Calculating generator function in canonical transformation

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To calculate the generator function for a canonical transformation, an example using a type 2 generator function F2(q,P,t) is provided, where new canonical variables are expressed in terms of old variables. The transformation involves spatial coordinates and momenta, specifically x = x' + Vt and px = mV + px'. The relationships for the generator function are established through partial derivatives, leading to the integration of the equation. The final form of the generator function is F2 = (mV + px')x - Vtpx', which accounts for the necessary adjustments in momentum transformation. This example illustrates the process of determining the generator function in classical mechanics.
Sourabh N
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I'm searching for an example of how to find out generator function for a canonical transformation, when new canonical variables are given in terms of old variables. Any help is greatly appreciated.
 
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See any classical mechanics textbook, such as Goldstein.
 
Let us take, for example, a generator function of type 2 (please see goldstein), i.e , F2(q,P,t) function of the old coordinates (q) and the new momenta (P), and consider the following non-relativistic transformation:

x=x'+Vt , x'- new (spacial) coordinate
t=t' , (time remains the same)

as you can see, you will also have to know how the momenta transforms (in order to determine F2)

px=mV+px' , px' new momentum (px'=P, if you prefer)

For the type 2 generator function,

p=dF2/dq (partial derivative)
Q=dF2/dP (partial derivative)

So, all you have to do is integrate, i.e

px is your old momentum, therefore F2 = (mV+px')x + A
now let's determine A,

"A" is not a constant because if you take dF2/dP it won't equal the new momenta. There's still a " - Vtpx' "
lacking in the equation (with the minus sign included).
Well if that quatity is missing , all we have to do is add it to the equation (A=-Vtpx').

So here it is, the F2(q,P,t) generator function for this non-relativistic transformation is:

F2 = (mV+px')x - Vtpx'

I down know if there is another (better) way to do it, but I hope it helps.

Best regards

Rico B.
 
This has been discussed many times on PF, and will likely come up again, so the video might come handy. Previous threads: https://www.physicsforums.com/threads/is-a-treadmill-incline-just-a-marketing-gimmick.937725/ https://www.physicsforums.com/threads/work-done-running-on-an-inclined-treadmill.927825/ https://www.physicsforums.com/threads/how-do-we-calculate-the-energy-we-used-to-do-something.1052162/
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