14.3: The Poisson Brackets for the Orbital Elements

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.