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:

enter image description here

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?


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 '20 at 20:22
  • $\begingroup$ In the $p=m\dot{q}$, then you will get the transform $p\rightarrow P$. $\endgroup$ – NDewolf Aug 25 '20 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 '20 at 21:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.