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This is somewhat related to another question I asked at this link.

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 $\mathbb{R}^{2n}$. Let's let $(\pmb{x},\dot{\pmb{x}})$ be some coordinate representation for $(\text{x},\mathbf{v})\in TQ$ and $(\pmb{q},\pmb{p})$ be some coordinate representation for $(\text{x},\mathbf{p})\in T^*Q$ (in general, $\pmb{q}\neq\pmb{x}$). That is, there are some smooth coordinate charts, $(TQ,\varphi)$ and $(T^*Q,\vartheta)$ (not necessarily defined globally):

$$ \varphi:TQ \to \mathbb{R}^{2n} \quad,\quad (\pmb{x},\dot{\pmb{x}}) = \varphi(\text{x},\mathbf{v}) \quad \leftrightarrow \quad (\text{x},\mathbf{v}) = \varphi^{-1}(\pmb{x},\dot{\pmb{x}}) \\ \vartheta:T^*Q \to \mathbb{R}^{2n} \quad,\quad (\pmb{q},\pmb{p}) = \vartheta(\text{x},\mathbf{p}) \quad \leftrightarrow \quad (\text{x},\mathbf{p}) = \vartheta^{-1}(\pmb{q},\pmb{p}) $$

But, typically, the ``Lagrangian state space'' coordinates, $(\pmb{x},\dot{\pmb{x}})$, are diffeomorphic to the phase space coordinates, $(\pmb{q},\pmb{p})$; there is some $\psi : \mathbb{R}^{2n}\to \mathbb{R}^{2n}$ (not necessarily defined globally) such that: $$ \psi : \mathbb{R}^{2n}\to \mathbb{R}^{2n} \quad, \quad (\pmb{q},\pmb{p}) = \psi(\pmb{x},\dot{\pmb{x}}) \quad \leftrightarrow \quad (\pmb{x},\dot{\pmb{x}}) = \psi^{-1}(\pmb{q},\pmb{p}) $$

So this means that $(\pmb{x},\dot{\pmb{x}})$ can also be mapped to $T^*Q$ and $(\pmb{q},\pmb{p})$ can be mapped to $TQ$:

$$ (\text{x},\mathbf{v}) = \varphi^{-1} \circ \psi^{-1}(\pmb{q},\pmb{p}) \quad \leftrightarrow \quad (\pmb{q},\pmb{p}) = \psi \circ \varphi(\text{x},\mathbf{v}) \\ (\text{x},\mathbf{p}) = \vartheta^{-1} \circ \psi(\pmb{x},\dot{\pmb{x}}) \quad \leftrightarrow \quad (\pmb{x},\dot{\pmb{x}}) = \psi^{-1} \circ \vartheta(\text{x},\mathbf{p}) $$

A.) So it seems that $TQ$ and $T^*Q$, which are supposedly very distinct spaces, can both be described using the same coordinates. This seems wrong to me for some reason. Is it?

B.) (this is my main question) In particular, let's take $\pmb{x}=\pmb{q}$ as the same coordinates for $Q$. The canonical symplectic form on $T^*Q$ is given in symplectic coordinates $(\pmb{q},\pmb{p})$ by $$ \omega = \mathbf{d}q^i \wedge \mathbf{d}p_i \; \in \Omega^2(T^*Q) $$ We have a coordinate transformation $(\pmb{q},\dot{\pmb{q}})\mapsto (\pmb{q},\frac{\partial L }{\partial \dot{\pmb{q}}}) = (\pmb{q},\pmb{p}) $ such that the above can be written in the $(\pmb{q},\dot{\pmb{q}})$ coordinates as

$$ \omega = \mathbf{d}q^i \wedge \mathbf{d}p_i \;=\; a_{ij} \mathbf{d}q^i \wedge \mathbf{d}q^j \,+\, b_{ij}\mathbf{d}q^i \wedge \mathbf{d}\dot{q}^j $$

where $a_{ij} = \frac{\partial^2 L}{\partial q^j \partial\dot{q}^i}$ and $b_{ij} = \frac{\partial^2 L}{\partial \dot{q}^j \partial\dot{q}^i}$. I have seen several books say that the above expression for $\omega$ is actually a totally different 2-form that belongs to $\Omega^2(TQ)$ rather than $\Omega^2(T^*Q)$. Why?

C.) So when working directly with coordinates and equations of motion in $\mathbb{R}^{2n}$, how do we know if we are working with coordinate representations of $TQ$ or $T^*Q$? If you start with coordinates on $TQ$, then do some transformation(s) in coordinate space, how can you tell if you have ``accidentally'' jumped to $T^*Q$, or vice versa?

edit: I know that $(\pmb{x},\dot{\pmb{x}})$ are not symplectic coordinates for $T^*Q$. But it seems like, if they are valid coordinates for $TQ$, then they are also valid coordinates for $T^*Q$.

J Peterson
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  • @MarianoSuárez-Álvarez I suppose I mean that some 2n-tuple of coordinates which are mapped to a point on $TQ$ can also be mapped to a point on $T^Q$ (under different mappings). For a particle on $Q=\mathbb{E}^2$, we can consider $(\pmb{q},\dot{\pmb{q}})=(r,\phi,\dot{r},\dot{\phi})$ and $(\pmb{q},\pmb{p})= (r,\phi ,p_r, p_{\phi}) \equiv (r,\phi, m\dot{r}, mr^2\dot{\phi})$. Both sets of coordinates specify the exact same state, which is an abstract point on $TQ$ or $T^Q$. But it seems that either coordinates can be used for (mapped to) either space. – J Peterson Dec 07 '22 at 19:02
  • But yes, my question is not posed well because I'm still not 100% sure what I'm confused about. The main reason I am asking is for question B in my post, if you can offer any clarity on it. – J Peterson Dec 07 '22 at 19:04
  • ahh ok, I'm aware of the Legendre transformation, but only as the "passive'' coordinate transformation on $\mathbb{R}^{2n}$ that gave in my post. It seems I need to learn the "active'' coordinate-free view of it as a map $TQ \to T^*Q$. That would indeed clarify things. – J Peterson Dec 07 '22 at 19:23
  • @MarianoSuárez-Álvarez YES! That is my confusion. I learned all of analytical mechanics (Lagrangian and Hamiltonian formulations) only in terms of coordinates on some $\mathbb{R}^m$. I'm now trying to teach myself the geometric view on manifolds. But, since ''everything is a function of everything'' on $\mathbb{R}^{m}$, it is not clear which things I learned in the past are coordinate representations for $TQ$ and which are coordinate representations for $T^*Q$. – J Peterson Dec 07 '22 at 20:41
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    based on my answer to your other question, the answers here should hopefully be obvious. For (A), since the spaces are different, the coordinates are different (since coordinates are by definition functions defined on a subset of that space), though as I mentioned to you, you’re pulling back the same base coordinates on $Q$ to $TQ$ and $T^Q$, but people often abuse notation here. For (B), you take the symplectic form $\omega\in \Omega^2(T^Q)$, and you consider the pullback by the Fiber-derivative, $\omega_L=(\mathbf{F}L)^*\omega\in \Omega^2(TQ)$. Hence, they’re completely different 2-forms. – peek-a-boo Jan 04 '23 at 00:40
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    (which by the way is none other than the exterior derivative of the “momentum 1-form” I told you about in the other answer, $d\mu_L=d(\mathbf{F}L^(\theta))=(\mathbf{F}L)^(d\theta)=(\mathbf{F}L)^\omega$, since exterior derivatives commute with pullbacks). For (C), you just need to read the surrounding context, and you need some experience to decipher and get used to the occasional sloppiness and abuse in notation and language. But if you’re using $(q,\dot{q})$, you’re in $TQ$‘s coordinates, while if you use $(q,p)$, you’re in $T^Q$‘s coordinates. – peek-a-boo Jan 04 '23 at 00:41

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