# In Hamilton-Jacobi theory, how is the new coordinate $Q$ time-independent when Hamilton's principal function separates?

Following the notation in Goldstein, the solution to the Hamilton-Jacobi equation is the generating function $$S$$ for a canonical transformation from old variables $$(q,p)$$ to new variables $$(Q,P)$$ where the new Hamiltonian is $$K=0$$ and the new momentum is the integration constant $$P=\alpha$$. Therefore, $$\dot{Q}=\dot{P}=0$$. If the old Hamiltonian is not a function of time, the solution may be expressed as

$$S(q,\alpha,t)=W(q,\alpha) - \alpha t.\tag{1}$$

What's confusing me is that, by definition, $$S$$ is a type 2 generating function where the new variables are constant in time. However, the new coordinate $$Q$$ from the type 2 generating function above is expressed as

$$Q=\frac{\partial S}{\partial P} = \frac{\partial S}{\partial \alpha} = \frac{\partial W}{\partial \alpha} - t.\tag{2}$$ Therefore giving

$$\dot{Q} = -1 \neq 0.\tag{3}$$

What am I missing here?

I will assume you are only talking about the case of $$1$$ degree of freedom (or else, your notation is problematic).

Generally, it is always a good reflex to study a specific case when you have two conflicting arguments. Take for example a free particle with $$H = \frac{p^2}{2}$$, solving the Hamilton-Jacobi equation, you get: $$S = \sqrt{2P}q-Pt$$ from which $$P = \frac{p^2}{2} \\ Q = \frac{q}{p}-t \\$$ and since $$\dot p = 0\\ \dot q = p$$ you have $$\dot Q = 0$$.

As you can see, your mistake was in the last step $$Q = \left(\frac{\partial W}{\partial P}\right)_q-t \\ \dot Q = -1$$ Remember that $$W$$ is a function of $$q,P$$, so there is an implicit time dependence. Using the chain rule, you get the extra terms: $$\dot Q = \left(\frac{\partial^2 W}{\partial P^2}\right)_q\dot P+\frac{\partial^2 W}{\partial q\partial P}\dot q-1$$ The first one is trivial by construction, since $$\dot P = 0$$, but the second one will cancel the $$-1$$.

To prove this fortuitous cancellation, the fastest route is to refer to the general proof of the HJE. In this simple case, there is also the following shortcut starting from the HJE: $$H\left(\frac{\partial W}{\partial q},q\right) = P$$ and by applying $$\left(\frac{\partial}{\partial P}\right)_q$$, you get: $$\left(\frac{\partial H}{\partial p}\right)_q\frac{\partial^2 W}{\partial P\partial q} = 1 \\ \dot q \frac{\partial^2 W}{\partial q\partial P} = 1 \\$$ hence the result. Notice that using $$p = \left(\frac{\partial W}{\partial q}\right)_P$$ and using $$\dot H = 0$$, you get from the HJE $$\dot P = 0$$.

Hope this helps and tell me if you need more details.

• Got it, thanks a lot! And yes, was keeping to 1 DOF for simplicity.
– Matt
Commented Jun 13, 2022 at 17:37

The culprit of OP's question seems to be that Goldstein considers 2 different generators for a type-2 canonical transformation (CT) in Hamilton-Jacobi theory:

1. Hamilton's principal function $$S(q,P,t)$$. Here the Kamiltonian $$K\equiv 0$$ is zero. Therefore all the new phase space variables $$(Q_S^i,P_j)$$ are constants of motion (COM).

2. Hamilton's characteristic function $$W(q,P)$$. The latter assumes that the Hamiltonian $$H(q,p)$$ has no explicit time dependence, and that the Kamiltonian $$K(P)$$ does not depend on the new coordinates and time $$t$$. In fact Goldstein in section 10.3 effectively assumes that the Kamiltonian $$K(P)=P_1$$ is the 1st new momentum $$P_1$$. As a consequence the 1st new position $$Q^1_W$$ is identified as time $$t$$ (up to a possible shift) rather than a COM.

When both methods apply, under the identification $$S(q,P,t)~=~W(q,P)-P_1t,\tag{A}$$ we conclude that the new positions in the 2 pictures are related as $$Q_S^i ~=~ \frac{\partial S}{\partial P_i} ~\stackrel{(A)}{=}~\frac{\partial W}{\partial P_i} -\delta^i_1 t~=~Q_W^i-\delta^i_1 t .\tag{B}$$

TL;DR: The main point is that the 2 CTs produce slightly different sets of new positions $$Q_S^i$$ and $$Q_W^i$$, where we for clarity have decorated them with a subscript $$S$$ and $$W$$, respectively.