Thanks for this super clear explanation of HJ. When I was graduating N year ago, this was a question during my exam and I hadn't the faintest idea on what it was :) Now I finally understood it, although it seems more of a beautiful thing rather than a useful one. In fact, I actually never used HJ in any practical application.
Very clear and enjoyable presentation. Note that in "Tensors, Differential Forms and Variational Principles; David Lovelock and Hanno Rund" (chapter 6 again coincidentally) provide a different solution to the HJ eq for the harmonic oscillator namely S(t, q) = -1/2 w q^2 tan (w t + Phi) with H(q, p) = 1/2 p^2 +1/2 w q^2. The difference is attributed to the fact that the equation in the video entails a choice of the integration parameter alpha which corresponds directly to total energy E and as such restricts oneself to a particular member of the one parameter (alpha) family of possible extremal curves (determined by initial conditions). As such the S(q, alpha, t) only reflects the value of S on one particular extremal, where S(t, q) has the correct value in the whole of R2, for all extremals, without the need to parameterize. On the extremal corresponding to specific initial conditions it is easily verified that the solution from Lovelock and Rund equals the solution in this video. Note also the comment in Lovelock and Rund that S should be a function of S(t, q) and should not be represented as a function S(t, q, parameters) where parameters merely serve to label individual members of the family of extremals.
I think it makes sense that the Hamilton-Jacobi theorem is nonunique, given the dependence of the generating function of the Lagrangian, which is also nonunique. Several Lagrangians would imply several generating functions. It's probably also worth keeping in mind that all of this business around the constraints for generating functions being canonical to begin with is built on a specific example of a nonunique Lagrangian where the function is added to the orginal Lagrangian and is of the form dF/dt. Maybe there are other functions that could work as canonical, which will have differing constraints, and then perhaps the constraints would then be different, and then the generating function that gives us back equillibrium points along all the points in ${p} \times {q}$ could be something other than the action...
If the GF is the action, and the path taken is the one s.t. dS/dt = 0, then according to HJ: S = W - tE => E = W/t - S/t, for the interval t of interest. Taking differential intervals then, wouldn't: => E = dW/dt - dS/dt = dW/dt Meaning that W would essentially be a sum of total energies kind of term? EDIT: in the time-independent case ofc.
I am curious about the connection between the action integral and Hamilton's principle function. If I was doing the functional derivative of the action I am treating the action integral as a map between a set of possible paths and the real numbers, whereas here Hamilton's principle function is a map from the R^d to R. How can I reconcile these two ideas? I am also curious as well, for a given Hamiltonian or initial conditions on the motion, does that imply boundary conditions/initial conditions on the solution to the PDE given by the HJE?
I do not see how S2(t) disappeared on the left hand side (18 mins in). It looks like they should be there but then divide both sides by S2 squared, and then set to constant.
We have a PDE for S(q,t) at 15:51, and then we assume the solution has the form S(q,t) = S1(q) + S2(t). If we substitute S(q,t) = S1(q) + S2(t) into the PDE, we get the equation at 17:46, which just involves two total derivatives. For instance, on the left hand side, the only derivative is ∂S(q,t)/∂q = ∂[S1(q) + S2(t)]/∂q = ∂S1(q)/∂q + ∂S2(t)/∂q = dS1(q)/dq + 0. The only way the resulting equation at 17:46 can hold is if both the left-hand-side and right-hand-side are equal to a constant, which we call alpha1. Does it make sense now?
So the "independent Hamiltonian example" at 52:50, is not necessarily totally decoupled, since V_theta term has r, the last term has r and phi, the "feeling like decouped in the sequencial solving procedure" can also be seen as independent, right?
It's an example of a "separable" Hamiltonian, not necessarily independent, but I think you're right. It's a "sort of" independence due to how it is solved in the Hamilton-Jacobi approach.
Finding the best transformation for what purpose? There are only two handfuls of integrable Hamiltonians and they have all been worked out a long time ago. Is this class working towards perturbation theory?
The best introductory level discourse I ever came across!
Wow, thanks for the high praise and for watching!
i've gone through about 1/2 of the length of this lecture and it's beautiful. Thanks for posting it.
This series and the series on Lagrangian and 3D Rigid Body Dynamics have been invaluable to me, thank you very much professor!
Byron, Thank you so much! I'm glad you appreciate it. Be sure to share with others you will find it useful.
Thanks for this super clear explanation of HJ. When I was graduating N year ago, this was a question during my exam and I hadn't the faintest idea on what it was :) Now I finally understood it, although it seems more of a beautiful thing rather than a useful one. In fact, I actually never used HJ in any practical application.
Very clear and enjoyable presentation. Note that in "Tensors, Differential Forms and Variational Principles; David Lovelock and Hanno Rund" (chapter 6 again coincidentally) provide a different solution to the HJ eq for the harmonic oscillator namely S(t, q) = -1/2 w q^2 tan (w t + Phi) with H(q, p) = 1/2 p^2 +1/2 w q^2. The difference is attributed to the fact that the equation in the video entails a choice of the integration parameter alpha which corresponds directly to total energy E and as such restricts oneself to a particular member of the one parameter (alpha) family of possible extremal curves (determined by initial conditions). As such the S(q, alpha, t) only reflects the value of S on one particular extremal, where S(t, q) has the correct value in the whole of R2, for all extremals, without the need to parameterize. On the extremal corresponding to specific initial conditions it is easily verified that the solution from Lovelock and Rund equals the solution in this video. Note also the comment in Lovelock and Rund that S should be a function of S(t, q) and should not be represented as a function S(t, q, parameters) where parameters merely serve to label individual members of the family of extremals.
I think it makes sense that the Hamilton-Jacobi theorem is nonunique, given the dependence of the generating function of the Lagrangian, which is also nonunique. Several Lagrangians would imply several generating functions. It's probably also worth keeping in mind that all of this business around the constraints for generating functions being canonical to begin with is built on a specific example of a nonunique Lagrangian where the function is added to the orginal Lagrangian and is of the form dF/dt. Maybe there are other functions that could work as canonical, which will have differing constraints, and then perhaps the constraints would then be different, and then the generating function that gives us back equillibrium points along all the points in ${p} \times {q}$ could be something other than the action...
I'm glad to have you in my life - Thank you!
Thank you. I'm glad you appreciate the videos.
If the GF is the action, and the path taken is the one s.t. dS/dt = 0, then according to HJ: S = W - tE
=> E = W/t - S/t, for the interval t of interest. Taking differential intervals then, wouldn't:
=> E = dW/dt - dS/dt = dW/dt
Meaning that W would essentially be a sum of total energies kind of term?
EDIT: in the time-independent case ofc.
I am curious about the connection between the action integral and Hamilton's principle function. If I was doing the functional derivative of the action I am treating the action integral as a map between a set of possible paths and the real numbers, whereas here Hamilton's principle function is a map from the R^d to R. How can I reconcile these two ideas? I am also curious as well, for a given Hamiltonian or initial conditions on the motion, does that imply boundary conditions/initial conditions on the solution to the PDE given by the HJE?
Much better explanation than Landau. Thank you.
Wow, thank you. That's high praise.
Thank you professor Rose, your video helped me a lot!
That is great news. You're very welcome.
Amazing video! This helped understand part of Schrödingers paper for qm mechanics
Glad it helped! You might like this other video which mentions a classical "uncertainty principle" th-cam.com/video/hfeKkZBtZTI/w-d-xo.html
I do not see how S2(t) disappeared on the left hand side (18 mins in). It looks like they should be there but then divide both sides by S2 squared, and then set to constant.
We have a PDE for S(q,t) at 15:51, and then we assume the solution has the form S(q,t) = S1(q) + S2(t). If we substitute S(q,t) = S1(q) + S2(t) into the PDE, we get the equation at 17:46, which just involves two total derivatives. For instance, on the left hand side, the only derivative is ∂S(q,t)/∂q = ∂[S1(q) + S2(t)]/∂q = ∂S1(q)/∂q + ∂S2(t)/∂q = dS1(q)/dq + 0. The only way the resulting equation at 17:46 can hold is if both the left-hand-side and right-hand-side are equal to a constant, which we call alpha1. Does it make sense now?
at 47:00 the integration should be with respect to dr (not dq)
You're right. Thanks for the correction!
So the "independent Hamiltonian example" at 52:50, is not necessarily totally decoupled, since V_theta term has r, the last term has r and phi, the "feeling like decouped in the sequencial solving procedure" can also be seen as independent, right?
It's an example of a "separable" Hamiltonian, not necessarily independent, but I think you're right. It's a "sort of" independence due to how it is solved in the Hamilton-Jacobi approach.
Finding the best transformation for what purpose? There are only two handfuls of integrable Hamiltonians and they have all been worked out a long time ago. Is this class working towards perturbation theory?
The best Video on this topic!
Nice!
Thankyou for your generosity. Teacher
This is the theory which was used by Erwin Schrodinger to solve H-atom as an eigenvalue problem.
best explanation online
Thank you!
great
Thank you so much!
HJE = sorcery