Classical Mechanics formulations (Part 2)

Last time I mentioned looking at two functions, M(p,\dot p) and J(\dot q,\dot p) that would ideally carry all the data needed to solve certain types of problems in classical mechanics. In the case of M, I conjectured that equations analogous to the Lagrange equations ought to exist and hold, namely,
\displaystyle \frac{d}{dt}\frac{\partial M}{\partial \dot p_i} = \frac{\partial M}{\partial p_i}.
If we got lucky, we could recover q and \dot q from those partials. While my approach on that front (prior to writing this post) has been… heuristic at best, I have found evidence that indicates M = L with the coordinates transformed. At least in some situations. In my attempt to present this evidence yesterday afternoon without complete notes, I was unable to recapture this evidence on the fly. There is probably a good lesson in that, even if nothing else is gained from this exercise.

In the comments for Part 1, Dr. Baez suggested looking at the one-dimensional Lagrangian L(q,\dot q) = \frac{1}{2}m\dot q\cdot\dot q - V(q), for which the equations of motion are well known. Acting on this suggestion over the weekend, during a bout of insomnia while running a fever, I wove together the evidence I found so elusive yesterday with a skein of half-baked yarn. I probably could mix more metaphors in there if I tried, but the point is, the following paragraph may need to be taken with a liberal grain of salt.

At first I was uncertain how to approach an ‘easy’ V(q), like a gravitational potential, so I took a simple spring potential, V(q) = \frac{1}{2}kq\cdot q. So that’s already a point in favor of feverish-insomnia me over at-the-whiteboard-without-notes me, who couldn’t even write down the potential correctly. The equation of motion is m\ddot q = -kq, from which we can find q(t) = A\sin(\sqrt{\frac{k}{m}}t+B). That can be simplified by choosing units where A=1 and a suitable time translation for which B=0, but I did not originally make those simplifications, so I won’t here. \dot q(t) = A\sqrt{\frac{k}{m}}\cos(\sqrt{\frac{k}{m}}t+B) = \frac{p}{m}, and \ddot q(t) = -A\frac{k}{m}\sin(\sqrt{\frac{k}{m}}t+B) = \frac{\dot p}{m}. Noticing those two outside equalities, \dot q = \frac{p}{m}, and \ddot q = \frac{\dot p}{m}, along with the equation of motion itself, m\ddot q = -kq, right off the bat would have saved me from having to do any of these contortions in t. There goes that point I gave to feverish-insomnia me.

In particular, L(q,\dot q) = \frac{1}{2}(m\dot q\cdot\dot q - kq\cdot q) can be rewritten by converting q to \ddot q via the equation of motion, and from there \dot q can be written in terms of p, so \ddot q can be written in terms of \dot p. Presto:
\displaystyle L = \frac{p\cdot p}{2m} - \frac{\dot p\cdot\dot p}{2k}
without any reference to t. But what happens when we treat this as M and write the Lagrange-analogy equations?
\displaystyle \frac{d}{dt}\frac{\partial L}{\partial \dot p} = \frac{d}{dt}\left(-\frac{\dot p}{k}\right) = \frac{d}{dt}(q), because -kq = m\ddot q = \dot p. Similarly,
\displaystyle \frac{\partial L}{\partial p} = \frac{p}{m} = \dot q,
and indeed, \frac{dq}{dt} = \dot q.

So that worked out for a simple spring. And it wasn’t just tenuous dreamstuff it was made of, after all. Bolstered by that success, I continued to the ‘simpler’ case of constant gravity. Here, V(q) = mgq, where g is the acceleration due to gravity, and is constant. Again, I took the cheap route of writing things in terms of t, but this time I was unable to fold all the ts back into an expression with only p,\dot p as variables. So that’s the bad news for this one. The good news is that if you take L(p,\dot p,q) and apply the partials, you do get \frac{dq}{dt} = \dot q again. Unfortunately, this requires more data than the original formulations, as the Hamiltonian approach gets by with a proper subset of those coordinate variables. I had thought I had gotten away from requiring q, but looking at what I wrote in my delirious state, that really was a phantasm of night.

It would appear this pseudo Lagrangian approach can work for certain situations, most likely ones whose equations of motion are second order homogeneous differential equations such that q can be solved for in terms of \dot q, \ddot q, which can be converted to p,\dot p. This seems likely at least for the given kinetic energy that was used here. A DHO (damped harmonic oscillator) might be worth looking at in the future, as those provide a generic linear second order homogeneous differential equation of motion.

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