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A point transformation $ Q(q,t)$ only modifies the location coordinates, while a canonical transformation, in general, also modifies the momentum coordinates $Q(q,p,t)$.

In the context of the Lagrangian formalism, it is usually argued that we have absolute freedom to perform point transformations. I was wondering if this freedom carries over to the Hamiltonian formalism.

Specifically, do arbitrary point transformations leave the form of Hamilton's equations unchanged? In other words, are transformations in phase space which only involve the location coordinates $q$ necessarily canonical?

The defining condition for canonical coordinates is $$ \{Q,P\} = 1 $$ Therefore, we can check for a point transformation $$ Q = Q(q)$$ $$ P = p$$ if this condition is fulfilled \begin{align} \{Q,P\} & = \frac{\partial Q}{\partial q} \frac{\partial P}{\partial p} - \frac{\partial Q}{\partial p} \frac{\partial P}{\partial q} \\ &= \frac{\partial Q(q)}{\partial q} \frac{\partial p}{\partial p} - \frac{\partial Q(q)}{\partial p} \frac{\partial p}{\partial q} \\ &= \frac{\partial Q(q)}{\partial q} \, . \end{align} In general, I think, this is not equal to one or any other constant and therefore point transformations aren't canonical.

But, for example, in Vol. 1 of Landau-Lifshitz they write:

"Since Lagrange's equations are unchanged by the [point transformation $Q_i = Q_i(q,t)$], Hamilton's equations are also unchanged."

This seems like a reasonable argument. However, the calculation above shows that does not seem to be the case. Is Landau-Lifshitz wrong or did I make a mistake in my reasoning?

I think it would be really strange if point transformations weren't canonical because in almost all Classical Mechanics textbook some like the following is mentioned:

[S]ince there are twice as many canonical variables (q,p) as generalized coordinate, the set of possible transformations is considerably larger. This is one of the advantages of the canonical formalism. (Calkin, Lagrangian and Hamiltonian Mechanics)

E.g. to quote once more Landau-Lifshitz

"The enlargement of the class of possible transformations is one of the important advantages of the Hamiltonian treatment."

jak
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2 Answers2

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  1. Yes, a point transformation
    $$q^i \longrightarrow Q^i~=~f^i(q,t)\tag{1}$$ is a canonical transformation (CT) with type 2 generating function $$ F_2(q,P,t)~=~\sum_{i=1}^nf^i(q,t)P_i, \tag{2}$$ cf. Ref. 1.

  2. Alternatively, it is a straightforward exercise to show that a point transformation (1) when extended to the phase space/cotangent bundle $$ (q^i,p_j) \longrightarrow (Q^i,P_j)~=~\left(Q^i(q,t), \sum_{k=1}^n\frac{\partial q^k}{\partial Q^j}p_k\right) \tag{3}$$ is a symplectomorphism. i.e. it preserves the fundamental/canonical Poisson bracket relations. Here it is important to realize that the canonical momenta $p_j$ transform as components of a covector/1-form, cf. e.g. this & this Phys.SE posts.

References:

  1. H. Goldstein, Classical Mechanics; eq. (9.26).
Qmechanic
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Understand that the defining condition for a canonical transformation is that Hamilton's equations remain unchanged.

For a point transformation, you can indeed define $q \rightarrow Q(q)$. But then the momentum, which is defined from Lagrangian formalism as

$$ p = \frac{\partial \mathcal{L}}{\partial \dot{q}} $$

must change also, so that

$$ P = \frac{\partial \mathcal{L}}{\partial \dot{Q}} \neq p. $$

It was your assumption that $P=p$ defines a point transformation that was false; one doesn't deal with momentum as a variable in Lagrangian mechanics, so when we discuss point transformations this is often not mentioned.

I imagine you could successfully show that this represented a canonical transformation, but using the fact that a symplectic Jacobian implies a canonical transformation might be easier than trying to use Poisson brackets.

flevinBombastus
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