Elliptic expansions

The well-known Bessel functions of the first kind, $J_n(x)$, where $n$ is an integer, are defined as the Fourier coefficients in the expansion of $\exp(\,{\rm i}\,x\,\sin\phi)$:

$${\rm e}^{\,{\rm i}\,x\,\sin\phi}\equiv \sum_{n=-\infty,\infty}J_n(x)\,{\rm e}^{\,{\rm i}\,n\,\phi}.$$ (A.116)

It follows that

$$J_n(x) = \frac{1}{2\pi}\oint {\rm e}^{-{\rm i}\,(n\,\phi-x\,\sin\phi)}\,d\phi = \frac{1}{\pi}\int_0^\pi \cos\left(x\,\sin\phi-n\,\phi\right)\,d\phi$$ (A.117)

(Gradshteyn and Ryzhik 1980a). The Taylor expansion of $J_n(x)$ about $x=0$ is

$$J_n(x) = \left(\frac{x}{2}\right)^n\sum_{k=0,\infty} \frac{(-x^{\,2}/4)^k}{k!\,(n+k)!}$$ (A.118)

for $n\geq 0$ (Gradshteyn and Ryzhik 1980b). Moreover,

$$J_{-n}(x) =(-1)^n\,J_n(x),$$ (A.119)

$$J_{-n}(-x) = J_n(x).$$ (A.120)

In particular,

$$J_0(x) = 1-\frac{x^{\,2}}{4} + {\cal O}(x^{\,4}),$$ (A.121)

$$J_1(x) =\frac{x}{2}-\frac{x^{\,3}}{16} + {\cal O}(x^{\,5}),$$ (A.122)

$$J_2(x) =\frac{x^{\,2}}{8}+{\cal O}(x^{\,4}),$$ (A.123)

$$J_3(x) =\frac{x^{\,3}}{48}+{\cal O}(x^{\,5}).$$ (A.124)

Let us write

$${\rm e}^{\,{\rm i}\,E} = \sum_{n=-\infty,\infty} A_n\,{\rm e}^{\,{\rm i}\,n\,{\cal M}},$$ (A.125)

where $E$ is the eccentric anomaly, and ${\cal M}$ the mean anomaly, of a Keplerian elliptic orbit. (See Section 4.11.) It follows that

$$A_n = \frac{1}{2\pi}\oint {\rm e}^{\,{\rm i}\,(E-n\,{\cal M})}\, d{\cal M}.$$ (A.126)

Integrating by parts, we obtain

$$A_n = \frac{1}{2\pi\,n}\oint{\rm e}^{\,{\rm i}\,(E-n\,{\cal M})} \,dE.$$ (A.127)

However, according to Equation (4.59), the relationship between the eccentric and the mean anomalies is

$$E-e\,\sin E={\cal M},$$ (A.128)

where $e$ is the orbital eccentricity. Hence,

$$A_n = \frac{1}{2\pi\,n} \oint {\rm e}^{-{\rm i}\,[(n-1)\,E - n\,e\,\sin E]}\,dE.$$ (A.129)

Comparison with Equation (A.117) reveals that

$$A_n = \frac{J_{n-1}(n\,e)}{n}.$$ (A.130)

For the special case $n=0$, L'Hopital's rule, together with Equations (A.119) and (A.122), yields

$$A_0 = e\,J_{-1}'(0)=-e\,J_1'(0) = -\frac{e}{2},$$ (A.131)

where $'$ denotes a derivative.

The real part of Equation (A.125) gives

$$\cos E = A_0 + \sum_{n=1,\infty}(A_n+A_{-n})\,\cos(n\,{\cal M})$$

$$=-\frac{e}{2} + \sum_{n=1,\infty} \left[\frac{J_{n-1}(n\,e)-J_{-n-1}(-n\,e)}{n}\right]\cos (n\,{\cal M})$$

$$= -\frac{e}{2} + \sum_{n=1,\infty}\left[\frac{J_{n-1}(n\,e)-J_{n+1}(n\,e)}{n}\right]\cos(n\,{\cal M}),$$ (A.132)

where use has been made of Equations (A.120), (A.130), and (A.131). Likewise, the imaginary part of Equation (A.125) yields

$$\sin E = \sum_{n=1,\infty}\left[\frac{J_{n-1}(n\,e)+ J_{n+1}(n\,e)}{n}\right]\sin(n\,{\cal M}).$$ (A.133)

It follows from Equations (A.121)–(A.124) that

$$\sin E = \left(1-\frac{e^{\,2}}{8}\right)\sin {\cal M}+\frac{e}{2}\left(1-\frac{e^{\,2}}{3}\right)\sin 2{\cal M} +\frac{3\,e^{\,2}}{8}\,\sin 3{\cal M}$$

$$\phantom{=}+ \frac{e^{\,3}}{3}\,\sin 4{\cal M} + {\cal O}(e^{\,4}),$$ (A.134)

$$\cos E =-\frac{e}{2} +\left(1-\frac{3\,e^{\,2}}{8}\right)\cos {\cal M}+\frac{e}{2}\left(1-\frac{2\,e^{\,2}}{3}\right)\cos 2{\cal M}$$

$$\phantom{=} +\frac{3\,e^{\,2}}{8}\,\cos 3{\cal M}+ \frac{e^{\,3}}{3}\,\cos 4{\cal M}+ {\cal O}(e^{\,4}).$$ (A.135)

Hence, from Equation (A.128),

$$E = {\cal M} + e\,\sin{\cal M} + \frac{e^{\,2}}{2}\,\sin 2{\cal M} + \frac{e^{\,3}}{8}\left(3\,\sin 3{\cal M}-\sin{\cal M}\right)$$

$$\phantom{=}+\frac{e^{\,4}}{6}\left(2\,\sin 4{\cal M}-\sin 2{\cal M}\right) + {\cal O}(e^5).$$ (A.136)

According to Equation (4.69),

$$\frac{r}{a} = 1-e\,\cos E,$$ (A.137)

where $r$ is is the radial distance from the focus of the orbit and $a$ is the orbital major radius. Thus,

$$\frac{r}{a} =1-e\,\cos{\cal M} -\frac{e^{\,2}}{2}\left(\cos 2{\cal M}-1\right) - \frac{3\,e^{\,3}}{8}\left( \cos 3{\cal M}- \cos{\cal M}\right)$$

$$\phantom{=} - \frac{e^{\,4}}{3}\left(\cos 4{\cal M}-\cos 2{\cal M}\right)+{\cal O}(e^5).$$ (A.138)

Equations (4.39) and (4.67) imply that

$$\frac{d\theta}{d{\cal M}} = (1-e^{\,2})^{1/2}\left(\frac{a}{r}\right)^2,$$ (A.139)

where $\theta$ is the true anomaly. Hence, it follows from Equations (A.128) and (A.137), and the fact that $\theta=0$ when ${\cal M}=0$, that

$$\theta = (1-e^{\,2})^{1/2}\int_0^{\cal M} \left(\frac{dE}{d{\cal M}}\right)^2\,d{\cal M}.$$ (A.140)

From Equation (A.136),

$$\frac{dE}{d{\cal M}} =1+e\left(1-\frac{e^{\,2}}{8}\right)\cos{\cal M} + e^{\,2}\left(1-\frac{e^{\,2}}{3}\right) \cos 2{\cal M} +\frac{9\,e^{\,3}}{8}\,\cos 3{\cal M}$$

$$\phantom{=}+\frac{4\,e^{\,4}}{3}\,\cos 4{\cal M} + {\cal O}(e^5).$$ (A.141)

Thus,

$$\left(\frac{dE}{d{\cal M}}\right)^2 = 1+\frac{e^{\,2}}{2}+\frac{3\,e^{\,4}}{8} + 2\,e\left(1+\frac{3\... ...al M} + \frac{5\,e^{\,2}}{2}\left(1+\frac{2\,e^{\,2}}{15}\right) \cos 2{\cal M}$$

$$\phantom{=}+\frac{13\,e^{\,3}}{4}\,\cos 3{\cal M} + \frac{103\,e^{\,4}}{24}\,\cos 4{\cal M} +{\cal O}(e^5),$$ (A.142)

and

$$\theta = {\cal M} + 2\,e\,\sin{\cal M} + \frac{5\,e^{\,2}}{4}\,\sin 2{\c... ... e^{\,3}\left( \frac{13}{12}\,\sin 3{\cal M} - \frac{1}{4}\,\sin{\cal M}\right)$$

$$\phantom{=}+e^{\,4}\left(\frac{103}{96}\,\sin 4{\cal M} - \frac{11}{24}\,\sin 2{\cal M}\right)+{\cal O}(e^5).$$ (A.143)