Time dependent canonical transformation

  • #1
LCSphysicist
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Homework Statement:
.
Relevant Equations:
.
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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Answers and Replies

  • #2
anuttarasammyak
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Homework Statement:: .
Relevant Equations:: .

But, if we add t′=λ2t, the question says that the transformation is not canonical anymore.
[tex]t' = \lambda^2 t [/tex] 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 ?
 
  • #3
LCSphysicist
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[tex]t' = \lambda^2 t [/tex] 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.
 
  • #4
anuttarasammyak
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Mutual changes in scale of coordinate and momentum, i.e.
[tex]P=\frac{p}{\lambda},Q=\lambda q[/tex]
keep {P,Q}=1 but I am afraid
[tex]P=\frac{p}{\lambda}, Q=\lambda q, T(=t')=\lambda^2 t[/tex]
are not compatible except ##\lambda = \pm 1##.
 
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