# Why does time-independent Hamiltonian not depend on angle variable?

In Landau and Lifshitz Mechanics, $\S50$ Canonical variables a time-independent Hamiltonian is considered, and a canonical transformation is done such that adiabatic invariant $I$ becomes the new momentum. Then the angle variable is found as

$$w=\frac{\partial S_0(q,I;\lambda)}{\partial I},$$

where $S_0$ is abbreviated action (and generating function for the canonical transformation), $q$ is old position variable and $\lambda$ is a constant parameter.

Now L&L say:

Since the generating function $S_0(q,I;\lambda)$ does not depend explicitly on time, the new Hamiltonian $H'$ is just $H$ expressed in terms of the new variables. In other words, $H'$ is the energy $E(I)$, expressed as a function of the action variable. Accordingly, Hamilton's equations in canonical variables are $$\dot I=0,\;\;\;\dot w=\frac{\mathrm dE(I)}{\mathrm dI}.\tag{50.4}$$

Now my question is: why does $H'=E$ not depend also on $w$? $H$ does in general depend on old position variable $q$ (even if it does not depend explicitly on time), so why shouldn't $H'$ depend on angle variable?

## 1 Answer

This is more or less an exercise in chasing definitions. The adiabatic invariant $I$ is defined as $$I\equiv \oint p \frac{\mathrm{d}q}{2\pi}\tag{49.7}$$ where the integral is taken over the path for given $E$ and $\lambda$. The external parameter $\lambda(t)$ is a slowly varying function of time $t$ in $\S49$, but is assumed to be a constant in $\S50$. Let us suppress the role of $\lambda$ in what follows to keep formulas simple. Then $I=f(E)$ is only a function of the energy $E$. Combined with the fact that the Hamiltonian $H$ does not depend explicitly on time, it show that the Kamiltonian ($\equiv$ the new Hamiltonian) $$H^{\prime}~\equiv~ K~=~H~=~E~=~f^{-1}(I)$$ is only a function of $I$, which is also the new momentum.

• Ah I see now: $I$ can't depend on neither $p$ nor $q$ since they are integrated away, and thus $w$, depending on $q$, can't somehow appear in the final expression for $I$. Thus, as $I=f(E)$, $E$ also can't have $w$ in its definition. – Ruslan May 28 '16 at 16:46