Worked examples — Cyclic coordinates — corresponding conservation law

2.1.8 · D3 · Physics › Analytical Mechanics › Cyclic coordinates — corresponding conservation law

Agar koi symbol naya lage, hum usse zero se banate hain. Sirf yeh ingredients yaad rakhein:

• $L$ = Lagrangian, ek akela number jo aap positions aur velocities se har instant par compute karte ho (banaya gaya hai Euler–Lagrange Equations mein — master equation $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}-\frac{\partial L}{\partial q}=0$ jis par hum is poore page mein rely karte hain).
• $q_k$ = ek generalized coordinate — koi bhi number jo configuration pin karta hai (ek distance, ek angle, kuch bhi).
• ${\dot{q}}_k$ = uski rate of change (dot ka matlab hai "per second").
• $p_k\equiv \frac{\partial L}{\partial {\dot{q}}_k}$ = conjugate momentum (from Generalized Momentum — general "oomph" jo plain coordinate ke liye $m\dot{x}$ ban jaati hai). Symbol $\partial /\partial {\dot{q}}_k$ ka matlab hai "sirf ${\dot{q}}_k$ ko wiggle karo, baaki sab frozen rakho, aur dekho $L$ kitni tezi se change hoti hai."

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Scenario matrix

Is topic ka har problem in aath cells mein se ek mein baithta hai. Neeche ki figure visual map hai: aath coloured tiles, friendly cases (top-left) se trap cases (bottom-right) tak arrange ki gayi hain, har ek par us example ka tag hai jo usse cover karta hai. Table se pehle ise padho — colour code fixed hai aur baad ki har figure mein reuse hota hai, taaki aap hamesha dekh sako ki ek picture kis "family" se belong karti hai: blue = clean cyclic case, green = mixed cyclic/non-cyclic, yellow = edge case (zero ya limiting input), red = trap.

Alt-text / caption: Aath coloured tiles ka 2×4 grid. Blue tiles (A, B) clean cyclic cases hain — ek momentum conserved. Green tiles (C, D) cyclic aur non-cyclic coordinates mix karte hain. Yellow tiles (E, F) edge cases hain — zero aur limiting inputs. Red tiles (G, H) traps hain — bilkul bhi cyclic coordinate nahi, aur ek velocity-coupling jo momentum mein chhup jaati hai. Har tile us example ka naam likhti hai (Ex 1–Ex 8) jo wahan rehta hai.

Cell	Isme kya distinctive hai	Example
A. Linear cyclic	ek straight-line coordinate $x$ absent hai → linear momentum conserved	Ex 1
B. Angular cyclic	ek angle $\varphi$ absent hai → angular momentum conserved	Ex 2
C. Mixed: ek cyclic, ek nahi	coordinate-by-coordinate decide karna hoga; force ka sign matter karta hai	Ex 3
D. Ek saath do cyclic	do conserved momenta simultaneously (ignorable $x$ aur $z$)	Ex 4
E. Degenerate / zero input	conserved momentum zero ke barabar hai — yeh kya force karta hai?	Ex 5
F. Limiting behaviour	ek parameter → 0 ya → ∞ jaane do aur dekho conservation law survive karti hai	Ex 6
G. Real-world word problem	physical words ko Lagrangian mein translate karo, phir dhundo kya conserved hai	Ex 7
H. Exam twist (trap)	$\dot{q}$ present hai par $q$ absent hai, ya ek coupling term jo lagta hai $q$ dependence jaisa par hai nahi	Ex 8

Har cell A–H neeche cover ki gayi hai. Jab bhi koi example milta hai, upar ke map par matching colour ki tile dekho — us tile ki border colour us example ki apni figure ki border colour hai, taaki aath cells aur aath worked cases ek-to-one line up ho sakein.

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

Forecast: Aage padhne se pehle — kya $x$ conserved hai, ya $p_x$? Andaza lagao kaunsi quantity fixed rehti hai.

1. Check karo ki $x$ cyclic hai ya nahi. $\frac{\partial L}{\partial x}$ compute karo. Kyunki $L$ mein koi bare $x$ nahi hai, yeh $0$ hai. To $x$ cyclic hai. Yeh step kyun? Poora method is sawaal se shuru hota hai "kya coordinate khud appear karta hai?" — $\dot{x}$ ka appear karna usse disqualify nahi karta.
2. Conjugate momentum form karo. $p_x=\frac{\partial L}{\partial \dot{x}}=m\dot{x}$. Yeh step kyun? Sirf $\frac{1}{2}m{\dot{x}}^2$ mein $\dot{x}$ hai; $\dot{x}$ ke w.r.t. uska derivative $m\dot{x}$ hai.
3. Conservation state karo. Kyunki $\partial L/\partial x=0$ hai, Euler–Lagrange deta hai ${\dot{p}}_x=0$, to $p_x=m\dot{x}$ constant hai. Symmetry se $p_y=m\dot{y}$ bhi constant hai.

Figure kya dikhati hai: red dots equal time-steps par particle hain; har dot par yellow arrow $p_x=m\dot{x}$ hai. Arrows ko left-to-right trace karo — har ek ki same length aur direction hai. Woh unchanging arrow hi woh conservation law hai jo aapne abhi derive ki; blue dashed line woh straight-line path hai jo yeh produce karta hai.

Alt-text / caption: Ek particle jo paanch equal time-steps par ek straight dashed blue path ke saath plot ki gayi hai, har ek par ek identical horizontal yellow momentum arrow hai — visually confirm karta hai ki p_x = m·x-dot kabhi nahi badlata.

Verify: $p_x$ ki units: $\text{kg}\cdot \text{m/s}$ = ordinary momentum. ✔ Sanity: ek free particle constant velocity par drift karta hai, to $m\dot{x}$ zaroor fixed hona chahiye.

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Cell B — Angular cyclic coordinate (blue tile, matrix ke top mein)

Forecast: Kya $\varphi$ constant hai, ya $m{r}^2\dot{\varphi }$? (Dhyan do: $\varphi$ badhta rehta hai jaise planet orbit karta hai.)

1. Cyclic check. $\frac{\partial L}{\partial \varphi }=0$ — koi bare $\varphi$ appear nahi karta ($V$ sirf $r$ par depend karta hai). To $\varphi$ cyclic hai. Yeh step kyun? Potential $V(r)$ koi $\varphi$ nahi chhupata; kinetic term sirf $\dot{\varphi }$ use karta hai. Chahe $k$ kuch bhi ho, $-k/r$ ko $\varphi$ ke respect mein differentiate karne par zero milta hai.
2. Conjugate momentum. Sirf $\frac{1}{2}m{r}^2{\dot{\varphi }}^2$ mein $\dot{\varphi }$ hai. Differentiate karo: $p_{\varphi }=m{r}^2\dot{\varphi }$. Yeh step kyun? $\frac{\partial }{\partial \dot{\varphi }}\left(\frac{1}{2}m{r}^2{\dot{\varphi }}^2\right)=m{r}^2\dot{\varphi }$; doosre terms mein koi $\dot{\varphi }$ nahi to woh drop ho jaate hain.
3. Interpret karo. $p_{\varphi }=m{r}^2\dot{\varphi }$ angular momentum $L_z$ hai — conserved. Yeh step kyun? $r^2\dot{\varphi }$ woh rate hai jis par centre-to-particle line area sweep karti hai (do se divide karo); fixed $m{r}^2\dot{\varphi }$ ka matlab hai area constant rate par sweep ho raha hai — yeh hi Kepler's second law hai. Central Force Motion mein yeh akela conserved number poore orbit problem ko ek one-dimensional radial equation mein reduce kar deta hai.
4. Numbers: $p_{\varphi }=2\cdot 3^2\cdot 0.5=9$ (in ${\text{kg⋅m}}^2/\text{s}$).

Figure kya dikhati hai: blue ellipse orbit hai, yellow dot force ka centre hai. Green aur red wedges equal time intervals mein sweep hote hain — ek centre ke paas (fast, short $r$) aur ek door (slow, long $r$). Unka area equal hai precisely kyunki $m{r}^2\dot{\varphi }$ pinned hai: jab $r$ shrink hota hai, $\dot{\varphi }$ ko product fixed rakhne ke liye badhna padta hai.

Alt-text / caption: Ek yellow force-centre ke around elliptical orbit, jisme do equal-area shaded wedges hain (centre ke paas green, door red) jo equal times mein sweep hote hain — conserved p_phi = m·r²·phi-dot ka geometric chehra.

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

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Cell C — Ek cyclic, ek nahi (sign ka dhyan rakhna) (matrix ka green tile)

Forecast: Ek coordinate par force hai aur ek par nahi. Kaunsa kaunsa hai?

1. Horizontal. $\frac{\partial L}{\partial x}=0$ → $x$ cyclic → $p_x=m\dot{x}=$ const. Yeh step kyun? Gravity ka term $-mgy$ mein koi $x$ nahi hai, to kuch bhi horizontally push nahi karta.
2. Vertical. $\frac{\partial L}{\partial y}=-mg\ne 0$ → $y$ cyclic nahi. Euler–Lagrange: ${\dot{p}}_y=\frac{\partial L}{\partial y}=-mg$. Yeh step kyun? $y$ par generalized "force" $\partial L/\partial y$ hai, aur woh nonzero hai — do coordinates ke beech ka poora fark yahi hai.
3. Numbers: ${\dot{p}}_y=-mg=-1\cdot 9.8=-9.8\text{N}$. To $p_y$ har second $9.8\text{kg⋅m/s}$ girta hai.

Figure kya dikhati hai: blue parabola ke saath, green arrows $p_x$ hain — sab identical (cyclic). Red arrows $p_y$ hain — pehle lamba aur upar, top par zero hota hai, phir neeche badhta hai. Sirf green arrows dekh ke aap kabhi nahi jaan paate ki gravity exist karti hai; yahi "cyclic" dikhta hai.

Alt-text / caption: Ek projectile ka parabolic arc jisme teen points par ek unchanging green horizontal momentum arrow hai (p_x conserved) aur ek red vertical arrow jo steadily shrink hoti hai aur reverse karti hai (p_y gravity ke under change hoti hai).

Verify: $p_x$ constant "projectile ki horizontal velocity kabhi nahi badlati" se match karta hai. ✔ ${\dot{p}}_y=-mg$ $y$-direction mein Newton's second law hai. ${\dot{p}}_y$ ki units: ${\text{kg⋅m/s}}^2=\text{N}$. ✔

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Cell D — Ek saath do cyclic coordinates (matrix ka green tile)

Forecast: Sum karne se pehle count guess karo.

1. Teeno test karo. $\partial L/\partial x=0$, $\partial L/\partial z=0$, lekin $\partial L/\partial y=-mg\ne 0$. Yeh step kyun? Single potential term $-mgy$ sirf $y$ mention karta hai, $x$ aur $z$ ko untouched chodta hai — do directions jahan koi force nahi.
2. Conserved momenta. $p_x=m\dot{x}=$ const aur $p_z=m\dot{z}=$ const. Do cyclic coordinates → do conservation laws. Yeh step kyun? Har cyclic coordinate independently apna $\dot{p}=0$ deta hai; woh interfere nahi karte.
3. Odd one out. ${\dot{p}}_y=-mg$, to $p_y$ conserved nahi hai.

Verify: 3-D mein ek projectile jisme $y$ along gravity hai, poore horizontal plane $(x,z)$ mein freely drift karta hai — exactly do conserved linear momenta. ✔ Count un directions ki sankhya se match karta hai jahan koi force nahi.

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Cell E — Degenerate case: conserved momentum zero ke barabar (matrix ka yellow tile)

Forecast: Agar $p_{\varphi }$ zero se shuru hota hai aur conserved hai… kya particle kabhi swirl karna shuru kar sakta hai?

1. Abhi ki value. $p_{\varphi }=m{r}^2\dot{\varphi }=m{r}^2\cdot 0=0$. Yeh step kyun? Initial $\dot{\varphi }=0$ ko conserved quantity mein plug karo taaki uski fixed value padh sako.
2. Hamesha ki value. Kyunki $\varphi$ cyclic hai, $p_{\varphi }$ constant hai. Woh $0$ par shuru hua, to $p_{\varphi }=0$ hamesha ke liye. Yeh step kyun? Conservation sirf acche non-zero numbers par apply nahi hoti — exactly $0$ ki value bilkul utni hi rigidly preserve hoti hai.
3. Consequence. $m{r}^2\dot{\varphi }=0$ jahan $m\ne 0$, $r\ne 0$ hota hai, woh force karta hai ki $\dot{\varphi }=0$ hamesha: particle ek straight radial line par gira (ya chadha) — koi orbit kabhi develop nahi hoti. Yeh step kyun? $\dot{\varphi }$ ke saath nonzero factors ka product tab hi zero ho sakta hai jab $\dot{\varphi }$ khud zero ho, to angle kabhi change nahi hoga.

Verify: Ek zero conserved quantity phir bhi ek conservation law hai — yeh motion ko ek line par pin karti hai. Physically: aap ek central force se spontaneously angular momentum acquire nahi kar sakte. ✔ Ex 2 ka special case jahan number $0$ set kiya gaya hai.

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Cell F — Limiting behaviour (matrix ka yellow tile)

Forecast: Agar $p_{\varphi }$ change nahi ho sakta par ${\sin }^2\theta \to 0$ ho raha hai, to $\dot{\varphi }$ ko kya karna chahiye?

1. Cyclic check. $L$ mein koi bare $\varphi$ nahi → $\varphi$ cyclic → $p_{\varphi }=m{R}^2{\sin }^2\theta \dot{\varphi }$ constant. Yeh step kyun? Sirf $\dot{\varphi }$ (through ${\sin }^2\theta {\dot{\varphi }}^2$) appear karta hai, $\varphi$ khud kabhi nahi; longitude ka sphere par koi preferred value nahi hoti.
2. $\dot{\varphi }$ ke liye solve karo. $\dot{\varphi }=\frac{p_{\varphi }}{m{R}^2{\sin }^2\theta }$. Yeh step kyun? Conserved quantity ko rearrange karo taaki spin rate isolate ho sake aur hum dekh sakein ki $\theta$ change hone par woh kya karta hai.
3. Limit. Jaise $\theta \to 0$, ${\sin }^2\theta \to 0$, to (fixed nonzero $p_{\varphi }$ ke liye) $\dot{\varphi }\to \infty$: particle pole ke paas tezi se tezi se whirl karta hai. Yeh step kyun? Numerator $p_{\varphi }$ locked hai, to denominator shrink karna quotient ko force karta hai upar jaane ko — mathematically ek skater ke arms andar kheenchne ki chhaya.
4. Numbers trend check karte hain: $R=1,m=1,p_{\varphi }=0.25$. $\theta =30°$ par, ${\sin }^2\theta =0.25$, to $\dot{\varphi }=0.25/0.25=1$. $\theta =90°$ (equator) par, $\dot{\varphi }=0.25/1=0.25$. Pole ke paas tezi se. ✔

Figure kya dikhati hai: yellow curve $\dot{\varphi }$ hai jo polar angle ke against plot ki gayi hai. Use equator (right) se pole (left) ki taraf follow karo: woh dheere chadha hai, phir vertical dotted line ki taraf $\theta =0$ par shoot karta hai. Green dot calm equator value $0.25$ mark karta hai; red annotation blow-up mark karta hai, jo upar diya caveat remind karta hai ki yeh sirf ek chart artefact hai.

Alt-text / caption: Phi-dot vs polar angle theta (pole se degrees mein) ka graph. Yellow curve equator par low aur flat hai, jahan ek green dot value 0.25 mark karta hai, aur theta equals zero par ek vertical dotted asymptote ki taraf tezi se rise karti hai, red mein label hai — kyunki conserved p_phi phi-dot ko upar force karta hai jaise sin-squared-theta shrink hota hai, halanki yeh blow-up sirf pole par ek coordinate-singularity artefact hai.

Verify: Conservation ${\sin }^2\theta \dot{\varphi }$ product ko fixed force karta hai; chhota ${\sin }^2\theta$ bada $\dot{\varphi }$ demand karta hai. Same physics angular momentum conservation jaisa. ✔

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

Forecast: Yahan bilkul bhi free $\varphi$ nahi hai — kya yeh ek trick hai? Kaunsa coordinate cyclic ho sakta hai?

1. Free coordinate identify karo. Akela free coordinate $r$ hai; angle externally driven hai, to yeh cyclic test ke liye available nahi hai. Yeh step kyun? Cyclic-coordinate theorem sirf un coordinates par apply hota hai jinhe system freely vary karne ke liye free hai — motor-imposed $\varphi (t)$ ek constraint hai, dynamical variable nahi.
2. Kya $r$ cyclic hai? $\frac{\partial L}{\partial r}=mr{\Omega }^2\ne 0$ ($r\ne 0$ ke liye). To $r$ cyclic nahi hai — yahan $p_r$ ka koi conservation nahi. Yeh step kyun? $\frac{1}{2}m{r}^2{\Omega }^2$ term mein ek bare $r$ hai; woh bare $r$ exactly ek real generalized force hai jo bead ko bahar dhakelta hai.
3. Equation ki jagah kya deta hai. $p_r=\frac{\partial L}{\partial \dot{r}}=m\dot{r}$, aur Euler–Lagrange deta hai ${\dot{p}}_r=\frac{\partial L}{\partial r}=mr{\Omega }^2$. Yeh step kyun? $L$ ko $\dot{r}$ ke w.r.t. differentiate karo momentum paane ke liye, phir $r$ ke w.r.t. force paane ke liye — standard do derivatives, ab ek nonzero force ke saath. Numbers: ${\dot{p}}_r=0.5\cdot 2\cdot 4^2=16\text{N}$ — bead bahar accelerate karta hai.
4. Lesson. Ek word problem design ho sakta hai taaki kuch bhi cyclic na ho; phir theorem honestly kehta hai "koi free lunch nahi," aur aap poore Euler–Lagrange machinery par fall back karte ho.

Verify: ${\dot{p}}_r=mr{\Omega }^2=0.5\cdot 2\cdot 16=16\text{N}$, outward centrifugal-type force $m{\Omega }^{2}r$. Units ${\text{kg⋅m/s}}^2=\text{N}$. ✔ Bead bahar fly karta hai — ek spinning disk par frictionless radial wire ke liye correct hai.

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Cell H — Exam twist (trap) (matrix ka red tile)

Forecast: Coupling term $b\dot{x}\dot{y}$ coordinates mix karta hai — kya yeh cyclicity todta hai?

1. Cyclic test — sirf bare coordinates dhundho. $\partial L/\partial x=0$ aur $\partial L/\partial y=0$: na $x$ aur na $y$ appear karta hai (sirf unki velocities). To dono cyclic hain — student diagnosis mein sahi hai. Yeh step kyun? Cyclicity coordinate ke baare mein hai, velocity ya velocity-couplings ke baare mein kabhi nahi; $b\dot{x}\dot{y}$ mein koi bare $x$ ya $y$ nahi hai.
2. Conjugate momenta (subtle part). Carefully differentiate karo: $p_x=\frac{\partial L}{\partial \dot{x}}=m\dot{x}+b\dot{y},p_y=\frac{\partial L}{\partial \dot{y}}=m\dot{y}+b\dot{x}.$ Yeh step kyun? Coupling $b\dot{x}\dot{y}$ mein $\dot{x}$ hai, to yeh $p_x$ mein contribute karta hai; aur isme $\dot{y}$ bhi hai, to yeh $p_y$ mein bhi contribute karta hai. Trap hai lazily $p_x=m\dot{x}$ likhna.
3. Dono conserved hain. ${\dot{p}}_x=\partial L/\partial x=0$ aur ${\dot{p}}_y=\partial L/\partial y=0$: to $m\dot{x}+b\dot{y}$ aur $m\dot{y}+b\dot{x}$ dono har ek constant hain. Yeh step kyun? Euler–Lagrange $\dot{p}=$ (vanishing $q$-derivative) set karta hai, to har momentum ki rate of change zero hai.
4. Numeric check. $m=1,b=0.5,\dot{x}=2,\dot{y}=4$ ke saath: $p_x=1\cdot 2+0.5\cdot 4=4$; $p_y=1\cdot 4+0.5\cdot 2=5$.

Verify: $x,y$ mein dono partials vanish hote hain, to dono momenta constant hain — diagnosis sahi thi lekin $p$ ke formula ko coupling chahiye thi. $p_x=4$, $p_y=5$. ✔ Dono Noether's theorem (har continuous symmetry ek conserved current deti hai) aur Hamiltonian formalism (jahan yeh $p$'s independent variables ban jaate hain) insist karte hain ki aap $p$ poore $L$ se compute karo, coupling included.

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Recall Quick self-test

Ex 6 mein, $\dot{\varphi }$ pole par kyun blow up karta hai? ::: Kyunki $p_{\varphi }=m{R}^2{\sin }^2\theta \dot{\varphi }$ fixed hai jabki ${\sin }^2\theta \to 0$ hota hai, to $\dot{\varphi }$ ko without bound badhna padta hai — lekin yeh sirf ek coordinate-singularity artefact hai; true speed finite rehti hai. Ex 8 mein, $p_x$ kya hai? ::: $p_x=m\dot{x}+b\dot{y}$, na ki sirf $m\dot{x}$ — coupling contribute karta hai. Ex 5 mein, $p_{\varphi }$ hamesha ke liye kaunsi value leta hai? ::: Zero, kyunki woh zero par shuru hua aur conserved hai.

Connections

• Parent topic — woh rule jise yeh examples exercise karte hain.
• Euler–Lagrange Equations — master equation jise hum tab differentiate karte hain jab ek coordinate cyclic nahi hota (Cells C, G).
• Generalized Momentum — woh quantity jo hum conserve karte hain; Cell H uska poora form velocity coupling ke saath dikhata hai.
• Central Force Motion — Cells B, E, F ka ghar; conserved $p_{\varphi }$ orbit ko ek 1-D radial problem mein reduce kar deta hai.
• Noether's Theorem — woh deep statement ki har continuous symmetry ek conserved quantity deti hai, jiski simplest case hamara cyclic rule hai.
• Hamiltonian Mechanics — jahan cyclic coordinates aapko variables bilkul drop karne dete hain (Routhian reduction).
• Conservation of Energy — time-translation cousin, deliberately ek cyclic-coordinate result nahi.