# Is the definition of a canonical transformation symmetric in the specification of old and new coordinates?

Consider the following transformation:

$$q=P^\alpha \cos(\beta Q)$$

$$p=P^\alpha \sin(\beta Q)$$

for $$\alpha=1/2$$ and $$\beta=2$$.

Now, by convention, one takes $$(q,p)$$ to be the old coordinates and $$(Q, P)$$ to be the new coordinates such that if for a given transformation $$\{Q,P\}_{q,p}=\frac{\partial Q}{\partial q}\frac{\partial P}{\partial p}-\frac{\partial Q}{\partial p}\frac{\partial P}{\partial q}=1$$

then the given transformation is canonical transformation.

If now we need to check whether the above-given transformation is canonical one might try to make $$Q$$ and $$P$$ subjects in the above equations such that $$Q=Q(q,p)$$ and $$P=P(q,p)$$ and then just check the Poisson bracket.

As one can see making $$Q$$ and $$P$$ subjects in the above equations and then taking the partial derivatives is an ugly job.

My friend suggested to me that we can instead simply check $$\{q,p\}_{Q, P}=1$$ and if it is true then the transformation is canonical since the system doesn't care which way we make the transformation.

Though this suggestion sounds plausible to me, I am not entirely convinced of it since I do not see how the "system doesn't care which way we make the transformation" argument justifies this since as far as I can remember in some cases an inverse transformation is not possible. (zero Jacobian determinant)

Is my friend's argument correct?

If yes, is there a better way (so that my dumb brain understands) of seeing it?

Canonical transformations preserve the volume and orientation in phase space, so the Jacobian is automatically unity. So your friend is correct.

1. Yes, the inverse of a symplectomorphism is also a symplectomorphism.

2. Here is perhaps a more down-to-Earth explanation using 2 local coordinate systems: If $$Z^I~=~f^I(z),\qquad I~\in~\{1,\ldots,2n\},$$ is a time-independent symplectomorphism, then by definition $$\{Z^I,Z^J\}_{z}~=~\{z^I,z^J\}_{z}, \qquad I,J~\in~\{1,\ldots,2n\},$$ or equivalently, the Jacobian matrix $$M^I{}_J=\frac{\partial Z^I}{\partial z^J}$$ is a symplectic matrix. The inverse coordinate transformation $$f^{-1}$$ exists in a local neighborhood due to the inverse function theorem. One may show that $$f^{-1}$$ is in fact a symplectomorphism since the inverse Jacobian matrix is also a symplectic matrix. $$\Box$$

References:

1. H. Goldstein, Classical Mechanics, 2nd edition; Section 9.3.

2. H. Goldstein, Classical Mechanics, 3rd edition; Section 9.4.

• Is the statement "the inverse of a symplectomorphism is also a symplectomorphism" in layman terms isomorphic to ""system doesn't care which way we make the transformation"?
– Lost
Commented Nov 14, 2021 at 16:03
• Since if that is true then I can instead try to understand my friend's statement which seems like a lot easier explanation than this advanced stuff!
– Lost
Commented Nov 14, 2021 at 16:05
• I updated the answer. Commented Nov 14, 2021 at 18:08