# numerically integrating a trajectory in polar coordinates

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. $$r\sin(\theta), r\cos(\theta)$$): This "orbit" is not numerical error. Increasing the time step by two orders of magnitude just makes it cleaner: (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?

• 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. – GRB Oct 2 at 0:03
• 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. – David Oct 2 at 0:16
• Your H is just a line element $ds^2$ – Eli Oct 2 at 7:21
• @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. – GRB Oct 2 at 9:42

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.

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

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.

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