What the book demonstrates

In No-Nonsense Classical Mechanics, the author spends some time discussing how point transformations in configuration space correspond with canonical transformations in phase space:

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Specifically, the author demonstrates via this proof that a point transformation $q \mapsto Q = Q(q)$ implies that $p \mapsto P = \frac{\partial q}{\partial Q} p$.

How is this not a counter-example?

I don't doubt the proof, but I'm having trouble understanding this intuitively. For example, let us suppose we're in the context of a falling ball, where we can demonstrate via the Lagrangian that $p = m \dot{q}$. It seems to me that if $\dot{q} \mapsto \frac{\partial Q}{\partial q} \dot{q}$, then $p = m \dot{q}$ will then get mapped to

$$ P = m \left( \frac{\partial Q}{\partial q} \dot{q} \right) = \frac{\partial Q}{\partial q} \left( m \dot{q} \right) = \frac{\partial Q}{\partial q} p $$

and of course this contradicts the idea that $p \mapsto P = \frac{\partial q}{\partial Q} p$. What is wrong with my reasoning?


2 Answers 2


TL;DR: $p=m\dot{q}$ and $P=m\dot{Q}$ are typically not both true.

Perhaps an example is in order.

  • Example: Consider a coordinate scaling $$Q~=~ \lambda q,$$ where $\lambda\in\mathbb{R}\backslash \{0\}$ is a non-zero constant. Consider the Lagrangian $$L~=~\frac{m}{2}\dot{q}^2-V(q)~=~\frac{m}{2\lambda^2}\dot{Q}^2-V(Q/\lambda).$$ Then $$p ~=~\frac{\partial L}{\partial \dot{q}}~=~m \dot{q},$$
    while $$P ~=~\frac{\partial L}{\partial \dot{Q}}~=~\frac{m}{\lambda^2}\dot{Q},$$ so that $$P~=~\lambda^{-1} p.$$

See also e.g. this & this related Phys.SE posts.


If $\dot{Q}=\frac{\partial Q}{\partial q}\dot{q}$ then $\dot{q}=\frac{\partial q}{\partial Q}\dot{Q}$ and by entering this in your formula you get the correct result.

  • $\begingroup$ What do you mean? Entering this into which formula? $\endgroup$
    – George
    Aug 25, 2020 at 20:22
  • $\begingroup$ In the $p=m\dot{q}$, then you will get the transform $p\rightarrow P$. $\endgroup$
    – NDewolf
    Aug 25, 2020 at 20:54
  • $\begingroup$ You started from the assumption $P=m\dot{Q}$, which is as @Qmechanic noted not necessarily true. $\endgroup$
    – NDewolf
    Aug 25, 2020 at 21:03

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