}\sin \zeta \cos \left(2{\theta }_{0}\right)+8\sin \zeta \sin \left(2{\theta }_{0}\right)\left(-3{e}_{7}\sin \left(2{\theta }_{0}\right)+9{e}_{1}\cos \left(2{\theta }_{0}\right)+11{e}_{1}\right)-\sqrt{2}{e}_{9}{e}_{50}\kappa ,\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{e}_{65} & = & {e}_{10}\kappa \left(2{e}_{63}\left(\sqrt{2}{e}_{9}-{e}_{13}\kappa \right)-54\kappa \left({e}_{13}\sin \left(2{\theta }_{0}\right)-2{e}_{9}^{2}{e}_{16}\right){}^{2}\right)\\ & & +\ 2\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right){}^{2}\left(6\left(2{e}_{9}^{2}-9{e}_{10}\right){\sin }^{2}\left(2{\theta }_{0}\right)+\left({e}_{9}^{4}-6{e}_{9}^{2}{e}_{10}\right)\cos \left(2{\theta }_{0}\right)\right)\\ & & -\ 48\sqrt{2}\left(6{e}_{10}-{e}_{9}^{2}\right)\kappa \left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right)\sin {\theta }_{0}\cos {\theta }_{0}\left({e}_{9}^{2}{e}_{16}-{e}_{13}\sin {\theta }_{0}\cos {\theta }_{0}\right),\\ {e}_{66} & = & 6{e}_{9}{e}_{62}\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)\left(-16{e}_{5}{e}_{9}\cos \zeta \sin \left(2{\theta }_{0}\right)+{e}_{64}\sin \zeta +8{e}_{49}{\cos }^{2}\zeta \right),\\ {e}_{67} & = & -12{e}_{5}\cos \left(\zeta -2{\theta }_{0}\right)-12{e}_{5}\cos \left(\zeta +2{\theta }_{0}\right)+\sqrt{2}{e}_{45}\kappa \sin \zeta -40{e}_{5}\cos \zeta -24{e}_{7}{e}_{9}\sin \left(2{\theta }_{0}\right)+24{e}_{1}{e}_{9}\cos \left(2{\theta }_{0}\right)+8{e}_{1}{e}_{9},\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{e}_{68} & = & -6\left(\cos \alpha \cos \zeta \cos \left(2{\theta }_{0}\right)+\sin \left(2{\theta }_{0}\right)\left({e}_{6}\sin \zeta -{e}_{2}\right)+{e}_{8}\sin \zeta \cos \left(2{\theta }_{0}\right)\right),\\ {e}_{69} & = & 2395\cos \left(2{\theta }_{0}\right)+\displaystyle \frac{3}{8}\left(5812\cos \left(4{\theta }_{0}\right)+2040\cos \left(6{\theta }_{0}\right)+243\cos \left(8{\theta }_{0}\right)+2585\right),\\ {e}_{70} & = & \sin \left(2{\theta }_{0}\right)\left(105\cos \left(2{\theta }_{0}\right)-18\cos \left(4{\theta }_{0}\right)+65\right),\\ {e}_{71} & = & \displaystyle \frac{1}{2}\sin \left(2{\theta }_{0}\right)\left(300\cos \left(2{\theta }_{0}\right)-9\cos \left(4{\theta }_{0}\right)+61\right),\\ {e}_{72} & = & \displaystyle \frac{1}{4}\sin \left(2{\theta }_{0}\right)\left(18111\cos \left(2{\theta }_{0}\right)+6120\cos \left(4{\theta }_{0}\right)+729\cos \left(6{\theta }_{0}\right)+11200\right),\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{e}_{73} & = & -2\sin \left(2{\theta }_{0}\right)\left(-3\left({e}_{4}-4{e}_{3}{e}_{7}\right)\cos \left(2{\theta }_{0}\right)+{e}_{3}{e}_{7}-{e}_{1}{e}_{8}\right),\\ {e}_{74} & = & \displaystyle \frac{1}{8}\left(16882{e}_{1}\sin \left(2{\theta }_{0}\right)+17502{e}_{1}\sin \left(4{\theta }_{0}\right)+6138{e}_{1}\sin \left(6{\theta }_{0}\right)+729{e}_{1}\sin \left(8{\theta }_{0}\right)\right.\\ & & +\ \left.19160{e}_{7}\cos \left(2{\theta }_{0}\right)+17436{e}_{7}\cos \left(4{\theta }_{0}\right)+6120{e}_{7}\cos \left(6{\theta }_{0}\right)+729{e}_{7}\cos \left(8{\theta }_{0}\right)+7755{e}_{7}\right),\\ {e}_{75} & = & {e}_{1}\sin \zeta \left(301\sin \left(2{\theta }_{0}\right)+60\sin \left(4{\theta }_{0}\right)+9\sin \left(6{\theta }_{0}\right)\right),\\ {e}_{76} & = & \left(-\sqrt{2}{e}_{13}{e}_{9}^{2}\kappa +3\sqrt{2}{e}_{10}{e}_{13}\kappa +2{e}_{9}^{3}-12{e}_{10}{e}_{9}\right)\sin \left(2{\theta }_{0}\right)+6\sqrt{2}{e}_{10}{e}_{16}{e}_{9}^{2}\kappa ,\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{e}_{77} & = & -32{e}_{5}{e}_{9}^{5}\cos \zeta \sin \left(2{\theta }_{0}\right)+\sin \zeta \left(4\left({e}_{7}{e}_{69}+{e}_{1}{e}_{72}\right)\sin \zeta -16\sqrt{2}{e}_{73}{e}_{9}^{5}\kappa +{e}_{75}\right)+4{e}_{74}{\cos }^{2}\zeta ,\\ {e}_{78} & = & 3{e}_{10}\left(2{e}_{13}^{2}{e}_{9}^{2}{\kappa }^{2}-5{e}_{13}^{2}{\kappa }^{2}-4\sqrt{2}{e}_{13}{e}_{9}^{3}\kappa +5\sqrt{2}{e}_{13}{e}_{9}\kappa +4{e}_{9}^{4}\right)-{e}_{9}^{4}\left({e}_{13}^{2}{\kappa }^{2}-2\sqrt{2}{e}_{13}{e}_{9}\kappa +2{e}_{9}^{2}\right),\\ {e}_{79} & = & 36{e}_{16}{e}_{9}\left(\sqrt{2}{e}_{13}\kappa +{e}_{9}\right)\sin \left(2{\theta }_{0}\right)-27{e}_{13}{\sin }^{2}\left(2{\theta }_{0}\right)+{e}_{9}^{3}\left(-54\sqrt{2}{e}_{16}^{2}\kappa +\sqrt{2}{e}_{13}{e}_{68}\kappa -2{e}_{9}{e}_{68}\right),\\ {e}_{80} & = & 6{e}_{68}{e}_{9}^{9}\sin \left(2{\theta }_{0}\right)+9{e}_{16}{e}_{9}^{7}\left(3\left(\cos \left(4{\theta }_{0}\right)-5\right)-20\cos \left(2{\theta }_{0}\right)\right)+8{e}_{16}{e}_{9}^{11}+{e}_{13}{e}_{71}{e}_{9}^{5},\\ {e}_{81} & = & 5{e}_{9}^{5}\kappa \left(2{e}_{9}^{2}{e}_{16}-{e}_{13}\sin \left(2{\theta }_{0}\right)\right)\left(-72{e}_{16}{e}_{9}^{2}\sin \left(2{\theta }_{0}\right)+60{e}_{13}\cos \left(2{\theta }_{0}\right)-9{e}_{13}\cos \left(4{\theta }_{0}\right)+4{e}_{68}{e}_{9}^{4}+45{e}_{13}\right)\\ & & +\ {e}_{80}\left(2{e}_{13}\kappa -2\sqrt{2}{e}_{9}\right)-4\sqrt{2}{e}_{9}^{9}{\rm{t}}6\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right)\left({e}_{9}^{2}{e}_{16}-{e}_{13}\sin {\theta }_{0}\cos {\theta }_{0}\right),\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{e}_{82} & = & 9{e}_{13}\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right){\cos }^{2}\left(2{\theta }_{0}\right)+15{e}_{13}\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right)\cos \left(2{\theta }_{0}\right)+{e}_{9}{e}_{79},\\ {e}_{83} & = & 4{e}_{9}^{5}\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right){}^{2}\left(-162{e}_{9}^{2}{\sin }^{3}\left(2{\theta }_{0}\right)+{e}_{9}^{6}\sin \left(2{\theta }_{0}\right)-18{e}_{9}^{4}\sin \left(4{\theta }_{0}\right)+6{e}_{10}{e}_{70}\right)+12\sqrt{2}\left({e}_{9}^{2}-6{e}_{10}\right){e}_{82}{e}_{9}^{5}\kappa \sin \left(2{\theta }_{0}\right)\\ & & +\ 36\sqrt{2}{e}_{9}^{5}\kappa \left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)\left(2{e}_{9}^{2}{e}_{16}-{e}_{13}\sin \left(2{\theta }_{0}\right)\right)\left(6\left(2{e}_{9}^{2}-9{e}_{10}\right){\sin }^{2}\left(2{\theta }_{0}\right)+\left({e}_{9}^{4}-6{e}_{9}^{2}{e}_{10}\right)\cos \left(2{\theta }_{0}\right)\right)-3{e}_{10}{e}_{81}\kappa ,\\ {e}_{84} & = & 3{e}_{10}\left(4{e}_{13}^{2}{\kappa }^{2}-13\sqrt{2}{e}_{13}{e}_{9}\kappa +24{e}_{9}^{2}\right)-4{e}_{9}^{2}\left({e}_{13}^{2}{\kappa }^{2}-3\sqrt{2}{e}_{13}{e}_{9}\kappa +4{e}_{9}^{2}\right),\\ {e}_{85} & = & 12{e}_{16}{e}_{9}^{2}\kappa \left(-2{e}_{13}{e}_{9}^{2}\kappa +6{e}_{10}{e}_{13}\kappa +2\sqrt{2}{e}_{9}^{3}-15\sqrt{2}{e}_{10}{e}_{9}\right)\sin \left(2{\theta }_{0}\right)+18{e}_{10}{e}_{13}\kappa \left(\sqrt{2}{e}_{9}-{e}_{13}\kappa \right){\cos }^{2}\left(2{\theta }_{0}\right)\\ & & +\ 3{e}_{84}{\sin }^{2}\left(2{\theta }_{0}\right)+2{e}_{78}\cos \left(2{\theta }_{0}\right)+2{e}_{10}{e}_{9}^{4}\kappa \left(54{e}_{16}^{2}\kappa -{e}_{13}{e}_{68}\kappa +\sqrt{2}{e}_{9}{e}_{68}\right),\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{e}_{86} & = & 3{e}_{9}^{6}\left({e}_{10}{e}_{77}\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right){}^{2}-6{e}_{85}\left(-16{e}_{5}{e}_{9}\cos \zeta \sin \left(2{\theta }_{0}\right)+{e}_{64}\sin \zeta +8{e}_{49}{\cos }^{2}\zeta \right)\right),\\ {e}_{87} & = & 36{e}_{16}{e}_{9}\left(\sqrt{2}{e}_{13}\kappa +{e}_{9}\right)\sin \left(2{\theta }_{0}\right)-27{e}_{13}{\sin }^{2}\left(2{\theta }_{0}\right)+{e}_{9}^{3}\left(-54\sqrt{2}{e}_{16}^{2}\kappa +\sqrt{2}{e}_{13}{e}_{68}\kappa -2{e}_{9}{e}_{68}\right),\\ {e}_{88} & = & -\displaystyle \frac{4\sqrt{2}\kappa \left(9{e}_{13}\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right){\cos }^{2}\left(2{\theta }_{0}\right)+15{e}_{13}\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right)\cos \left(2{\theta }_{0}\right)+{e}_{9}{e}_{87}\right)}{{e}_{9}^{4}\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right){}^{2}},\\ {e}_{89} & = & \cos \left(2{\theta }_{0}\right)\left(6{e}_{8}\kappa -3\sqrt{2}{e}_{9}\sin \zeta \right)-\sqrt{2}{e}_{9}\sin \zeta +10{e}_{8}\kappa ,\\ {e}_{90} & = & {e}_{89}\sin \alpha \cos \phi +{e}_{12}\cos \alpha -\sqrt{2}{e}_{9}\cos \zeta \left(-3{e}_{8}\sin \left(2{\theta }_{0}\right)+3{e}_{3}\cos \left(2{\theta }_{0}\right)+{e}_{3}\right),\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{e}_{91} & = & \sin \alpha \cos \phi \left(\cos \left(2{\theta }_{0}\right)\left(6{e}_{8}\kappa -3\sqrt{2}{e}_{9}\sin \zeta \right)-\sqrt{2}{e}_{9}\sin \zeta +10{e}_{8}\kappa \right)\\ & & +\ {e}_{12}\cos \alpha -\sqrt{2}{e}_{9}\cos \zeta \left(-3{e}_{8}\sin \left(2{\theta }_{0}\right)+3{e}_{3}\cos \left(2{\theta }_{0}\right)+{e}_{3}\right),\\ {e}_{92} & = & \sqrt{2}{e}_{9}^{2}\sin \zeta \sin \left(4{\theta }_{0}\right)+6{e}_{9}\kappa \sin \left(4{\theta }_{0}\right)\sin \left(2{\theta }_{0}\right)\left({e}_{3}-{e}_{6}+\sin \phi \sin \phi ^{\prime} \right)+4{e}_{3}{e}_{9}\kappa {\sin }^{2}\left(2{\theta }_{0}\right)-3\sqrt{2}\sin \zeta {\sin }^{3}\left(2{\theta }_{0}\right),\\ {e}_{93} & = & \left(3{e}_{10}-{e}_{9}^{2}\right)\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)\sin \left(2{\theta }_{0}\right)-6{e}_{9}{e}_{10}\left(\sqrt{2}{e}_{9}{e}_{16}\kappa -\sin \left(2{\theta }_{0}\right)\right),\\ {e}_{94} & = & -\displaystyle \frac{1}{2}{e}_{92}\cos \alpha \csc {\theta }_{0}\sec {\theta }_{0}+{e}_{17}\sin \alpha \cos \phi +\sqrt{2}{e}_{18}\cos \zeta ,\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{e}_{95} & = & -{e}_{25}\sin \zeta -24{e}_{9}^{3}\kappa \cos \left(4{\theta }_{0}\right)\left({e}_{6}-\sin \phi \sin \phi ^{\prime} \right)-24{e}_{3}{e}_{9}^{3}\kappa {\sin }^{2}\left(2{\theta }_{0}\right)+24{e}_{3}{e}_{9}^{3}\kappa {\cos }^{2}\left(2{\theta }_{0}\right)+8{e}_{3}{e}_{9}^{3}\kappa \cos \left(2{\theta }_{0}\right),\\ {e}_{96} & = & \sqrt{2}{e}_{9}^{2}\sin \zeta \sin \left(4{\theta }_{0}\right)+6{e}_{9}\kappa \sin \left(4{\theta }_{0}\right)\sin \left(2{\theta }_{0}\right)\left({e}_{3}-{e}_{6}+\sin \phi \sin \phi ^{\prime} \right)+4{e}_{3}{e}_{9}\kappa {\sin }^{2}\left(2{\theta }_{0}\right)-3\sqrt{2}\sin \zeta {\sin }^{3}\left(2{\theta }_{0}\right),\\ {e}_{97} & = & -\displaystyle \frac{1}{2}{e}_{96}\cos \alpha \csc {\theta }_{0}\sec {\theta }_{0}+{e}_{17}\sin \alpha \cos \phi +\sqrt{2}{e}_{18}\cos \zeta ,\\ {e}_{98} & = & -54{e}_{9}^{2}{e}_{10}\left(\sin \left(2{\theta }_{0}\right)-\sqrt{2}{e}_{9}{e}_{16}\kappa \right){}^{2}+12{e}_{9}\left({e}_{9}^{2}-3{e}_{10}\right)\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)\sin \left(2{\theta }_{0}\right)\left(\sin \left(2{\theta }_{0}\right)-\sqrt{2}{e}_{9}{e}_{16}\kappa \right)\\ & & +\ \displaystyle \frac{3}{2}{e}_{9}{e}_{10}\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)\left(4\sqrt{2}{e}_{22}{e}_{9}^{3}\kappa -20\cos \left(2{\theta }_{0}\right)-3\cos \left(4{\theta }_{0}\right)-9\right)\\ & & +\ \left({e}_{9}^{2}-3{e}_{10}\right)\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right){}^{2}\left({e}_{9}^{2}\cos \left(2{\theta }_{0}\right)+6{\sin }^{2}\left(2{\theta }_{0}\right)\right),\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{e}_{99} & = & \left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)\left(6{e}_{9}{e}_{10}\left(\sqrt{2}{e}_{9}{e}_{16}\kappa -\sin \left(2{\theta }_{0}\right)\right)-\left(3{e}_{10}-{e}_{9}^{2}\right)\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)\sin \left(2{\theta }_{0}\right)\right),\\ {e}_{100} & = & \left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right){}^{2}\left({e}_{23}\sin \alpha \cos \phi -{e}_{95}\cos \alpha +\sqrt{2}{e}_{24}\cos \zeta \right),\\ {e}_{101} & = & 18{e}_{9}\left(\sin \left(2{\theta }_{0}\right)-\sqrt{2}{e}_{9}{e}_{16}\kappa \right){}^{2}-\displaystyle \frac{1}{2}\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)\left(4\sqrt{2}{e}_{22}{e}_{9}^{3}\kappa -20\cos \left(2{\theta }_{0}\right)-3\cos \left(4{\theta }_{0}\right)-9\right),\\ {e}_{102} & = & -\displaystyle \frac{1}{4}\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right){}^{2}\left(-16\sqrt{2}{e}_{16}{e}_{9}^{5}\kappa +301\sin \left(2{\theta }_{0}\right)+60\sin \left(4{\theta }_{0}\right)+9\sin \left(6{\theta }_{0}\right)\right)-24{e}_{9}{e}_{101}\left(\sqrt{2}{e}_{9}{e}_{16}\kappa -\sin \left(2{\theta }_{0}\right)\right)\\ & & +\ 15{e}_{9}\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)\left(\sqrt{2}{e}_{9}{e}_{16}\kappa -\sin \left(2{\theta }_{0}\right)\right)\left(4\sqrt{2}{e}_{22}{e}_{9}^{3}\kappa -20\cos \left(2{\theta }_{0}\right)-3\cos \left(4{\theta }_{0}\right)-9\right),\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{e}_{103} & = & -\displaystyle \frac{1}{2}\sin (2\alpha )\csc \alpha \left({e}_{33}\sin \zeta +16{e}_{9}^{5}\kappa \left(6\sin \left(4{\theta }_{0}\right)\left({e}_{3}-{e}_{6}+\sin \phi \sin \phi ^{\prime} \right)+{e}_{3}\sin \left(2{\theta }_{0}\right)\right)\right)-{e}_{34}\sin \alpha \cos \phi +\sqrt{2}{e}_{35}\cos \zeta ,\\ {e}_{104} & = & \sqrt{2}{e}_{9}^{2}\sin \zeta \sin \left(4{\theta }_{0}\right)+6{e}_{9}\kappa \sin \left(4{\theta }_{0}\right)\sin \left(2{\theta }_{0}\right)\left({e}_{3}-{e}_{6}+\sin \phi \sin \phi ^{\prime} \right)+4{e}_{3}{e}_{9}\kappa {\sin }^{2}\left(2{\theta }_{0}\right)-3\sqrt{2}\sin \zeta {\sin }^{3}\left(2{\theta }_{0}\right),\\ {e}_{105} & = & -{e}_{25}\sin \zeta -24{e}_{9}^{3}\kappa \cos \left(4{\theta }_{0}\right)\left({e}_{6}-\sin \phi \sin \phi ^{\prime} \right)-24{e}_{3}{e}_{9}^{3}\kappa {\sin }^{2}\left(2{\theta }_{0}\right)+24{e}_{3}{e}_{9}^{3}\kappa {\cos }^{2}\left(2{\theta }_{0}\right)+8{e}_{3}{e}_{9}^{3}\kappa \cos \left(2{\theta }_{0}\right),\\ {e}_{106} & = & -\displaystyle \frac{1}{2}{e}_{104}\cos \alpha \csc {\theta }_{0}\sec {\theta }_{0}+{e}_{17}\sin \alpha \cos \phi +\sqrt{2}{e}_{18}\cos \zeta ,\\ {e}_{107} & = & -54{e}_{9}^{2}{e}_{10}\left(\sin \left(2{\theta }_{0}\right)-\sqrt{2}{e}_{9}{e}_{16}\kappa \right){}^{2}+12{e}_{9}\left({e}_{9}^{2}-3{e}_{10}\right)\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)\sin \left(2{\theta }_{0}\right)\left(\sin \left(2{\theta }_{0}\right)-\sqrt{2}{e}_{9}{e}_{16}\kappa \right)\\ & & +\ \displaystyle \frac{3}{2}{e}_{9}{e}_{10}\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)\left(4\sqrt{2}{e}_{22}{e}_{9}^{3}\kappa -20\cos \left(2{\theta }_{0}\right)-3\cos \left(4{\theta }_{0}\right)-9\right)\\ & & +\ \left({e}_{9}^{2}-3{e}_{10}\right)\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right){}^{2}\left({e}_{9}^{2}\cos \left(2{\theta }_{0}\right)+6{\sin }^{2}\left(2{\theta }_{0}\right)\right),\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{e}_{108} & = & 6{e}_{9}{e}_{10}\left(\sqrt{2}{e}_{9}{e}_{16}\kappa -\sin \left(2{\theta }_{0}\right)\right)-\left(3{e}_{10}-{e}_{9}^{2}\right)\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)\sin \left(2{\theta }_{0}\right),\\ {e}_{109} & = & 2\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right){}^{3}\left(-54{e}_{9}^{2}{\sin }^{3}\left(2{\theta }_{0}\right)+{e}_{9}^{6}\sin \left(2{\theta }_{0}\right)-9{e}_{9}^{4}\sin \left(4{\theta }_{0}\right)+3{e}_{10}{e}_{32}\right),\\ {e}_{110} & = & -18{e}_{9}\left({e}_{9}^{2}-3{e}_{10}\right){e}_{101}\left(2{e}_{9}-\sqrt{2}{e}_{13}\kappa \right)\sin \left(2{\theta }_{0}\right)\\ & & +\ 36{e}_{9}\left({e}_{9}^{2}-3{e}_{10}\right)\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right){}^{2}\left(\sqrt{2}{e}_{9}{e}_{16}\kappa -\sin \left(2{\theta }_{0}\right)\right)\left({e}_{9}^{2}\cos \left(2{\theta }_{0}\right)+6{\sin }^{2}\left(2{\theta }_{0}\right)\right)+3{e}_{9}{e}_{10}{e}_{102}+{e}_{109},\\ {e}_{111} & = & 18{e}_{9}{e}_{108}\left(\sqrt{2}{e}_{13}\kappa -2{e}_{9}\right){}^{2}\left({e}_{23}\sin \alpha \cos \phi -{e}_{105}\cos \alpha +\sqrt{2}{e}_{24}\cos \zeta \right).\end{array}\end{eqnarray*}$$

The series coefficients appearing in Equation (24) are

$$\begin{eqnarray}\begin{array}{rcl}{a}_{0} & = & \displaystyle \frac{{r}_{{\rm{e}}}}{12c\kappa }\left(-2\sqrt{3}{{cg}}_{2}\kappa {k}_{0}+12c\kappa {k}_{0}+\displaystyle \frac{\mathrm{log}\left(2+\sqrt{3}\right){g}_{7}}{{r}_{e}}-3{g}_{3}\cos {\theta }_{0}-2\sqrt{3}{g}_{2}\omega +2\sqrt{3}{g}_{2}{\omega }_{0}\right.\\ & & +\ \left.{g}_{4}+{g}_{5}+{g}_{6}+{g}_{8}+12(\omega -{\omega }_{0}\right),\end{array}\end{eqnarray} \tag{ B13 }$$

$$\begin{eqnarray}&&{a}_{1}=-\displaystyle \frac{{r}_{e}\left(2\sqrt{3}\sin \left(2{\theta }_{0}\right)\left(c\kappa {k}_{0}+\omega -{\omega }_{0}\right)-{e}^{{g}_{2}}{g}_{12}{\sin }^{3}\left({\theta }_{0}\right)+{g}_{11}\sin {\theta }_{0}+{g}_{10}\right){e}^{-{g}_{2}}}{4{{cg}}_{1}\kappa },\end{eqnarray} \tag{ B14 }$$

$$\begin{eqnarray}&&{a}_{2}=\displaystyle \frac{{r}_{{\rm{e}}}\left({g}_{16}+{g}_{17}+{g}_{18}+{g}_{19}+{g}_{20}+{g}_{21}\right){e}^{-2{g}_{2}}}{8{{cg}}_{1}^{3}\kappa },\end{eqnarray} \tag{ B15 }$$

$$\begin{eqnarray}&&{a}_{3}=\displaystyle \frac{\left(2\sqrt{3}{r}_{e}\left(-{{cg}}_{28}\kappa {k}_{0}+{g}_{27}\omega +{g}_{28}{\omega }_{0}\right)-12{e}^{{g}_{2}}{g}_{23}{g}_{1}^{5}\kappa \omega {r}_{e}\sin \zeta +{e}^{{g}_{2}}\left(12{g}_{22}{g}_{1}^{5}\kappa \omega +{g}_{29}\right)\right){e}^{-{g}_{2}}}{72{{cg}}_{1}^{5}\kappa },\end{eqnarray} \tag{ B16 }$$

where

$$\begin{eqnarray*}\begin{array}{rcl}{g}_{1} & = & \sqrt{5+3\cos \left(2{\theta }_{0}\right)},\\ {g}_{2} & = & \mathrm{log}\left({g}_{1}+\sqrt{6}\cos {\theta }_{0}\right),\\ {g}_{3} & = & 4\kappa \omega \cos \alpha \cos \zeta {\sin }^{2}{\theta }_{0}+4\kappa \omega \sin \alpha \sin \zeta {\sin }^{2}{\theta }_{0}\cos \phi ^{\prime} +\sqrt{2}{g}_{1}\left(c\kappa {k}_{0}+\omega -{\omega }_{0}\right),\\ {g}_{4} & = & 12\kappa \omega \sin \zeta {\sin }^{3}{\theta }_{0}\sin \phi \sin \phi ^{\prime} ,\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{g}_{5} & = & -12\kappa \omega \cos \alpha \sin \zeta {\sin }^{3}{\theta }_{0}\cos \phi \cos \phi ^{\prime} ,\\ {g}_{6} & = & 12\kappa \omega \sin \alpha \cos \zeta {\sin }^{3}{\theta }_{0}\cos \phi ,\\ {g}_{7} & = & 2\sqrt{3}{r}_{{\rm{e}}}\left(c\kappa {k}_{0}+\omega -{\omega }_{0}\right),\\ {g}_{8} & = & \sqrt{3}\mathrm{log}(2)\left(c\kappa {k}_{0}+\omega -{\omega }_{0}\right),\\ {g}_{9} & = & 9\sqrt{2}\left(c\kappa {k}_{0}+\omega -{\omega }_{0}\right)-8{g}_{1}\kappa \omega \cos \alpha \cos \zeta -8{g}_{1}\kappa \omega \sin \alpha \sin \zeta \cos \phi ^{\prime} ,\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{g}_{10} & = & 12{e}^{{g}_{2}}{g}_{1}\kappa \omega {\sin }^{2}{\theta }_{0}\cos {\theta }_{0}\cos \phi \left(\sin \alpha \cos \zeta -\cos \alpha \sin \zeta \cos \phi ^{\prime} \right),\\ {g}_{11} & = & \sqrt{2}\left(2{g}_{1}+5{e}^{{g}_{2}}\right)\left(c\kappa {k}_{0}+\omega -{\omega }_{0}\right)+6{e}^{{g}_{2}}{g}_{1}\kappa \omega \sin \zeta \sin \left(2{\theta }_{0}\right)\sin \phi \sin \phi ^{\prime} +{e}^{{g}_{2}}{g}_{9}{\cos }^{2}{\theta }_{0},\\ {g}_{12} & = & 3\sqrt{2}\left(c\kappa {k}_{0}+\omega -{\omega }_{0}\right)-4{g}_{1}\kappa \omega \cos \alpha \cos \zeta -4{g}_{1}\kappa \omega \sin \alpha \sin \zeta \cos \phi ^{\prime} ,\\ {g}_{13} & = & 3\sqrt{2}\left(5{e}^{2{g}_{2}}-2{g}_{1}^{2}\right)\left(c\kappa {k}_{0}+\omega -{\omega }_{0}\right)-7{e}^{2{g}_{2}}{g}_{1}^{3}\kappa \omega \cos \alpha \cos \zeta -7{e}^{2{g}_{2}}{g}_{1}^{3}\kappa \omega \sin \alpha \sin \zeta \cos \phi ^{\prime} ,\\ {g}_{14} & = & \sin \zeta \left(\sin \phi \sin \phi ^{\prime} -\cos \alpha \cos \phi \cos \phi ^{\prime} \right)+\sin \alpha \cos \zeta \cos \phi ,\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{g}_{15} & = & \sqrt{2}{e}^{{g}_{2}}\left(9{e}^{{g}_{2}}{\sin }^{2}\left(2{\theta }_{0}\right)+2{g}_{1}^{3}+25{e}^{{g}_{2}}\right)\left(c\kappa {k}_{0}+\omega -{\omega }_{0}\right)+27\sqrt{2}{e}^{2{g}_{2}}{\cos }^{2}\left(2{\theta }_{0}\right)\left(c\kappa {k}_{0}+\omega -{\omega }_{0}\right)\\ & & +\ 60\sqrt{2}{e}^{2{g}_{2}}\cos \left(2{\theta }_{0}\right)\left(c\kappa {k}_{0}+\omega -{\omega }_{0}\right)-4{g}_{13}{\sin }^{2}{\theta }_{0},\\ {g}_{16} & = & 4{e}^{{g}_{2}}\left(3\sqrt{3}{\cos }^{2}\left(2{\theta }_{0}\right)\left(c\kappa {k}_{0}+\omega -{\omega }_{0}\right)-2{e}^{{g}_{2}}{g}_{1}^{3}\kappa \omega {\cos }^{3}{\theta }_{0}\left(\sin \alpha \sin \zeta \cos \phi ^{\prime} +\cos \alpha \cos \zeta \right)\right),\\ {g}_{17} & = & 4\left(5\sqrt{3}{e}^{{g}_{2}}\cos \left(2{\theta }_{0}\right)\left(c\kappa {k}_{0}+\omega -{\omega }_{0}\right)+{g}_{1}^{3}\kappa \sin {\theta }_{0}\left(\sqrt{3}{{ck}}_{0}\sin {\theta }_{0}+6{e}^{2{g}_{2}}{g}_{14}\omega {\cos }^{2}{\theta }_{0}\right)\right),\\ {g}_{18} & = & {g}_{15}\cos {\theta }_{0}-4{g}_{1}^{3}{\sin }^{2}{\theta }_{0}\left(3{e}^{2{g}_{2}}\kappa \omega \sin \alpha \cos \zeta \sin {\theta }_{0}\cos \phi +\sqrt{3}\left({\omega }_{0}-\omega \right)\right),\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{g}_{19} & = & 6\kappa \left(\sqrt{3}c\left({g}_{1}+{e}^{{g}_{2}}\right){k}_{0}{\sin }^{2}\left(2{\theta }_{0}\right)+2{e}^{2{g}_{2}}{g}_{1}^{3}\omega \cos \alpha \sin \zeta {\sin }^{3}{\theta }_{0}\cos \phi \cos \phi ^{\prime} \right),\\ {g}_{20} & = & -12{\sin }^{2}{\theta }_{0}\cos {\theta }_{0}\left(3\sqrt{2}{{ce}}^{2{g}_{2}}\kappa {k}_{0}\cos \left(2{\theta }_{0}\right)-2\sqrt{3}\left({g}_{1}+{e}^{{g}_{2}}\right)\left(\omega -{\omega }_{0}\right)\cos {\theta }_{0}\right),\\ {g}_{21} & = & -3{e}^{2{g}_{2}}\sin {\theta }_{0}\left(4{g}_{1}^{3}\kappa \omega \sin \zeta {\sin }^{2}{\theta }_{0}\sin \phi \sin \phi ^{\prime} +3\sqrt{2}\left(\omega -{\omega }_{0}\right)\sin \left(4{\theta }_{0}\right)\right),\\ {g}_{22} & = & {r}_{{\rm{e}}}\cos \zeta \left(6\sin \alpha {\cos }^{3}{\theta }_{0}\cos \phi -21\sin \alpha {\sin }^{2}{\theta }_{0}\cos {\theta }_{0}\cos \phi +20\cos \alpha \sin {\theta }_{0}{\cos }^{2}{\theta }_{0}-7\cos \alpha {\sin }^{3}{\theta }_{0}\right),\\ {g}_{23} & = & -\displaystyle \frac{3}{4}\cos \alpha \left(\cos {\theta }_{0}-9\cos \left(3{\theta }_{0}\right)\right)\cos \phi \cos \phi ^{\prime} -20\sin \alpha \sin {\theta }_{0}{\cos }^{2}{\theta }_{0}\cos \phi ^{\prime} \\ & & +\ 7\sin \alpha {\sin }^{3}{\theta }_{0}\cos \phi ^{\prime} -6{\cos }^{3}{\theta }_{0}\sin \phi \sin \phi ^{\prime} +21{\sin }^{2}{\theta }_{0}\cos {\theta }_{0}\sin \phi \sin \phi ^{\prime} ,\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{g}_{24} & = & -2\sin {\theta }_{0}\left(861\cos \left(2{\theta }_{0}\right)+252\cos \left(4{\theta }_{0}\right)+27\cos \left(6{\theta }_{0}\right)+604\right),\\ {g}_{25} & = & \displaystyle \frac{1}{2}\sin \left(2{\theta }_{0}\right)\left(60\cos \left(2{\theta }_{0}\right)+9\cos \left(4{\theta }_{0}\right)+155\right),\\ {g}_{26} & = & 18{e}^{{g}_{2}}{\cos }^{2}\left(2{\theta }_{0}\right)+30{e}^{{g}_{2}}\cos \left(2{\theta }_{0}\right)+12{\sin }^{2}{\theta }_{0}\left(3\left(2{g}_{1}+{e}^{{g}_{2}}\right){\cos }^{2}{\theta }_{0}+{g}_{1}^{3}\right)+\sqrt{6}{g}_{1}^{2}\cos {\theta }_{0}\left({e}^{{g}_{2}}{g}_{1}+24{\sin }^{2}{\theta }_{0}\right),\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{g}_{27} & = & -\sqrt{6}{g}_{1}^{5}\sin {\theta }_{0}+{e}^{-{g}_{2}}{g}_{1}\sin {\theta }_{0}\left(\sqrt{6}{g}_{1}+6\cos {\theta }_{0}\right)\left(2\sqrt{6}{g}_{1}^{3}\cos {\theta }_{0}+60\cos \left(2{\theta }_{0}\right)+9\left(\cos \left(4{\theta }_{0}\right)+3\right)\right)\\ & & +\ {e}^{-2{g}_{2}}{g}_{26}{g}_{1}\sin {\theta }_{0}\left(\sqrt{6}{g}_{1}+6\cos {\theta }_{0}\right)-3{g}_{25},\\ {g}_{28} & = & \sqrt{6}{g}_{1}^{5}\sin {\theta }_{0}-{e}^{-{g}_{2}}{g}_{1}\sin {\theta }_{0}\left(\sqrt{6}{g}_{1}+6\cos {\theta }_{0}\right)\left(2\sqrt{6}{g}_{1}^{3}\cos {\theta }_{0}+60\cos \left(2{\theta }_{0}\right)+9\left(\cos \left(4{\theta }_{0}\right)+3\right)\right)\\ & & -\ {e}^{-2{g}_{2}}{g}_{26}{g}_{1}\sin {\theta }_{0}\left(\sqrt{6}{g}_{1}+6\cos {\theta }_{0}\right)+3{g}_{25},\\ {g}_{29} & = & 3\sqrt{2}{g}_{24}{r}_{{\rm{e}}}\left(c\kappa {k}_{0}+\omega -{\omega }_{0}\right).\end{array}\end{eqnarray*}$$

Consider the integral appearing in Equation (27),

$$\begin{eqnarray}&&{I}_{0}=\underset{-\infty }{\overset{+\infty }{\displaystyle \int }}{{\rm{e}}}^{i({c}_{0}+{c}_{1}\theta +{c}_{2}{\theta }^{2}+{c}_{3}{\theta }^{3}}d\theta .\end{eqnarray} \tag{ C1 }$$

By changing the variable of integration θ = (x/l) + m and defining the constants $l=\sqrt[3]{{c}_{3}}$ and m = − c₂/(3c₃), we obtain

$$\begin{eqnarray}&&\underset{-\infty }{\overset{+\infty }{\displaystyle \int }}{{\rm{e}}}^{i({c}_{0}+{c}_{1}\theta +{c}_{2}{\theta }^{2}+{c}_{3}{\theta }^{3})}d\theta =U\underset{-\infty }{\overset{+\infty }{\displaystyle \int }}{e}^{i\left(zx+{x}^{3}\right)}{dx},\end{eqnarray} \tag{ C2 }$$

where

$$\begin{eqnarray*}&&z=\displaystyle \frac{1}{\sqrt[3]{{c}_{3}}}\left({c}_{1}-\displaystyle \frac{{c}_{2}^{2}}{3{c}_{3}}\right),\quad U=\displaystyle \frac{1}{\sqrt[3]{{c}_{3}}}{e}^{i\left({c}_{0}+\tfrac{2{c}_{2}^{3}}{27{c}_{3}^{2}}-\tfrac{{c}_{1}{c}_{2}}{3{c}_{3}}\right)}.\end{eqnarray*}$$

For $\mathrm{Im}(z)=0$, we know

$$\begin{eqnarray}&&{j}_{0}=\underset{-\infty }{\overset{\infty }{\displaystyle \int }}{{\rm{e}}}^{i({zx}+{x}^{3})}{dx}=\displaystyle \frac{\pi }{\sqrt[3]{3}}\left[\left(1-\displaystyle \frac{\sqrt{{z}^{2}}}{z}\right){\rm{Ai}}\left(-\displaystyle \frac{\sqrt{{z}^{2}}}{\sqrt[3]{3}}\right)+\left(1+\displaystyle \frac{\sqrt{{z}^{2}}}{z}\right){\rm{Ai}}\left(\displaystyle \frac{\sqrt{{z}^{2}}}{\sqrt[3]{3}}\right)\right],\end{eqnarray} \tag{ C3 }$$

where Ai (z) is an entire Airy function of z with no branch cut discontinuities, and

$$\begin{eqnarray}&&{j}_{1}=\mathop{\mathop{\displaystyle \int }\limits^{+\infty }}\limits_{-\infty }x\,{{\rm{e}}}^{i({zx}+{x}^{3})}{dx}=-i\displaystyle \frac{2\pi }{\sqrt[3]{{3}^{2}}}{{\rm{Ai}}}^{{\rm{{\prime} }}}\left(\displaystyle \frac{z}{\sqrt[3]{3}}\right),\end{eqnarray} \tag{ C4 }$$

where Ai$^{\prime} (z)$ is the derivative of the Airy function Ai (z). Therefore, we have

$$\begin{eqnarray}&&{I}_{0}=U\,{j}_{0}.\end{eqnarray} \tag{ C5 }$$

By differentiating Equation (C2) on both sides with respect to c₁, we obtain

$$\begin{eqnarray}\begin{array}{rcl}{I}_{1} & = & \underset{-\infty }{\overset{+\infty }{\displaystyle \int }}\theta {{\rm{e}}}^{i({c}_{1}\theta +{c}_{2}{\theta }^{2}+{c}_{3}{\theta }^{3})}d\theta =\displaystyle \frac{U}{\sqrt[3]{{c}_{3}}}\underset{-\infty }{\overset{+\infty }{\displaystyle \int }}\left(x-\displaystyle \frac{{c}_{2}}{3\sqrt[3]{{c}_{3}^{2}}}\right){e}^{i\left(zx+{x}^{3}\right)}{dx}\,\\ & = & \ \displaystyle \frac{U}{\sqrt[3]{{c}_{3}}}\left({j}_{1}-\displaystyle \frac{{c}_{2}}{3\sqrt[3]{{c}_{3}^{2}}}{j}_{0}\right).\end{array}\end{eqnarray} \tag{ C6 }$$

Differentiation of Equation (C2) on both sides with respect to c₂ gives

$$\begin{eqnarray}\begin{array}{rcl}{I}_{2} & = & \underset{-\infty }{\overset{+\infty }{\displaystyle \int }}{\theta }^{2}{{\rm{e}}}^{i({c}_{1}\theta +{c}_{2}{\theta }^{2}+{c}_{3}{\theta }^{3})}d\theta =\displaystyle \frac{U}{3{c}_{3}}\underset{-\infty }{\overset{+\infty }{\displaystyle \int }}\left(\displaystyle \frac{2{c}_{2}^{2}}{3{c}_{3}}-{c}_{1}-\displaystyle \frac{2{c}_{2}}{\sqrt[3]{{c}_{3}}}x\right){e}^{i\left(zx+{x}^{3}\right)}{dx}\,\\ & = & \ \displaystyle \frac{U}{3{c}_{3}}\left[\left(\displaystyle \frac{2{c}_{2}^{2}}{3{c}_{3}}-{c}_{1}\right){j}_{0}-\displaystyle \frac{2{c}_{2}}{\sqrt[3]{{c}_{3}}}{j}_{1}\right].\end{array}\end{eqnarray} \tag{ C7 }$$

Next, by differentiating Equation (C2) on both sides with respect to c₃, we get

$$\begin{eqnarray}\begin{array}{rcl}{I}_{3} & = & \underset{-\infty }{\overset{+\infty }{\displaystyle \int }}{\theta }^{3}{{\rm{e}}}^{i({c}_{1}\theta +{c}_{2}{\theta }^{2}+{c}_{3}{\theta }^{3})}d\theta =\displaystyle \frac{U}{9\sqrt[3]{{c}_{3}^{7}}}\underset{-\infty }{\overset{+\infty }{\displaystyle \int }}\left(\displaystyle \frac{9{c}_{1}{c}_{2}{c}_{3}-4{c}_{2}^{3}}{3\sqrt[3]{{c}_{3}^{2}}}+(4{c}_{2}^{2}-3{c}_{1}{c}_{3})x\right){e}^{i\left(zx+{x}^{3}\right)}{dx}\,\\ & = & \ \displaystyle \frac{U}{9\sqrt[3]{{c}_{3}^{7}}}\left[\displaystyle \frac{(9{c}_{1}{c}_{2}{c}_{3}-4{c}_{2}^{3})}{3\sqrt[3]{{c}_{3}^{2}}}{j}_{0}+(4{c}_{2}^{2}-3{c}_{1}{c}_{3}){j}_{1}\right].\end{array}\end{eqnarray} \tag{ C8 }$$

By differentiating the position vector r presented in Equation (8) with respect to r_e , θ, and ϕ, we get

$$\begin{eqnarray}&&\displaystyle \frac{{\rm{\partial }}{\boldsymbol{r}}}{{\rm{\partial }}{r}_{{\rm{e}}}}{dr}_{{\rm{e}}}={\sin }^{2}\theta \,{dr}_{{\rm{e}}}\,{\hat{r}}_{{\rm{e}}},\,\displaystyle \frac{{\rm{\partial }}{\boldsymbol{r}}}{{\rm{\partial }}\theta }d\theta =ds\,\hat{b},\,{\rm{a}}{\rm{n}}{\rm{d}}\,\displaystyle \frac{{\rm{\partial }}{\boldsymbol{r}}}{{\rm{\partial }}\phi }d\phi ={r}_{e}{\sin }^{3}\theta \,d\phi \,\hat{\phi },\end{eqnarray} \tag{ D1 }$$

where we have chosen ${\hat{r}}_{{\rm{e}}}=\hat{r}.$ The tangent to the field line is $\hat{b}$, and the arc length ${ds}=| {\boldsymbol{b}}| d\theta =({r}_{{\rm{e}}}/\sqrt{2})\sin \theta \sqrt{5+3\cos (2\theta )}\,d\theta .$ Then the infinitesimal line element is given by $d{\boldsymbol{l}}={\sin }^{2}\theta \,{{dr}}_{{\rm{e}}}\,\hat{r}+{ds}\,\hat{b}+r\sin \theta \,d\phi \,\hat{\phi }.$

The infinitesimal area element, which is shown by the shaded region in the $({\hat{r}}_{{\rm{e}}},\hat{b})$-plane in Figure 2, is given by

$$\begin{eqnarray}&&d{\boldsymbol{A}}=\displaystyle \frac{{r}_{{\rm{e}}}}{\sqrt{2}}{\sin }^{3}\theta \,\sqrt{5+3\cos (2\theta )}\,{{dr}}_{{\rm{e}}}\,d\theta ({\hat{r}}_{{\rm{e}}}\times \hat{b}).\end{eqnarray} \tag{ D2 }$$

Since ${\hat{r}}_{{\rm{e}}}$ and $\hat{b}$ are nonorthogonal, we have

$$\begin{eqnarray}&&{\hat{r}}_{{\rm{e}}}\times \hat{b}=\displaystyle \frac{\sqrt{2}\sin \theta }{\sqrt{5+3\cos (2\theta )}}\hat{\phi }.\end{eqnarray} \tag{ D3 }$$

So, we get

$$\begin{eqnarray}&&d{\boldsymbol{A}}={r}_{{\rm{e}}}{\sin }^{4}\theta \,{{dr}}_{{\rm{e}}}\,d\theta \,\hat{\phi }.\end{eqnarray} \tag{ D4 }$$

Therefore, the infinitesimal volume element, as shown in Figure 3, is given by

$$\begin{eqnarray}\begin{array}{rcl}{dV} & = & {r}_{e}{\sin }^{3}\theta \,d\phi \,(d{\boldsymbol{A}}\cdot \hat{\phi })\\ & = & {r}_{{\rm{e}}}^{2}{\sin }^{7}\theta \,{{dr}}_{{\rm{e}}}\,d\theta \,d\phi .\end{array}\end{eqnarray} \tag{ D5 }$$

We can also obtain dV by estimating the Jacobian:

$$\begin{eqnarray*}&&J=\left|\displaystyle \frac{\partial (X,Y,Z)}{\partial ({r}_{{\rm{e}}},\theta ,\phi )}\right|={r}_{{\rm{e}}}^{2}{\sin }^{7}\theta .\end{eqnarray*}$$