(Chapter 2, p. 68, Problem 24) From Goldstein Classical Mechanics.


The one-dimensional harmonic oscillator has the Lagrangian $L=m\dot{x}^2/2-kx^2/2.$ Suppose you did not know the solution to the motion, but realized that the motion must be periodic and therefore could be described by a Fourier series of the form $$x(t) =\sum_{j=0}a_j\cos(j\omega t)$$ (taking $t=0$ at a turning point) where $\omega$ is the (unknown) angular frequency of the motion. This representation for $x(t)$ defines a many-parameter path for the system point in configuration space. (That is, a set of (infinite) parameters $a_0, a_1, a_2,\ldots$ defines a path $x = x(t)$). Consider the action integral $I$ between $t = 0$ and $t = T$, where $T = \frac{2\pi}{\omega}$. Show that $I$ is an extremum for nonvanishing $x$ only if $a_j = 0$ for $j\ne1$ , and only if $\omega^2 = \frac{k}{m} $. (Hint: calculate $I$ and take partial derivatives $\frac{\partial I}{\partial a_i} $, which must be 0. Since we are assuming that $x(t)$ repeats with the period $T$, the $j = 1$ term must be nonzero.)

All I know is $\partial I=0$ from Hamilton's principle (where $I = \int Ldt)$ and $L = T - V$, where $T =$ Kinetic energy and $V =$ potential energy. The trouble comes when trying to square a Fourier series (twice) inside the Lagrangian which we much integrate over and take the partial derivatives of with respect to several $a_j's$


I) The Fourier series

$$\tag{1} x(t) ~=~\sum_{j\in\mathbb{N}_0}a_j\cos(j\omega t) ,\qquad \omega~>~0,$$

is a change of variables from paths $t\mapsto x(t)$ to Fourier coefficients $(a_j)_{j\in\mathbb{N}_0}$. The Lagrangian

$$\tag{2} L~=~\frac{m}{2}\dot{x}^2-\frac{k}{2}x^2, \qquad m,k~>~0,$$

for a harmonic oscillator leads to an action

$$\tag{3} I~=~\int_{t_1}^{t_2}\! dt ~L ~=~\int_{0}^{\frac{2\pi}{\omega}}\! dt ~L ~\stackrel{(1)+(2)}{=}~ \frac{m\omega}{4}\sum_{j\in\mathbb{N}}j^2a_j^2-\frac{k}{4\omega}\sum_{j\in\mathbb{N}_0} a_j^2.$$

II) The stationary condition for the action (3) is

$$\tag{4} \forall j\in\mathbb{N}_0: ~~0~=~\frac{\partial I}{\partial a_j} ~\stackrel{(3)}{=}~ \left(\frac{m\omega}{2}j^2-\frac{k}{2\omega}\right)a_j,$$

which implies that

$$\tag{5} \forall j\in\mathbb{N}_0: ~~ a_j~=~0 \quad\vee \quad\sqrt{\frac{k}{m\omega^2}}~=~j.$$

In other words: At most one variable $a_j$ is non-zero in eq. (5).

  1. If $j_0:=\sqrt{\frac{k}{m\omega^2}}\in\mathbb{N}$, there exists a one-dimensional stationary line, where only $a_{j_0}$ is non-zero. In fact, the entire $a_{j_0}$ dependence drops out of the action (3).

  2. If $\sqrt{\frac{k}{m\omega^2}}\notin\mathbb{N}$, there exists only one stationary point, the trivial configuration, where all the $a_j$ are zero.

III) Finally, we would like to address why the exercise formulation seems to favor the stationary solution $\sqrt{\frac{k}{m\omega^2}}=1$. Eq. (1) implies the "boundary conditions"

$$\tag{6} x(t)~\stackrel{(1)}{=}~x(-t) \quad\text{and} \quad x(t)~\stackrel{(1)}{=}~x(t+\frac{2\pi}{\omega}).$$

Note that all stationary points (5) has $a_0=0$, and note that there is a minus in front of the $a_0^2$ term in the action (3). Hence without further boundary conditions, the stationary points (5) are not minimum points for the action. It seems natural to additionally impose that the average

$$\tag{7} \frac{2\pi}{\omega}a_0~\stackrel{(1)}{=}~\int_{0}^{\frac{2\pi}{\omega}}\! dt~x(t)~=~0$$

should vanish. Let us assume (7) from now on. Then a stationary line with $\sqrt{\frac{k}{m\omega^2}}= 1$ is a minimum for the action (3), while a stationary line with $\sqrt{\frac{k}{m\omega^2}}\in\mathbb{N}\backslash\{1\} $ is only a saddle. Concerning saddle points, see also my related Phys.SE answer here.


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.