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I have been using results from this paper in calculations. In sections 2.4 and 3.4 they perform a canonical transformation into new coordinates consisting of constants of motion. They then construct an SL$(2, \mathbb{R})$ algebra that acts on these coordinates.

The canonical transformation for Schwarzschild $(r, \phi, p_r, p_\phi) \rightarrow (T, \Phi, H, L)$ is given by \begin{align} d T &= \frac{Hr^4}{\left(r^2-2Mr\right)\sqrt{\mathcal{R}(r)}}d r \\ d \Phi &= d\phi - \frac{L}{\sqrt{\mathcal{R}(r)}} d r. \end{align} where \begin{equation}\label{eq:HamSchw} H = \sqrt{\left(1-\frac{2M}{r}\right)\left(\frac{p_\phi^2}{r^2}+ \left(1-\frac{2M}{r}\right)p_r^2\right).} \end{equation}

We can then define a generator \begin{equation} H_0 = (H- \tilde{H}(L)) T \end{equation} where $\tilde{H}(L)$ is the Hamiltonian of the null bound geodesics \begin{equation} \tilde{H}(L) = |L|/ \sqrt{27M^2}. \end{equation} Now the number of orbits for an unbound geodesic can be calculated through the winding number $w = \Delta \phi /(2\pi)$. It can be shown that $\Delta \phi$ in the limit of bound null geodesics is given by $$\Delta \phi = \ln(R_{\text{min}}^{-2})$$ where $R$ is defined through $r = 3M(1+R)$. In this case $$\{w, H_0\} = 1/(2\pi).$$

In section 2.4 the authors claim that the group action on the observable $w$ increases it to $w+1$ (see eq. (2.71)). I am confused on how this group action is defined, as I am not used to their conventions. How is the group action here defined in terms of Poisson brackets?

Geigercounter
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1 Answers1

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This should be familiar to you from the first example of generators of transformations we usually encounter: The bracket between position $q$ and momentum $p$ is constant as $\{q,p\} = 1$, and momentum generates translations $q\mapsto q + c, c\in \mathbb{R}$ as the corresponding one-parameter group action.

You have a constant Poisson bracket $\{w,H_0\}$, hence $H_0$ likewise generates translations $w\mapsto w+ c$ in $w$ as the corresponding group.

Both of these are special cases of the general idea that for any phase space function $f$, the group action generated by it on any other phase space function $g$ is simply given by $\partial_\phi (g\circ\exp(\phi f)) = \{g, f\}$ where $\exp(\phi f), \phi \in \mathbb{R}$ is the flow of the Hamiltonian vector field of $f$. See also this answer of mine.

ACuriousMind
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