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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?

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1 Answer

up vote 5 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 $t$.

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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
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