# Time dependent canonical transformation

LCSphysicist
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.

PhDeezNutz and Delta2

Gold Member
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 ?

PhDeezNutz
LCSphysicist
$$t' = \lambda^2 t$$ 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.

Gold Member
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##.

Last edited: