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Considering a classical, non-damped 1D harmonic oscillator (e.g. mass $m$ oscillating along $x$-axis attached to spring with constant $k$) -- described by Hamiltonian (for constant energy $E$) $$E=p^2/2m + kx^2/2,$$ which system follows an elliptical trajectory in corresponding phase space $(p, x)$:

I want to formulate an equation for this phase space trajectory in polar form for an ellipse (origin at ellipse center), having semi-major/minor axes $a/b$, using the standard polar form:

$$r(θ)=\frac{ab}{\sqrt{(b\cos(θ))^2 + (a\sin(θ))^2}}$$

I am stumped as to what "a" and "b" would be in terms of "m", "k" and "E". Can you suggest what they might be?

It seems that, in general, the square of the radius of the ellipse would be in units of energy (kg-m^2/s^2), so I'm trying to find what "a" and "b" would be such that r^2(theta) would thus be in units of energy. Or am I incorrect in assuming that $r^2(\theta)$ would be in units of energy?

Any insights/suggestions appreciated.

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  • $\begingroup$ Al ellipse doesn't have a 'center': it has two focal points. Similarly, it has no radius... $\endgroup$
    – Gert
    Jul 26, 2021 at 17:05
  • $\begingroup$ OK, my mistake for not describing correctly. I meant to convey as described here (with "center" on semi-major axis half way twixt focal points): en.wikipedia.org/wiki/Ellipse#Polar_form_relative_to_center $\endgroup$
    – Dave
    Jul 26, 2021 at 17:18
  • $\begingroup$ OK, that's been edited. $\endgroup$
    – Gert
    Jul 26, 2021 at 17:34

1 Answer 1

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There isn't much to suggest. Simply write your equation in terms of dimensionless quantities/ratios: $$ 1= \frac{x^2}{2E/k}+ \frac{p^2}{2Em}~~\leadsto \\ a=\sqrt{2E/k}, ~~b=\sqrt{2Em}. $$ This is to say now, that the semimajor axis, a, is in units of length and the semi minor one, b, in units of momentum. Assuming neither vanishes or blows up, if you measure x in units of a, and p in units of b, you really have a circle.

If you measure space-amplitudes and momenta in fixed units, then a and b are also dimensionless numbers times these same units; this is called nondimensionalization. Giving them the same names in the polar form, $$r(θ)=\frac{1}{\sqrt{\cos(θ)^2/a^2 + \sin(θ)^2/b^2}}~~.$$

It is then evident that $r(0)= a$, and $r(\pi/2)= b$. There is no point in labelling the axes of the ellipse with differently dimensioned units, since quantities such as r are defined a bit oxymoronically, and would confuse readers who did not follow the above discussion. So, nondimensionalization is the simplest option.


Edit on comment:

No, the eccentricity is not meaningful, since the units of x and those of p are different. You may choose to measure x in m, and p in kg m/s, which fixes a=A m and b= B kg m/s, in which case the pure numerical eccentricity would be $$ e=\sqrt {1-B^2/A^2}~, $$ but if you chose to measure x in 1/2 m, then A would be twice the previous A and the eccentricity would be different. The above expression for r has dimensionless numbers A and B instead of a and b once you've fixed your units.

So, rescaling of the abscissa or the ordinate adjust the circle to an ellipse of arbitrary eccentricity. Conversely, the generic oscillator phase-space trajectory is always a circle, in natural units, as the standard non-dimensionalized picture yields, and bothering with ellipses is conceptually a do-nothing machine.

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  • $\begingroup$ I think I follow your answer, in that if I make the variable substitutions, for example, X=x*sqrt(k/2E) and P=p/sqrt(2Em) I have the circle P^2+X^2=1. In the equation for r(theta) above, I was mistaken in thinking that the units of a and b had to be the same in order to be able to add "cos^2(theta)/a^2" and "sin^2(theta)/b^2". Is there some meaningful measure of the eccentricity (sqrt(1-b^2/a^2)) of the ellipse 1=x^2/a^2 + p^2/b^2 , even if a and b do not have the same units? Thank you again for your help. $\endgroup$
    – Dave
    Jul 26, 2021 at 23:56

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