An excellent introduction to Action-Angle variables beginning around 43:00. This is a very clear lecture which clarifies a lot of concepts on this difficult topic.
@@Unexpectedthings007 many students do hardwork apart from iitans... Just because of limited seats opportunities didn't go for many dedicated talented students... But nptel through these lectures filling the gap
Regarding the discussion at about 33 minutes. I understand why the transformation must be invertible, but making it volume preserving seems to me to be frosting on the cake and not necessary. If one can solve the new system and invert the solutions, who cares about the volume in the new coordinates? Anyone have an explanation or comment?
From what I understood in class (really not an expert, I'm having the course right now) it has to do with the thing that it must preserve uncertainty. If you have little uncertainty in your data, you want it to stay like that. But I insist, I'm far from an expert, maybe you should ask in stackexchange
It has to do with the fact that given hamiltonian and hamiltons equations, we can track the system perfectly in phase space for given initial conditions, i.e. each initial condition should correspond to a single trajectory at any instant of time which means volume should be preserved.
You need the transformation to be volume preserving because you need the transformation to also preserve Hamilton's equations of motion. Any given Hamiltonian preserved the phase space volume and if after the transformation, you end up with a Hamiltonian H' that doesn't preserve volume element in the phase space, then you won't get your usual Hamilton's equations.
Think of it in this way: in 3 dimensional space, two planes can intersect to form a one-dimensional line. Even if they are not planes, and more complicated surfaces, they will intersect to form a one-dimensional trajectory. In 2n dimensional phase space, each constant of motion corresponds to a 2n-1 dimensional hyper-plane. Hence, 2n-1 of these hyper-planes intersect to form a on-dimensional phase trajectory.
An excellent introduction to Action-Angle variables beginning around 43:00. This is a very clear lecture which clarifies a lot of concepts on this difficult topic.
thanks, i am exactly looking for the Action-Angle variables
Thanks me too
These lectures are highly insightful. Thank you Prof Balakrishnan
I like how he incorporates Poisson brackets with Hamiltonians right from the start.
iit students are indeed lucky ones , great lecture
Hardworking bro hardworking
@@Unexpectedthings007 many students do hardwork apart from iitans... Just because of limited seats opportunities didn't go for many dedicated talented students... But nptel through these lectures filling the gap
This professor is excellent. He really understands the topic and is a great teacher.
Brilliant lecture series. Cleared up so many of my doubts from these textbooks...
Best teacher I have ever seen
Long live the legend
Very nice to learn.
Organised and knowledgeable
can any one post a good link to study Hamilton jacobi theory
i just found out that jacobian is another type of possion bracket ..or its just lie algebra .. :O
Regarding the discussion at about 33 minutes. I understand why the transformation must be invertible, but making it volume preserving seems to me to be frosting on the cake and not necessary. If one can solve the new system and invert the solutions, who cares about the volume in the new coordinates? Anyone have an explanation or comment?
From what I understood in class (really not an expert, I'm having the course right now) it has to do with the thing that it must preserve uncertainty. If you have little uncertainty in your data, you want it to stay like that. But I insist, I'm far from an expert, maybe you should ask in stackexchange
It has to do with the fact that given hamiltonian and hamiltons equations, we can track the system perfectly in phase space for given initial conditions, i.e. each initial condition should correspond to a single trajectory at any instant of time which means volume should be preserved.
You need the transformation to be volume preserving because you need the transformation to also preserve Hamilton's equations of motion. Any given Hamiltonian preserved the phase space volume and if after the transformation, you end up with a Hamiltonian H' that doesn't preserve volume element in the phase space, then you won't get your usual Hamilton's equations.
Great lecturer!
Can anyone plz explain why a 2n dimnetional space needs 2n-1 constants of motion🤔😓??
If you have 2n constant of motion then you find a point instead of a trajectory
Because we need to have an one-dimensional phase-space trajectory.
Think of it in this way: in 3 dimensional space, two planes can intersect to form a one-dimensional line. Even if they are not planes, and more complicated surfaces, they will intersect to form a one-dimensional trajectory. In 2n dimensional phase space, each constant of motion corresponds to a 2n-1 dimensional hyper-plane. Hence, 2n-1 of these hyper-planes intersect to form a on-dimensional phase trajectory.
Unless he learnt well he can't explain like this he burnt many nights mid night oil to come to this position.
Karka kasadara is best axiom for IITM but Sanskrit slogan no use at all.
tooo tough lecture