I have been thinking recently about how one could introduce and motivate elementary mechanics starting from the hamiltonian (or lagrangian, but that is not mostly what I'm thinking about) point of view, without going through Newton's laws. I don't claim this makes sense or is a good idea — consider it an idle intellectual exercise.

Introducing dynamics using Newton's 2nd law is relatively straightforward, because the idea of pushing or pulling on something to change its motion makes intuitive sense. From there, there is a definite path to understanding momentum, energy, and conservation laws, and I think this at least opens the door to explaining Hamilton's equations.

However, without Newton's 2nd law, it seems like there is no way to motivate Hamilton's equations. Worse than that, there doesn't even seem to be any way to introduce the momentum, on which the whole thing is based. It seems the best one can do is write down Hamilton's equations by fiat in terms of this strange variable $p$, then try to understand what they tell us.

My specific question is: if you were forced to explain momentum (specifically in the context of hamiltonian mechanics) to someone who didn't know newtonian mechanics, and you couldn't say anything about forces or Newton's laws, how would you do it? I realize this is a pretty open-ended question, and the answer might just be "this is a pointless waste of time," but I am hoping others have thought about this before and may have some insights I don't.

  • $\begingroup$ In quantum mechanics, a momentum $p$ is the rate of change of phase as you adjust the conjugate position variable $x$. (In classical mechanics, the same statement is true except it's the rate of change of the "on-shell action".) So you could drop all mention of dynamics involving differential equations entirely, and focus on the path integral / phase / action. $\endgroup$ – knzhou May 23 at 21:43
  • $\begingroup$ The only problem is that while this approach alone might satisfy some math student who likes formal-sounding statements, it'll leave them with absolutely no physical intuition... $\endgroup$ – knzhou May 23 at 21:44
  • $\begingroup$ With respect to the Hamiltonian formalism, position and momenta are on symmetric footing (you can perform a canonical transformation that takes momenta to positions and positions to momenta). So I don't see a way of explaining what momentum is from a Hamiltonian formalism viewpoint. Of course, unless you specify a particular Hamiltonian that might break this symmetry. $\endgroup$ – Dvij D.C. May 23 at 23:37

I have never actually seen it done like this, but without introducing generalised coordinates (for which I think vector notations are better) one could use conservation of momentum as the basis for understanding Newtonian mechanics. This encompasses Newton's third law and leads to a natural definition of Force via his second.

I would regard this as a more fundamental approach because the classical idea of Force is not as natural in either general relativity or quantum mechanics. One can define Force in general relativity, but it is really easier to work with conservation of momentum which is embodied in Einstein's equation. Conservation of momentum can be proven for interactions in quantum field theory using only relativistic principles, but the classical idea of force is not useful in quantum theory.

| cite | improve this answer | |

Well, you could start by introducing the momentum as the generator of translations, as it is a well know characteristic. For that you need to have a Hamiltonian $H(q,p)$ and Hamilton's equation of motion:

$$\frac{dQ}{dt} = \{Q,H\}$$

Where $Q$ is any observable and $\{A,B\}$ denote Poisson's brackets:

$$\{A,B\} = \frac{\partial A}{\partial q} \frac{\partial B}{\partial p} - \frac{\partial B}{\partial q} \frac{\partial A}{\partial p}$$

This is the most general form of Hamilton's equations, which may be kinda like Noether's Theorem for finding conserved currents here in the Hamiltonian formalism.

Hey, I know what you're going to say. You're gonna say it's pointless to use a thing that depends on the momentum to define it. However, you must note that this equation up there is a generalization; at the principle of the theory, only two really were there, pretty much as axioms:

$$\frac{dp}{dt} = - \frac{\partial H}{\partial q}$$

$$\frac{dq}{dt} = \frac{\partial H}{\partial p}$$

Which are the most know forms of Hamilton's equation, the one's that create the basis of the theory. However, they can also be understood as a way of defining $q$ and $p$ (and don't you dare ask where those equations come from).

Ok, having a closer look at the first of those you notice that, for a conserved $p$, $H$ must be invariant under spatial translations (assuming you've already defined $q$ as space, for god's sake). This property makes $p$ the generator of translations in $H$.

Therefore, it seems reasonable to build a theory using $q$ and the generator of translations in $q$, which is $p$, the momentum. If there wasn't any generator of this kind in $H$, the theory would not involve translations and, thus, $q$ would need to be a constant, making the theory really boring (and useless). From this you can already imagine $p$ being the genesis of the movement; something deeply connected to the dynamics of the position $q$.

The first equation is also telling you that $p$ depends on a function of $q$, let's call it $\frac{\partial H}{\partial q} = F(q)$. So, in crude terms, the second equation is telling you that "how much your system is translating/moving" depends on where you are, which might lead you to think of $F(q)$ as a way of restricting and directing the system's dynamics, like a force would do. So, even though we don't require forces at all it ended up appearing in the theory, as you can see However, they are nowhere close to being the center of the theory --- that's $q$ and $p$ places.

Now, we summarize what we know about the first equation: it is saying that $p$ generates translations in $H$, directing the trajectory of the system, and is restricted by a function $\frac{\partial H}{\partial q} = F(q)$.

Finally, looking at the second equation, we may notice that the lhs is already well defined, since the position variable must be predetermined. The rhs, though, must be a constant or a function of momentum, $u(p) = \frac{\partial H}{\partial p}$, and therefore this is an equation defining the momentum, $\frac{d q}{dt} = u(p)$. Actually, because the lhs is the velocity, you can see that the momentum is somehow proportional to the velocity already. Because momentum is proportional to the velocity, you already know that $F(q)$ is proportional to the acceleration.

A summary of the variable $p$ thus is:

  • It generates and maintains translations/movement in $H(q, p)$ (this will lead to inertia)
  • Its evolution is dictated by a function $F(q) = \frac{ \partial H}{\partial q}$ which is proportional to the acceleration (this will lead to Newton's law)
  • It is proportional to the velocity, not the acceleration (this will lead to the actual definition of momentum, like $p=mv$ for a free system).

So yes, you can introduce momentum without using Newtonian mechanics, even though it might cost you some math interpretation skills and some creativity (read hand-waving) to attain physical meaning to the concepts.

Obs: the first two Hamilton's equations come from the Lagrangian formalism, which then have much deeper origins, like the action of a system, if you are wondering.

Moreover: it is even easier to correlate momentum with translation in the Lagrangian formalism, since Noether's theorem is "native" of there, thus allowing you to calculate each conserved current explicitly given a certain transformation, including the spatial translations that characterize momentum.

Curiosity: interpretations of momentum exist way before Newton's laws, dating all the way back to Aristotle, followed by the Theory of Impetus, the baby of the 17th century classical mechanics. However, you probably won't find Aristotle messing around with derivatives or translation generators --- he (and with he I mean "ancient" physicists) rather work mostly through the intuitive association of momentum with movement and the tendency to remain at it (inertia, how we call it nowadays). Also, they used to call it impetus instead of momentum, elucidating it's connection to inertia.

Edit: So... I am kind of new here, so sorry for any organization mistakes, but I do hope this ends up being useful.

| cite | improve this answer | |

Not the answer you're looking for? Browse other questions tagged or ask your own question.