# 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 $t$s 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.