Cyclic coordinates — corresponding conservation law: Question bank

2.1.8 · D5 · Physics › Analytical Mechanics › Cyclic coordinates — corresponding conservation law

Traps par attack karne se pehle, hum woh do pictures set up karte hain jinpar neeche ke har sawaal ka daromadar hai, taaki koi bhi symbol anjana na lage.

Definition Lagrangian aur woh coordinates jo hum baar baar use karenge

Lagrangian $L$ ek single number hai jo system ki positions aur velocities se bana hai; ise "score sheet" $L = (kinetic energy) - (potential energy)$ samjho jo Nature motion decide karne ke liye padhti hai.

$m$ particle ki mass hai.
Plane mein, $(x, y)$ sadharan left–right / up–down coordinates hain. Dot ka matlab hai "rate of change", toh $\overset{x}{˙}$ sideways speed hai.
Polar coordinates $(r, ϕ)$ usi plane ko alag tarike se describe karte hain: $r$ origin se doori hai aur $ϕ$ ek reference direction se ghoom ke naapa gaya angle hai. Neeche wali picture dekho.

Is page par do Lagrangians baar baar aate hain:

Plane mein free particle: $L = \frac{1}{2} m (\overset{x}{˙}^{2} + \overset{y}{˙}^{2})$ .
Polar form mein central force: $L = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{ϕ}^{2}) - V (r)$ , jahan $V (r)$ sirf doori $r$ par depend karne wala potential hai.

Figure — Cyclic coordinates — corresponding conservation law

Figure dekho: wahi point $P$ $(x, y)$ ya $(r, ϕ)$ se pin kiya gaya hai. Angle $ϕ$ woh cheez hai jo "cycle" karti hai — ise $0$ se $2 π$ tak ghuma do aur tum wapis wohin pahunch jaate ho jahan se shuru kiya tha. Yahi periodicity $ϕ$ ko ek natural cyclic coordinate banati hai, aur yeh bhi warn karti hai ki $ϕ$ aur $ϕ + 2 π$ same physical jagah hain (ek branch subtlety jis par hum Edge cases mein wapas aate hain).

Ab yeh char anchors apne dimaag mein rakho — neeche ke har trap inhi mein se kisi ek ko poke karta hai:

Recall Char anchors (agar topic bhool gaye to open karo)

Cyclic ka matlab hai ki coordinate $q_{k}$ $L$ mein absent hai: $\partial L / \partial q_{k} = 0$ . Uski velocity $\overset{q}{˙}_{k}$ phir bhi appear ho sakti hai.
Jo conserved hota hai woh conjugate momentum $p_{k} = \partial L / \partial \overset{q}{˙}_{k}$ hai — coordinate khud nahi.
Engine hai Euler–Lagrange equation: $\overset{p}{˙}_{k} = \partial L / \partial q_{k}$ . Right side zero ⇒ left side ka rate of change zero.
$L$ mein time ka absent hona ek alag theorem hai — woh energy (Hamiltonian) conserve karta hai, dekho Conservation of Energy aur Hamiltonian Mechanics.

True or false — justify karo

L = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{ϕ}^{2}) - V (r)

: coordinate

ϕ

cyclic hai chahe

\dot{ϕ}

L

mein appear kare.

True. "Cyclic" coordinate

ϕ

ke khud absent hone ke baare mein hai; yahan

ϕ

kabhi appear nahi karta, sirf

\dot{ϕ}

karta hai, toh

\partial L / \partial ϕ = 0

aur

p_{ϕ}

conserved hai.

Agar

q_{k}

cyclic hai, toh

q_{k}

khud time mein constant rehta hai.

False. Jo constant rehta hai woh conjugate momentum

p_{k} = \partial L / \partial \overset{q}{˙}_{k}

hai. Central-force motion mein

ϕ

hamesha ke liye badhta rehta hai, phir bhi

m r^{2} \dot{ϕ}

fixed rehta hai.

Agar velocity

\overset{q}{˙}_{k}

L

mein missing hai, toh

q_{k}

cyclic hai.

False. Cyclic ke liye coordinate

q_{k}

ka absent hona zaroori hai, uski velocity ka nahi. Missing

\overset{q}{˙}_{k}

ek alag (degenerate) situation hai.

Ek cyclic coordinate ka conjugate momentum hamesha constant rehta hai, chahe baaki coordinates kuch bhi kar rahe hon.

True.

\overset{p}{˙}_{k} = \partial L / \partial q_{k} = 0

tab identity hold karta hai jab

q_{k}

absent ho, chahe baaki coordinates kitni bhi wildly evolve karein — wahi constancy bilkul theorem ka content hai.

Agar

L

explicitly time par depend nahi karta, toh koi linear momentum conserved hai.

False. Time-independence energy/Hamiltonian conserve karta hai, jo ek alag theorem hai. Yeh directly kisi bhi conjugate momentum

p_{k}

ke baare mein kuch nahi kehta.

Free particle ke liye

L = \frac{1}{2} m (\overset{x}{˙}^{2} + \overset{y}{˙}^{2})

, dono

p_{x}

aur

p_{y}

conserved hain.

True. Na

x

y

L

mein appear karta hai, toh dono cyclic hain aur dono

m \overset{x}{˙}

aur

m \overset{y}{˙}

constant hain — plane mein poori translational symmetry.

x

par depend karne wala potential

V (x)

add karna

p_{x}

ki conservation kabhi destroy nahi kar sakta.

False. Jab

V

x

par depend karta hai,

\partial L / \partial x = - V^{'} (x) \neq = 0

generally, toh

x

ab cyclic nahi raha aur

p_{x}

change hota hai. Symmetry (aur uska law) potential se tod di jaati hai.

Ek cyclic angular coordinate ka conjugate momentum hamesha physical angular momentum

L_{z}

hota hai.

Mostly, lekin dhyan se padho. Yeh ek plane mein genuine rotation ke liye angular momentum ke barabar hai (

p_{ϕ} = m r^{2} \dot{ϕ}

); generally

p_{ϕ}

jo bhi

\partial L / \partial \dot{ϕ}

evaluate karta hai wahi hota hai, jo physical angular momentum se tab milta hai jab

ϕ

sach mein ek rotation parametrize kare.

Ek charged particle in a magnetic field ke liye,

\partial L / \partial q_{k} = 0

guarantee karta hai ki mechanical momentum

m \overset{q}{˙}_{k}

conserved hai.

False. Velocity-dependent potential ke saath conserved conjugate momentum

p_{k} = m \overset{q}{˙}_{k} + q A_{k}

hai (mechanical plus field part), na ki sirf

m \overset{q}{˙}_{k}

. Poori story ke liye Edge cases dekho.

Error dhundo

"Kyunki

ϕ

L = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{ϕ}^{2}) - V (r)

mein cyclic hai, angular velocity

\dot{ϕ}

constant hai."

Error: conserved quantity

p_{ϕ} = m r^{2} \dot{ϕ}

hai,

\dot{ϕ}

nahi. Jab

r

chota hota hai,

m r^{2} \dot{ϕ}

fixed rakhne ke liye

\dot{ϕ}

badhna chahiye (yahi precisely Kepler's second law hai).

"Gravity projectile par act karti hai, toh koi bhi momentum conserved nahi ho sakta."

Error: gravity sirf vertically act karti hai, toh

x

cyclic rehta hai aur

p_{x} = m \overset{x}{˙}

conserved rehta hai. Sirf vertical momentum change hota hai; horizontal momentum untouched rehta hai.

q_{k}

cyclic hai matlab generalized force

\partial L / \partial q_{k}

q_{k}

ke along nonzero hai."

Error: bilkul ulta hai. Cyclic ka matlab

\partial L / \partial q_{k} = 0

hai, yaani

q_{k}

ke along koi generalized force nahi — wahi vanishing force isliye hai kyunki momentum ko change karne ke liye kuch nahi hai.

"Kyunki

x

cyclic hai, hum

x

ke liye Euler–Lagrange equation drop kar sakte hain; yeh koi information nahi deta."

Error: yeh sabse useful information deta hai — ek first integral

p_{x} = const

. Hum ise drop nahi karte; hum ise problem reduce karne ke liye use karte hain (yahi Routhian reduction ka basis hai Hamiltonian Mechanics mein).

"Time

t

L

mein absent hai, toh

t

ek cyclic coordinate hai aur uska conjugate momentum

p_{t}

conserved hai."

Error:

t

evolution parameter hai, generalized coordinate nahi — differentiate karne ke liye koi

\dot{t}

nahi hai. Time-independence Hamiltonian conserve karta hai, energy-function argument ke through, cyclic-coordinate wale se nahi.

"Bead-on-sphere

L = \frac{1}{2} m R^{2} (\dot{θ}^{2} + sin^{2} θ \dot{ϕ}^{2})

mein

θ

cyclic hai kyunki yeh ek angle hai."

Error:

θ

explicitly

sin^{2} θ

ke through appear karta hai, toh

\partial L / \partial θ \neq = 0

aur

θ

cyclic nahi hai. Sirf

ϕ

(jo kabhi appear nahi karta) cyclic hai. Angle hona na zaroori hai na sufficient. (Edge cases mein neeche bead-on-sphere figure dekho.)

Why questions

"Lagrangian ka

q_{k}

ki parwah na karna" "kuch conserved hai" mein kaise translate hota hai?

Agar

L

tab bhi unchanged hai jab tum

q_{k}

shift karo, toh

q_{k}

ke along push karne ke liye koi generalized force

\partial L / \partial q_{k}

nahi hai; kuch change karne ke liye nahi hone par, us direction mein momentum

p_{k}

frozen ho jaata hai. Symmetry ⇒ conservation (dekho Noether's Theorem).

Conserved quantity

q_{k}

ke conjugate momentum kyun hai, koi aur momentum kyun nahi?

Kyunki coordinate

q_{k}

ke liye Euler–Lagrange padhta hai

\overset{p}{˙}_{k} = \partial L / \partial q_{k}

jahan

p_{k} = \partial L / \partial \overset{q}{˙}_{k}

. Right side zero karne par exactly wahi

p_{k}

freeze hota hai aur koi nahi; yeh pairing equation mein built-in hai.

p_{ϕ} = m r^{2} \dot{ϕ}

mein

r^{2}

(sirf

m \dot{ϕ}

kyun nahi) kyun hai?

Kyunki

p_{ϕ} = \partial L / \partial \dot{ϕ}

hai, aur

\dot{ϕ}

wala sirf term

\frac{1}{2} m r^{2} \dot{ϕ}^{2}

hai; differentiate karne par

m r^{2} \dot{ϕ}

milta hai.

r^{2}

genuine angular momentum ki units aur geometry carry karta hai.

Time-independence ek alag theorem kyun hai cyclic spatial coordinate se?

Ek cyclic coordinate coordinate shift karne ke under symmetry hai aur uska conjugate momentum conserve karta hai. Time-independence clock shift karne ke under symmetry hai aur energy function

h

conserve karta hai — alag symmetry, alag conserved object, alag manipulation se prove hota hai.

Ek cyclic coordinate phir bhi fully solvable equation of motion kyun chhod deta hai, chahe

q_{k}

"gayab" ho gaya ho?

Conserved

p_{k} = const

tumhe constant aur baaki coordinates ke terms mein

\overset{q}{˙}_{k}

algebraically solve karne deta hai, phir ise remaining equations mein feed karo — ignored coordinate ek first integral ke source ke roop mein reappear hota hai, problem ke degrees of freedom reduce ho jaate hain.

Do coordinates ek saath cyclic kyun ho sakte hain, aur usse kya faida hota hai?

L

ke kai coordinates se independent hone par koi rok nahi; har independent absence apna conserved conjugate momentum deta hai. Har conserved momentum ek first integral hai jo problem ki effective dimension ek se kam karta hai.

Ek coordinate non-cyclic lagta hua bhi phir bhi conserved momentum kyun carry kar sakta hai (gauge trap)?

Velocity-dependent potentials (magnetic fields) ke saath, gauge change

A \to A + \nabla χ

L

mein ek total time derivative add karta hai aur

q_{k}

ko explicitly appear kara sakta hai, symmetry chhupa deta hai. Conserved

p_{k}

unchanged rehta hai; sirf mechanical aur field parts mein uski split aur uski visibility gauge ke saath shift hoti hai.

Edge cases

Agar

\partial L / \partial q_{k} = 0

sirf ek instant par hold kare, har time ke liye nahi?

Toh

q_{k}

cyclic nahi hai aur

p_{k}

conserved nahi hai. "Cyclic"

q_{k}

mein ek identity hai —

L

har jagah

q_{k}

se free hona chahiye, sirf jahan

\overset{p}{˙}_{k}

momentarily zero ho wahan nahi.

Plane mein free particle: kya angular momentum conserved hai chahe uski trajectory ek straight line ho, aur kis point ke baare mein?

Haan, chosen origin ke baare mein. Angular momentum origin-dependent hai; us origin par centred polar coordinates mein free-particle

L

mein

ϕ

cyclic hai, toh

p_{ϕ} = m r^{2} \dot{ϕ}

conserved hai. Ek straight line ka kisi bhi fixed point ke baare mein constant angular momentum hota hai, lekin value is baat par depend karti hai ki kaunsa point choose kiya.

Agar

q_{k}

cyclic hai lekin corresponding

\overset{q}{˙}_{k} = 0

t = 0

(particle us direction mein momentarily still hai), toh kya coordinate hamesha wahi rehta hai?

Haan us special case ke liye:

p_{k} = 0

conserved hai, toh

\overset{q}{˙}_{k}

zero rehta hai (assuming

p_{k}

\overset{q}{˙}_{k}

ka simple multiple hai). Lekin yeh ek boundary value hai, general feature nahi — nonzero initial

p_{k}

q_{k}

ko hamesha ke liye chalata rehta hai.

Maano

L

mein koi bhi coordinate appear nahi karta, sirf velocities hain (e.g. free particle). Theorem kya kehta hai?

Har coordinate cyclic hai, toh har conjugate momentum conserved hai. Yeh maximally symmetric case hai: sab momenta constants of motion hain aur trajectory straight-line, uniform hai.

Central motion mein,

r

ke turning points par (jahan

\overset{r}{˙} = 0

), kya

p_{ϕ}

phir bhi conserved hai?

Haan, bilkul.

p_{ϕ} = m r^{2} \dot{ϕ}

har time ke liye conserved hai chahe

\overset{r}{˙}

kuch bhi kare;

r

ke turning points

ϕ

-symmetry se irrelevant hain. Isliye particle equal areas sweep karta hai chahe radially speed up aur slow down ho (dekho Central Force Motion).

Kyunki

ϕ

periodic hai (

ϕ

aur

ϕ + 2 π

same physical point hain), kya "branch cut"

ϕ = 2 π

par uski cyclicity ya

p_{ϕ}

ki conservation break hoti hai?

Nahi. Cyclicity ek statement hai ki

L

mein

ϕ

nahi hai, jo is par unaffected hai ki hum branch ko kaise label karein;

p_{ϕ} = m r^{2} \dot{ϕ}

velocity

\dot{ϕ}

par depend karta hai, jo

ϕ

2 π

past wind hone par smooth rehta hai. Branch cut ek bookkeeping artifact hai, physical discontinuity nahi — actually periodicity precisely isliye hai kyunki

ϕ

cyclic hai.

Velocity-dependent potential (charge

q

in magnetic vector potential

A

L = \frac{1}{2} m \dot{r}^{2} + q \dot{r} \cdot A (r)

. Agar

x

L

mein absent hai, toh kya conserved hai?

Canonical momentum

p_{x} = \partial L / \partial \overset{x}{˙} = m \overset{x}{˙} + q A_{x}

, na ki mechanical

m \overset{x}{˙}

. Field part

q A_{x}

difference carry karta hai. Ek gauge choose karna jahan

A

x

par depend kare

x

ko explicitly appear kara sakta hai aur symmetry chhupa sakta hai, phir bhi physics (conserved

p_{x}

) added total derivative tak gauge-invariant rehta hai.

Agar

L

q_{k}

par sirf ek total time derivative ke through depend kare, e.g.

L \to L^{'} = L + \frac{d}{d t} f (q_{k}) = L + \frac{\partial f}{\partial q _{k}} \overset{q}{˙}_{k}

Equations of motion total time derivative se unchanged rehti hain, toh sirf is tarah appear karne wala coordinate abhi bhi effectively cyclic hai. Concretely naya momentum hai

p_{k}^{'} = \partial L^{'} / \partial \overset{q}{˙}_{k} = p_{k} + \partial f / \partial q_{k}

;

d / d t

lene aur

\partial L^{'} / \partial q_{k} = \partial L / \partial q_{k} + \frac{d}{d t} \frac{\partial f}{\partial q _{k}}

use karne par, extra pieces Euler–Lagrange mein cancel ho jaate hain, toh

\overset{p}{˙}_{k}^{'} = \overset{p}{˙}_{k}

. Conserved quantity sirf constant-along-motion term

\partial f / \partial q_{k}

se shift hoti hai; conservation law survive karta hai.

Ek coordinate generalized coordinates ke ek set mein cyclic hai lekin variables change karne ke baad explicitly appear karta hai. Kaunsa conclusion sahi hai?

Dono consistent hain: cyclicity coordinate choice ki property hai, absolute nahi. Ek clever choice jo ek coordinate ko cyclic banaye ek conserved momentum reveal karti hai; ek poor choice use chhupa deti hai. Underlying symmetry (Noether's Theorem ke through) coordinate-independent hai chahe "cyclic" label nahi ho.

Bead-on-sphere

L = \frac{1}{2} m R^{2} (\dot{θ}^{2} + sin^{2} θ \dot{ϕ}^{2})

: kaunsa angle cyclic hai, aur geometry kya dikhati hai?

Azimuthal

ϕ

(polar axis ke around spin) cyclic hai — yeh kabhi appear nahi karta — toh

p_{ϕ} = m R^{2} sin^{2} θ \dot{ϕ}

conserved hai. Polar

θ

cyclic nahi hai kyunki

sin^{2} θ

\dot{ϕ}^{2}

term ko multiply karta hai. Sphere figure dekho: poori picture ko vertical axis ke around rotate karne se kuch nahi badalta, jo exactly

ϕ

-symmetry hai.

Connections

Euler–Lagrange Equations — yahan ka har "kyun" $\overset{p}{˙}_{k} = \partial L / \partial q_{k}$ par trace back hota hai.
Generalized Momentum — woh object jo sealed hota hai.
Noether's Theorem — coordinate-free reason ki symmetry variable change se survive karti hai.
Central Force Motion — $ϕ$ -cyclic, $p_{ϕ}$ -conserved traps ka ghar.
Conservation of Energy / Hamiltonian Mechanics — time-translation analogue aur reduction machinery.