4

I asked a similar question on the physics SE at this link but did not really get an answer so I'll ask here. My question there is a bit more detailed.

From what I have recently learned, in classical mechanics, the configuration of some mechanical systems (say, a double pendulum) can be described by a point on an n-dimensional smooth configuration manifold, $Q$. The dynamics are then often studied using Lagrangian mechanics on $TQ$, or using Hamiltonian mechanics on $T^*Q$, both of which have coordinate representations in $\mathbb{R}^{2n}$.

Long before I ever heard the word manifold, I was "doing" dynamics using some 2n "state space" variables in $\mathbb{R}^{2n}$. I know now that when using state space variables of the form $(\pmb{x},\dot{\pmb{x}})$, I'm really working with some coordinate representation of $(\text{x},\mathbf{v})\in TQ$. I could transform to some other coordinates $(\pmb{q},\dot{\pmb{q}})$ but, contrary to what I used to think, this is not actually a different state space but rather just different coordinates for the same $(\text{x},\mathbf{v})\in TQ$ (right?).
Similarly, when using state space variables of the form $(\pmb{x},\pmb{p}_{x})$ (phase space coordinates), I'm really working with some coordinate representation of $(\text{x},\mathbf{p})\in T^*Q$. I could transform to some other $(\pmb{q},\pmb{p}_{q})$ but again this is not a different phase space but just different coordinates for the same $(\text{x},\mathbf{p})\in T^*Q$ (right?).

My Question: When working directly with coordinates in $\mathbb{R}^{2n}$, we also commonly use more general "state space" coordinates which are not necessarily of the lagrangian form, $(\pmb{x},\dot{\pmb{x}})$, nor of the canonical/symplectic form $(\pmb{x},\pmb{p})$, but are rather just some 2n variables that fully define the state (for example, the classic Keplerian orbit element for the two-body problem). Are such coordinates on $\mathbb{R}^{2n}$ actually a coordinate representation for some ``state space manifold''? Or put differently, it seems $TQ$ and $T^*Q$ are two particular types of a "state space manifold" for which the coordinates and equations of motion have certain properties, but is there a more general manifold on which we can study the dynamics?


note: I'm aware that in the Hamiltonian formulation we can use some canonical transformation such that the coordinates $(\pmb{x},\pmb{p})$ are not necessarily split into "position level" coordinates and "velocity-level" coordinates. The Delaunay variables for the two-body problem are an example of this. So this would be an example of "jumbled" coordinates on $\mathbb{R}^{2n}$ which are actually just still coordinates for the same $T^*Q$.

3 Answers3

5

Everything you said on your third paragraph is correct: you're just dealing with different coordinates for $TQ$ and $T^*Q$.

But what about state spaces which are not $T^*Q$ for any manifold $Q$? You're looking for symplectic manifolds: a pair $(M,\omega)$ where $\omega$ is a non-degenerate closed $2$-form. The main example is $(T^*Q, -{\rm d}\lambda)$, where $\lambda$ is the $1$-form on $T^*Q$ taking $Z\in T_{(x,p)}(T^*Q)$ to $\lambda_{(x,p)}(Z) = p({\rm d}\pi_{(x,p)}(Z))$, where $\pi\colon T^*Q\to Q$ is the bundle projection. Now, Darboux's theorem says that around any point in $M$ there are coordinates $(q^1,\ldots, q^n,p_1,\ldots, p_n)$ such that $$\omega = {\rm d}q^1\wedge {\rm d}p_1+\cdots + {\rm d}q^n\wedge {\rm d}p_n,$$so this situation is locally indistinguishable from working in $T^*Q$.

The literature on symplectic geometry and its relations with classical mechanics is extremely extensive. For a very quick introduction, my own notes might be helpful. I have included in the bibliography links for whatever I could find legally available online, so it might be a good starting point.

Ivo Terek
  • 77,665
  • 1
    Oh I actually just started reading these notes the other week. Your website has been helpful to me. Re your answer: So I had thought that, in mechanics, $T^* Q$ is the symplectic manifold for the system and the setting for all Hamiltonian formulations. If some coordinates in $\mathbb{R}^{2n}$ obey the coordinate representation of Hamilton's equations, then they are actually coordinates for $T^* Q$, or is that wrong? But we sometimes use coordinates in $\mathbb{R}^{2n}$ that do not obey Hamilton's equations. Are they still somehow coordinates for $T^* Q$? Or some other symplectic manifold? – J Peterson Nov 12 '22 at 20:12
  • 1
    The cotangent bundle of $\Bbb R^n$ is $T^(\Bbb R^n) \cong \Bbb R^n\times (\Bbb R^n)^ \cong \Bbb R^{2n}$ and $T^Q$ is a $2n$-dimensional manifold, but we don't look at arbitrary coordinate systems on $T^Q$, just the ones induced from coordinates in $Q$ from the following procedure: given $(q^1,\ldots, q^n)$ coordinates on $Q$, take $(x,p) \in T^*Q$ with $x$ in the coordinate domain, and write $p$ as a linear combination of ${\rm d}q^i\big|_x$. [...] – Ivo Terek Nov 12 '22 at 20:40
  • 1
    [...] Then $$(x,p) \mapsto \left(q^1(x),\ldots, q^n(x), p\left(\frac{\partial}{\partial q^1}\bigg|_x\right),\ldots, p\left(\frac{\partial}{\partial q^n}\bigg|_x\right)\right)$$defines coordinates on $(q^1,\ldots, q^n,p_1,\ldots, p_n)$ on the inverse image of the coordinate domain under $\pi$. The $p_i$'s are special, working as linear coordinates on each cotangent space. If you take completely arbitrary coordinates on $T^*Q$, the canonical symplectic form won't have the special expression displayed in the original answer. – Ivo Terek Nov 12 '22 at 20:41
  • 1
    And taking a quick look at the Phys.SE post, I think what may be throwing you off is the identification $(\Bbb R^n)^\ast \cong \Bbb R^n$, which is not natural and relies on the standard Riemannian metric (the usual dot product) in $\Bbb R^n$. When you think of $p=mv$, $p$ lives in $T^Q$ and $v$ lives in $TQ$, the mass is actually an isomorphism $m\colon TQ\to T^Q$ given by a Riemannian metric in $Q$. Position and velocity are the same thing up to isomorphism only. – Ivo Terek Nov 12 '22 at 20:48
  • ohhh something you said may be the key to my understanding but permit me one more clarifying question: So can any state space coordinates $\pmb{z}\in\mathbb{R}^{2n}$ be seen as coordinates for $T^Q$, but only for certain coordinates (symplectic/Darboux coordinates) will the symplectic 2-form have the ''nice'' feature that its matrix representation is given by the usual matrix $J$? Can all coordinate EOMs $\dot{\pmb{z}}=\pmb{f}(\pmb{z})$ be seen as the coordinate representation of $\tfrac{\text{d}}{\text{d}t} \text{z} = \omega^{-1}(\text{d}H,\cdot)$, for $\text{z}\in T^Q$? [...] – J Peterson Nov 12 '22 at 21:06
  • [...] but only for symplectic coordinates will the coordinate EOMs — the coordiante representation of $\tfrac{\text{d}}{\text{d}t} \text{z} = \omega^{-1}(\text{d}H,\cdot)$ — take the form of Hamilton's equations? – J Peterson Nov 12 '22 at 21:08
  • 1
    First question: yes, absolutely. Coordinates for which $\omega$ look nice are usually called "canonical" by physicists. And yes, once a Hamiltonian function $H\colon M\to \Bbb R$ has been fixed on an arbitrary symplectic manifold, the Hamiltonian vector field $X_H$ is defined by ${\rm d}H = \omega(X_H,\cdot)$. The EOM are $\dot{z}(t) = X_H(z(t))$. This is manifestly coordinate free. But if you choose canonical coordinates, then $$X_H = \sum_i\frac{\partial H}{\partial p_i}\frac{\partial}{\partial q^i} - \frac{\partial H}{\partial q^i}\frac{\partial}{\partial p_i}.$$ – Ivo Terek Nov 12 '22 at 22:29
  • 1
    And so an integral curve described as $t\mapsto (q(t),p(t))$ is characterized by $$\dot{q}^i = \frac{\partial H}{\partial p_i}\quad\mbox{and}\quad \dot{p}_i = -\frac{\partial H}{\partial q^i}.$$If you're not using canonical coordinates, Hamilton's equations become a mess. For example, see item (b) of Exercise 16 on page 24 of my notes. Canonical coordinates for the usual symplectic form are not canonical for the magnetic one, and so you get extra terms. In that particular situation things are still well-behaved, but may quickly become unyieldy. – Ivo Terek Nov 12 '22 at 22:31
  • 1
    Now, about whether every equation $\dot{z} = f(z)$ can be seen (up to a coordinate change) as the coordinate expression for Hamilton's equations? My knee-jerk reaction is "hell no", it would be the same as saying that every first order system of ODEs with an even number of equations is Hamiltonian. It is simply too strong. – Ivo Terek Nov 12 '22 at 22:34
2

My understanding on this is as follow. In general, you can think of points of a manifold $M$ as states of a "system", not necessarily a classical mechanical system. It can be a biological system (think of the Lotka-Volterra model) or any other type. You can introduce a dynamical law, for example fixing a vector field on $M$, and you have a (continuous) dynamical system. The vector field tells you where do you move when time goes on. You can use coordinate changes to express the manifold and the dynamical law in a more comfortable manner.

Having said that, Lagrangian mechanics and Hamiltonian mechanics are nothing but different ways to create, "automatically", dynamical systems that model phenomena of a certain type, that we call classical mechanical systems. They both provide you with a manifold (TQ and T^*Q, respectively) and a dynamical law (Euler-Lagrange equations, or Hamilton equations) that reflect the behaviour of the part of reality you are studying.

2

There seems to be three different questions here, here are some attempts at answering them:

Long before I ever heard the word manifold, I was "doing" dynamics using some 2n "state space" variables in $\mathbb{R}^{2n}$. I know now that when using state space variables of the form $(\pmb{x},\dot{\pmb{x}})$, I'm really working with some coordinate representation of $(\text{x},\mathbf{v})\in TQ$. I could transform to some other coordinates $(\pmb{q},\dot{\pmb{q}})$ but, contrary to what I used to think, this is not actually a different state space but rather just different coordinates for the same $(\text{x},\mathbf{v})\in TQ$ (right?).
Similarly, when using state space variables of the form $(\pmb{x},\pmb{p}_{x})$ (phase space coordinates), I'm really working with some coordinate representation of $(\text{x},\mathbf{p})\in T^*Q$. I could transform to some other $(\pmb{q},\pmb{p}_{q})$ but again this is not a different phase space but just different coordinates for the same $(\text{x},\mathbf{v})\in T^*Q$ (right?).

Whether or not one is changing the phase space by changing the coordinates is dependent on one's philosophical disposition, training and interests. E.g. for the purist you are merely using different letters to model same (or analogous) things, but for the practitioner changing the notation (let alone coordinates) can have serious ramifications in terms of the doing of the mathematics. To give two examples consider the following questions: (1) Are the sets $\{1,2,...,n\}$ and $\{0,1,...,n-1\}$ the same or not? (2) Is a function the same as its Fourier series?

When working directly with coordinates in $\mathbb{R}^{2n}$, we also commonly use more general "state space" coordinates wich are not necessarily of the lagrangian form, $(\pmb{x},\dot{\pmb{x}})$, nor of the canonical/symplectic form $(\pmb{x},\pmb{p})$, but are rather just some 2n variables that fully define the state (for example, the classic Keplerian orbit element for the two-body problem). Are such coordinates on $\mathbb{R}^{2n}$ actually a coordinate representation for some "state space manifold"?

They are, although it must be noted that calling the state space a "manifold" is more about how the coordinates transform. There are essentially two paradigms of how coordinates transforms (covariant/homologically/by pushforwards and contravariant/cohomologically/by pullbacks), and $TQ$ and $TQ^\ast$ exemplify these two paradigms. Of course as you pointed out, and connecting this to the paragraph above, the fact that $TQ$ and $TQ^\ast$ are transformable into eachother (or the correspondence between the Hamiltonian and Lagrangian formalism) is arguably the foundational reason why the mathematics of classical mechanics is so satisfying. With Riemannian or pseudo-Riemannian formalism there is a similar situation, but of a different flavor.

Said differently, just as the number $5$ on its own is nowhere near as important as how it relates to other numbers, the individual choices for variables are nowhere near as important as how they relate to other choices for variables.

Or put differently, it seems $TQ$ and $T^*Q$ are two particular types of a "state space manifold" for which the coordinates and equations of motion have certain properties, but is there a more general manifold on which we can study the dynamics?

From the perspective of modern dynamics, what the manifold or the state space is is secondary; once one has a coherent notion of a time evolution one can always come up with a state space. And indeed at times it becomes useful to consider the state space with a smooth structure, in which case one has a manifold as the state space. But even within the Hamiltonian paradigm a sophisticated enough time evolution may make it inconvenient to consider the state space as a manifold, at least in the sense you seem to take a manifold. E.g. for hydrodynamics it has proved to be useful to take the state space as the group of diffeomorphisms of a manifold, which is an "infinite-dimensional manifold" (in a precise sense) and the time evolution as the geodesic flow on this group appropriately defined.

A much more classical example is the geodesic flow on a (finite dimensional) compact Riemannian manifold $M$. One can indeed think of the geodesic flow as a dynamical system with state space $TM$; however due to a conservation law the state space is often taken as the unit tangent bundle $UTM$. This reduces the degrees of freedom by $1$, and more importantly $UTM$ is still a compact Riemannian manifold in and of itself, making it more convenient to consider as a state space.

See e.g. How a group represents the passage of time?, What are some interesting examples of non-classical dynamical systems? (Group action other than $\mathbb{Z}$ or $\mathbb{R}$), Discrete-time dynamical systems with variable state space dimensions (or output space dimensions), How does symbolic dynamics contribute to the study of dynamical systems?, Uniqueness of a homomorphism in a projective limit dynamical system; showing that a projective limit system is compact - Is my proof correct? for various discussions along the lines of the modern perspective on dynamics.

Alp Uzman
  • 10,742