Worked examples — Cyclic coordinates — corresponding conservation law

If a symbol here is new, we build it from zero. Recall only these ingredients:

• $L$ = the Lagrangian, a single number you compute at each instant from positions and velocities (built in Euler–Lagrange Equations — the master equation $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}-\frac{\partial L}{\partial q}=0$ that we lean on all page).
• $q_k$ = a generalized coordinate — any number that pins down the configuration (a distance, an angle, whatever).
• ${\dot{q}}_k$ = its rate of change (the dot means "per second").
• $p_k\equiv \frac{\partial L}{\partial {\dot{q}}_k}$ = the conjugate momentum (from Generalized Momentum — the general "oomph" that reduces to $m\dot{x}$ for a plain coordinate). The symbol $\partial /\partial {\dot{q}}_k$ means "wiggle only ${\dot{q}}_k$, hold everything else frozen, and ask how fast $L$ changes."

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The scenario matrix

Every problem this topic can throw sits in one of eight cells. The figure below is the visual map: eight coloured tiles arranged from the friendly cases (top-left) to the trap cases (bottom-right), each tagged with the example that covers it. Read it before the table — the colour code is fixed and reused on every later figure, so you can always see which "family" a picture belongs to: blue = clean cyclic case, green = mixed cyclic/non-cyclic, yellow = edge case (zero or limiting input), red = trap.

Alt-text / caption: A 2×4 grid of eight coloured tiles. Blue tiles (A, B) are the clean cyclic cases — one momentum conserved. Green tiles (C, D) mix cyclic and non-cyclic coordinates. Yellow tiles (E, F) are edge cases — zero and limiting inputs. Red tiles (G, H) are the traps — no cyclic coordinate at all, and a velocity-coupling that hides in the momentum. Each tile names the example (Ex 1–Ex 8) that lives there.

Cell	What makes it distinctive	Example
A. Linear cyclic	a straight-line coordinate $x$ is absent → linear momentum conserved	Ex 1
B. Angular cyclic	an angle $\varphi$ absent → angular momentum conserved	Ex 2
C. Mixed: one cyclic, one not	must decide coordinate-by-coordinate; sign of the force matters	Ex 3
D. Two cyclic at once	two conserved momenta simultaneously (ignorable $x$ and $z$)	Ex 4
E. Degenerate / zero input	conserved momentum equals zero — what does that force?	Ex 5
F. Limiting behaviour	let a parameter → 0 or → ∞ and watch the conservation law survive	Ex 6
G. Real-world word problem	translate physical words into a Lagrangian, then find what's conserved	Ex 7
H. Exam twist (trap)	$\dot{q}$ present but $q$ absent, or a coupling term that looks like $q$ dependence but isn't	Ex 8

Every cell A–H is covered below. As you meet each example, glance back at the tile of the matching colour on the map above — that tile's border colour is the border colour of the example's own figure, so the eight cells and the eight worked cases line up one-to-one.

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Cell A — Linear cyclic coordinate (blue tile, top-left of the matrix)

Forecast: Before reading on — is $x$ conserved, or $p_x$? Guess which quantity stays fixed.

1. Check if $x$ is cyclic. Compute $\frac{\partial L}{\partial x}$. Since no bare $x$ sits in $L$, it is $0$. So $x$ is cyclic. Why this step? The whole method starts by asking "does the coordinate itself appear?" — the velocity $\dot{x}$ appearing does not disqualify it.
2. Form the conjugate momentum. $p_x=\frac{\partial L}{\partial \dot{x}}=m\dot{x}$. Why this step? Only $\frac{1}{2}m{\dot{x}}^2$ contains $\dot{x}$; its derivative w.r.t. $\dot{x}$ is $m\dot{x}$.
3. State conservation. Because $\partial L/\partial x=0$, Euler–Lagrange gives ${\dot{p}}_x=0$, so $p_x=m\dot{x}$ is constant. By symmetry $p_y=m\dot{y}$ is also constant.

What the figure shows: the red dots are the particle at equal time-steps; the yellow arrow at each dot is $p_x=m\dot{x}$. Trace the arrows left-to-right — every one has the same length and direction. That unchanging arrow is the conservation law you just derived; the blue dashed line is the straight-line path it produces.

Alt-text / caption: A particle plotted at five equal time-steps along a straight dashed blue path, each carrying an identical horizontal yellow momentum arrow — visually confirming p_x = m·x-dot never changes.

Verify: Units of $p_x$: $\text{kg}\cdot \text{m/s}$ = ordinary momentum. ✔ Sanity: a free particle drifts at constant velocity, so $m\dot{x}$ must be fixed.

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Cell B — Angular cyclic coordinate (blue tile, top of the matrix)

Forecast: Is it $\varphi$ that's constant, or $m{r}^2\dot{\varphi }$? (Watch: $\varphi$ keeps growing as the planet orbits.)

1. Cyclic check. $\frac{\partial L}{\partial \varphi }=0$ — no bare $\varphi$ appears ($V$ depends only on $r$). So $\varphi$ is cyclic. Why this step? The potential $V(r)$ hides no $\varphi$; the kinetic term uses only $\dot{\varphi }$. Whatever $k$ is, differentiating $-k/r$ with respect to $\varphi$ gives zero.
2. Conjugate momentum. Only $\frac{1}{2}m{r}^2{\dot{\varphi }}^2$ holds $\dot{\varphi }$. Differentiate: $p_{\varphi }=m{r}^2\dot{\varphi }$. Why this step? $\frac{\partial }{\partial \dot{\varphi }}\left(\frac{1}{2}m{r}^2{\dot{\varphi }}^2\right)=m{r}^2\dot{\varphi }$; the other terms have no $\dot{\varphi }$ so they drop.
3. Interpret. $p_{\varphi }=m{r}^2\dot{\varphi }$ is the angular momentum $L_z$ — conserved. Why this step? $r^2\dot{\varphi }$ is (twice) the rate at which the line from centre to particle sweeps out area; a fixed $m{r}^2\dot{\varphi }$ therefore means area is swept at a constant rate — this is Kepler's second law. In Central Force Motion this single conserved number is what reduces the whole orbit problem to a one-dimensional radial equation.
4. Numbers: $p_{\varphi }=2\cdot 3^2\cdot 0.5=9$ (in ${\text{kg⋅m}}^2/\text{s}$).

What the figure shows: the blue ellipse is the orbit, the yellow dot the centre of force. The green and red wedges are swept in equal time intervals — one near the centre (fast, short $r$) and one far away (slow, long $r$). They have equal area precisely because $m{r}^2\dot{\varphi }$ is pinned: when $r$ shrinks, $\dot{\varphi }$ must grow to keep the product fixed.

Alt-text / caption: An elliptical orbit around a yellow force-centre, with two equal-area shaded wedges (green near the centre, red far away) swept in equal times — the geometric face of conserved p_phi = m·r²·phi-dot.

Verify: Units: $\text{kg}\cdot {\text{m}}^2\cdot {\text{s}}^{-1}$ = angular momentum. ✔

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Cell C — One cyclic, one not (mind the sign) (green tile of the matrix)

Forecast: One coordinate has a force on it and one doesn't. Which is which?

1. Horizontal. $\frac{\partial L}{\partial x}=0$ → $x$ cyclic → $p_x=m\dot{x}=$ const. Why this step? Gravity's term $-mgy$ has no $x$ in it, so nothing pushes horizontally.
2. Vertical. $\frac{\partial L}{\partial y}=-mg\ne 0$ → $y$ not cyclic. Euler–Lagrange: ${\dot{p}}_y=\frac{\partial L}{\partial y}=-mg$. Why this step? The generalized "force" on $y$ is $\partial L/\partial y$, and it is nonzero — this is the whole difference between the two coordinates.
3. Numbers: ${\dot{p}}_y=-mg=-1\cdot 9.8=-9.8\text{N}$. So $p_y$ drops by $9.8\text{kg⋅m/s}$ every second.

What the figure shows: along the blue parabola, the green arrows are $p_x$ — all identical (cyclic). The red arrows are $p_y$ — long and up early, shrinking to zero at the top, then growing downward. Watching only the green arrows you'd never know gravity exists; that's what "cyclic" looks like.

Alt-text / caption: A projectile's parabolic arc with, at three points, an unchanging green horizontal momentum arrow (p_x conserved) and a red vertical arrow that steadily shrinks and reverses (p_y changing under gravity).

Verify: $p_x$ constant matches "horizontal velocity of a projectile never changes." ✔ ${\dot{p}}_y=-mg$ is Newton's second law in the $y$-direction. Units of ${\dot{p}}_y$: ${\text{kg⋅m/s}}^2=\text{N}$. ✔

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Cell D — Two cyclic coordinates at once (green tile of the matrix)

Forecast: Guess the count before summing.

1. Test all three. $\partial L/\partial x=0$, $\partial L/\partial z=0$, but $\partial L/\partial y=-mg\ne 0$. Why this step? The single potential term $-mgy$ mentions only $y$, leaving $x$ and $z$ untouched — two directions with no force.
2. Conserved momenta. $p_x=m\dot{x}=$ const and $p_z=m\dot{z}=$ const. Two cyclic coordinates → two conservation laws. Why this step? Each cyclic coordinate independently gives its own $\dot{p}=0$; they don't interfere.
3. The odd one out. ${\dot{p}}_y=-mg$, so $p_y$ is not conserved.

Verify: A projectile in 3-D with gravity along $y$ drifts freely in the whole horizontal plane $(x,z)$ — exactly two conserved linear momenta. ✔ Count matches the number of directions with no force.

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Cell E — Degenerate case: conserved momentum equals zero (yellow tile of the matrix)

Forecast: If $p_{\varphi }$ starts at zero and is conserved… can the particle ever start swirling?

1. Value now. $p_{\varphi }=m{r}^2\dot{\varphi }=m{r}^2\cdot 0=0$. Why this step? Plug the initial $\dot{\varphi }=0$ into the conserved quantity to read off its fixed value.
2. Value forever. Since $\varphi$ is cyclic, $p_{\varphi }$ is constant. It started at $0$, so $p_{\varphi }=0$ for all time. Why this step? Conservation doesn't only apply to nice non-zero numbers — a value of exactly $0$ is preserved just as rigidly.
3. Consequence. $m{r}^2\dot{\varphi }=0$ with $m\ne 0$, $r\ne 0$ forces $\dot{\varphi }=0$ always: the particle falls (or climbs) on a straight radial line — no orbit ever develops. Why this step? A product of nonzero factors times $\dot{\varphi }$ can only be zero if $\dot{\varphi }$ itself is zero, so the angle can never start changing.

Verify: A zero conserved quantity is still a conservation law — it pins the motion to a line. Physically: you can't spontaneously acquire angular momentum from a central force. ✔ Special case of Ex 2 with the number set to $0$.

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Cell F — Limiting behaviour (yellow tile of the matrix)

Forecast: If $p_{\varphi }$ can't change but ${\sin }^2\theta \to 0$, what must $\dot{\varphi }$ do?

1. Cyclic check. No bare $\varphi$ in $L$ → $\varphi$ cyclic → $p_{\varphi }=m{R}^2{\sin }^2\theta \dot{\varphi }$ constant. Why this step? Only $\dot{\varphi }$ (through ${\sin }^2\theta {\dot{\varphi }}^2$) appears, never $\varphi$ itself; longitude has no preferred value on a sphere.
2. Solve for $\dot{\varphi }$. $\dot{\varphi }=\frac{p_{\varphi }}{m{R}^2{\sin }^2\theta }$. Why this step? Rearrange the conserved quantity to isolate the spin rate so we can watch it as $\theta$ changes.
3. Limit. As $\theta \to 0$, ${\sin }^2\theta \to 0$, so (for fixed nonzero $p_{\varphi }$) $\dot{\varphi }\to \infty$: the particle whirls faster and faster near the pole. Why this step? The numerator $p_{\varphi }$ is locked, so shrinking the denominator forces the quotient up — the mathematical shadow of a skater pulling their arms in.
4. Numbers check the trend: $R=1,m=1,p_{\varphi }=0.25$. At $\theta =30°$, ${\sin }^2\theta =0.25$, so $\dot{\varphi }=0.25/0.25=1$. At $\theta =90°$ (equator), $\dot{\varphi }=0.25/1=0.25$. Faster near the pole. ✔

What the figure shows: the yellow curve is $\dot{\varphi }$ plotted against the polar angle. Follow it from the equator (right) toward the pole (left): it climbs gently, then shoots up toward the vertical dotted line at $\theta =0$. The green dot marks the calm equator value $0.25$; the red annotation marks the blow-up, which the caveat above reminds us is only a chart artefact.

Alt-text / caption: A graph of phi-dot versus polar angle theta (in degrees from the pole). The yellow curve is low and flat at the equator, where a green dot marks the value 0.25, and rises ever more steeply toward a vertical dotted asymptote at theta equals zero, labelled in red — because the conserved p_phi forces phi-dot upward as sin-squared-theta shrinks, though this blow-up is only a coordinate-singularity artefact at the pole.

Verify: Conservation forces the ${\sin }^2\theta \dot{\varphi }$ product fixed; smaller ${\sin }^2\theta$ demands bigger $\dot{\varphi }$. Same physics as conservation of angular momentum. ✔

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Cell G — Real-world word problem (red tile of the matrix)

Forecast: There's no free $\varphi$ at all here — is that a trick? Which coordinate could be cyclic?

1. Identify the free coordinate. The only free coordinate is $r$; the angle is externally driven, so it is not available for the cyclic test. Why this step? The cyclic-coordinate theorem applies only to coordinates the system is free to vary — a motor-imposed $\varphi (t)$ is a constraint, not a dynamical variable.
2. Is $r$ cyclic? $\frac{\partial L}{\partial r}=mr{\Omega }^2\ne 0$ (for $r\ne 0$). So $r$ is not cyclic — no conservation of $p_r$ here. Why this step? The $\frac{1}{2}m{r}^2{\Omega }^2$ term contains a bare $r$; that bare $r$ is exactly a real generalized force pushing the bead outward.
3. What the equation gives instead. $p_r=\frac{\partial L}{\partial \dot{r}}=m\dot{r}$, and Euler–Lagrange gives ${\dot{p}}_r=\frac{\partial L}{\partial r}=mr{\Omega }^2$. Why this step? Differentiate $L$ w.r.t. $\dot{r}$ to get the momentum, then w.r.t. $r$ to get the force driving it — the standard two derivatives, now with a nonzero force. Numbers: ${\dot{p}}_r=0.5\cdot 2\cdot 4^2=16\text{N}$ — the bead accelerates outward.
4. Lesson. A word problem can be designed so nothing is cyclic; then the theorem honestly tells you "no free lunch," and you fall back on the full Euler–Lagrange machinery.

Verify: ${\dot{p}}_r=mr{\Omega }^2=0.5\cdot 2\cdot 16=16\text{N}$, the outward centrifugal-type force $m{\Omega }^{2}r$. Units ${\text{kg⋅m/s}}^2=\text{N}$. ✔ The bead flies outward — correct for a frictionless radial wire on a spinning disk.

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Cell H — Exam twist (the trap) (red tile of the matrix)

Forecast: The coupling term $b\dot{x}\dot{y}$ mixes the coordinates — does that break cyclicity?

1. Cyclic test — look only for bare coordinates. $\partial L/\partial x=0$ and $\partial L/\partial y=0$: neither $x$ nor $y$ appears (only their velocities). So both are cyclic — the student is right on the diagnosis. Why this step? Cyclicity is about the coordinate, never the velocity or velocity-couplings; $b\dot{x}\dot{y}$ has no bare $x$ or $y$.
2. Conjugate momenta (the subtle part). Differentiate carefully: $p_x=\frac{\partial L}{\partial \dot{x}}=m\dot{x}+b\dot{y},p_y=\frac{\partial L}{\partial \dot{y}}=m\dot{y}+b\dot{x}.$ Why this step? The coupling $b\dot{x}\dot{y}$ contains $\dot{x}$, so it contributes to $p_x$; and it contains $\dot{y}$, so it also contributes to $p_y$. The trap is lazily writing $p_x=m\dot{x}$.
3. Both conserved. ${\dot{p}}_x=\partial L/\partial x=0$ and ${\dot{p}}_y=\partial L/\partial y=0$: so $m\dot{x}+b\dot{y}$ and $m\dot{y}+b\dot{x}$ are each constant. Why this step? Euler–Lagrange sets $\dot{p}=$ (the vanishing $q$-derivative), so each momentum's rate of change is zero.
4. Numeric check. With $m=1,b=0.5,\dot{x}=2,\dot{y}=4$: $p_x=1\cdot 2+0.5\cdot 4=4$; $p_y=1\cdot 4+0.5\cdot 2=5$.

Verify: Both partials in $x,y$ vanish, so both momenta are constant — the diagnosis was right but the formula for $p$ needed the coupling. $p_x=4$, $p_y=5$. ✔ Both Noether's theorem (every continuous symmetry gives a conserved current) and the Hamiltonian formalism (where these $p$'s become independent variables) insist you compute $p$ from the full $L$, coupling included.

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Recall Quick self-test

In Ex 6, why does $\dot{\varphi }$ blow up at the pole? ::: Because $p_{\varphi }=m{R}^2{\sin }^2\theta \dot{\varphi }$ is fixed while ${\sin }^2\theta \to 0$, so $\dot{\varphi }$ must grow without bound — but this is only a coordinate-singularity artefact; the true speed stays finite. In Ex 8, what is $p_x$? ::: $p_x=m\dot{x}+b\dot{y}$, not just $m\dot{x}$ — the coupling contributes. In Ex 5, what value does $p_{\varphi }$ take forever? ::: Zero, because it started at zero and is conserved.

Connections

• Parent topic — the rule these examples exercise.
• Euler–Lagrange Equations — the master equation we differentiate the moment a coordinate is not cyclic (Cells C, G).
• Generalized Momentum — the quantity we conserve; Cell H shows its full form with velocity coupling.
• Central Force Motion — home of Cells B, E, F; conserved $p_{\varphi }$ reduces the orbit to a 1-D radial problem.
• Noether's Theorem — the deep statement that every continuous symmetry gives one conserved quantity, of which our cyclic rule is the simplest case.
• Hamiltonian Mechanics — where cyclic coordinates let you drop variables entirely (Routhian reduction).
• Conservation of Energy — the time-translation cousin, deliberately not a cyclic-coordinate result.