Program Listing for File InclinationEccentricity_pms2.md

↰ Return to documentation for file (InclinationEccentricity_pms2.md)

Calculation of the Pm,s Coefficients {#InclinationEccentricity_pms2}
=============================================

We need only \f$m=0\f$ and \f$m=\pm2\f$.
Clearly:
\f{eqnarray*}{
    p_{m,s}&=&\int_0^{2\pi/\omega}
        \frac{a^3\exp(-im\Delta \phi(t))}{r^3(t)}e^{i s \omega t}dt\\
    &=& \int_0^{2\pi/\omega}
        a^3\exp(-im\phi_0)\frac{\cos(m\phi(t))-i\sin(m\phi(t))}{r^3(t)}
        e^{i s \omega t}dt
\f}
For \f$m=0\f$:
\f{eqnarray*}{
    p_{0,s}&=& \int_{0}^{2\pi}
        \frac{\exp(i s (u-e\sin u))}{\omega (1-e\cos u)^2} du\\
    &=& \frac{1}{\omega}
        \int_{0}^{2\pi} \frac{\exp(i s (u-e\sin u))} {(1-e\cos u)^2} du
\f}
From \f$1/(1-x)^2=\sum_{k=0}^\infty (k+1)x^k\f$:
\f{eqnarray*}{
    p_{0,s}&=&\frac{1}{\omega}\int_0^{2\pi}
        \left[\cos u + i \sin u\right]^s
        \left[\sum_{l=0}^\infty \frac{(-ise)^l\sin^l u}{l!}\right]
        \left[\sum_{p=0}^\infty (p+1) e^p\cos^p u\right]\\
    &=&\frac{1}{\omega}\int_0^{2\pi}
        \left[\sum_{k=0}^s {s \choose k} i^k\cos^{s-k} u \sin^k u\right]
        \left[\sum_{l=0}^\infty \frac{(-ise)^l\sin^l u}{l!}\right]
        \left[\sum_{p=0}^\infty (p+1) e^p\cos^p u\right]\\
    &=&\frac{1}{\omega}\sum_{k=0}^s \sum_{l=0}^\infty \sum_{p=0}^\infty
            (-1)^l i^{k+l} (p+1) {s \choose k} \frac{s^l}{l!} e^{l+p}
            I_{k+l,s-k+p}\\
    &=&\frac{1}{\omega}\sum_{n=0}^\infty \left[\sum_{k=0}^s {s \choose k}
            \sum_{l=0}^n (-1)^l i^{k+l} (n-l+1) \frac{s^l}{l!}
            I_{k+l,s-k+n-l}\right]e^n

\f}
with
\f[
    I_{m,n} \equiv \int_0^{2\pi} \sin^m u \cos^n u
\f]

To find an expression for \f$m=\pm2\f$ consider:
\f{eqnarray*}{
    Q_{p,q}&\equiv&
        \int_{0}^{2\pi} \frac{\exp[i s (u-e\sin u)]} {(1-e\cos u)^4} \sin^p u
                        \cos^q udu\\
        &=& \sum_{\lambda=0}^\infty \frac{(-ise)^\lambda}{\lambda!}
            \int_{0}^{2\pi}
                \frac{\sin^{\lambda+p} u \cos^q u
                    \left(\cos u + i\sin u\right)^s}{(1-e\cos u)^4}

\f}
Using: \f$1/(1-x)^4=\sum_{k=0}^\infty {{k+3} \choose 3} x^k\f$
\f{eqnarray*}{
    Q_{p,q}&=&\sum_{\lambda=0}^\infty \frac{(-ise)^\lambda}{\lambda!}
            \sum_{k=0}^\infty {k+3 \choose 3} e^k
            \sum_{\sigma=0}^{s} {s \choose \sigma} i^\sigma
                I_{\lambda+\sigma+p, s-\sigma+k+q}\\
    Q_{p,q}&=&\sum_{n=0}^\infty \left[
            \sum_{\lambda=0}^n \frac{(-is)^\lambda}{\lambda!}
            {n-\lambda+3 \choose 3}
            \sum_{\sigma=0}^{s} {s \choose \sigma} i^\sigma
                I_{\lambda+\sigma+p, s+n+q-\sigma-\lambda}\right] e^n
\f}

[here](@ref InclinationEccentricity_pms1) we show:
\f{eqnarray*}{
    p_{\pm2,s}&=&\exp\left(\mp 2i\phi_0\right) p_{0,s} -\\
        &&{}-\frac{\exp(\mp 2i\phi_0)(1-e^2)}{\omega}
        \int_{0}^{2\pi} \frac{\exp[i s (u-e\sin u)]} {(1-e\cos u)^4} du+\\
        &&{}+\frac{\exp(\mp 2i\phi_0)(1-e^2)}{\omega}
        \int_{0}^{2\pi} \frac{\exp[i s (u-e\sin u)]\cos 2u} {(1-e\cos u)^4}
        du\mp\\
        &&{}\mp i\frac{\exp(\mp 2i\phi_0)\sqrt{1-e^2}}{\omega}
        \int_{0}^{2\pi} \frac{\exp[i s (u-e\sin u)]\sin 2u} {(1-e\cos u)^4}
        du\pm\\
        &&{}\pm i\frac{2e\exp(\mp 2i\phi_0)\sqrt{1-e^2}}{\omega}
        \int_{0}^{2\pi} \frac{\exp[i s (u-e\sin u)]\sin u} {(1-e\cos u)^4}
        du\\
    &=&\exp\left(\mp 2i\phi_0\right) p_{0,s} -\\
        &&{}-\frac{\exp(\mp 2i\phi_0)(1-e^2)}{\omega}Q_{0,0}+\\
        &&{}+\frac{\exp(\mp 2i\phi_0)(1-e^2)}{\omega}(Q_{0,0}-2Q_{2,0})\mp\\
        &&{}\mp i\frac{\exp(\mp 2i\phi_0)\sqrt{1-e^2}}{\omega}2Q_{1,1}\pm\\
        &&{}\pm i\frac{2e\exp(\mp 2i\phi_0)\sqrt{1-e^2}}{\omega}Q_{1,0}\\
    &=&\exp\left(\mp 2i\phi_0\right) \left[p_{0,s} -
        2\frac{(1-e^2)Q_{2,0} \pm i\sqrt{1-e^2}(Q_{1,1} - e Q_{1,0})}{\omega}
        \right]

\f}

According to [this page](http://planetmath.org/taylorexpansionofsqrt1x)
\f[
    \sqrt{1-e^2}=1-\frac{e^2}{2}-
        \sum_{n=2}^\infty\frac{(2n-3)!}{2^{2n-2} n! (n-2)!} e^{2n}
\f]

Also not that:
\f{eqnarray*}{
    I_{m,n}&=&I_{m,n-2}-I_{m+2,n-2}=I_{m,n-2}-\frac{m+1}{n-1}I_{m,n}\\
    \Rightarrow I_{m,n}&=&\frac{n-1}{n+m}I_{m,n-2}
\f}
Similarly:
\f{eqnarray*}{
    I_{m,n}&=&I_{m-2,n}-I_{m-2,n+2}=I_{m-2,n}-\frac{n+1}{m-1}I_{m,n}\\
    \Rightarrow I_{m,n}&=&\frac{m-1}{n+m}I_{m-2,n}
\f}
Further \f$I_{m,1}=I_{1,n}=0\f$, so \f$I_{m,n}\f$ is non-zero only if both m
and n are even.

Finally, \f$I_{m,n}=I_{n,m}\f$.