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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?

Qmechanic
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physicscircus
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1 Answers1

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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.

References:

  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.

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$^1$ 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.

Qmechanic
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