QPL 2015

On July 17 I gave a talk at Oxford in the Quantum Physics and Logic 2015 conference.  It was recently released on the Oxford Quantum Group youtube channel.  There were a bunch of other really cool talks that week.  In my talk I refer to Pawel Sobocinski’s Graphical Linear Algebra tutorial on Monday and Tuesday, as well as Sean Tull’s talk on Categories of relations as models of quantum theory.  There are benefits to speaking near the end of the conference.

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.

Classical Mechanics formulations (Part 1)

Given fixed starting position and time, q_0 and t_0, and fixed ending position and time, q_1 and t_1, when a particle travels along a path of ‘least’ action, it obeys the Euler-Lagrange equations for some Lagrangian, L:
\displaystyle \frac{d}{dt}\frac{\partial L}{\partial \dot{q}^i} = \frac{\partial L}{\partial q^i}
which is a fancy way of saying the time derivative of p_i (momentum) is equal to F_i (force). The natural variables for this Lagrangian approach are q (position) and \dot{q} (velocity), from which p and F are built.

Given the right conditions, we can recast this formulation into the Hamiltonian approach, where q^i and p_i are the natural variables, and several nice things happen. By ‘right conditions’ I mean conditions that would allow us to return to the Lagrangian approach. One of the nice things that happen is position and momentum can be seen to be conjugate to each other in a way that is impossible for position and velocity. Indeed, Hamilton’s equations illustrate this conjugacy nicely:
\displaystyle \frac{dp_i}{dt} = -\frac{\partial H}{\partial q^i}\\ \frac{dq^i}{dt} = \frac{\partial H}{\partial p_i}
where H(q^i,p_i) is the total energy function.

If p and q are so similar, could we not recast this yet again, but in terms of p and \dot p = F, much like the Lagrangian approach was in terms of q and \dot q? Or if we are feeling zealous with reformulations, why not consider mechanics recast in terms of \dot q and F, like a derived Hamiltonian approach? These are some questions that occurred to me near the end of last quarter in Dr. Baez’ Classical Mechanics (Math 241) course, when he went over how to transition from the Lagrangian approach to the Hamiltonian approach.

At first blush, these ‘new’ approaches seem to give us less information than the old ones. To wit, in Newtonian mechanics, p = m \dot q, so any information about the absolute position appears to be lost to a constant of integration, especially for the \dot q, F approach. But it is worse than this. For each particle being considered, the constant of integration may be different. So not only do we lose absolute position, we also lose relative position. This limits our considerations to situations where potential energy is zero, unless something very nice happens that would allow us to recover q.

The nice thing about mathematics versus physics, is that I can cast aside the difficulties of whether or not things are physically relevant, as long as they make sense mathematically. So I will set aside any complaints about possible non-utility and forge ahead. I have not actually looked that far ahead yet, but I suspect the first ‘new’ approach will be somewhat similar to the Lagrangian approach, via the analogy between q and p. That is, I suspect there will be some function, M, analogous to the Lagrangian, L, such that:
\displaystyle \frac{d}{dt}\frac{\partial M}{\partial \dot{p}_i} = \frac{\partial M}{\partial p_i}
where \frac{\partial M}{\partial \dot{p}_i} and \frac{\partial M}{\partial p_i} will be interesting quantities, perhaps even position and velocity.

The second ‘new’ approach will almost certainly be lossy, but I suspect it will follow a pattern similar to the Hamilton equations. For convenience, I will write v:=\dot q for these:
\displaystyle \frac{dF_i}{dt} = -\frac{\partial J}{\partial v^i}\\ \frac{dv^i}{dt} = \frac{\partial J}{\partial F_i}
where J is something analogous to total energy, and should have units of power / time.

So I leave, for now, with some things to ponder, and some guesses as to the direction they will lead.