Exercises — Cyclic coordinates — corresponding conservation law

What we check: does $x$ appear? does $y$ appear? Neither $x$ nor $y$ shows up in $L$ (only their velocities $\dot{x},\dot{y}$ do). So $\partial L/\partial x=0$ and $\partial L/\partial y=0$. Answer: ==both $x$ and $y$ are cyclic==. The flat plane "looks the same" wherever you stand — that translational sameness is exactly the cyclic property.

What we check: the term $-\frac{1}{2}k{x}^2$ contains $x$ explicitly. So $\partial L/\partial x=-kx\ne 0$ (unless $x=0$). Answer: ==$x$ is NOT cyclic==. The spring's potential energy changes as you move, so there is a genuine force $-kx$. No missing coordinate ⇒ no free conservation law here.

What we check: $r$ appears (inside $r^2{\dot{\varphi }}^2$ and inside $V(r)$), so $r$ is not cyclic. The angle $\varphi$ appears only through its velocity $\dot{\varphi }$ — the coordinate $\varphi$ itself is absent. Answer: ==$\varphi$ is cyclic, $r$ is not==. See Central Force Motion.

Step (WHAT): differentiate $L$ with respect to $\dot{x}$, freezing $\dot{y}$. Only $\frac{1}{2}m{\dot{x}}^2$ contains $\dot{x}$; its derivative is $m\dot{x}$. $p_x=\frac{\partial L}{\partial \dot{x}}=m\dot{x}.$ Why conserved: $x$ cyclic (L1.1) ⇒ ${\dot{p}}_x=0$ ⇒ $p_x=m\dot{x}=\text{const}$. This is ordinary linear momentum. Translational symmetry ⇒ momentum conservation.

Step: only the term $\frac{1}{2}m{r}^2{\dot{\varphi }}^2$ contains $\dot{\varphi }$. Differentiate: $\frac{1}{2}m{r}^2\cdot 2\dot{\varphi }=m{r}^2\dot{\varphi }$. $p_{\varphi }=\frac{\partial L}{\partial \dot{\varphi }}=m{r}^2\dot{\varphi }=\text{const}.$ Identify: this is the angular momentum about the axis perpendicular to the plane of motion — we write it $J_z$ (the letter $J$ chosen to avoid clashing with $L$, which on this whole page means the Lagrangian). $J_z$ measures "how much rotation-oomph" the particle carries about the origin: mass $\times$ radius${}^2$ $\times$ angular rate. Its constancy is Kepler's second law (equal areas in equal times).

$x$: $\partial L/\partial x=0$ ⇒ cyclic ⇒ $p_x=m\dot{x}=\text{const}$ (horizontal momentum frozen). $y$: $\partial L/\partial y=-mg\ne 0$ ⇒ not cyclic. Euler–Lagrange gives ${\dot{p}}_y=\partial L/\partial y=-mg$, so $p_y=m\dot{y}$ decreases steadily — gravity is pumping momentum out vertically. Answer: ==horizontal $p_x$ conserved; vertical $p_y$ changes at rate $-mg$.==

Tool: $p_{\varphi }=m{r}^2\dot{\varphi }=\ell$ is the same number at both radii. With $m=1$: $r_1^2{\dot{\varphi }}_1=r_2^2{\dot{\varphi }}_2\Rightarrow (2{)}^2(3)=(6{)}^2{\dot{\varphi }}_2\Rightarrow 12=36{\dot{\varphi }}_2.$ $\boxed{{\dot{\varphi }}_2=\frac{12}{36}=\frac{1}{3}\approx \mathrm{0.333.}}$ What it looks like (figure s01): the shaded area swept per second is the same at $r_1$ and $r_2$; at large radius the planet crawls, at small radius it whips around.

Step: from $p_{\varphi }=m{r}^2\dot{\varphi }$ we solve $\dot{\varphi }=\frac{p_{\varphi }}{m{r}^2}$. Square and multiply by $mr$: $mr{\dot{\varphi }}^2=mr{\left(\frac{p_{\varphi }}{m{r}^2}\right)}^2=\frac{p_{\varphi }^2}{m{r}^3}.$ Why useful: the radial equation becomes $m\ddot{r}=\frac{p_{\varphi }^2}{m{r}^3}-\frac{dV}{dr}$ — an equation in $r$ alone, because the cyclic coordinate has been eliminated. This is exactly the "effective potential" reduction of Central Force Motion and the spirit of Routhian reduction in Hamiltonian Mechanics.

Cyclic check: $\varphi$ does not appear in $L$ (only $\dot{\varphi }$ does, and ${\sin }^2\theta$ depends on $\theta$, not $\varphi$). So $\partial L/\partial \varphi =0$ ⇒ cyclic. Momentum: $p_{\varphi }=\partial L/\partial \dot{\varphi }=m{R}^2{\sin }^2\theta \dot{\varphi }=\text{const}$. At $\theta =90^{\circ }$: $\sin 90^{\circ }=1$, so ${\sin }^2\theta =1$. With $m{R}^2=1$ and $p_{\varphi }=\frac{1}{2}$: $\frac{1}{2}=(1)(1)\dot{\varphi }\Rightarrow \boxed{\dot{\varphi }=\frac{1}{2}=\mathrm{0.5.}}$ Near the pole ($\theta \to 0$), ${\sin }^2\theta \to 0$, so $\dot{\varphi }$ must blow up to keep $p_{\varphi }$ fixed — the bead spins furiously near the top. That degenerate limit is the geometric warning of the next level.

Build $L$: the kinetic energy $T$ (energy of motion) is the sum for both beads, $T=\frac{1}{2}{m}_1{\dot{x}}_1^2+\frac{1}{2}{m}_2{\dot{x}}_2^2$; the potential energy of the spring depends only on the stretch $x_2-x_1$: $L=T-V=\frac{1}{2}{m}_1{\dot{x}}_1^2+\frac{1}{2}{m}_2{\dot{x}}_2^2-\frac{1}{2}k((x_2-x_1)-{\ell }_0{)}^2.$ Spot the problem: neither $x_1$ nor $x_2$ is cyclic alone, because both appear in the spring term. Clever variables: let the centre of mass $X=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$ and the separation $s=x_2-x_1$. Then $L=\frac{1}{2}M{\dot{X}}^2+\frac{1}{2}\mu {\dot{s}}^2-\frac{1}{2}k(s-{\ell }_0{)}^2,M=m_1+m_2,\mu =\frac{m_1{m}_2}{M}.$ Read off: $X$ is now absent from $L$ ⇒ cyclic ⇒ $p_X=M\dot{X}=\text{const}=\text{total linear momentum}.$ The internal coordinate $s$ carries the spring force; the whole two-body system drifts with constant total momentum. This is the analytical-mechanics face of "internal forces cannot move the centre of mass."

Build $L$: $L=T-V=\frac{1}{2}m({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2)-mgz.$ Cyclic check, all three:

• $\partial L/\partial x=0$ ⇒ $x$ cyclic ⇒ $p_x=m\dot{x}=\text{const}$.
• $\partial L/\partial y=0$ ⇒ $y$ cyclic ⇒ $p_y=m\dot{y}=\text{const}$.
• $\partial L/\partial z=-mg\ne 0$ ⇒ $z$ not cyclic; ${\dot{p}}_z=-mg$, so $p_z=m\dot{z}$ falls steadily. Answer: ==two conserved momenta $p_x,p_y$ (the horizontal plane), none vertical.== Symmetry is broken only along gravity's arrow.

First, the object we need — the energy function. Define the energy function (also called the Jacobi integral or, once rewritten in momenta, the Hamiltonian): $h\equiv \dot{x}\frac{\partial L}{\partial \dot{x}}-L=\dot{x}{p}_x-L.$ In words: take each velocity times its conjugate momentum, add them up, and subtract $L$. For $L=\frac{1}{2}m{\dot{x}}^2-V(x)$ this gives $h=\dot{x}(m\dot{x})-(\frac{1}{2}m{\dot{x}}^2-V)=\frac{1}{2}m{\dot{x}}^2+V(x)$ — the total energy $T+V$. The key identity (why $h$ matters): differentiating $h$ along the motion and using Euler–Lagrange, one finds $\frac{dh}{dt}=-\frac{\partial L}{\partial t}.$ So if $L$ has no explicit time ($\partial L/\partial t=0$), then $dh/dt=0$ and $h$ is conserved. This statement — that time-translation symmetry conserves the energy function $h$ — is the energy-function / Beltrami identity theorem. (In one dimension it is exactly Beltrami's identity; in general it is the energy-function argument that leads into Hamiltonian Mechanics.) What is right in the student's claim: yes, a conservation law does follow from "$L$ does not depend on time," and the conserved quantity is the energy $h$. What is wrong: $t$ is not a generalized coordinate — there is no velocity $\dot{t}$ and hence no conjugate momentum $p_t\equiv \partial L/\partial \dot{t}$. The cyclic-coordinate theorem does not apply to $t$ at all; energy conservation comes from the separate energy-function/Beltrami theorem above, driven by time-translation symmetry rather than a missing spatial coordinate. See Conservation of Energy. This is exactly "Mistake C" in the parent note Cyclic coordinates — corresponding conservation law.

Conservation is intact: $p_{\varphi }=m{R}^2{\sin }^2\theta \dot{\varphi }$ is exactly constant for all valid motion; the divergence of $\dot{\varphi }$ is precisely what keeps the product fixed as ${\sin }^2\theta \to 0$. Nothing about $\varphi$ being cyclic breaks. Physical resolution — the pole is walled off. Look at the kinetic energy stored in the $\varphi$-motion: $T_{\varphi }=\frac{1}{2}m{R}^2{\sin }^2\theta {\dot{\varphi }}^2=\frac{1}{2}\frac{p_{\varphi }^2}{m{R}^2{\sin }^2\theta }.$ As $\theta \to 0$ this grows without bound (it behaves like $1/{\sin }^2\theta$). A bead has only finite total energy, so it can never actually reach the pole while carrying nonzero $p_{\varphi }$ — the $1/{\sin }^2\theta$ term acts like an infinite energy wall (an "effective potential barrier") near $\theta =0$. The trajectory therefore turns around at some minimum ${\theta }_{\min }>0$ set by energy balance. When can the pole be reached? Only if $p_{\varphi }=0$ (pure meridian, "straight over the top" motion): then $\dot{\varphi }=0$ throughout, no barrier, and the bead can pass through $\theta =0$. So the divergence is the conservation law protecting itself, not failing — it forbids exactly the trajectories that would violate finite energy.

(a) Translation symmetry ⇒ linear momentum. If shifting a coordinate $x\to x+a$ (for every constant $a$) leaves $L$ unchanged, then $x$ is cyclic and Noether's Theorem's conserved current is exactly the conjugate momentum $p_x=m\dot{x}$ — ordinary linear momentum. (b) Rotation symmetry ⇒ angular momentum. If rotating an angle $\varphi \to \varphi +\alpha$ (for every constant $\alpha$) leaves $L$ unchanged, then $\varphi$ is cyclic and Noether's conserved current is $p_{\varphi }=m{r}^2\dot{\varphi }$ — the angular momentum $J_z$ of L2.2. Why Noether is strictly more general: the cyclic-coordinate rule only fires when a single coordinate literally disappears from $L$. Noether's theorem covers any continuous symmetry, including ones where no single $q_k$ vanishes but a combination of coordinates is preserved — e.g. rotations expressed in Cartesian coordinates (where $x$ and $y$ both appear yet the mixture is symmetric), Lorentz boosts, and gauge transformations. Cyclic coordinates are the easy, directly-visible corner of Noether's far larger landscape.

• L1: can you spot the absent coordinate at a glance? ::: Look for the bare $q_k$ missing from $L$; ignore ${\dot{q}}_k$.
• L2: can you write and name $p_k=\partial L/\partial {\dot{q}}_k$? ::: Yes — linear for $x$, angular $m{r}^2\dot{\varphi }$ for $\varphi$.
• L3: can you use conservation to eliminate a coordinate? ::: Solve for $\dot{\varphi }=p_{\varphi }/(m{r}^2)$ and substitute.
• L4: can you invent variables to create a cyclic coordinate? ::: Centre-of-mass coordinate for two-body systems.
• L5: can you separate the momentum theorem from the energy theorem? ::: Space symmetry (cyclic coord) vs time symmetry (energy function $h$) — different theorems.