Time dependent canonical transformation

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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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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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