In the ICTP lectures of Y. Grossman: Standard Model 1, in about minute 54:00, he leaves an informal homework for the students. He ask to find the symmetry related to the conservation of the amplitude of the simple classical harmonic oscillator.

I thought about it and it is a bit obvious from Newton's equations. A harmonic oscillator is defined (with unit constants) as $$\ddot{x}=-x$$ and the solution is $x(t)=A\cos(t+\phi)$, where $A$ is a constant amplitude and $\phi$ a phase.

Guess 1

Replacing the solution in the equation for the energy gives


thus the amplitude is conserved because it is proportional to the energy, which is conserved (no dissipative forces).

Thus as a first guess, the symmetry is just the same as the symmetry for the energy conservation: time translation symmetry.

Guess 2

The equation for the amplitude is $$A^2=2E=x^2+\dot{x}^2$$ which is the equation for a circle. Any rotation in phase space, will leave this quantity the same. So I can try the following infinitesimal transformation $\delta x = \dot{x}\epsilon$ and $\delta \dot{x}=-x\epsilon$, write the Lagrangian and input all this in the machinery of Noether's theorem. In the end, I found that $A^2$ is conserved. This is not a surprise because I started from the equation of a circle. (This can also be verified using inverse Noether theorem to confirm that it is the good infinitesimal transformation). Thus the symmetry is the invariance by rotations in phase space.

Which is it?

The first guess seems trivial and it is probably what it was meant but I am unable to confirm. The second guess is a bit of an artifact because I basically started from the equation of Hamiltonian times a constant, but the symmetry is a bit different that what I was expecting from guess 1. What is the symmetry related to that question? Is there another way to independently define the amplitude which makes its connection to the Hamiltonian less straightforward?

I just watched again the part where Grossman asks the question, and he says more precisely "what is the symmetry that guarantees that the period does not depend on the amplitude?". I suppose that it is related to what I have already written here. A much more naive argument would be to say that the period $\omega^{-1}$ has to depend on $k$ (spring constant), $m$ (mass), and $A$, but by dimensional analysis $k$ and $m$ suffice $\omega\propto\sqrt{k/m}$, so what is the symmetry there?

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    $\begingroup$ Nice question. Are we sure that there can just be one symmetry that leads to energy conservation ? It looks like for the SHO there could be two. $\endgroup$
    – Kurt G.
    Commented Jul 12, 2021 at 14:07
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    $\begingroup$ If I had to guess, I suspect that it's the second symmetry: the second symmetry is a rotation in phase space, and the angular frequency is in some sense "conjugate" to the angle in phase space. But I don't have the knowledge at my fingertips to formalize this. $\endgroup$ Commented Jul 12, 2021 at 14:10
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    $\begingroup$ The fact that the amplitude doesn't depend on the period is not a conservation law, so why would you expect it to be related to a symmetry? $\endgroup$
    – Javier
    Commented Jul 12, 2021 at 17:49
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    $\begingroup$ @Javier It can be interpreted as a conservation law for $\varphi - \omega t$ where $\varphi$ is the phase angle -- see Qmechanic's answer. $\endgroup$
    – nanoman
    Commented Jul 12, 2021 at 22:34
  • 2
    $\begingroup$ Please make your post one cohesive question rather than tacking edits on at the end. There is an edit history available for those who are interested. $\endgroup$ Commented Jul 13, 2021 at 1:04

6 Answers 6

  1. The most clean derivation is to go to the Hamiltonian formulation. Then the conserved charge is the Hamiltonian $$H~=~\frac{p^2}{2m}+\frac{kq^2}{2}\tag{A}$$ (basically the square of the amplitude $A$), and it generates the infinitesimal symmetry transformation in phase space $$\begin{align}\delta q~=~&\epsilon \{q ,H\}, \cr \delta p~=~&\epsilon \{p ,H\}, \cr \delta t~=~&0. \end{align}\tag{B} $$ This is essentially OP's 2nd guess.

  2. However, the symmetry transformation in Noether's theorem is not unique, cf. e.g. my Phys.SE answer here. For energy conservation (OP's 1st guess), this is e.g. demonstrated in my Phys.SE answer here.

  3. Let us go to angle-action variables $$\begin{align} \varphi~:=~&\arg(p+im\omega q)~\sim~\varphi+2\pi, \cr J~:=~&\frac{1}{2\pi}\oint p \mathrm{d}q. \cr \{\varphi,J\}~=~&1. \end{align}\tag{C}$$ Then the SHO Hamiltonian (A) becomes $$H~=~\omega J,\quad\text{where}\quad \omega~:=~\sqrt{k/m}.\tag{D}$$

  4. The fact that the period doesn't depend on the amplitude is encoded by the constant of motion $$Q~=~\varphi -\omega t.\tag{E}$$ It generates the infinitesimal quasi-symmetry transformation
    $$\begin{align} \delta \varphi~=~&\epsilon \{\varphi ,Q\}~=~0, \cr \delta J~=~&\epsilon \{J ,Q\}~=~-\epsilon, \cr \delta t~=~&0.\end{align} \tag{F} $$

  • $\begingroup$ Your Hamiltonian answer has inspired my Lagrangian one. $\endgroup$
    – nanoman
    Commented Jul 13, 2021 at 11:30

For the simple harmonic oscillator the two symmetry transformations are the same:

Time translation takes $$ \left(\begin{array}{c}x\\ \dot x\end{array}\right)=A\left(\begin{array}{c}\cos(\omega\,t+\phi)\\ -\sin(\omega\,t+\phi)\end{array}\right)\quad\mbox{ to }\quad \left(\begin{array}{c}x'\\ \dot x'\end{array}\right)=A\left(\begin{array}{c}\cos(\omega\,t_0+\omega\,t+\phi)\\ -\sin(\omega\,t_0+\omega\,t+\phi)\end{array}\right)\,. $$ A clockwise rotation in phase space takes $(x,\dot x)$ to $$ A\left(\begin{array}{cc}\cos\alpha & \sin\alpha\\ -\sin\alpha &\cos\alpha\end{array}\right)\left(\begin{array}{c}\cos(\omega\,t+\phi)\\ -\sin(\omega\,t+\phi)\end{array}\right)= A\left(\begin{array}{c}\cos(\alpha+\omega\,t+\phi)\\ -\sin(\alpha+\omega\,t+\phi)\end{array}\right)\,. $$ Edit (after watching the lecture): Grossman is saying that for the harmonic oscillator $\ddot x=-\omega^2x$ the period does not depend on the amplitude, and for a real pendulum $\ddot x=-\omega^2\sin x$ it does. He is also saying that energy is always conserved (for SHO and real pendulum). So far so good. His question to the students is (for a real pendulum): "what is the symmetry that guarantees that the period does not depend on the amplitude?" My pedestrian answer: If we want the amplitude $A$ to be choosen freely, regardless of the period $\omega$, and still get a solution to the EOM this means nothing else than the EOM must be linear. Then only the SHO $\ddot x=-x$ survives. And what is the symmetry? I am still thinking. One comment was made that the independece of the amplitude from the period is not a conservation law, and not necessarily directly related to a symmetry.

  • $\begingroup$ Do you see any relation between the period and the amplitude? $\endgroup$
    – Mauricio
    Commented Jul 12, 2021 at 15:32
  • $\begingroup$ Just watched a bit of Grossman's lecture. He is saying that for the harmonic oscillator the period does not depend on the amplitude, and for a real pendulum $\ddot x=\sin x$ it does. He is also saying that energy is always conserved. So far so good. As far as I can tell: His homework question seems : for a --general pendulum-- "what is the symmetry that guarantees that the period does not depend on the amplitude?" I am a bit confused now about the formulation of your question because it doesn't distinguish the SHO form the real pendulum. $\endgroup$
    – Kurt G.
    Commented Jul 12, 2021 at 17:46
  • $\begingroup$ Maybe you aré right and that would deserve another separate post. Do you have an idea by the way (for the general pendulum)? $\endgroup$
    – Mauricio
    Commented Jul 12, 2021 at 19:23
  • $\begingroup$ Not yet. Only that the real pendulum becomes a SHO when $x$ is small. More tomorrow. $\endgroup$
    – Kurt G.
    Commented Jul 12, 2021 at 19:38
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    $\begingroup$ I shall take Qmechanic's answer as a motivation to learn, on the basis of the real pendulum, Noether's theorem (thoroughly), action angle variables and quasi symmetry. There are other answers, too, worth reading carefully. Let me now edit mine adding some more pedestrian approach. $\endgroup$
    – Kurt G.
    Commented Jul 14, 2021 at 3:21

Let's write the harmonic oscillator Lagrangian as $L = \frac{1}{2} (\dot{x}^2 - \omega^2 x^2)$.

what is the symmetry that guarantees that the period does not depend on the amplitude?

It is natural to guess that this symmetry should involve a transformation that varies the amplitude. Now, we can simply write $\delta x = \epsilon x$, where $\epsilon$ is an infinitesimal parameter, and this spatial rescaling is a symmetry of the equation of motion $\ddot{x} = -\omega^2 x$, which is enough to draw the desired conclusion. Any solution can be transformed into a solution with a different amplitude but the same period.

But this is not good enough. The spirit of the question is to find a symmetry of the action and a corresponding Noether conservation law that enforces the desired property; $\delta x = \epsilon x$ is not a symmetry of the action and does not correspond to a conservation law. Let's try to overcome this within the Lagrangian framework.

To find such a symmetry, the key step is that we can use a function of the state in place of the constant $\epsilon$. Now, I'm going to cheat slightly to identify the right function: Qmechanic's answer has already indicated (in a Hamiltonian formalism) that the relevant transformation changes the amplitude so as to alter the energy $H$ by an additive constant. Because $H$ is a homogeneous polynomial in $x$, when we change $x$ by an amount proportional to $x$, we change $H$ by an amount proportional to $H$. So to change $H$ by a constant, we must compensate by dividing the amount of rescaling of $x$ by $H$. This leads to $$\delta x = \frac{\epsilon x}{2H} = \frac{\epsilon x}{\dot{x}^2 + \omega^2 x^2},$$ where of course $\epsilon$ now has different dimensions (energy).

Is this a symmetry of the action? Let's evaluate: $$\begin{split} \delta L &= \dot{x}\, \delta\dot{x} - \omega^2 x\, \delta x = \frac{d}{dt}(\dot x\, \delta x) - \ddot{x}\, \delta x - \omega^2 x\, \delta x\\ &= \frac{d}{dt}(\dot x\, \delta x) - \frac{\epsilon(x\ddot{x} + \omega^2 x^2)}{2H} = \frac{d}{dt}(\dot x\, \delta x) - \frac{\epsilon(x\ddot{x} - \dot{x}^2)}{2H} - \epsilon\\ &= \frac{d}{dt}(\dot x\, \delta x) + \epsilon(B - 1). \end{split}$$ The change in $L$ includes a portion that is expressible as a total time derivative, and a portion that is not: the weird term $B \equiv -(x\ddot{x} - \dot{x}^2)/(\dot{x}^2 + \omega^2 x^2)$.

However, $B$ is "like" a total time derivative in that the equation-of-motion contribution derived from it appears to vanish identically, i.e., plugging and chugging gives $$\frac{\partial B}{\partial x} - \frac{d}{dt} \frac{\partial B}{\partial\dot{x}} + \frac{d^2}{dt^2} \frac{\partial B}{\partial\ddot{x}} \equiv 0.$$ The caveat is that we have to assume that $H \ne 0$ (otherwise we would divide by zero and $B$ would be undefined).

Chugging through Noether's theorem, the explicit total time derivative we extracted in $\delta L$ above cancels out, and the "conservation law" is expressed as $$B - 1 = 0.$$ Indeed, on a physical trajectory where $\ddot{x} = -\omega^2 x$, we can directly simplify $B$ to $1$.

What is going on? Since $B$ is not a total time derivative, $\int dt\, B$ does not depend solely on the initial and final states. However, the calculation above shows that $\int dt\, B$ is unchanged by infinitesimal variations of the trajectory. The resolution is that $\int dt\, B$ is a topological invariant.

The trajectory can be plotted as a curve in the $(\dot{x}, \omega x)$ plane. If this is treated as a Euclidean plane, then $\omega B$ is the angular velocity around the origin. So $(\omega/2\pi) \int dt\, B$ is the number of "orbits" or "cycles", which is well-defined for an arbitrary trajectory in the punctured plane with the origin $(\dot{x}, \omega x) = (0, 0)$ excluded by the constraint $H \ne 0$.

While $B$ is not a total time derivative, we can cheat and relate it to the time derivative of a multivalued function of the state: the phase angle $\varphi$ in the $(\dot{x}, \omega x)$ plane, which is well-defined up to a multiple of $2\pi$. The specific multiple of $2\pi$ can be determined by following the specific continuous trajectory provided it obeys $H \ne 0$.

Now, with $B = \dot{\varphi}/\omega$, the "conservation law" becomes $$\frac{\dot{\varphi}}{\omega} - 1 = 0,$$ which integrates to $$\varphi - \omega t = \mathrm{const}.$$ This says that the phase advances at the fixed rate $\omega$ (resulting in the fixed period $2\pi/\omega$) for all physical trajectories -- in particular, independent of the amplitude.

  • $\begingroup$ Thanks for this alternative solution! I love how your version is self contained and it is easier to share. $\endgroup$
    – Mauricio
    Commented Jul 13, 2021 at 13:41

First I transformed the second-order differential equation to a first-order differential equation.

With $x=y_1~$ and $~\dot x=y_2$ you obtain

$$\dot{\vec{y}}=\underbrace{\begin{bmatrix} 0 & 1 \\ -1 & 0 \\ \end{bmatrix}}_{Q}\,\vec y$$

The eigenvalue $~\pm i~$ of the matrix Q are proportional to the period of $x(t)$ . The eigenvalues wouldn't change if you transformed the matrix Q with 2D orthogonal transformation $~R$.

$$R=\left[ \begin {array}{cc} \cos \left( \alpha \right) &-\sin \left( \alpha \right) \\ \sin \left( \alpha \right) &\cos \left( \alpha \right) \end {array} \right] $$

Thus $Q\mapsto R^T\,Q,R$

The solution is

$$\vec y(t)=\Re\left(A\,\vec v_1 e^{i\,t}+B\,\vec v_2\,e^{-i\,t}\right)$$

where $\vec v_i$ are the constant eigenvectors.

for $Q\mapsto i\,\left(T^{-1}\,Q\,T\right)$ where $T=\left[\vec v_1~,\vec v_2\right]$, the eigenvalues also don't change.

And because the matrix Q is constant, the eigenvalues (period) don't depend on the amplitude.


Guess 1 notes energy conservation of each solution implies amplitude conservation of each solution.

Guess 2 notes the solution set is closed under rotation in phase space because such rotation connects the solutions of a given amplitude.

Guess 1 is the only one to present an argument relevant to associating a system of the action to a conservation law; as guessed, the energy (Hamiltonian, call it what you will) is conserved.

Guess 2 concerns a separate concern, namely the fact that specifying a phase is a form of spontaneous symmetry breaking. (Sure, people usually have something like the Higgs boson in mind when they use that jargon, but this is also a valid use of the term.)

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    $\begingroup$ Do you see any relation with the period? $\endgroup$
    – Mauricio
    Commented Jul 12, 2021 at 15:32
  • $\begingroup$ @Mauricio If we no longer impose $\omega=1$ for nondimensionalization, the two guesses respectively note $E:=(\omega^2x^2+\dot{x}^2)/2$ is conserved and $t\mapsto\delta t$ preserve $E$. So the conserved quantity depends on $\omega$ and not just $A=\sqrt{2E}/\omega$, but we can regard either as what's conserved for a given choice of $\omega$. $\endgroup$
    – J.G.
    Commented Jul 12, 2021 at 15:36

Only the first guess is a symmetry because only there did you have to use the equations of motion to claim that $\delta x = \dot{x} \delta t$ left the amplitude invariant.

The second guess is a redundancy in how you specify the energy functional. You can think of it as a co-ordinate transformation. But since it acts on $x \equiv q$ and $\dot{x} \equiv p$ independently, its form is required to be canonical. From the global form of the transformation \begin{equation} \begin{pmatrix}Q \\ P\end{pmatrix} = \begin{pmatrix}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix}q \\ p\end{pmatrix}, \end{equation} we can check that \begin{align} \frac{\partial Q}{\partial p} &= -\frac{\partial q}{\partial P} \\ \frac{\partial Q}{\partial q} &= \frac{\partial p}{\partial P} \end{align} and similarly for $(q, Q) \leftrightarrow (p, P)$ which means that Hamilton's equations are preserved.


The thing about amplitude and frequency is that they are two completely different types of parameters. The first is set by the initial state (after which it's conserved) while the second is a fixed property of the system. So a question about why they are independent can be interpreted as "why are arbitrary amplitudes allowed". To me, this follows from the fact that $A$ can be expressed using only $x$ and $\dot{x}$, which are the degrees of freedom left unspecified by a second order ODE.

Asking which conservation law is responsible for the existence of this (or any other) free parameter in the solution sounds misleading. One can count the number of free parameters by checking how many time derivatives appear in the equations of motion but this will in general be larger than the number of independent conserved quantities. Even in problems like this one where the free parameters and conserved quantities are equal in number (known as integrable systems), I don't see what would make one way to pair them up better than the other.

  • $\begingroup$ Do you see any relation with the period? $\endgroup$
    – Mauricio
    Commented Jul 12, 2021 at 15:31
  • $\begingroup$ Have you spent time thinking about whether the answers to the first part of your question increase your own ability to answer the second? $\endgroup$ Commented Jul 12, 2021 at 15:38
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    $\begingroup$ @Mauricio Regarding relation b/w amplitude and time period: In adiabatic systems, the total energy for SHM can be written as $E=nI\omega$ where $2\pi I=\int p.dq$ is an adiabatic invariant and n is the winding number for this integral. From your guess $A^2 \sim E$. Can you see the relation? This might help: en.wikipedia.org/wiki/…. You should look for action angle variables in clas mech $\endgroup$
    – paul230_x
    Commented Jul 13, 2021 at 6:59

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