Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

In Arnold's Mathematical Methods of Classical Mechanics, he derives the Hamilton-Jacobi equation (HJE) using a generating function $S_1(Q, q)$ to get

$$ H\left(\frac{\partial S_1(Q, q)}{\partial q}, q, t \right) ~=~ K(Q, t). $$

However, this is different from what I've seen in other physics texts. For example, Goldstein uses the generating function $S_2(q, P, t)$ to get the equation

$$ H\left(\frac{\partial S_2(q, P, t)}{\partial q}, q, t\right) ~=~ - \frac{\partial S_2(q, P, t)}{\partial t}. $$

Why is there this difference? Are the two equations saying the same thing?

share|cite|improve this question

1 Answer 1

up vote 6 down vote accepted

The main points are:

  1. We are studying a Canonical Transformation (CT) $$(q,p) \longrightarrow (Q,P) $$ from old canonical coordinates $(q,p)$ and Hamiltonian $H(q,p,t)$ to new canonical coordinates $(Q,P)$and Kamiltonian $K(Q,P,t)$.

  2. $S_1(q,Q,t)$ is a so-called type 1 generating function of the CT.

  3. $S_2(q,P,t)$ is a so-called type 2 generating function of the CT.

  4. The two types of generating function are connected via a Legendre transformation $$S_2(q,P,t)-S_1(q,Q,t)~=~Q^i P_i. $$

  5. For all four types of generating functions hold that $$K-H~=~\frac{\partial S_i}{\partial t},\qquad i~=~1,2,3,4. $$

  6. Goldstein, Classical Mechanics, uses $S_2(q,P,t)$ in the treatment of Hamilton-Jacobi equation. Goldstein assumes that the Kamiltonian $K=0$ vanishes identically.

  7. Arnold, Mathematical Methods of Classical Mechanics, uses $S_1(q,Q,t)$ in Section 47 and $S_2(q,P,t)$ in Section 48. Arnold assumes (among other things) that $S_1(q,Q,t)$ does not depend explicitly on $t$.

share|cite|improve this answer
The way I understand it, the unknown in both PDEs is the generating function. Does the $K$ somehow turn into $-\partial S_2 / \partial t$ if you switch from type 1 to type 2? – Alan C Nov 1 '12 at 21:58
I updated the answer. – Qmechanic Nov 1 '12 at 22:24
What do you mean by depending explicitly on $t$? Is there a way for it to depend implicitly on $t$? And does this have to do with the fact that Arnold's definition of canonical requires that $H(q, p, t) = K(Q, P, t)$? (since this would imply that $K - H \equiv 0$. – Alan C Nov 1 '12 at 23:35
There is also implicit time-dependence via the canonical variables. – Qmechanic Nov 1 '12 at 23:40
Thank you for your help! – Alan C Nov 1 '12 at 23:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.