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So I've reduced my problem to not being sure how to integrate a trajectory in polar coordinates. Suppose I have a free particle and I express its Hamiltonian thus:

$H =\eta_{ij}P^iP^j,$

where $\eta_{ij}$ is the flat-space metric in 2D polar coordinates:

$ \eta_{ij} = \begin{pmatrix} 1 & 0\\0&r^2 \end{pmatrix}$.

And the $P^i$ vector contains the momenta: $P_i = (P_r, P_\theta)$

This gives a Hamiltonian of

$H = P_r^2 + r^2P_\theta^2$.

(I know this isn't the most straightforward way to do this, but it generalizes to my problem in a useful way– I need to determine a first-order system for my equations of motion via the Hamiltonian given a metric). Getting the equations of motion gives

$\dot{r} = 2P_r, \dot{P_r} = -2P_\theta^2 r,$

$\dot{\theta} = 2P_\theta r^2, \dot{P_\theta} = 0$.

Now, when I plug this into my integrator (a simple RK4 implementation where the initial condition is $(r,P_r,\theta,P_\theta)$), I get bizarre plots (this is the particle's trajectory, the axes are $x$ and $y$ (i.e. $r\sin(\theta), r\cos(\theta)$): enter image description here

This "orbit" is not numerical error. Increasing the time step by two orders of magnitude just makes it cleaner: enter image description here (interestingly, the initial condition for these plots was $r = 10, P_r = 1, \theta = \pi/3, P_\theta = 1$)

What's going on here? It should be a straight line, right?

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  • $\begingroup$ What does the plot represent? What are the axis labels? I suspect what I see here is just approximation errors piling up, but I need some more elements to be sure. $\endgroup$ – GRB Oct 2 at 0:03
  • $\begingroup$ The plot is the particle's trajectory in $x$ and $y$ $(r\sin(\theta), r\cos(\theta))$. It's not numerical error, as the orbital shape is stable under increasing the precision. $\endgroup$ – David Oct 2 at 0:16
  • $\begingroup$ Your H is just a line element $ds^2$ $\endgroup$ – Eli Oct 2 at 7:21
  • $\begingroup$ @David If you increase the precision and the orbit shape changes, then it's exactly what I define as numerical error. As for the circular shape of the orbit, try to put $P_\theta = 0$ as initial condition. $\endgroup$ – GRB Oct 2 at 9:42
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Your Hamiltonian doesn't mean what you think it means, because the canonical momenta are the covariant components of the momentum, not the contravariant ones.

If we start with the Lagrangian

$$L = g_{ij} \dot{q}^i \dot{q}^j$$

we can straightforwardly find that the canonical momenta are

$$p_i = 2 g_{ij} \dot{q}^j.$$

You can see by the index position on the RHS that it makes sense for $p_i$ to have a downstairs index, and that (modulo a normalization) it's nothing more than the covariant version of the velocity. After some index gymnastics we arrive at the Hamiltonian

$$H = g^{ij}p_i p_j.$$

Explicitly for your example, we have $L = \dot{r}^2 + r^2 \dot{\theta}^2$ and $H = p_r^2 + p_\theta^2/r^2$. If you use this Hamiltonian you should get the right trajectories.

There is, however, a subtlety here: My Hamiltonian and yours are equal! After all, index position doesn't matter in a contraction: $g^{ij}p_i p_j$ and $g_{ij} p^i p^j$ are the same. The problem, though, is that the $p^i$ are not canonical variables, so Hamilton's equations don't apply. To use the Hamiltonian formalism you need to use the covariant momentum.

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  • $\begingroup$ Aaaaahhh! thank you! That clears up a lot. $\endgroup$ – David Oct 2 at 2:28

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