Calculating generator function in canonical transformation

Click For Summary
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
Messages
634
Reaction score
0
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.
 
Physics news on Phys.org
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.
 
For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

Similar threads

  • · Replies 2 ·
Replies
2
Views
541
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K