2

The following is taken nearly verbatim from section 1.1.2.1 Free Energy Computations: A Mathematical Perspective by Mathias Rousset, Gabriel Stoltz and Tony Lelievre. An excerpt, in which the part I am quoting from is contained, can be viewed for free here.

Consider microscopic systems composed of $N$ particles described by positions $q := (q_k)_{k = 1}^{N} \in D$ ($D$ is the configuration space) and momenta $p := (p_k)_{k = 1}^{N} \in \mathbb{R}^{3 N}$. The vector $(q, p)$ is the microscopic state or the configuration of the system. [...]

In the framework of statistical physics, macroscopic quantities of interest are written as averages over thermodynamic ensembles, which are probability measures on all the admissible microscopic configurations: $$ \mathbb{E}_{\mu}(A) := \int_{T^* D} A(q, p) \mu(dq \; dp). $$ In this expression, $A$ is an observable. [...] As $p \in \mathbb{R}^{3 N}$, the cotangent space $T^* D$ from above can be identified with $D \times \mathbb R^{3 N}$.

My question. Why does the notion of a cotangent space come into play here? If $q \in D$ and $p \in \mathbb{R}^{3 N}$, why isn't the set that is integrated over simply $D \times \mathbb{R}^{3 N}$, but a priori the cotangent space of $D$?

ViktorStein
  • 5,024
  • 1
    I looked briefly through the table of contents of that book and get the impression that it is not necessary to understand all about symplectic geometry on the phase space to keep reading on. Correct me if I am wrong. A hint why cotangential space is used is here. In short: $dq$ and $dp$ are differential forms (covector fields). What I find a bit confusing is their notation $T^*D$. That they identify it with $D\times \mathbb R^{3N}\subset \mathbb R^{3N}\times \mathbb R^{3N}= \mathbb R^{6N}$ does make sense though. – Kurt G. Apr 06 '22 at 06:45
  • @KurtG. Thank you for the interest in my question. You are right that the answer to my question will not be relevant for the understanding the book, this question is posed purely out of interest. Unfortunately, my background in differential geometry is very weak, but at least I see that 1-forms on $M$ are sections of the cotangent bundle and there the cotangent space comes into play. I will look into the canonical symplectic form you linked. Just knowing the term "symplectic geometry on the phase space" helps me a lot, since googling it provided me with further resources. Thank you! – ViktorStein Apr 06 '22 at 10:42
  • 1
    To learn that matter I would probably start with the survey arXiv:0708.1585v1. – Kurt G. Apr 06 '22 at 10:47
  • I'm not a physicist, but physically the phase space is the cotangent bundle because velocities transform like vectors and momenta transform like covectors under change of coordinates. – Andrew D. Hwang Apr 07 '22 at 02:19
  • I don't have references to suggest offhand; the fact the velocities transform like vectors is essentially the chain rule, and momenta are real-valued functions of velocities, i.e., covectors, along the lines of Kurt's comments. (The related post listed by the site might be useful?) – Andrew D. Hwang Apr 07 '22 at 18:45

0 Answers0