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.

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2 thoughts on “Classical Mechanics formulations (Part 1)

  1. Hi! Why don’t you try working out your first new approach in an example (or in general)? It could be interesting. You could do the case of a system with just one variable q, with Lagrangian equal to \frac{1}{2} m \dot{q}^2 - V(q). Then the equations of motion are familiar, and you could try to see what function M gives those equations using your ‘modified Lagrange equation’.

    • A cursory look at that Lagrangian suggests the modified Lagrange equations I conjectured still hold, with M=L, even. I took a naive look at V(q) for an ideal spring, and then for constant gravity. Basically I took everything (position, velocity, momentum and force) as functions of t, and then checked to see if the modified Lagrange equations held. I had a couple of choices for this naive approach for V(\dot p) in the constant gravity case; one made the M equation work, and one that did not (\frac{1}{2}\ddot q_0 \dot p vs. \frac{1}{2m}\dot p\cdot\dot p).
      I’ll be more thoroughly checking my work later – this comment is the fruit of thinking through the fog of insomnia while running a fever.

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