14.3: The Poisson Brackets for the Orbital Elements
- Page ID
- 6877
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)A worked example is in order. From Equations 14.2.7 and 14.2.8, we see that the Poisson brackets are defined by
\[\{ A_i , A_k \}_{α_j , β_j} = \sum_j \left( \frac{\partial A_i}{\partial α_j} \frac{\partial A_k}{\partial β_j} - \frac{\partial A_i}{\partial β_j} \frac{\partial A_k}{\partial α_j} \right) . \label{14.3.1} \]
The \(A_i\) are the orbital elements.
For our example, we shall calculate \(\{ Ω , i \}\) and we write out the sum in full:
\[\begin{align} \{ Ω , i \} &= \sum_j \left( \frac{\partial Ω}{\partial α_j} \frac{\partial i}{\partial β_j} - \frac{\partial Ω}{\partial β_j} \frac{\partial i}{\partial α_j} \right) \\[4pt] &= \frac{\partial Ω}{\partial α_1} \frac{\partial i}{\partial β_1} + \frac{\partial Ω}{\partial α_2} \frac{\partial i}{\partial β_2} + \frac{\partial Ω}{\partial α_3} \frac{\partial i}{\partial β_3} - \frac{\partial Ω}{\partial β_1} \frac{\partial i}{\partial α_1} - \frac{\partial Ω}{\partial β_2} \frac{\partial i}{\partial α_2} - \frac{\partial Ω}{\partial β_3} \frac{\partial i}{\partial α_3}. \label{14.3.2} \end{align}\]
Refer now to Equations 10.11.27 and 29, and we find
\[\{ Ω, i \} = 0 + 0 + 0 - 0 + \frac{1}{α_3 \sqrt{1 - α_2^2 / α_3^2}} - 0 . \label{14.3.3} \]
Finally, referring to Equations 10.11.20 and 21, we obtain
\[ \{ Ω , i \} = \frac{1}{\sqrt{GMm^2 a (1-e^2). \sin i}} . \label{14.3.4}\]
Proceeding in a similar manner for the others, we obtain
\[\begin{align} \{ a , T \} &= - \frac{2a^2}{GMm}, \label{14.3.5} \\[4pt] \{ e , T \} &= - \frac{a(1-e^2)}{GMme}, \label{14.3.6} \\[4pt] \{ i , ω \} &= \frac{1}{\sqrt{GMm^2 a (1-e^2). \tan i}} . \label{14.3.7} \\[4pt] \{ e , ω \} &= - \frac{\sqrt{1-e^2}}{em \sqrt{GMa}}, \label{14.3.8} \end{align}\]
In addition, we have, of course,
\[ \{ i , Ω \} = - \{ Ω , i \} , \{ T , a \} = - \{ a , T \} , \{ T , e \} = - \{ e , T \} \text{ and } \{ ω , i \} = - \{ i , ω \} .\]
All other pairs are zero.