Condition for a time-dependent transformation to be a canonical transformation

Asked Jul 27 '21 at 10:14

Active Jul 27 '21 at 11:14

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I'm looking for a sufficient condition to determine if a given transformation is a canonical transformation. I have found two conditions, but they are only valid for the case that the transformation is is time-independent (i.e., a restricted canonical transformation); if $F$ is the generating function,

$$pdq-PdQ=dF \tag{1}$$

$$\frac{\partial Q}{\partial q} = \frac{\partial P}{\partial p} \quad;\quad \frac{\partial Q}{\partial p} = -\frac{\partial P}{\partial q} \tag{2}$$

Is there a general condition for determining if an arbitrary transformation is canonical? For example, how would we deterimine if the following transformation is canonical?

$$Q=q$$ $$P=p+\frac{\partial \Phi(q,t)}{\partial q}$$

With $\Phi(q,t)$ an arbitrary funcion of the generalized coordinate $q$ and the time.

edited Jul 27 '21 at 11:14

asked Jul 27 '21 at 10:14

Invenietis

2

Possible duplicate: Which transformations are canonical? – Qmechanic Jul 27 '21 at 10:20
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Your example preserves PBs. – Cosmas Zachos Jul 27 '21 at 15:09
@Cosmas Zachos This example is indeed supposed to be a canonical transformation, but, in $\frac{\partial Q}{\partial q} \frac{\partial P}{\partial p}-\frac{\partial Q}{\partial p} \frac{\partial P}{\partial q}=1-\frac{\partial^2 \Phi(q,t)}{\partial q^2}$, wouldn't $\frac{\partial^2 \Phi(q,t)}{\partial q^2}$ need to be zero? (Thanks for the PB trick) – Invenietis Jul 28 '21 at 19:09
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No. Your second term vanishes by dint of dQ/dp=0. – Cosmas Zachos Jul 28 '21 at 20:56

Condition for a time-dependent transformation to be a canonical transformation

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