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Let the configuration space of a single "point particle" be the one-dimensional affine space $\mathbb{A}^1 \cong \mathbb{R}$, with a chosen linear coordinate chart identifying some distinguished point $O \in \mathbb{A}^1$ with $0 \in \mathbb{R}$. (So basically this is a point particle confined to a line -- so we should be able to analyze it using Newtonian mechanics with "scalar", as in 1d-vector, quantities.)

The phase space will then be the cotangent bundle of $\mathbb{A}^1$, which will be a trivial bundle isomorphic to $$\mathbb{R} \times (\mathbb{R})^\ast \cong \mathbb{R} \times \mathbb{R} = \mathbb{R}^2 = \mathbb{R} \oplus \mathbb{R}.\tag{1}$$ One of the $\mathbb{R}$ "factors" corresponds to (a coordinate chart of) the configuration space (denoted "$q$"), and the other to the cotangent spaces $T_q^\ast \mathbb{A}^1$ of canonical momenta (denoted "p").

I read that symplectic geometry considers the space $\mathbb{R}^{2n}$ as $\mathbb{R}^n \oplus (\mathbb{R}^n)^\ast$ (like in the standard Hamiltonian picture), while complex geometry considers the space $\mathbb{R}^{2n}$ as $\mathbb{R}^n \oplus i\mathbb{R}^n = \mathbb{C}^n$, with these two interpretations being suitably isomorphic and consistent for $\mathbb{R}^{2n}$, and more generally in the context of any Kahler manifold.

I am not claiming that every symplectic manifold corresponding to a (classical) phase space in the Hamiltonian formalism is a Kahler manifold, I am just pointing out that this is so at least in these very special cases, like $\mathbb{A}^1$ above.

So consider the position $q(t)$ of a single point particle in $\mathbb{A}^1$ -- then we can treat its Hamiltonian phase space as $\mathbb{C}$. (More generally, the phase space corresponding to $N$ point particles on $\mathbb{A}^1$ would be isomorphic to $\mathbb{C}^N$.)

Questions:

  1. Given a momentum "vector" $p$ in the cotangent space $T_q^\ast \mathbb{A}^1$ with base point $q \in \mathbb{A}^1$, corresponding to $(q,p)$, or more suggestively $q + p i$, what is the linear function(al) $T_q \mathbb{A}^1 \to \mathbb{R}$ that it corresponds to? $p: \dot{q}^a \mapsto m \dot{q}^a$ for some fixed scalar $m$ corresponding to the mass $m$ that the given choice of $p$ "assigns" to the particle?

  2. Given two different time points $t_1, t_2$, and a given trajectory $$(q(t), p(t)) \overset{def}{=} (q + pi)(t):\mathbb{R} \to T^\ast \mathbb{A}^1 \cong \mathbb{C},\tag{2}$$ we can consider the two complex numbers $$q(t_1) + ip(t_1) \overset{def}{=} q_1 + p_1 i \overset{def}{=} z_1\tag{3}$$ and $$q(t_2) + i p(t_2) \overset{def}{=} q_2 + p_2 i \overset{def}{=} z_2.\tag{4}$$ Then we can define their "complex inner product" $$z_1 \bar{z_2}= (q_1 q_2 + p_1 p_2) + (p_1 q_2 - p_2 q_1)i ,\tag{5}$$ with the real part being their dot product considered as vectors in $\mathbb{R}^2$ and the imaginary part the (signed) area of the parallelogram spanned by $z_1$ and $z_2$ in the complex plane. Is there a physical interpretation to either the real part or the imaginary part? A physical interpretation to either the real part or the imaginary part being conserved as $t_2$ (and/or $t_1$) changes?

My understanding is that the imaginary part of the complex inner product in this case should correspond to the canonical/tautological symplectic 2-form associated with the symplectic manifold $T_q^\ast \mathbb{A}^1$. I can see how it is an alternating bilinear form at least.

By Liouville's theorem, the value of the symplectic form is supposed to be preserved over time for trajectories corresponding to a conservative system. (That is my poor understanding at least -- I am trying to improve that understanding by asking this question.) So would this correspond to $$0 = \frac{d}{dt_2}(p_1 q_2 - p_2 q_1) = p_1 \frac{d}{dt_2}q_2 - q_1 \frac{d}{dt_2}p_2 \quad \iff \quad p_1 \frac{d}{dt_2}q_2 = q_1 \frac{d}{dt_2}p_2~?\tag{6}$$ Being really generous with myself this "sort of" looks like the Liouville equation, but obviously isn't. Did I go wrong with trying to identify the tautological 2-form with the imaginary part of the complex inner product? Or with the interpretation of Liouville's theorem as saying that this 2-form's value is constant over time?

Possibly related questions (although most likely not): here, here and here.

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  1. Suggestion: It seems that instead of looking at 1 (complex) coordinate $z=q+ip$ at $N$ different times $t_1,\ldots, t_N$, as OP does in eqs. (3-6), it would be more fruitful to look at $N$ (complex) coordinates $z^1,\ldots, z^n$, at the same time $t$.

  2. Then the imaginary part $${\rm Im}(z^{1\ast} z^2)~=~q^1p_2-q^2p_1$$ of OP's complex inner product (5) has an interpretation as angular momentum, and is closely related with the Wronskian $$W(q^1,q^2) ~=~q^1\dot{q}^2-q^2\dot{q}^1 $$ if we identify $p_j=m\dot{q}^j,$ cf. e.g. this Phys.SE post and links therein.

  3. The canonical symplectic 2-form $$\omega~=~\sum_{j=1}^n\mathrm{d}p_j\wedge\mathrm{d}q^j ~=~\frac{1}{2}{\rm Im}\sum_{j=1}^n\mathrm{d}z^{j\ast}\wedge\mathrm{d}z^j $$ sets the stage for Liouville's theorem.

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  • $\begingroup$ Sorry for the slow response -- this is very helpful! I didn't know about the Wronskian at all. The linked to question is very relevant and very helpful in that regard too. Your reformulation of the question also seems to address the question I had about Liouville's theorem, e.g. the only 2-form on $\mathbb{A}^1$ is trivial (can't "cheat" by considering two different $(p_i, q_i)$ and saying they correspond to "different times"), and you're correct that Liouville's theorem doesn't say anything about time invariance of $\omega$, just that the Lie derivative corresponding to the Hamiltonian is 0. $\endgroup$ Commented Nov 5 at 1:33
  • $\begingroup$ (I.e. thinking of $dp_1 \land dq^1$ as the antisymmetrization of $dp_1 \otimes dq^1$ when $n=1$.) I admit I'm still confused about the interpretation of momentum as a covector (if I can't figure it out even in this simplest possible case, then I'm a little pessimistic about figuring it out more generally), BUT that question has effectively already been covered many times already on this site, and I just need to re-read and review those answers until I understand how to apply them in the simplest possible (1D straight line "Newtonian" free particle) context. $\endgroup$ Commented Nov 5 at 1:35

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