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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?

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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 $$ p_i ~=~\frac{\partial S}{\partial q^i} ~=~\text{function of } (q,\alpha,t).\tag{1}$$ Granted pertinent rank conditions, this relation (1) can in principle can be solved for $\alpha$, which then becomes a function of $(q,p,t)$.

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

  4. 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.

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$^1$ Despite the name, a complete solution to a PDE is not a general solution!

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

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  • $\begingroup$ 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

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