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The harmonic oscillator action variable $J_{a} (E) = E / ω$ and $J_{a}^{'} (E) = T / (2 π)$

Example

Let $m > 0$ , $k > 0$ , $U (q) = \frac{1}{2} k q^{2}$ on $J_{0} = R$ , and for $E > 0$ set $a (E) : = - \sqrt{\frac{2 E}{k}}, b (E) : = \sqrt{\frac{2 E}{k}}, ω : = \sqrt{\frac{k}{m}}, T : = \frac{2 π}{ω} .$ Then the datum $(J_{0}, U, m, a, b)$ satisfies the hypotheses of The action variable of a two-simple-turning-point family is $C^{1}$ with derivative $T (E) / (2 π)$ , the period quadrature over $2 π$ on every bounded open interval $(E_{0}, E_{1})$ with $0 < E_{0} < E_{1} < \infty$ . These restrictions define the same action on overlaps, hence together give a $C^{1}$ function on $(0, \infty)$ ; the action variable of The one-dimensional action variable of a family of two-turning-point energy levels is $J_{a} (E) = \frac{E}{ω} (E > 0),$ and $J_{a}^{'} (E) = \frac{1}{ω} = \frac{T}{2 π} (E > 0),$ in agreement with the general theorem, whose period quadrature is $T (E) = \sqrt{2 m} \int_{a (E)}^{b (E)} d q / \sqrt{E - U (q)} = 2 π / ω$ .

Facts & Assumptions

Given: Positive $m, k$ , the quadratic potential $U (q) = \frac{1}{2} k q^{2}$ , the turning-point functions $\pm \sqrt{2 E / k}$ , and the frequency $ω = \sqrt{k / m}$ .

[L1]

Action variable and theorem: $J_{a} (E) = \frac{1}{π} \int_{a (E)}^{b (E)} \sqrt{2 m (E - U (q))} d q$ , and under the two simple turning points of this datum, $J_{a}^{'} (E) = T (E) / (2 π)$ with $T (E) = \sqrt{2 m} \int_{a (E)}^{b (E)} d q / \sqrt{E - U (q)}$ (The one-dimensional action variable of a family of two-turning-point energy levels, The action variable of a two-simple-turning-point family is $C^{1}$ with derivative $T (E) / (2 π)$ , the period quadrature over $2 π$ ).

[L2]

Substitution with $q = \sqrt{2 E / k} \sin ϑ$ on $[- π / 2, π / 2]$ , where $\cos ϑ \geq 0$ , is a differentiable monotone change of variable, and $\cos^{2} ϑ = \frac{1}{2} (1 + \cos 2 ϑ)$ from the addition formula; the Taylor-free identities $\sin^{2} + \cos^{2} = 1$ and the endpoint values $\sin (\pm π / 2) = \pm 1$ , $\cos (\pm π / 2) = 0$ hold (Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ , The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The addition formulas for sine and cosine, Parity and the Pythagorean identity for sine and cosine, Signs, monotonicity intervals, and ranges of sine and cosine).

[L3]

$\sqrt{2 E / k} \cdot \sqrt{k / 2} = \sqrt{E}$ and $\sqrt{2 m E} \cdot \sqrt{E} = \sqrt{2 m} E$ ; for $E > 0$ the square roots are the positive ones (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

Verification

technique · direct

1.1givenL1algebra

Fix any bounded energy interval $(E_{0}, E_{1})$ with $0 < E_{0} < E_{1} < \infty$ . Restrict $a, b$ to that interval. The datum satisfies the hypotheses of the theorem: $U$ is $C^{3}$ with $U^{'} (q) = k q$ , so $U^{'} (a (E)) = - k \sqrt{2 E / k} < 0 < k \sqrt{2 E / k} = U^{'} (b (E))$ , and $a (E) < b (E)$ ; moreover $\frac{1}{2} k q^{2} < E$ exactly for $a (E) < q < b (E)$ , and $U (a (E)) = U (b (E)) = E$ . Every $E > 0$ lies in such an interval, for example $(E / 2, 2 E)$ , and the defining integrals agree on overlaps; thus the theorem gives a $C^{1}$ action on all of $(0, \infty)$ .

2.1step 1.1L2L3algebra

Computation of the action variable: substituting $q = \sqrt{2 E / k} \sin ϑ$ by [L2] and using $\frac{1}{2} k q^{2} = E \sin^{2} ϑ$ , $J_{a} (E) = \frac{1}{π} \int_{- π / 2}^{π / 2} \sqrt{2 m E \cos^{2} ϑ} \sqrt{\frac{2 E}{k}} \cos ϑ d ϑ = \frac{\sqrt{2 m E}}{π} \sqrt{\frac{2 E}{k}} \int_{- π / 2}^{π / 2} \cos^{2} ϑ d ϑ,$ because $\sqrt{\cos^{2} ϑ} = \cos ϑ \geq 0$ on the interval. With $\cos^{2} ϑ = \frac{1}{2} (1 + \cos 2 ϑ)$ the integral is $π / 2$ , so by [L3] $J_{a} (E) = \frac{\sqrt{2 m E} \sqrt{2 E / k}}{2} = \frac{2 E \sqrt{m / k}}{2} = \frac{E}{ω} .$

2.2step 1.1L2L3algebra

Period quadrature: put $A = \sqrt{2 E / k}$ . For $- π / 2 < c < d < π / 2$ the substitution $q = A \sin ϑ$ lies strictly between the turning points, where the integrand is continuous and $\cos ϑ > 0$ . Hence [L2] gives $\int_{A \sin c}^{A \sin d} d q / \sqrt{E - k q^{2} / 2} = \sqrt{2 / k} (d - c)$ , since $\sqrt{E \cos^{2} ϑ} = \sqrt{E} \cos ϑ$ there. Taking the independent limits $c ↓ - π / 2$ , $d ↑ π / 2$ proves convergence at both endpoints and gives $\int_{a (E)}^{b (E)} d q / \sqrt{E - k q^{2} / 2} = π \sqrt{2 / k}$ . Thus $T (E) = \sqrt{2 m} π \sqrt{2 / k} = 2 π / ω$ .

3.1step 1.1step 2.1step 2.2L1algebra∎

Differentiating the closed form of step 2.1, $J_{a}^{'} (E) = 1 / ω$ for every $E > 0$ ; by step 2.2 this equals $T (E) / (2 π)$ , consistent with the general theorem [L1], which the datum satisfies by step 1.1.

Depends on

Dependency tree · 2 levels

Sources

David Tong, Classical Dynamics (University of Cambridge), §4.5.1 and §4.5.3, printed pp. 105-111: the harmonic oscillator action variable and the period relation $\mathrm dE/\mathrm dI=\omega$

The harmonic oscillator action variable Ja(E)=E/ωJa​(E)=E/ω and Ja′(E)=T/(2π)Ja′​(E)=T/(2π)