I am reading goldstein there is some comment I don't understand. Consider the following hamiltonian $$H = \frac{p^2}{2m} + \frac{kq^2}{2}$$, which can be rewritten as follows $$H = \frac{1}{2m}(p^2 + m^2\omega^2q^2)$$.

"This form of the hamiltonian, as sum of two squares, suggests a transformation in which $H$ is cyclic in the new coordinates."

I don't see why is that the case why does it suggest a transformation in which $H$ is cyclic?

  • $\begingroup$ is that because we can somehow relate to sin and cos or is there some other reason ? $\endgroup$ – Illustionisttt. Dec 10 '15 at 5:21

I don't have enough reputation to comment, but since this is a simple homework problem, let me give you a hint. You know that the total energy $H$ is conserved. You've seen that $H$ can be written in the following form:

$$H = a (X^2 + b Y^2) $$

where $a$ and $b$ are some constants and $X,Y$ can be identified with $p$ and $q$. Let's suppose that $b=1$, which can be achieved by rescaling $Y$. For simplicity we can also set $a=1$ for now. Suppose that the energy of your particle is $E$. Then what are the possible coordinates $(X,Y)$ that your particle can have? Which shape does this describe?

You should find that polar coordinates will be very useful. So your guess about sine and cosine is on the mark.

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  • $\begingroup$ Welcome. short and precise answer. $\endgroup$ – HolgerFiedler Dec 10 '15 at 5:44

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