2
$\begingroup$

I'm studying action-angle coordinates and I've come across two distinct, yet correct, definitions for the new generalized momenta, $\text{J}_k$:

$$\text{J}_k \equiv \oint p_k dq_k $$

and

$$\text{J}_k \equiv \frac{1}{2\pi} \oint p_k dq_k .$$

I'm trying to understand how they relate.

$\endgroup$
  • 2
    $\begingroup$ one includes a factor of $2\pi$ and the other one does not? it is just about conventions, there is no physics behind that factor (besides the fact that it simplifies the harmonic oscillator) $\endgroup$ – AccidentalFourierTransform Nov 28 '16 at 17:40
  • $\begingroup$ Doesn't it have to do with normalization or something? I'm trying to understand why one has the factor and the other does not. $\endgroup$ – Phy09 Nov 28 '16 at 17:42
  • $\begingroup$ So, I can use either one for any case and I'll obtain the same frequencies, same equations of motion etc? But wouldn't the new generalized momenta, $\text{J}_k$, differ by a factor of $2\pi$ depending on which definition I used? $\endgroup$ – Phy09 Nov 28 '16 at 17:46
0
$\begingroup$

Imagine you have the harmonic oscillator

$$ H(q, p) = \frac{p^2}{2} + \frac{q^2}{2} \tag{1} $$

In phase space, the are enclosed by a solution of Eq. (1) is

$$ A = \pi\times\sqrt{2}\times \sqrt{2} = 2\pi $$

For this orbit, the action is

$$ J = \frac{1}{2\pi}\oint {\rm d}q\; p = \frac{1}{2\pi}A = 1 $$

That is the reason behind the "$2\pi$", but it is absolutely not required. The action $J$ will have the same meaning even if this factor is removed

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.