# 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:

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 $$P~=~\lambda^{-1} p.$$

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.
• In the $p=m\dot{q}$, then you will get the transform $p\rightarrow P$. – NDewolf Aug 25 '20 at 20:54
• You started from the assumption $P=m\dot{Q}$, which is as @Qmechanic noted not necessarily true. – NDewolf Aug 25 '20 at 21:03