Generalized momenta and generalized forces — Hinglish notes

2.1.7 · Physics › Analytical Mechanics

1. Setup: generalized coordinates

YEH form kyun? Kyunki constraints bake-in hain: ek pendulum bob ka $r = (ℓ sin θ, - ℓ cos θ)$ hota hai — ek coordinate $θ$ , na ki do Cartesian wale jo $x^{2} + y^{2} = ℓ^{2}$ se bandhe hain.

Generalized velocity hai $\overset{q}{˙}_{j} = d q_{j} / d t$ . Chain rule se, $v_{i} = \frac{d r _{i}}{d t} = \sum_{j} \frac{\partial r _{i}}{\partial q _{j}} \overset{q}{˙}_{j} + \frac{\partial r _{i}}{\partial t} .$

2. Generalized force — first principles se

Sabse clean definition virtual work se aati hai. Socho ki har particle ko ek infinitesimal virtual amount $δ r_{i}$ se displace karo jo constraints ke saath consistent ho, fixed time par. Applied forces $F_{i}$ dwara kiya gaya work hai

$δ W = \sum_{i} F_{i} \cdot δ r_{i} .$

Ab $δ r_{i}$ ko generalized coordinates mein express karo (time frozen hai, isliye $\partial / \partial t$ term nahi): $δ r_{i} = \sum_{j} \frac{\partial r _{i}}{\partial q _{j}} δ q_{j} .$

Substitute karo aur summation ka order swap karo: $δ W = \sum_{j} \equiv Q_{j} i \sum F_{i} \cdot \frac{\partial r _{i}}{\partial q _{j}} δ q_{j} .$

3. Generalized momentum — first principles se

ISE aise kyun define karte hain? Dekho Euler–Lagrange equation kya banti hai: $\frac{d}{d t} \frac{\partial L}{\partial q ˙ _{j}} - \frac{\partial L}{\partial q _{j}} = 0 ⟹ \overset{p}{˙}_{j} = \frac{\partial L}{\partial q _{j}} .$ Yeh Newton's second law disguise mein hai: (generalized) momentum ki rate of change = (generalized) force. $p_{j}$ ko $\partial L / \partial \overset{q}{˙}_{j}$ define karna precisely woh hai jo yeh sach banata hai.

Figure — Generalized momenta and generalized forces

4. Worked example: pendulum (full pipeline)

System: mass $m$ , rod length $ℓ$ , gravity $g$ , coordinate $θ$ .

Step 1 — position. $r = (ℓ sin θ, - ℓ cos θ)$ . Kyun: ek DOF, $θ$ constraint $∣ r ∣ = ℓ$ encode karta hai.

Step 2 — kinetic energy. $v = ℓ \dot{θ} (cos θ, sin θ)$ , isliye $∣ v ∣ = ℓ \dot{θ}$ , jo deta hai $T = \frac{1}{2} m ℓ^{2} \dot{θ}^{2}$ .

Step 3 — potential. $V = m g y = - m g ℓ cos θ$ .

Step 4 — Lagrangian. $L = \frac{1}{2} m ℓ^{2} \dot{θ}^{2} + m g ℓ cos θ$ .

Step 5 — generalized momentum. $p_{θ} = \frac{\partial L}{\partial θ ˙} = m ℓ^{2} \dot{θ} .$ Yeh step kyun? $p_{θ}$ pivot ke baare mein angular momentum hai (units kg·m²/s). ✓

Step 6 — generalized force. Conservative hai, isliye $Q_{θ} = - \partial V / \partial θ = - m g ℓ sin θ$ . Yeh step kyun? $θ$ ek angle hai ⇒ $Q_{θ}$ ek torque hai. Aur $- m g ℓ sin θ$ exactly gravity ka restoring torque hai. ✓

Step 7 — equation of motion. $\overset{p}{˙}_{θ} = Q_{θ} \Rightarrow m ℓ^{2} \ddot{θ} = - m g ℓ sin θ \Rightarrow \ddot{θ} = - \frac{g}{ℓ} sin θ .$ ✓ Pendulum equation, zero free-body diagrams se derive ki.

5. Forecast-then-Verify drill

Common mistakes

Recall Feynman: 12-year-old ko explain karo

Socho tum ek merry-go-round dhakka de rahe ho. Ise describe karne ke liye tum har plank ka $x, y$ track nahi karte — bas ek angle track karo. "Kitna hard dhakka diya" abhi bhi matter karta hai, lekin ab useful number seedha push ki jagah twist (torque) hai, aur simple speed ki jagah "kitna ghoom raha hai" (angular momentum). Lagrangian mechanics ka ek master recipe hai: energy likh do, aur jo bhi variable tune choose ki uske liye force-jaisi aur momentum-jaisi numbers khud nikl aati hain — slider ke liye push, spinner ke liye twist, automatically.

Flashcards

Generalized momentum

p_{j}

define karo.

p_{j} = \frac{\partial L}{\partial q ˙ _{j}}

, Lagrangian ka generalized velocity ke w.r.t. derivative.

Generalized force

Q_{j}

ko virtual work se define karo.

Q_{j} = \sum_{i} F_{i} \cdot \frac{\partial r _{i}}{\partial q _{j}}

, define kiya gaya taaki

δ W = \sum_{j} Q_{j} δ q_{j}

Conservative forces ke liye,

Q_{j} = ?

Q_{j} = - \frac{\partial V}{\partial q _{j}}

Angular coordinate torque kyun deta hai, force kyun nahi?

Kyunki

Q_{j} δ q_{j}

work ke barabar hona chahiye; agar

q_{j}

angle hai to

Q_{j}

ke units energy/radian = N·m = torque hain.

"Cancellation of dots" identity state karo.

\frac{\partial v _{i}}{\partial q ˙ _{j}} = \frac{\partial r _{i}}{\partial q _{j}}

, kyunki

v_{i}

\overset{q}{˙}_{j}

mein linear hai.

Cyclic (ignorable) coordinate kya hai aur iski consequence kya hai?

Woh jo

L

mein appear nahi karta; tab

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

, isliye

p_{j}

conserved hai.

Plane-polar

θ

ka generalized momentum

T = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{θ}^{2})

ke liye?

p_{θ} = m r^{2} \dot{θ}

= angular momentum.

p_{j}

hamesha

m \overset{q}{˙}_{j}

kyun nahi hota?

Kyunki

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

; units aur mass factors coordinate par depend karte hain (jaise pendulum ke liye

m ℓ^{2} \dot{θ}

Virtual work mein

\partial r_{i} / \partial t

kyun drop karte hain?

Virtual displacements frozen time par li jaati hain, isliye explicit time term vanish ho jaata hai.

Euler–Lagrange

p_{j}

aur

Q_{j}

ko kaise relate karta hai?

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

, yaani generalized momentum ki rate of change generalized "force" term ke barabar hai — Newton's law restate kiya gaya.

Connections

Lagrangian Mechanics — $L = T - V$ , $p_{j}$ aur $Q_{j}$ dono ka source object.
Euler-Lagrange Equation — equation $\overset{p}{˙}_{j} = \partial L / \partial q_{j}$ jo dono ko unify karti hai.
Generalized coordinates and constraints — jahan se $q_{j}$ aur $r_{i} (q, t)$ aate hain.
D'Alembert's Principle — virtual work foundation jo $Q_{j}$ define karta hai.
Noether's Theorem — cyclic coordinate ⇒ conserved $p_{j}$ ek special case ki tarah.
Hamiltonian Mechanics — $p_{j}$ ek independent variable ban jaata hai; Legendre transform.
Angular momentum — $p_{θ}$ special case.