Transformation of Lagrange brackets

The most common set of orbital elements used to parameterize Keplerian orbits consists of the major radius, $a$; the mean longitude at epoch, $\skew{5}\bar{\lambda}_0$; the eccentricity, $e$; the inclination (relative to some reference plane), $I$; the longitude of the perihelion, $\varpi $; and the longitude of the ascending node, ${\mit \Omega }$. (See Section 4.12.) The mean orbital angular velocity is $n=(\mu/a^3)^{1/2}$ [see Equation (4.117)].

Consider how a particular Lagrange bracket transforms under a rotation of the coordinate system $X$, $Y$, $Z$ about the $Z$-axis (if we look along the axis). We can write

$\displaystyle [p,q] = \frac{\partial(X,\dot{X})}{\partial (p,q)}+ \frac{\partial(Y,\dot{Y})}{\partial (p,q)}+ \frac{\partial(Z,\dot{Z})}{\partial (p,q)},$ (G.41)


$\displaystyle \frac{\partial (a,b)}{\partial (c,d)}\equiv \frac{\partial a}{\pa...
... b}{\partial d} - \frac{\partial a}{\partial d}\,\frac{\partial b}{\partial c}.$ (G.42)

Let the new coordinate system be $x', y', z'$. A rotation about the $Z$-axis though an angle ${\mit \Omega }$ brings the ascending node to the $x'$-axis. See Figure 4.6. The relation between the old and new coordinates is (see Section A.6)

$\displaystyle X$ $\displaystyle = \cos{\mit\Omega}\,x' - \sin{\mit\Omega}\,y',$ (G.43)
$\displaystyle Y$ $\displaystyle =\sin{\mit\Omega}\,x' + \cos{\mit\Omega}\,y',$ (G.44)
$\displaystyle Z$ $\displaystyle =z'.$ (G.45)

The partial derivatives with respect to $p$ can be written

$\displaystyle \frac{\partial X}{\partial p}$ $\displaystyle = A_1\,\cos{\mit\Omega} - B_1\,\sin{\mit\Omega},$ (G.46)
$\displaystyle \frac{\partial Y}{\partial p}$ $\displaystyle = B_1\,\cos{\mit\Omega} +A_1\,\sin{\mit\Omega},$ (G.47)
$\displaystyle \frac{\partial \dot{X}}{\partial p}$ $\displaystyle = C_1\,\cos{\mit\Omega} - D_1\,\sin{\mit\Omega},$ (G.48)
$\displaystyle \frac{\partial \dot{Y}}{\partial p}$ $\displaystyle = D_1\,\cos{\mit\Omega} +C_1\,\sin{\mit\Omega},$ (G.49)


$\displaystyle A_1$ $\displaystyle = \frac{\partial x'}{\partial p} - y'\,\frac{\partial {\mit\Omega}}{\partial p},$ (G.50)
$\displaystyle B_1$ $\displaystyle = \frac{\partial y'}{\partial p} + x'\,\frac{\partial {\mit\Omega}}{\partial p},$ (G.51)
$\displaystyle C_1$ $\displaystyle = \frac{\partial \dot{x}'}{\partial p} - \dot{y}'\,\frac{\partial {\mit\Omega}}{\partial p},$ (G.52)
$\displaystyle D_1$ $\displaystyle = \frac{\partial \dot{y}'}{\partial p} + \dot{x}'\,\frac{\partial {\mit\Omega}}{\partial p}.$ (G.53)

Let $A_2$, $B_2$, $C_2$, and $D_2$ be the equivalent quantities obtained by replacing $p$ by $q$ in the preceding equations. It thus follows that

$\displaystyle \frac{\partial(X,\dot{X})}{\partial (p,q)}$ $\displaystyle =
(A_1\,C_2-A_2\,C_1)\,\cos^2{\mit\Omega} + (B_1\,D_2-B_2\,D_1)\,\sin^2{\mit\Omega}$    
  $\displaystyle \phantom{=}+ (-A_1\,D_2-B_1\,C_2+A_2\,D_1+B_2\,C_1)\,\sin{\mit\Omega}\,\cos{\mit\Omega},$ (G.54)
$\displaystyle \frac{\partial(Y,\dot{Y})}{\partial (p,q)}$ $\displaystyle =
(B_1\,D_2-B_2\,D_1)\,\cos^2{\mit\Omega} + (A_1\,C_2-A_2\,C_1)\,\sin^2{\mit\Omega}$    
  $\displaystyle \phantom{=} + (A_1\,D_2+B_1\,C_2-A_2\,D_1-B_2\,C_1)\,\sin{\mit\Omega}\,\cos{\mit\Omega}.$ (G.55)


$\displaystyle [p,q] = A_1\,C_2-A_2\,C_1 + B_1\,D_2-B_2\,D_1+ \frac{\partial(Z,\dot{Z})}{\partial(p,q)}.$ (G.56)


$\displaystyle A_1\,C_2-A_2\,C_1$ $\displaystyle = \left(\frac{\partial x'}{\partial p} - y'\,\frac{\partial {\mit...{x}'}{\partial p} - \dot{y}'\,\frac{\partial{\mit\Omega}}{\partial p}\right)$    
  $\displaystyle =\frac{\partial (x',\dot{x}')}{\partial (p,q)}$    
  $\displaystyle \phantom{=} +\left(-y'\,\frac{\partial \dot{x}'}{\partial q}+\dot...
...c{\partial \dot{x}'}{\partial p}\right)\frac{\partial{\mit\Omega}}{\partial q}.$ (G.57)


$\displaystyle B_1\,D_2-B_2\,D_1$ $\displaystyle = \left(\frac{\partial y'}{\partial p} + x'\,\frac{\partial {\mit...
...\dot{y}'}{\partial p} +\dot{x}'\,\frac{\partial{\mit\Omega}}{\partial p}\right)$    
  $\displaystyle =\frac{\partial (y',\dot{y}')}{\partial (p,q)} +\left(x'\,\frac{\...
...c{\partial \dot{y}'}{\partial p}\right)\frac{\partial{\mit\Omega}}{\partial q}.$ (G.58)


$\displaystyle [p,q]' = \frac{\partial(x',\dot{x}')}{\partial (p,q)}+ \frac{\partial(y',\dot{y}')}{\partial (p,q)}+ \frac{\partial(z',\dot{z}')}{\partial (p,q)}.$ (G.59)

Because $Z=z'$ and $\dot{Z} = \dot{z}'$, it follows that

$\displaystyle [p,q]$ $\displaystyle = [p,q]' + \left(x'\,\frac{\partial \dot{y}'}{\partial q}+ \dot{y...
...'\,\frac{\partial y'}{\partial q}\right)\frac{\partial{\mit\Omega}}{\partial p}$    
  $\displaystyle \phantom{=}-\left(x'\,\frac{\partial \dot{y}'}{\partial p}+ \dot{...
...'\,\frac{\partial y'}{\partial p}\right)\frac{\partial{\mit\Omega}}{\partial q}$    
  $\displaystyle = [p,q]' + \frac{\partial({\mit\Omega}, x'\,\dot{y}'-y'\,\dot{x}')}{\partial (p,q)}.$ (G.60)


$\displaystyle x'\,\dot{y}'-y'\,\dot{x}' = h\,\cos\,I = [\mu\,a\,(1-e^{\,2})]^{1/2}\,\cos\,I \equiv {\cal G},$ (G.61)

because the left-hand side is the component of the angular momentum per unit mass parallel to the $z'$-axis. Of course, this axis is inclined at an angle $I$ to the $z$-axis, which is parallel to the angular momentum vector. Thus, we obtain

$\displaystyle [p,q] = [p,q]' + \frac{\partial({\mit\Omega}, {\cal G})}{\partial (p,q)}.$ (G.62)

Consider a rotation of the coordinate system about the $x'$-axis. Let the new coordinate system be $x''$, $y''$, $z''$. A rotation through an angle $I$ brings the orbit into the $x''$-$y''$ plane. See Figure 4.6. Let

$\displaystyle [p,q]'' = \frac{\partial(x'',\dot{x}'')}{\partial (p,q)}+ \frac{\...
...'',\dot{y}'')}{\partial (p,q)}+ \frac{\partial(z'',\dot{z}'')}{\partial (p,q)}.$ (G.63)

By analogy with the previous analysis,

$\displaystyle [p,q]' = [p,q]'' + \frac{\partial(I,y''\,\dot{z}''-z''\,\dot{y}'')}{\partial(p,q)}.$ (G.64)

However, $z''$ and $\dot{z}''$ are both zero, because the orbit lies in the $x''$-$y''$ plane. Hence,

$\displaystyle [p,q]'=[p,q]''.$ (G.65)

Consider, finally, a rotation of the coordinate system about the $z''$-axis. Let the final coordinate system be $x$, $y$, $z$. A rotation through an angle $\varpi-{\mit\Omega}$ brings the perihelion to the $x$-axis. See Figure 4.6. Let

$\displaystyle [p,q]''' = \frac{\partial(x,\dot{x})}{\partial (p,q)}+ \frac{\partial(y,\dot{y})}{\partial (p,q)}.$ (G.66)

By analogy with the previous analysis,

$\displaystyle [p,q]'' = [p,q]''' + \frac{\partial(\varpi-{\mit\Omega},x\,\dot{y}-y\,\dot{x})}{\partial (p,q)}.$ (G.67)


$\displaystyle x\,\dot{y}-y\,\dot{x} = h= [\mu\,a\,(1-e^{\,2})]^{1/2}\equiv H,$ (G.68)

so, from Equations (G.62) and (G.65),

$\displaystyle [p,q] = [p,q]''' + \frac{\partial(\varpi-{\mit\Omega},H )}{\partial (p,q)}
+ \frac{\partial({\mit\Omega}, {\cal G})}{\partial (p,q)}.$ (G.69)

It thus remains to calculate $[p,q]'''$.

The coordinates $x=r\,\cos \theta$ and $y=r\,\sin \theta$—where $r$ represents radial distance from the Sun, and $\theta $ is the true anomaly—are functions of the major radius, $a$, the eccentricity, $e$, and the mean anomaly, ${\cal M}=\skew{5}\bar{\lambda}_0-\varpi+ n\,t$. Because the Lagrange brackets are independent of time, it is sufficient to evaluate them at ${\cal M}=0$; that is, at the perihelion point. It is easily demonstrated from Equations (4.86) and (4.87) that

$\displaystyle x$ $\displaystyle = a\,(1-e)+ {\cal O}({\cal M}^{\,2}),$ (G.70)
$\displaystyle y$ $\displaystyle = a\,{\cal M}\left(\frac{1+e}{1-e}\right)^{1/2}+{\cal O}({\cal M}^{\,3}),$ (G.71)
$\displaystyle \dot{x}$ $\displaystyle = -a\,n\,\frac{{\cal M}}{(1-e)^2}+{\cal O}({\cal M}^{\,3}),$ (G.72)
$\displaystyle \dot{y}$ $\displaystyle = a\,n\left(\frac{1+e}{1-e}\right)^{1/2}+{\cal O}({\cal M}^{\,2})$ (G.73)

at small ${\cal M}$. Hence, at ${\cal M}=0$,

$\displaystyle \frac{\partial x}{\partial a}$ $\displaystyle = 1-e,$ (G.74)
$\displaystyle \frac{\partial x}{\partial e}$ $\displaystyle = -a,$ (G.75)
$\displaystyle \frac{\partial y}{\partial (\skew{5}\bar{\lambda}_0-\varpi)}$ $\displaystyle = a\left(\frac{1+e}{1-e}\right)^{1/2},$ (G.76)
$\displaystyle \frac{\partial\dot{x}}{\partial(\skew{5}\bar{\lambda}_0-\varpi)}$ $\displaystyle =- \frac{a\,n}{(1-e)^2},$ (G.77)
$\displaystyle \frac{\partial\dot{y}}{\partial a}$ $\displaystyle = -\frac{n}{2}\left(\frac{1+e}{1-e}\right)^{1/2},$ (G.78)
$\displaystyle \frac{\partial\dot{y}}{\partial e}$ $\displaystyle = a\,n\,(1+e)^{-1/2}\,(1-e)^{-3/2},$ (G.79)

because $n\propto a^{-3/2}$. All other partial derivatives are zero. Because the orbit in the $x$, $y$, $z$ coordinate system only depends on the elements $a$, $e$, and $\skew{5}\bar{\lambda}_0-\varpi$, we can write

$\displaystyle [p,q]'''$ $\displaystyle =\frac{\partial(a,e)}{\partial(p,q)}\left[\frac{\partial(x,\dot{x})}{\partial(a,e)}+\frac{\partial(y,\dot{y})}{\partial(a,e)}\right]$    
  $\displaystyle \phantom{=}+\frac{\partial(e,\skew{5}\bar{\lambda}_0-\varpi)}{\pa...
  $\displaystyle \phantom{=}+\frac{\partial(\skew{5}\bar{\lambda}_0-\varpi,a)}{\pa...
...+\frac{\partial(y,\dot{y})}{\partial(\skew{5}\bar{\lambda}_0-\varpi,a)}\right].$ (G.80)

Substitution of the values of the derivatives evaluated at ${\cal M}=0$ into this expression yields

$\displaystyle \frac{\partial(x,\dot{x})}{\partial(a,e)}+\frac{\partial(y,\dot{y})}{\partial(a,e)}$ $\displaystyle = 0,$ (G.81)
$\displaystyle \frac{\partial(x,\dot{x})}{\partial(e,\skew{5}\bar{\lambda}_0-\varpi)}+\frac{\partial(y,\dot{y})}{\partial(e,\skew{5}\bar{\lambda}_0-\varpi)}$ $\displaystyle = 0,$ (G.82)
$\displaystyle \frac{\partial(x,\dot{x})}{\partial(\skew{5}\bar{\lambda}_0-\varpi,a)}+\frac{\partial(y,\dot{y})}{\partial(\skew{5}\bar{\lambda}_0-\varpi,a)}$ $\displaystyle =\frac{a\,n}{2},$ (G.83)


$\displaystyle [p,q]'''= \frac{\partial(\skew{5}\bar{\lambda}_0-\varpi,a)}{\part...
...2\,a^{1/2}} = \frac{\partial(\skew{5}\bar{\lambda}_0-\varpi,L)}{\partial(p,q)},$ (G.84)

where $L=(\mu\,a)^{1/2}$. Hence, from Equation (G.69), we obtain

$\displaystyle [p,q] = \frac{\partial(\skew{5}\bar{\lambda}_0-\varpi,L)}{\partia...
...H )}{\partial (p,q)}
+ \frac{\partial({\mit\Omega}, {\cal G})}{\partial (p,q)}.$ (G.85)