Time dependent canonical transformation

AI Thread Summary
The discussion centers on the conditions for a transformation to be canonical, specifically examining the transformation defined by p = λP and Q = λq. While this transformation maintains the Poisson bracket condition {Q, P} = 1, adding the time transformation t' = λ²t renders the overall transformation non-canonical. The confusion arises because the equations of motion remain unchanged despite the transformation's non-canonical status. It is clarified that mutual scaling of coordinates and momenta can preserve the canonical nature only under specific conditions, particularly when λ equals ±1. The participants express concern about potentially misapplying the canonical transformation criteria in future problems.
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Homework Statement
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Relevant Equations
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THe question is pretty simple. I was doing an exercise, in which $$p = \lambda P, Q = \lambda q$$ is a canonical transformation.

We can check it by $$\{Q,P \} = 1$$

But, if we add $$t' = \lambda ^2 t$$, the question says that the transformation is not canonical anymore.

I am a little confused, since the equations of motion remain the same.

So two question:

Why the second transformation is not canonical? And,
When can we use ##\{Q,P\}=1## to check if it is canonical? SInce in the second transformation we still have the same Poisson bracket, but it is not canonical anymore, i am afraid i have been using it unconsciously many times and by coincidence being right.
 
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Herculi said:
Homework Statement:: .
Relevant Equations:: .

But, if we add t′=λ2t, the question says that the transformation is not canonical anymore.
t' = \lambda^2 t seems t=1 second corresponds to t'=##\lambda^2## second say one new second. Should p=1 kg m/s correspond to p'=##\lambda^2## kg m / new second ?
 
anuttarasammyak said:
t' = \lambda^2 t seems t=1 second corresponds to t'=##\lambda^2## second say one new second. Should p=1 kg m/s correspond to p'=##\lambda^2## kg m / new second ?
I am afraid i didn't get what you mean.
 
Mutual changes in scale of coordinate and momentum, i.e.
P=\frac{p}{\lambda},Q=\lambda q
keep {P,Q}=1 but I am afraid
P=\frac{p}{\lambda}, Q=\lambda q, T(=t')=\lambda^2 t
are not compatible except ##\lambda = \pm 1##.
 
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