Constraints using Lagrange multipliers — Hinglish notes

2.1.10 · Physics › Analytical Mechanics

WHY do we need this method?

WHY yeh hai: agar kabhi tumhe poochna ho "bead hoop se kab ude gi?" ya "tension kya hai?", to embedded method directly jawab nahi de sakti. Multiplier method de sakti hai.

HOW: d'Alembert's principle se derive karna

Hum foundation se shuru karte hain, d'Alembert's principle (applied + inertial forces ka virtual work zero hota hai):

$\sum_{k} (\frac{d}{d t} \frac{\partial T}{\partial q ˙ _{k}} - \frac{\partial T}{\partial q _{k}} - Q_{k}) δ q_{k} = 0.$

Conservative applied forces ke liye yeh ban jaata hai, $L = T - V$ ke saath:

$\sum_{k=1}^{n}\left(\frac{d}{dt}\frac{\partial L}{\partial \dot q_k}-\frac{\partial L}{\partial q_k}\right)\delta q_k = 0. \tag{1}$

Yeh bas "har bracket = 0" kyun nahi hai: agar $q_{k}$ independent hote, to har $δ q_{k}$ free hota, isliye har bracket zero ho jaata — ordinary Euler–Lagrange equations. Lekin constraint $f (q, t) = 0$ ke under, $δ q_{k}$ independent nahi hote. Fixed time par constraint ka variation lene par:

$\delta f = \sum_{k}\frac{\partial f}{\partial q_k}\,\delta q_k = 0. \tag{2}$

To (1) sirf unhi $δ q$ ke liye satisfy hona chahiye jo (2) ko maante hain. Abhi har bracket zero nahi set kar sakte.

(2) ko ek as-yet-unknown function $λ (t)$ se multiply karo aur (1) mein add karo:

$\sum_{k} (\frac{d}{d t} \frac{\partial L}{\partial q ˙ _{k}} - \frac{\partial L}{\partial q _{k}} - λ \frac{\partial f}{\partial q _{k}}) δ q_{k} = 0.$

Ab $λ$ ko aise choose karo ki ek dependent coordinate ka bracket zero ho jaaye. Baaki $δ q_{k}$ independent hain, isliye har bracket zero hona chahiye:

$m$ constraints $f_{a} = 0$ ke liye sum karo: RHS ban jaata hai $\sum_{a = 1}^{m} λ_{a} \partial f_{a} / \partial q_{k}$ .

Figure — Constraints using Lagrange multipliers

WORKED EXAMPLE 1 — Wire par bead / rolling disk, constraint force nikalna

Setup: Mass $m$ ka ek particle radius $R$ ke vertical hoop ke andar slide karta hai (polar coords $r, θ$ ). Constraint: $f = r - R = 0$ . Hoop se normal force $N$ nikalo.

Plane polar mein Lagrangian: $L = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{θ}^{2}) - m g r sin θ .$

Polar kyun? Kyunki constraint simply $r = R$ hai, jisse $\partial f / \partial r = 1$ , $\partial f / \partial θ = 0$ — clean.

$r$ -equation: $\frac{d}{d t} (m \overset{r}{˙}) - m r \dot{θ}^{2} + m g sin θ = λ \cdot 1.$ Yeh step kyun? RHS mein $\partial f / \partial r = 1$ use hoti hai.

Constraint $r = R \Rightarrow \overset{r}{˙} = 0, \overset{r}{¨} = 0$ apply karo: $- m R \dot{θ}^{2} + m g sin θ = λ .$

Yeh answer kyun hai: $λ =$ $r$ ke along $\partial L$ -imbalance = radial constraint force. To $N = λ = m g sin θ - m R \dot{θ}^{2} .$ Bead hoop se tab jaati hai jab $N = 0$ , yaani $m g sin θ = m R \dot{θ}^{2}$ . Yeh sawaal embedded method se directly answer nahi hota tha — yahi is method ka faida hai.

WORKED EXAMPLE 2 — Atwood-style: multiplier se rope tension

Setup: Radius $a$ , moment $I$ wala disk, pulley par mass wagera. Simplify karein: ek mass $m$ cylinder par lapti string se latki hai. $x$ = mass ka giraav, $ϕ$ = cylinder ka rotation. "No slipping" ka constraint: $f = x - a ϕ = 0.$

$L = \frac{1}{2} m \overset{x}{˙}^{2} + \frac{1}{2} I \dot{ϕ}^{2} + m g x .$

$x$ -equation: $\partial f / \partial x = 1$ $m\ddot x - mg = \lambda. \tag{i}$ $ϕ$ -equation: $\partial f / \partial ϕ = - a$ $I\ddot\phi = -a\lambda. \tag{ii}$

Yeh kyun? RHS $λ \partial f / \partial q$ hai.

Constraint: $\overset{x}{¨} = a \ddot{ϕ} \Rightarrow \ddot{ϕ} = \overset{x}{¨} / a$ . (ii) mein substitute karo: $\frac{I}{a} \frac{x ¨}{a} = - aλ \Rightarrow λ = - \frac{I x ¨}{a ^{2}} .$ (i) mein plug in karo: $m \overset{x}{¨} - m g = - \frac{I x ¨}{a ^{2}} \Rightarrow \overset{x}{¨} = \frac{m g}{m + I / a ^{2}} .$ Aur tension uski constraint force magnitude hai: $T = ∣ λ ∣ = \frac{I x ¨}{a ^{2}} = \frac{m g ( I / a ^{2} )}{m + I / a ^{2}} .$ $T = ∣ λ ∣$ kyun: $λ \partial f / \partial x = λ$ string se $x$ par generalized force hai = $- T$ (string upar kheenchti hai). Sign sirf direction track karta hai.

WORKED EXAMPLE 3 — Sphere incline par roll karti hai (friction nikalo)

Constraint (rolling, no slip): $f = x - R θ = 0$ . $L = \frac{1}{2} M \overset{x}{˙}^{2} + \frac{1}{2} I \dot{θ}^{2} + M g x sin α$ ke saath:

$x$ : $M \overset{x}{¨} - M g sin α = λ$
$θ$ : $I \ddot{θ} = - R λ$

Same algebra: $λ = - f_{friction}$ . Solid sphere $I = \frac{2}{5} M R^{2}$ deta hai $\overset{x}{¨} = \frac{5}{7} g sin α$ aur friction $∣ λ ∣ = \frac{2}{7} M g sin α$ . Multiplier hi static friction force hai.

Common mistake Common errors ko steel-man karna

Error 1: "Main independent coordinates AND multipliers dono use karunga." Lagta hai sahi hai kyunki kam coordinates simple lagte hain. Yeh galat kyun hai: multipliers ko zaroori hai ki constraint abhi bhi tumhare coordinates mein present ho. Agar tumne use already eliminate kar diya, to $\partial f / \partial q = 0$ aur $λ$ meaningless hai. Fix: jab constraint force chahiye, redundant coordinate rakho.

Error 2: Yeh sochna ki $λ$ ke fixed units hain ya yeh dimensionless hai. Kyun tempting lagta hai: yeh "ek number jaisa lagta hai." Sach: $λ \partial f / \partial q_{k}$ ke units generalized force ke hone chahiye, isliye $λ$ ke units $f$ par depend karte hain. Agar $f$ ke length units hain, to $λ$ ek force hai. Fix: hamesha $λ \partial f / \partial q$ se units padho.

Error 3: Constraint equation ko bhool jaana. Log $n$ EL equations likhte hain aur panic karte hain ki $λ$ undetermined hai. Fix: tumhare paas $n + 1$ unknowns hain aur $(n + 1)$ th equation ke liye constraint $f = 0$ chahiye.

Error 4: Non-holonomic velocity constraints ke liye $f (q) = 0$ likh kar multipliers use karna. Non-holonomic constraints $\sum a_{k} \overset{q}{˙}_{k} = 0$ integrable nahi hote; tumhe velocity form use karni hogi: $\frac{d}{d t} \frac{\partial L}{\partial q ˙ _{k}} - \frac{\partial L}{\partial q _{k}} = \sum_{a} λ_{a} a_{ak}$ , kisi $f$ ke gradient se nahi.

Recall Feynman: 12-saal ke bachche ko samjhao

Socho tum ek katori ke andar marble roll kar rahe ho. Katori marble ko surface par rokne ke liye dhakka deti hai — yahi "constraint force" hai. Aam taur par physics mein hum lazy hote hain aur bas kehte hain "marble katori par rehta hai," aur kabhi nahi sochte ki katori kitna zyada dhakka de rahi hai. Lagrange-multiplier trick ek chota sa honest helper number $λ$ hamare equations mein add karne jaisi hai jiska sirf ek kaam hai — har moment yeh batana ki kitna zyada katori dhakka de rahi hai. Jab woh dhakka zero ho jaata hai, marble surface se uchhal jaata hai!

Active Recall

Holonomic constraint ke under virtual displacements independent kyun nahi hote?

Kyunki

δ f = \sum_{k} (\partial f / \partial q_{k}) δ q_{k} = 0

unhe link karta hai; ek

δ q

baaki se determine hota hai.

Ek constraint ke saath modified Lagrange equation kya hai?

\frac{d}{d t} \frac{\partial L}{\partial q ˙ _{k}} - \frac{\partial L}{\partial q _{k}} = λ \frac{\partial f}{\partial q _{k}}

, plus

f = 0

Lagrange multiplier

λ

ka physical meaning kya hai?

Yeh generalized constraint force

Q_{k}^{(c)} = λ \partial f / \partial q_{k}

(tension, normal force, friction) ko scale karta hai.

m

constraints ke liye EL equation ka RHS kya hai?

\sum_{a = 1}^{m} λ_{a} \partial f_{a} / \partial q_{k}

n

coords aur 1 constraint ke liye kitni equations aur unknowns hain?

n

EL equations + 1 constraint equation

= n + 1

n

coordinates +

λ

ke liye.

Embedded (independent-coordinate) method mein constraint forces kyun gayab ho jaati hain?

Kyunki woh virtual work nahi karti (allowed motion ke perpendicular hoti hain), isliye jab constraints solve ho jaate hain to woh drop out ho jaati hain.

Circular hoop se bead leave karne ki condition kya hai?

Jab normal force

λ = m g sin θ - m R \dot{θ}^{2} = 0

ho.

Incline par rolling sphere mein

λ

kiske barabar hai?

Minus static friction force; solid sphere ke liye magnitude

\frac{2}{7} M g sin α

δ f

ko fixed time par (

δ

d

nahi) kyun liya jaata hai?

Virtual displacements instantaneous hote hain; virtual variation ke dauran

f

ki explicit

t

-dependence freeze rehti hai.

Non-holonomic constraints ke liye kya alag hai?

Velocity-form

a_{ak}

coefficients use karo: RHS

= \sum_{a} λ_{a} a_{ak}

, kyunki

\sum_{k} a_{ak} \overset{q}{˙}_{k} = 0

koi integrable

f (q, t) = 0

nahi hai.

Connections

Generalized Coordinates and Degrees of Freedom — jab hum constraint rakhte hain to hum kya trade karte hain.
d'Alembert's Principle and Virtual Work — woh foundation jisse humne derive kiya.
Euler-Lagrange Equations — unconstrained special case ( $λ = 0$ ).
Holonomic vs Non-holonomic Constraints — yeh method kab apply hota hai vs kab velocity form chahiye.
Normal Force and Tension as Constraint Forces — physical payoff.
KKT Conditions — optimization/inequality constraints mein same multiplier idea.