Noether's theorem — symmetry ↔ conservation law: Worked examples

2.1.9 · D3 · Physics › Analytical Mechanics › Noether's theorem — symmetry ↔ conservation law

Yeh page ek drill hai. Parent note ne machinery banai thi; yahan hum usse har tarah ke case pe apply karte hain, taaki exam ya homework ka koi bhi sawaal anjaana na lage. Hum sirf teen tools reuse karte hain, jo sab parent mein define hain:

Recall Teen tools jinpe hum rely karte hain (parent se)

Conjugate momentum $p_{i} \equiv \frac{\partial L}{\partial q ˙ _{i}}$ — yeh " $L$ ka slope, velocity $\overset{q}{˙}_{i}$ ki direction mein" hai.
Cyclic coordinate: agar $L$ mein $q_{k}$ nahi hai, toh $\frac{\partial L}{\partial q _{k}} = 0$ aur $p_{k}$ frozen hai.
Noether charge: symmetry $q_{i} \to q_{i} + ϵ δ q_{i}$ ke liye, jahan $δ L = ϵ \frac{d F}{d t}$ , conserved quantity hai $Q = \sum_{i} p_{i} δ q_{i} - F .$ Jab $F = 0$ toh yeh bas $\sum_{i} p_{i} δ q_{i}$ hai.

Recall Do aur results jo hum quote karenge (woh bhi parent se)

Euler–Lagrange equation (har coordinate ka equation of motion): $\frac{d}{d t} (\frac{\partial L}{\partial q ˙ _{i}}) - \frac{\partial L}{\partial q _{i}} = 0 ⟺ \frac{d p _{i}}{d t} = \frac{\partial L}{\partial q _{i}} .$ Right-hand form ko aise padho: "momentum ke change ki rate equals $L$ ka us coordinate pe kitna lean karna hai." Agar $L$ $q_{i}$ pe lean nahi karta (coordinate absent hai), toh momentum change nahi ho sakta.
Energy function (time-translation charge): $t \to t + ϵ$ ki conserved quantity hai $h = \sum_{i} p_{i} \overset{q}{˙}_{i} - L,$ aur iski rate identity follow karti hai $\frac{d h}{d t} = - \frac{\partial L}{\partial t}$ . Toh $h$ frozen hota hai exactly jab $L$ mein koi explicit time nahi hota.

Neeche sab kuch vault chain Lagrangian Mechanics → Euler-Lagrange Equation → bade teen Conservation of Energy, Conservation of Momentum, Angular Momentum se feed hota hai.

The scenario matrix

Noether problems ko ek grid ki tarah socho. Har column ek type of symmetry hai; har row ek type of trap hai (ek sign, ek zero, ek degenerate ya limiting case). Humara kaam har cell ko kam se kam ek baar hit karna hai.

Cell	Kyun tricky hai	Kaun sa example hit karta hai
A. Pure cyclic coordinate	turant pehchano " $L$ mein koi $q$ nahi"	Ex 1
*B. Symmetry present hai lekin $L$ lagta hai* coordinate pe depend karta hai**	$δ L = 0$ check karna zaroori, eyeballing se kaam nahi	Ex 2
C. Rotation → angular momentum, cross product ka sign	$δ r = \overset{n}{^} \times r$ sahi karna, $L_{z}$ ka sign**	Ex 3
D. Symmetry broken → conserved NAHI (degenerate case)	explicit time / position dependence conservation khatam kar deta hai	Ex 4
E. Boundary term $F \neq = 0$ (Galilean boost)	$F$ subtract karna zaroori hai, warna wrong charge	Ex 5
F. Canonical $p \neq = m v$ (velocity-dependent term)	charged particle in a field	Ex 6
G. Limiting / zero input	kya hota hai jab parameter $\to 0$ **	Ex 7
H. Real-world word problem	ek physical story ko symmetry mein translate karo	Ex 8
I. Exam-style twist (partial symmetry)	sirf ek component conserved hai	Ex 9
J. Pathological / edge cases	non-invertible $L$ , singular coordinate maps	Ex 10

Dus examples das cells ke liye. Forecast padho aur scroll karne se pehle guess karo.

Ex 1 — Cell A: pure cyclic coordinate

Worked example Polar coordinates mein free particle

Ek plane mein free particle, polar coordinates $(r, ϕ)$ mein likha gaya, $L = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{ϕ}^{2}) .$ Inspection se ek conserved quantity dhundho.

Forecast: $L$ mein kaun sa coordinate missing hai? Frozen quantity guess karo pehle padhne se.

$L$ mein missing coordinate scan karo. $L$ mein $r, \overset{r}{˙}, \dot{ϕ}$ hain — lekin koi bare $ϕ$ nahi. Toh $ϕ$ cyclic hai. Yeh step kyun? $ϕ$ ke liye Euler–Lagrange equation apply karo: $\frac{d}{d t} p_{ϕ} = \frac{\partial L}{\partial ϕ}$ . Kyunki $ϕ$ $L$ mein kabhi appear nahi karta, right side $0$ hai, toh $\frac{d}{d t} p_{ϕ} = 0$ . Missing coordinate spot karna hi symmetry (rotational) spot karna hai.
Conjugate momentum $p_{ϕ}$ compute karo. $p_{ϕ} = \frac{\partial L}{\partial ϕ ˙} = m r^{2} \dot{ϕ} .$ Yeh step kyun? Shift $ϕ \to ϕ + ϵ$ ke liye Noether charge $p_{ϕ} \cdot δ ϕ$ hai, jahan $δ ϕ = 1$ , yaani bas $p_{ϕ}$ .
Conservation law state karo. $m r^{2} \dot{ϕ} = const$ — yeh angular momentum $L_{z}$ hai.

Verify: Units: $kg \cdot m^{2} \cdot s^{- 1} = J\cdots$ , angular momentum ke liye sahi. Sanity: bhale hi $r$ $L$ mein appear karta hai, $r$ cyclic nahi hai (ek $r^{2} \dot{ϕ}^{2}$ term hai), toh $p_{r}$ conserved nahi hai — sirf $p_{ϕ}$ . Yahi asymmetry poora point hai.

Ex 2 — Cell B: symmetry ek coordinate ke peeche chupi hai jo lagta hai use ho raha hai

Worked example Spring se jude do particles

$L = \frac{1}{2} m \overset{x}{˙}_{1}^{2} + \frac{1}{2} m \overset{x}{˙}_{2}^{2} - \frac{1}{2} k (x_{1} - x_{2})^{2} .$ Dono coordinates $x_{1}$ aur $x_{2}$ $L$ mein appear karte hain. Kya kuch conserved hai?

Forecast: Koi bhi coordinate apne aap cyclic nahi hai. Kya iska matlab hai kuch bhi conserved nahi hai? Guess karo.

Translation symmetry try karo $δ x_{1} = 1, δ x_{2} = 1$ (dono particles ko equally shift karo). Yeh step kyun? Potential sirf difference $x_{1} - x_{2}$ pe depend karta hai. Dono ko same amount shift karne se $x_{1} - x_{2}$ unchanged rehta hai — yahi symmetry hai, bhale hi har coordinate individually appear karta ho.
$δ L = 0$ check karo. $δ L = \frac{\partial L}{\partial x _{1}} (1) + \frac{\partial L}{\partial x _{2}} (1) = - k (x_{1} - x_{2}) (1) - k (x_{2} - x_{1}) (1) = 0. ✓$ Yeh step kyun? Noether ko $δ L = 0$ chahiye (yahan $F = 0$ ). " $x_{1}$ appear karta hai" eyeballing galat conclusion dega ki koi symmetry nahi; honest check pakad leta hai.
Charge banao $Q = p_{1} δ x_{1} + p_{2} δ x_{2} = m \overset{x}{˙}_{1} + m \overset{x}{˙}_{2} = P_{total}$ .

Verify: $\frac{d P}{d t} = m \overset{x}{¨}_{1} + m \overset{x}{¨}_{2}$ . Equations of motion: $m \overset{x}{¨}_{1} = - k (x_{1} - x_{2})$ , $m \overset{x}{¨}_{2} = - k (x_{2} - x_{1}) = + k (x_{1} - x_{2})$ . Sum $= 0$ . Toh total momentum genuinely constant hai. ✓ (Internal spring center of mass ko push nahi kar sakta — Newton's third law symmetry ke roop mein reborn.)

Ex 3 — Cell C: rotation, cross-product sign

Worked example 2D mein central-force particle

$L = \frac{1}{2} m (\overset{x}{˙}^{2} + \overset{y}{˙}^{2}) - V (x^{2} + y^{2}) .$ Potential sirf distance $r = x^{2} + y^{2}$ pe depend karta hai. Rotational symmetry se conserved charge dhundho, signs ke baare mein careful rehte hue.

Forecast: Charge $+ (x \overset{y}{˙} - y \overset{x}{˙})$ hoga ya $- (x \overset{y}{˙} - y \overset{x}{˙})$ ? Abhi sign guess karo.

Figure kya dikhata hai: dashed lavender circle radius $r$ ka ek orbit hai; mint dots uspe sample points hain; har coral arrow us point ka displacement $δ r$ hai ek tiny rotation ke under. Notice karo ki har arrow circle ke along point karta hai, counter-clockwise sense mein, aur slate radius line ke perpendicular hai — yahi $δ r = \overset{z}{^} \times r = (- y, x)$ ka geometric meaning hai.

Infinitesimal rotation likho. $\overset{z}{^}$ ke baare mein small angle $ϵ$ ka rotation $r \to R r$ bhejta hai, giving $δ r = \overset{z}{^} \times r$ . Components mein (figure mein har coral arrow): $δ x = - y, δ y = + x .$ Yeh step kyun? Hume symmetry ki shape $δ q_{i}$ chahiye. Cross product $\overset{z}{^} \times r = (- y, x)$ ek point ki velocity hai jo counter-clockwise spin ho raha hai — yahi "poori picture rotate karo" locally matlab hai.
$δ L = 0$ check karo. Kinetic part: $δ \overset{x}{˙} = - \overset{y}{˙}, δ \overset{y}{˙} = \overset{x}{˙}$ , toh $m (\overset{x}{˙} δ \overset{x}{˙} + \overset{y}{˙} δ \overset{y}{˙}) = m (- \overset{x}{˙} \overset{y}{˙} + \overset{y}{˙} \overset{x}{˙}) = 0$ . Potential part: $r$ rotation se unchanged hai, toh $δ V = 0$ . Total $δ L = 0$ . ✓ Yeh step kyun? Rotational invariance tabhi real hai jab dono pieces invariant hon. Distance-only potentials automatically pass karte hain.
Charge banao. $Q = p_{x} δ x + p_{y} δ y = m \overset{x}{˙} (- y) + m \overset{y}{˙} (x) = m (x \overset{y}{˙} - y \overset{x}{˙}) = L_{z} .$ Toh sign positive $m (x \overset{y}{˙} - y \overset{x}{˙})$ hai.

Verify: $r = (1, 0)$ pe particle lo jo $v = (0, 1)$ se move kar raha hai (counter-clockwise), $m = 1$ . Toh $L_{z} = 1 \cdot (1 \cdot 1 - 0 \cdot 0) = + 1 > 0$ , counter-clockwise motion ke liye correctly positive — figure mein arrow direction se match karta hai. Units $kg\cdotm^{2} / s$ . ✓

Ex 4 — Cell D: symmetry BROKEN, kuch conserved nahi (degenerate case)

Worked example Driven oscillator — classic "energy conserved nahi" trap

$L = \frac{1}{2} m \overset{x}{˙}^{2} - \frac{1}{2} k x^{2} + x F_{0} cos (ω t) .$ Ek spring plus ek external push jo time ke saath vary karta hai. Kya energy conserved hai?

Forecast: Intro physics kehta hai "energy hamesha conserved hoti hai." Kya yahan hai? Check karne se pehle guess karo.

Time-translation symmetry test karo $t \to t + ϵ$ . Yeh symmetry tabhi hai jab $L$ mein koi explicit $t$ nahi ho. Yeh step kyun? Energy function $h = \sum_{i} p_{i} \overset{q}{˙}_{i} - L$ (upar box mein recalled) time-translation ka Noether charge hai, aur iski rate $\frac{d h}{d t} = - \frac{\partial L}{\partial t}$ hai. Koi time-symmetry nahi ⇒ $h$ ke frozen hone ki koi wajah nahi.
Explicit $t$ dhundho. Term $x F_{0} cos (ω t)$ mein $t$ directly hai. Toh $\frac{\partial L}{\partial t} = - x F_{0} ω sin (ω t) \neq = 0$ . Yeh step kyun? Yeh akela non-zero partial derivative poora verdict hai.
Energy ke change ki actual rate compute karo. Recall box ki identity use karte hue, $\frac{d h}{d t} = - \frac{\partial L}{\partial t} = + x F_{0} ω sin (ω t) \neq = 0.$ Energy conserved nahi hai: driver energy andar aur bahar pump karta hai.

Verify: Drive band karo, $F_{0} = 0$ . Toh $\partial L / \partial t = 0$ aur $d h / d t = 0$ — energy conservation wapas aa jati hai. Yeh sahi limiting behaviour hai: broken symmetry exactly tab heal hoti hai jab time-dependent term vanish ho. ✓

Ex 5 — Cell E: boundary term $F \neq = 0$ (Galilean boost)

Worked example Velocity boost ke under free particle

$L = \frac{1}{2} m \overset{x}{˙}^{2} .$ Boost $x \to x + ϵ t$ apply karo (poore system ko ek small extra velocity $ϵ$ do). Conserved charge dhundho — naive Noether formula jhooth bolega agar $F$ include nahi kiya.

Forecast: Guess karo: kya naive charge $p δ x$ apne aap conserved hoga, ya hume correction chahiye?

$δ x$ read off karo. $x \to x + ϵ t$ matlab $δ x = t$ (aur $δ \overset{x}{˙} = 1$ , kyunki $\frac{d}{d t} (ϵ t) = ϵ$ ). Yeh step kyun? Boost ki "shape" time ke saath badhti hai — yahi isse plain shift se alag karta hai.
$δ L$ compute karo — yeh ZERO nahi hai. $δ L = \frac{\partial L}{\partial x ˙} δ \overset{x}{˙} = m \overset{x}{˙} \cdot 1 = m \overset{x}{˙} = \frac{d}{d t} (m x) .$ Toh $δ L = \frac{d F}{d t}$ jahan $F = m x$ . Yeh step kyun? Action sirf ek boundary term se change hota hai, toh symmetry abhi bhi valid hai — lekin parent ka formula demand karta hai ki hum $F$ subtract karen.
Corrected charge banao. $Q = p δ x - F = (m \overset{x}{˙}) (t) - m x = m (\overset{x}{˙} t - x) .$

Verify: Free particle ke liye $\overset{x}{˙} =$ const $\Rightarrow x = x_{0} + \overset{x}{˙} t$ . Toh $Q = m (\overset{x}{˙} t - x_{0} - \overset{x}{˙} t) = - m x_{0} = const$ . ✓ Yeh literally starting position $x_{0}$ encode karta hai — boost symmetry "center of mass $t = 0$ pe kahan tha" conserve karta hai. Agar $F$ bhool jaate, toh naive $p δ x = m \overset{x}{˙} t$ linearly grow karta aur nonsense hota.

Ex 6 — Cell F: canonical $p \neq = m v$ (charged particle)

Worked example Uniform magnetic field mein charged particle

Vector potential $A_{x} = - \frac{1}{2} B y, A_{y} = \frac{1}{2} B x$ (uniform field $B \overset{z}{^}$ ) ke saath, $L = \frac{1}{2} m (\overset{x}{˙}^{2} + \overset{y}{˙}^{2}) + q (\overset{x}{˙} A_{x} + \overset{y}{˙} A_{y}) = \frac{1}{2} m (\overset{x}{˙}^{2} + \overset{y}{˙}^{2}) + \frac{q B}{2} (x \overset{y}{˙} - y \overset{x}{˙}) .$ Rotational symmetry se conserved charge dhundho — aur dhyan rakho " $p$ " ka matlab kya hai.

Forecast: Kya conserved quantity plain $m (x \overset{y}{˙} - y \overset{x}{˙})$ hogi, ya kuch extra $B$ -term ke saath? Guess karo.

Rotation invariance check karo. $δ x = - y, δ y = x$ ke under, kinetic term invariant hai (Ex 3) aur term $x \overset{y}{˙} - y \overset{x}{˙}$ khud $r \times v$ ka $z$ -component hai, jo rotation-invariant hai. Toh $δ L = 0$ . ✓ Yeh step kyun? $\overset{z}{^}$ ke along uniform field mein $\overset{z}{^}$ ke baare mein full rotational symmetry hai.
Canonical momenta use karo, $m \overset{x}{˙}$ nahi. $p_{x} = \frac{\partial L}{\partial x ˙} = m \overset{x}{˙} - \frac{q B}{2} y, p_{y} = \frac{\partial L}{\partial y ˙} = m \overset{y}{˙} + \frac{q B}{2} x .$ Yeh step kyun? Parent mein Mistake 2: Noether charge $p_{i} = \partial L / \partial \overset{q}{˙}_{i}$ use karta hai, jo yahan vector-potential piece carry karta hai.
Charge assemble karo. $Q = p_{x} δ x + p_{y} δ y = (m \overset{x}{˙} - \frac{q B}{2} y) (- y) + (m \overset{y}{˙} + \frac{q B}{2} x) (x)$ $= m (x \overset{y}{˙} - y \overset{x}{˙}) + \frac{q B}{2} (x^{2} + y^{2}) .$

Verify: $q = 0$ set karo: $Q \to m (x \overset{y}{˙} - y \overset{x}{˙}) = L_{z}$ , ordinary angular momentum — sahi limit. $B \neq = 0$ ke saath, conserved "canonical angular momentum" $\frac{q B}{2} r^{2}$ term gain karta hai, yahi wajah hai ki cyclotron orbits ka ek fixed guiding center hota hai. Units: har term $kg\cdotm^{2} / s$ hai (note $[q B] = kg/s$ , times $m^{2}$ ). ✓

Ex 7 — Cell G: limiting / zero input

Worked example Vanishing torque ki limit mein cyclic angle

Potential $L = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{ϕ}^{2}) - α ϕ$ mein ek particle, jahan $α$ ek small constant "torque" parameter hai. $p_{ϕ}$ ke conservation ko $α \to 0$ ke saath study karo.

Forecast: $α \neq = 0$ ke saath coordinate $ϕ$ $L$ mein appear karta hai. Kya $p_{ϕ}$ conserved hai? $α \to 0$ par kya hota hai?

Test karo ki $ϕ$ cyclic hai ya nahi. $L$ mein term $- α ϕ$ hai, toh $\frac{\partial L}{\partial ϕ} = - α \neq = 0$ ( $α \neq = 0$ ke liye). Cyclic nahi. Yeh step kyun? Coordinate ki koi bhi explicit appearance exact shift symmetry tod deti hai.
$ϕ$ ke liye Euler–Lagrange. $\frac{d}{d t} p_{ϕ} = \frac{\partial L}{\partial ϕ}$ use karte hue, $\frac{d}{d t} p_{ϕ} (m r^{2} \dot{ϕ}) = - α \Rightarrow \frac{d p _{ϕ}}{d t} = - α .$ Yeh step kyun? Yeh breakage ko quantify karta hai: $p_{ϕ}$ constant rate $- α$ pe drift karta hai, jaise ek constant applied torque.
Limit $α \to 0$ lo. Toh $\frac{d p _{ϕ}}{d t} \to 0$ , aur $p_{ϕ} = m r^{2} \dot{ϕ}$ phir se conserved ho jaata hai.

Verify: Integrate karo: $p_{ϕ} (t) = p_{ϕ} (0) - α t$ . $α = 0$ par yeh ek flat line (constant) hai — symmetry, aur uska conservation law, limit mein continuously recover hota hai. Yahi honest tarika hai "zero input" ke behave karne ka: koi discontinuity nahi, bas torque switch off ho raha hai. ✓

Ex 8 — Cell H: real-world word problem

Worked example Arms andar kheenchti ice-skater

Ek skater freely spin kar rahi hai (koi friction torque nahi). Woh arms andar kheenchti hai, moment of inertia $I_{0}$ se $I_{0} /2$ ho jaata hai. Starting angular speed $ω_{0} = 3 rad/s$ . Uski final angular speed kya hai, aur kaun si symmetry answer guarantee karti hai?

Forecast: Koi algebra karne se pehle uski final spin rate guess karo.

Symmetry identify karo. Ice frictionless hai aur space uske around empty hai: physics mein kuch bhi vertical ke baare mein rotation ki preferred direction pick nahi karta → $\overset{z}{^}$ ke baare mein rotational symmetry. Yeh step kyun? Noether: rotational symmetry ⇒ angular momentum $L_{z}$ conserved. Hum yahi law use karenge, "energy" nahi (woh arms kheenchne mein kaam karti hai, toh energy conserved nahi hai).
$L_{z} = I ω = const$ likho. $I_{0} ω_{0} = \frac{I _{0}}{2} ω_{f} .$ Yeh step kyun? $L_{z} = I ω$ Ex 1 ke single-particle Noether charge $m r^{2} \dot{ϕ}$ ka rigid-body version hai, poore body pe summed. Conservation matlab pehle ki value equals baad ki value.
$ω_{f}$ solve karo. $I_{0}$ cancel karo: $ω_{0} = \frac{1}{2} ω_{f}$ , toh $ω_{f} = 2 ω_{0} = 2 \times 3 = 6 rad/s .$ Yeh step kyun? Inertia ko aadha karne se spin double hona chahiye taaki product $I ω$ fixed rahe — familiar skater speed-up.

Verify: Angular momentum pehle $= I_{0} \cdot 3 = 3 I_{0}$ ; baad mein $= \frac{I _{0}}{2} \cdot 6 = 3 I_{0}$ — equal. ✓ Energy pe sanity: $K E = \frac{1}{2} I ω^{2}$ $\frac{1}{2} I_{0} (3)^{2} = 4.5 I_{0}$ se $\frac{1}{2} \cdot \frac{I _{0}}{2} (6)^{2} = 9 I_{0}$ ho jaata hai — yeh double ho gaya, extra energy uske muscles ne supply ki. Energy conserved nahi, angular momentum hai. Yahi difference Noether predict karta hai (sirf symmetric quantity frozen hoti hai). ✓

Ex 9 — Cell I: exam twist, sirf EK class of component conserved

Worked example Uniform gravitational field mein particle

$L = \frac{1}{2} m (\overset{x}{˙}^{2} + \overset{y}{˙}^{2} + \overset{z}{˙}^{2}) - m g z, \overset{z}{^} upar .$ Kaun se momentum components conserved hain, aur kaun se nahi? (Classic "partial symmetry" trap.)

Forecast: Guess karo $p_{x}, p_{y}, p_{z}$ mein se kitne conserved hain.

Figure kya dikhata hai: mint double-arrow ek horizontal shift hai — picture ko left/right slide karne se height energy shading (lavender bands) nahi badalti, toh horizontal translation genuine symmetry hai. Coral vertical double-arrow particle ko $m g z$ background ki ek alag shade pe le jaata hai, yaani $L$ change ho jaata hai — toh vertical translation symmetry nahi hai.

Har coordinate scan karo. $L$ mein $z$ hai ( $- m g z$ mein) lekin $x$ ya $y$ nahi. Yeh step kyun? Cyclic = "coordinate absent". Sirf woh coordinates jo $L$ mein absent hain conserved momenta dete hain.
$x, y$ ke liye conclude karo. Dono cyclic hain ⇒ $\frac{d}{d t} p_{i} = \frac{\partial L}{\partial q _{i}} = 0$ se, $p_{x} = m \overset{x}{˙} = const, p_{y} = m \overset{y}{˙} = const .$ Yeh step kyun? Horizontal translational symmetry gravity se unbroken hai, toh horizontal momentum conserved hai.
$z$ ke liye conclude karo. $\frac{\partial L}{\partial z} = - m g \neq = 0$ , toh $\frac{d p _{z}}{d t} = \frac{\partial L}{\partial z} = - m g \neq = 0.$ Vertical momentum conserved nahi — gravity vertical translation symmetry tod deti hai.

Verify: Yeh projectile motion hai: $\overset{x}{˙}, \overset{y}{˙}$ constant (horizontal plane mein flat trajectory), jabki $p_{z} = m \overset{z}{˙}$ rate $m g$ se decrease karta hai (yahi $F = - m g$ hai). $g = 0$ set karo aur teeno conserved ho jaate hain — free particle. Conserved momenta ki sankhya = unbroken translation directions ki sankhya. ✓

Ex 10 — Cell J: pathological / edge cases

Worked example Jab machinery khud wobble kare

Noether ki recipe do quiet cheezein assume karti hai: ki $L$ well-defined momenta deta hai, aur ki coordinate map smooth aur reversible hai. Do short cases dikhate hain kya hota hai jab woh assumptions fail hoti hain.

Forecast: Neeche har case mein, kya conserved charge abhi bhi exist karta hai? Padhne se pehle guess karo.

Case (i): non-invertible ("singular") Lagrangian. $L = \overset{x}{˙} y - \overset{y}{˙} x .$

Momenta compute karo. $p_{x} = \frac{\partial L}{\partial x ˙} = y$ , $p_{y} = \frac{\partial L}{\partial y ˙} = - x$ . Yeh step kyun? Inme $\overset{x}{˙}, \overset{y}{˙}$ bilkul bhi nahi hain — tum unhe velocities ke liye solve nahi kar sakte. Yahi exactly "non-invertible / singular $L$ " ka matlab hai: map (velocities) $\to$ (momenta) one-to-one nahi hai.
Kya abhi bhi symmetry charge hai? Rotation $δ x = - y, δ y = x$ : $δ L$ check karo. Pata chalta hai $δ L = 0$ (form rotation-invariant hai), aur Noether charge hai $Q = p_{x} δ x + p_{y} δ y = y (- y) + (- x) (x) = - (x^{2} + y^{2})$ . Yeh step kyun? Noether ka charge formula abhi bhi ek quantity output karta hai, lekin kyunki $L$ singular hai, yeh object ek constraint hai ( $p_{x} - y = 0$ , $p_{y} + x = 0$ ), ordinary evolution law nahi. Aise systems ko Hamiltonian Mechanics with constraints (Dirac's method) chahiye, naive $p = \partial L / \partial \overset{q}{˙}$ story nahi.
Lesson: formula survive karta hai, interpretation badal jaati hai — hamesha check karo ki momenta actually velocities pe depend karte hain ya nahi, "conserved" charge trust karne se pehle.

Case (ii): singular coordinate transformation. Ex 1 ke polar coordinates par wapas jaao. Origin $r = 0$ par, angle $ϕ$ undefined hai aur map $(r, ϕ) \to (x, y)$ invertible nahi hai (uska Jacobian $r$ vanish karta hai).

Kya toota? Conserved $p_{ϕ} = m r^{2} \dot{ϕ}$ motion ke saath perfectly conserved rehta hai — lekin tum $r = 0$ par $ϕ$ read off nahi kar sakte. Yeh step kyun? Exactly origin se guzarta particle $r = 0$ hai, toh $p_{ϕ} = m \cdot 0 \cdot \dot{ϕ} = 0$ ; coordinate singularity charge ko $0$ force karti hai wahan, jo center se seedhi line ke zero angular momentum carry karne ke consistent hai.
Lesson: coordinate singularity conservation law destroy nahi karti; sirf us chart ko motion describe karne ki ek buri jagah banati hai. Cartesian $x, y$ switch karo (har jagah smooth) aur $L_{z} = m (x \overset{y}{˙} - y \overset{x}{˙})$ finite aur conserved rehta hai.

Verify: Case (i): $\frac{d}{d t} (- (x^{2} + y^{2})) = - 2 (x \overset{x}{˙} + y \overset{y}{˙})$ . Is $L$ ke constrained equations of motion $\overset{x}{˙} = x, \overset{y}{˙} = y$ force nahi karte; balki constraints $x, y$ ko pin karte hain taaki $x \overset{x}{˙} + y \overset{y}{˙} = 0$ , $Q$ constant rakha jaaye — charge consistent hai. Case (ii): $r = 0$ par, $p_{ϕ} = 0$ aur Cartesian $L_{z} = m (x \overset{y}{˙} - y \overset{x}{˙})$ evaluated at $x = y = 0$ bhi $0$ hai; dono descriptions agree karte hain. ✓

Recall Self-test clozes

Driven oscillator energy tabhi conserve karta hai limit mein ::: $F_{0} \to 0$ (drive off), time-translation symmetry restore hoti hai. Boost $x \to x + ϵ t$ ke liye, conserved charge hai ::: $Q = m (\overset{x}{˙} t - x) = - m x_{0}$ , initial position. Charged particle ka conserved "angular momentum" extra term gain karta hai ::: $\frac{q B}{2} (x^{2} + y^{2})$ , kyunki $p_{i} = \partial L / \partial \overset{q}{˙}_{i}$ vector potential include karta hai. Uniform gravity mein conserved momenta hain ::: $p_{x}$ aur $p_{y}$ (horizontal), kabhi $p_{z}$ nahi. Agar momenta velocities contain nahi karte (singular $L$ ), Noether output hai ::: ek constraint, evolution law nahi — constrained Hamiltonian methods use karo.

In symmetries ki deeper structure Symmetry Groups & Lie Algebras aur Gauge Symmetry mein hai; energy charge Hamiltonian Mechanics se connect hota hai.