Noether's theorem — symmetry ↔ conservation law: Question bank

2.1.9 · D5 · Physics › Analytical Mechanics › Noether's theorem — symmetry ↔ conservation law

True or false — justify

Reflection (parity) ek symmetry hai, toh iska ek conserved Noether charge zaroor hona chahiye.

False. Noether ko ek continuous parameter

ϵ

chahiye jise differentiate kiya ja sake; reflection discrete hai, toh koi

d / d ϵ

nahi banta jisse charge banaya ja sake — yeh selection rules deta hai, conserved quantities nahi.

Agar Lagrangian

L

kisi transformation ke under change ho jaata hai, toh action invariant nahi ho sakta.

False.

L

total time derivative

ϵ d F / d t

se change ho sakta hai; tab

\int d L d t

sirf endpoint values hote hain, toh action phir bhi invariant rehta hai aur charge ban jaata hai

Q = \sum_{i} p_{i} δ q_{i} - F

Energy har isolated mechanical system mein conserved hoti hai.

False. Energy (function

h = \sum_{i} p_{i} \overset{q}{˙}_{i} - L

) tab conserved hoti hai jab

L

mein koi explicit time dependence nahi hoti. Time-dependent potential time-translation symmetry tod deta hai aur energy drift karne lagti hai.

Ek cyclic coordinate hamesha

m \overset{q}{˙}

quantity conserve karta hai.

False. Yeh canonical momentum

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

conserve karta hai, jo

m \overset{q}{˙}

ke barabar hota hai sirf simple free particles ke liye; magnetic field ke saath

p_{x} = m \overset{x}{˙} + q A_{x}

hota hai, aur curvilinear coordinates mein yeh bilkul alag dikhta hai.

Agar

L

explicitly time par depend nahi karta, toh

h = T + V

hamesha hota hai.

False.

h = T + V

tab hota hai jab kinetic energy

T

velocities ka quadratic function ho (Euler's theorem on homogeneous functions ke through). Time-dependent constraints ya velocity-linear terms yeh tod dete hain even jab

\partial L / \partial t = 0

ho.

Equations of motion ki har continuous symmetry ek Noether charge deti hai.

False. Noether's theorem ko action ki invariance chahiye (ya

L

ki boundary term tak), jo zyada strong hai. Kuch symmetries jo EL equations ki hain, action ko invariant nahi chhod'tein aur koi conserved charge nahi deti.

Ek free particle ke liye, linear aur angular momentum dono ek saath conserved hote hain.

True. Free-particle

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

mein koi

x, y

dependence nahi (translation symmetry →

p

) aur yeh rotationally invariant hai (rotation symmetry →

L_{z}

), toh teeno charges ek saath hold karte hain.

Spot the error

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

ka matlab hai

q_{k}

kabhi change nahi karta."

Coordinate

q_{k}

phir bhi evolve karta hai; jo constant hai woh hai uska conjugate momentum

p_{k}

. Ek cyclic coordinate kehta hai

L

q_{k}

par depend nahi karta, yeh nahi ki

q_{k}

frozen hai.

"Space-translation invariance ka matlab hai

L

bilkul bhi position par depend nahi karta."

Iska matlab hai

L

sirf relative positions

r_{a} - r_{b}

par depend karta hai, absolute origin par nahi. Center-of-mass coordinate cyclic hai; internal separations phir bhi

L

mein aa sakti hain.

"Time shift ke liye,

δ q_{i} = 0

kyunki time external hai, coordinate nahi."

Time translation ke liye trajectory aage carry hoti hai, toh

δ q_{i} = \overset{q}{˙}_{i}

hota hai. Ise (boundary term

F = L

ke saath) Noether mein daalne par exactly energy function

h

milta hai.

Q = \sum_{i} \overset{q}{˙}_{i} δ q_{i}

Noether charge hai."

Charge mein canonical momentum

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

use hota hai, velocity nahi:

Q = \sum_{i} p_{i} δ q_{i}

\overset{q}{˙}_{i}

use karne par mass aur koi bhi field terms drop ho jaate hain.

"Rotation deta hai

Q = \sum_{a} p_{a} \cdot (\overset{n}{^} \times r_{a})

, aur yahi final answer hai."

Yeh sahi hai lekin scalar-triple-product identity se simplify karna chahiye taaki

\overset{n}{^} \cdot \sum_{a} (r_{a} \times p_{a}) = \overset{n}{^} \cdot L_{total}

mile — jo reveal karta hai ki yeh

\overset{n}{^}

ke baare mein angular momentum hai.

"Humne Noether prove karne ke liye Euler–Lagrange use kiya, toh Noether sirf motion ke end par hold karta hai."

EL physical path par har instant par hold karta hai, toh substitution

\partial L / \partial q_{i} = \frac{d}{d t} (\partial L / \partial \overset{q}{˙}_{i})

throughout valid hai — charge

Q

saare

t

ke liye constant hai, sirf kisi special moment par nahi.

"Kyunki

δ L = \sum_{i} (\partial L / \partial q_{i} δ q_{i} + \partial L / \partial \overset{q}{˙}_{i} δ \overset{q}{˙}_{i})

chain rule se aaya, iske liye

δ q

small chahiye lekin

δ \overset{q}{˙} = \frac{d}{d t} δ q

nahi."

Relation

δ \overset{q}{˙}_{i} = \frac{d}{d t} δ q_{i}

essential hai — yahi Step 3 mein product rule recognize karne aur sab kuch ek

\frac{d}{d t}

mein collapse karne mein madad karta hai. Iske bina charge kabhi appear nahi hoga.

Why questions

Noether's theorem ko symmetry continuous kyun chahiye, koi bhi symmetry kyun nahi chalti?

Kyunki proof mein ek infinitesimal transformation

q_{i} \to q_{i} + ϵ δ q_{i}

liya jaata hai aur first order mein

ϵ

tak expand kiya jaata hai; tumhein ek smooth knob

ϵ \to 0

chahiye differentiate karne ke liye. Discrete symmetries mein aisa koi infinitesimal generator nahi hota.

"Symmetry" ek conserved quantity mein kyun translate hoti hai, kisi aur cheez mein kyun nahi?

Symmetry ka matlab hai

δ L = 0

(ya ek boundary term), aur algebra force karta hai

δ L = \frac{d}{d t} \sum_{i} p_{i} δ q_{i}

. Use zero set karne par kehta hai ek specific combination ka change rate zero hai — yaani yeh conserved hai.

Energy charge woh special kyun hai jise boundary term

F = L

ki zarurat hai?

Time shift ke under Lagrangian strictly invariant nahi hota; yeh apni total derivative

δ L = d L / d t

se change hota hai. Yeh ek boundary term hai, toh hum

F = L

subtract karte hain, jisse

h = \sum_{i} p_{i} \overset{q}{˙}_{i} - L

milta hai instead of naive

\sum_{i} p_{i} δ q_{i}

Ek magnetic (velocity-dependent) potential phir bhi momentum conservation respect kyun kar sakta hai jabki

p \neq = m \overset{x}{˙}

Conserved object canonical momentum

p_{x} = m \overset{x}{˙} + q A_{x}

hai. Translation invariance is poore expression ko conserve karta hai, even though mechanical momentum

m \overset{x}{˙}

akela conserved nahi hai.

Rotational symmetry

\overset{n}{^}

ke baare mein sirf component

\overset{n}{^} \cdot L

kyun conserve karta hai, poora

L

kyun nahi?

Symmetry usi ek axis

\overset{n}{^}

ke baare mein rotation ke under invariance hai. Noether tumhein charge us specific direction ke along deta hai jiske baare mein tumne rotate kiya; dusre components tab conserved hote hain jab woh rotations bhi symmetries hon.

Cyclic coordinate Noether's theorem ka "cheap" version kyun hai?

Ek cyclic coordinate precisely

q_{k}

mein translation symmetry hai (

q_{k} \to q_{k} + ϵ

shift karne se

L

unchanged rehta hai), toh Euler–Lagrange turant deta hai

d p_{k} / d t = 0

— poori Noether machinery ek line mein collapse ho jaati hai.

Edge cases

Noether charge ka kya hota hai jab parameter

ϵ

transformation mein appear karta hai lekin

δ L \neq = 0

aur yeh total time derivative nahi hai?

Koi conserved charge nahi hoga — transformation simply action ki symmetry nahi hai. Noether tab hi bolta hai jab

δ L = 0

δ L = ϵ d F / d t

ho.

Ek pendulum jiska support upar-neeche

y_{0} (t) = A cos ω t

se driven hai: kya energy conserved hai?

Nahi. Constraint mein explicit

t

hone se

\partial L / \partial t \neq = 0

ho jaata hai, time-translation symmetry tod'ta hai; mechanical energy oscillate karti hai jab drive energy andar-bahar pump karta hai.

Ek particle uniform gravitational field

V = m g y

mein

x

–

y

plane mein move kar raha hai: kaun se momenta survive karte hain?

L

mein koi

x

nahi, toh

p_{x} = m \overset{x}{˙}

conserved hai (horizontal translation symmetry), lekin

y

V

mein appear karta hai, toh

p_{y}

conserved nahi hai — gravity ek direction pick karti hai aur vertical translation tod'ti hai.

Ek perfectly free particle empty space mein rest par (

\overset{q}{˙}_{i} = 0

): kya rotation charge

L_{z} = m (x \overset{y}{˙} - y \overset{x}{˙})

phir bhi "conserved" hai?

Haan, trivially — yeh zero ke barabar hai aur zero rehta hai. Conservation ka matlab hai rate of change zero hai; jo charge constantly

0

hai woh ek perfectly valid conserved quantity hai (degenerate lekin legitimate case).

Agar kisi system mein symmetry hai lekin trajectory aisa hai ki

δ q_{i} = 0

everywhere ho (jaise exactly axis par baithe particle ka rotation), toh kya Noether fail hota hai?

Fail nahi hota; charge

Q = \sum_{i} p_{i} δ q_{i}

us trajectory ke liye sirf

0

hai aur

0

rehta hai. Theorem ki guarantee (time mein constant) phir bhi hold karti hai — constant sirf zero happen karta hai.

Kya ek symmetry jo sirf ek instant par hold kare (poore

t

ke liye nahi) ek conserved charge deti hai?

Nahi. Invariance poore path ke along hold karni chahiye taaki

\frac{d}{d t} Q = 0

saare

t

ke liye ho. Ek ek-instant coincidence se kuch conserved nahi milta.

Recall Har trap ki one-line summary

Noether ko ek continuous symmetry of the action chahiye (boundary terms allow karke), charge canonical momenta use karta hai, aur har conserved quantity tab tak survive karti hai jab tak uski specific symmetry unbroken ho — energy explicit time dependence se, $p_{y}$ gravity se, etc.