I have had this confusion for a while now. We solve the Hamilton Jacobi equation,

$$H+\frac{\partial S}{\partial t}=0$$

Say we get a solution $S(q,\alpha,t)$ where $\alpha$ is a constant of integration. The approach is then to identity $\alpha$ as the new momentum.

I have trouble understanding this, when we define $\alpha$ as the new momentum, is $\alpha(p,q,t)$? Is $\alpha$ a function of the old co-ordinates and time? My understanding is that $\alpha$ is a constant, a number which is determined by the initial conditions we give and we try to invert the solutions locally in HJ approach.

And what is the difference between a constant of integration and constant of motion?


Well, the logic is as follows:

  1. The HJ equation is a first-order non-linear PDE in $n+1$ variables $(q^1,\ldots q^n,t)$, which may in principle be solved using e.g. the method of characteristics. A complete$^1$ solution $S(q,\alpha,t)$ has $n$ non-trivial$^2$ constants of integration $\alpha=(\alpha_1, \ldots, \alpha_n)\in\mathbb{R}^n$.

  2. Hamilton's principal function $S(q,\alpha,t)$ is a type-2 generating function for a CT $(q,p,t)\to (Q,P,t)$, which (among other things) implies that $$ p_i ~=~\frac{\partial S}{\partial q^i} ~=~\text{function of } (q,\alpha,t).\tag{1}$$

  3. A complete$^1$ solution has by definition $$ \det\frac{\partial^2 S}{\partial q^i\alpha_j}~\neq~ 0, \tag{2}$$ so that the relation (1) can in principle be solved for $\alpha$, which then becomes a function of $(q,p,t)$.

  4. The constants of integration $\alpha$ are next identified with the new momenta $P$.

  5. The Kamiltonian $K\equiv 0$ vanishes identically, so that the new phase space variables $(Q,P)$ are constants of motion, cf. Kamilton's equations. The definition of a constant of motion in a Hamiltonian context is given in my Phys.SE answer here.


  1. H. Goldstein, Classical Mechanics; Section 10.1 first footnote.

  2. L.D. Landau & E.M. Lifshitz, Mechanics, vol. 1 (1976); $\S$47 footnote on p. 148.


$^1$ A complete solution to a 1st-order PDE is not a general solution [1,2], despite the name!

$^2$ There is also a trivial constant of integration $\alpha_0$ associated with a shift $S\to S+\alpha_0$, which we suppress.

  • $\begingroup$ Can we set $\alpha_1=p^{1}(0)$? I mean to ask if we can designate $\alpha_i$ with initial momentum? $\endgroup$ – Abhikumbale Apr 10 '18 at 16:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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