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What is the physical significance of coordinate trasformations in phase space that are canonical but do not transform $q$s and $p$s contravariantly w.r.t each other?

For instance, the following families of linear transformations are canonical (their Jacobian matrix is symplectic)

$$\begin{cases} P_1=\frac{p_1-q_1+a(p_2-q_2)}{\sqrt{2}} \\ P_2=\frac{(p_2-q_2)}{\sqrt{2}} \\ Q_1=\frac{(p_1+q_1)}{\sqrt{2}} \\ Q_2=\frac{p_2+q_2-a(p_1+q_1)}{\sqrt{2}} \\ \end{cases} $$

where $a$ is a real parameters.

They map for instance the Hamiltonian $H=\frac{1}{2}(P_1Q_1+P_2Q_2)^2$ into $H=\frac{1}{8}(p_1^2-q_1^2+p_2^2-q_2^2)^2$.

However, if I assume that $H$ is the Legendre transform of a Lagrangian $L(q_1,q_2,\dot{q_1},\dot{q_2})$, being $p_i=\frac{\partial L}{\partial \dot{q_i}}$, I also need $P=(M^t)^{-1}p$ ($M=\frac{\partial(Q_1,Q_2)}{\partial(q_1,q_2)}$ is the Jacobian). This forces the families of transformations above to the single linear transformation where $a=0$.

My questions now are:

  • what is the physical significance (if any) of the tranformations with $a\neq 0$?
  • why requiring the transformation to be canonical does not yield automatically $a=0$?

I am asking this because any transformation, even with $a\neq 0$, being symplectic, hence canonical, should guarantee that the evolution of a system $\dot x=J \nabla H(p,q)$ transforms into $\dot X=J \nabla H(P,Q)$.

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They map for instance the Hamiltonian $H=\frac{1}{2}(P_1Q_1+P_2Q_2)^2$ into $H=\frac{1}{8}(p_1^2-q_1^2+p_2^2-q_2^2)^2$.

Without the square in the Hamiltonian, this would be just a transformation to normal coordinates/normal modes, i.e., decomposing oscillations into a superposition of independent oscillators: $$ H=P_1Q_1+P_2Q_2\longrightarrow H=\frac{1}{2}(p_1^2-q_1^2+p_2^2-q_2^2) $$ As a non-quadratic Hamiltonian generally describes non-linear system, it is unlikely that there is some general meaning to it - the transformation is likely inspired by that of a quadratic Hamiltonian, in order to simplify the analysis.

Note that transformation to normal modes is closely related to those for classifying second-order curves/conic sections, so perhaps the meaning can be inferred from a similar transformation for fourth-order curves.

Finally, not that physical meaning/significance is a rather intuitive concept, related to a specific problem, the objective of study, etc. - it is not hard-coded in mathematical form used to describe the problem (which can be the same for very different system - mechanical, electromagnetic, atomic, etc.)

Related: What is physical meaning? [closed]

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  • $\begingroup$ thanks for your reply. my question on the "physical significance" was not on the trasformations themselves but related to the fact that with $a \neq 0$ the $p$s and $q$s do not transform contravariantly w.r.t. each other. so what is the relationship between canonical transformations that do not necessarily transform the positions and momenta contravariantly and those that also respect contravariance? when can we use the former and when only the latter? $\endgroup$ Commented Oct 30 at 13:09
  • $\begingroup$ my example trasformation comes from a classical mechanics exercise that required to find it so that $H(p,q)$ was transformed in $H(P,Q)$ as in my question. but contravariance of $p$ and $q$ was not explicitely required in the exercise. on the other hand, when the text shows that point coordinate transformations preserve the hamiltonian structure, the contravariance of $p$ w.r.t $q$ is used. and all the other examples the text gives satisfy contravariance. $\endgroup$ Commented Oct 30 at 13:09
  • $\begingroup$ @l4teLearner in principle, once we are dealing with the Hamiltonian, we forget about Lagrangian - your transformation with $a\neq 0$ may correspond to a different Lagrangian or an equivalent Lagrangian (since they are defined up to a full derivative of an arbitrary function.) From the point of view of electrodynamics it could be a gauge transformation. $\endgroup$
    – Roger V.
    Commented Oct 30 at 13:39

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