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I am reading the book Classical dynamics: a contemporary approach by J. V. José and E. J. Saletan. In chapter 6.1, after deducing Hamilton-Jacobi (HJ) equation, the following is stated:

This (obtaining the solution of the HJ equation) can be done only if the Hessian satisfies the condition $|\partial^2S/\partial q^\alpha\partial Q^\beta|\neq 0$, but recall that this is a condition already imposed on a Type 1 generator.

A Type 1 generating function is $F_1(q,Q,t)=S(q,Q,t)$. I don't understand why such condition is already imposed in this kind of generating function, this only happens to type 1, or it also works in type 2 and others? Also, why is this condition even needed?

The book doesn't explain any further, so I would like to understand this mathematical needs for solving HJ equation.

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  1. The full quote in Ref. 1 reads

    [...] In the meantime, assume that a solution for $S(q, Q, t)$ has been found. Because $K = 0$ by construction, the motion in the $(Q, P)$ coordinates is trivial: $$ Q^i(t) ~=~ \text{const}., \qquad P_i(t) = \text{const}.\tag{6.1b}$$ To find the motion in the $(q, p)$ coordinates, use Eqs. (5.121): $$p_i~=~\frac{\partial S}{\partial q^i}, \qquad P_i~=~-\frac{\partial S}{\partial Q^i}. \tag{6.2} $$ Invert the second of these to find the $q^i$ in terms of $(Q, P, t)$ and insert that into the first to find the $p_i$ in terms of $(Q, P, t).$ This can be done only if the Hessian satisfies the condition $$\det\frac{\partial^2 S}{\partial q^i\partial Q^j}~\neq~ 0,\tag{6.2c}$$ but recall that this is a condition already imposed on a Type 1 generator. The problem is thereby solved, since $q(t)$ and $p(t)$ have been obtained in terms of the $2n$ constants of the motion $(Q, P).$ [...]

  2. Note that Ref. 1 is not claiming that condition (6.2c) is needed to solve the Hamilton-Jacobi (HJ) equation wrt. $S$. Instead condition (6.2c) is needed to solve eq. (6.2b) wrt. $q^i$, cf. the inverse function theorem.

  3. In the rest of this answer, we would like to show that condition (6.2c) is automatically imposed on a Type 1 generator. To this end, let us introduce the collective notation $$z^I~\equiv~(q^i,p_i), \qquad Z^I~\equiv~(Q^i,P_i), \qquad \mathbb{Q}^I~\equiv~(q^i,Q^i), \qquad \mathbb{P}_I~\equiv~(p_i,P_i), $$ $$ i~\in~\{1,\ldots,n\},\qquad I~\in~\{1,\ldots,2n\}. \tag{1}$$

  4. For notational simplicity, the time parameter $t$ is implicitly implied in what follows. So for instance a canonical transformation (CT) $$Z^I~=~G^I(z,t)\tag{2}$$ is written as $$Z^I~=~G^I(z) ~=~\begin{pmatrix} g^i(z) \\ \star \end{pmatrix}, \qquad Q^I~=~g^i(z),\tag{3}$$ with the parametric $t$-dependence implicitly implied, and so forth. (The star $\star$ indicates that the precise form of this sector will play no role in what follows.)

  5. If the CT has a type 1 generating function $F_1(\mathbb{Q})$, it means that $$\mathbb{P}_I~=~ H_I(\mathbb{Q}) ~=~\begin{pmatrix} h_i(\mathbb{Q}) \\ \star \end{pmatrix},\qquad p_i~=~h_i( \mathbb{Q})~=~\frac{\partial F_1(\mathbb{Q})}{\partial q^i}.\tag{4}$$

  6. For fixed $q$, we conclude that the functions $g$ and $h$ are each other's inverse functions. More precisely $$ Q^i~=~g^i(q, h(q,Q)), \qquad p_i~=~h_i(q, g(q,p)).\tag{5}$$ We conclude that the matrices $$ \frac{\partial g^i(z)}{\partial p_j} \quad \text{and}\quad \frac{\partial h_i(\mathbb{Q})}{\partial Q^j} ~=~\frac{\partial^2 F_1(\mathbb{Q})}{\partial q^i\partial Q^j} \tag{6}$$ are each other's inverse matrices, and in particular have non-zero determinants, which is the sought-for eq. (6.2c).

References:

  1. J.V. Jose & E.J. Saletan, Classical dynamics: a contemporary approach; p.244 eq. (5.122b) & p. 286.
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  • $\begingroup$ Solving the HJ equation means obtaining $q(t)$ and $p(t)$. I realize from your answer that the elements of the matrix $\partial^2 F_1/\partial{q^i}\partial Q^j$ need not to be zero in order to invert $Q(q,p)$ to $q(Q,p)$. With $p(q,Q)$ already given by the derivative of the generating function, the problem is then solved. What I still don't see, is why it is also needed that the determinant is not zero. $\endgroup$ – Saavestro Jan 2 '17 at 19:45
  • $\begingroup$ I would say that it is shown in the last line. Could you be more specific? $\endgroup$ – Qmechanic Jan 2 '17 at 21:18
  • $\begingroup$ I see it is clear that the determinant is non-zero in this case. But I don't see why this is a requirement for obtaining solutions of the HJ equation. What would happen if it is zero? It has something to do with the Cramer's Rule from linear algebra? or is there anything else? I heard that the HJ equation has solutions only if it is separable, but I'm not sure, and don't know if maybe this is related somehow. $\endgroup$ – Saavestro Jan 3 '17 at 19:51
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Jan 3 '17 at 21:16

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