Generalized momenta and generalized forces: Question bank

2.1.7 · D5 · Physics › Analytical Mechanics › Generalized momenta and generalized forces

Definition Is page par use hone wale symbols (pehle yeh padho)

$q_{j}$ — ek generalized coordinate: ek number (length jaise $r$ , ya angle jaise $θ$ ) jo configuration pin down karne mein help karta hai. $\overset{q}{˙}_{j} = d q_{j} / d t$ iska generalized velocity hai.
$r_{i}$ — particle $i$ ki ordinary Cartesian position, jo function $r_{i} (q_{1}, \dots, q_{n}, t)$ ke roop mein likhi hoti hai.
$T$ — kinetic energy, $T = \frac{1}{2} \sum_{i} m_{i} ∣ v_{i} ∣^{2}$ (hamesha positive, velocities se bani).
$V$ — potential energy, positions ka function (aur shayad $t$ ka bhi); gravity near Earth ke liye $V = m g y$ .
$L$ — Lagrangian, defined as $L = T - V$ (dhyan rakho minus sign hai, $T + V$ nahi).
$δ W$ — virtual work: applied forces jo kaam karte woh agar ek imagined tiny displacement $δ r_{i}$ ho jo frozen time par, constraints ke saath consistent ho. Chhota $δ$ (na ki $d$ ) yeh flag hai ki "imagined, time fixed rakha gaya hai".

Running example coordinate systems: ek pendulum ek angle $θ$ use karta hai rod length $ℓ$ ke saath (toh $r = (ℓ sin θ, - ℓ cos θ)$ ); polar coordinates mein radius $r$ aur angle $θ$ hota hai (toh $T = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{θ}^{2})$ ).

Key objects jinke baare mein tum reason karoge: generalized momentum $p_{j} = \partial L / \partial \overset{q}{˙}_{j}$ , generalized force $Q_{j} = \sum_{i} F_{i} \cdot \partial r_{i} / \partial q_{j}$ , aur Euler–Lagrange statement $\overset{p}{˙}_{j} = \partial L / \partial q_{j}$ .

Neeche teen pictures hain jo tumhara visual scaffolding hain — is bank ka har abstract phrase (" $\partial r_{i} / \partial q_{j}$ ", "figure-skater effect", "virtual displacement") wahan draw kiya gaya hai. Jab bhi koi line abstract lage, unhe dekh lo.

Figure — Generalized momenta and generalized forces

True or false — justify

Generalized momentum hamesha mass times velocity hota hai,

p_{j} = m \overset{q}{˙}_{j}

False.

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

; ek angular coordinate ke liye

p_{θ} = m ℓ^{2} \dot{θ}

ka unit kg·m²/s hota hai. "

m

" aur yahan tak ki dimensions bhi depend karte hain ki tumne kaun sa coordinate choose kiya.

Ek angle ke conjugate generalized force ek torque hota hai.

True. Kyunki

Q_{j} δ q_{j}

energy honi chahiye, aur

δ θ

dimensionless hai (radians), toh

Q_{θ}

ke units energy = N·m = torque hone chahiye.

Agar ek Lagrangian mein

q_{j}

par koi explicit dependence nahi hai, toh

\overset{q}{˙}_{j}

constant hoti hai.

False. Yeh momentum

p_{j}

hai jo conserved hota hai, velocity nahi.

p_{θ} = m r^{2} \dot{θ}

constant ke liye,

\dot{θ}

badlega jab

r

badlega (figure-skater effect jo figure 3 mein draw hai).

Ek conservative system ke liye time-independent potential ke saath, generalized force poori tarah se

Q_{j} = - \partial V / \partial q_{j}

se capture hoti hai.

True, yeh statement sirf applied conservative force ke baare mein hai. Lekin dhyan rakho: yeh

Q_{j}

hamesha equation of motion ka poora right-hand side nahi hoti, kyunki

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

\partial T / \partial q_{j}

term bhi le sakta hai (centrifugal). "Complete generalized force" aur "complete driving term" alag claims hain — inhe alag rakho.

Canonical momentum

p_{j}

physical

x y z

space mein ek vector hai.

False. Har

p_{j}

ek single scalar number hai jo coordinate

q_{j}

se attached hai. Yeh configuration/phase space mein rehta hai, real 3D space mein nahi — dekho Hamiltonian Mechanics.

Agar do systems ki kinetic energy

T

same hai, toh unke generalized momenta same hain.

False.

p_{j}

depend karta hai ki $T$ , $\overset{q}{˙}_{j}$ ke saath kaise vary karta hai, na ki sirf

T

ki value par. Do systems ek instant par numerically same

T

share kar sakte hain lekin alag

\partial T / \partial \overset{q}{˙}_{j}

ho sakti hai.

Ek cyclic (ignorable) coordinate ka matlab hai ki coordinate problem mein kahin bhi appear nahi karta.

False. Iska matlab hai ki

q_{j}

L

mein absent hai (toh

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

); iska velocity

\overset{q}{˙}_{j}

typically

L

mein appear karta rehta hai. Isi liye

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

meaningful aur conserved hai.

Generalized momentum hamesha us direction mein physical linear momentum ke barabar hota hai.

False. Tabhi jab

q_{j}

ek Cartesian coordinate ho aur

T

mein koi cross-terms na hon. Magnetic field mein,

p_{x} = m \overset{x}{˙} + q A_{x}

— canonical momentum, kinetic momentum

m \overset{x}{˙}

se alag hota hai.

Agar ek coordinate cyclic hai, toh iska conjugate momentum conserved hota hai, chahe Lagrangian explicitly time par depend kare.

True. Cyclicity sirf

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

ke baare mein hai; explicit

t

-dependence us derivative ko affect nahi karta, toh

\overset{p}{˙}_{j} = 0

phir bhi hold karta hai. (Jo explicit time-dependence ke under fail hota hai woh energy ka conservation hai, cyclic

p_{j}

ka nahi.)

Spot the error

"Rotating wire par bead par koi applied force nahi hai, toh

Q_{r} = 0

aur

m \overset{r}{¨} = 0

Error: equation of motion hai

m \overset{r}{¨} = m r ω^{2}

. Centrifugal term

\partial T / \partial r

se aata hai (

q

-dependence of kinetic energy se), kisi

V

se nahi. "Koi applied force nahi" ka matlab "koi force-like term nahi" nahi hota.

p_{θ} = m ℓ^{2} \dot{θ}

, aur pendulum ke liye

θ

cyclic hai, toh angular momentum conserved hai."

Error: pendulum ke liye

V = - m g ℓ cos θ

θ

par depend karta hai, toh

θ

cyclic nahi hai. Isliye

\overset{p}{˙}_{θ} = - m g ℓ sin θ \neq = 0

; angular momentum conserved nahi hai.

"Virtual work se

Q_{j}

nikalne ke liye main system ko

δ r_{i} = \sum_{j} \frac{\partial r _{i}}{\partial q _{j}} δ q_{j} + \frac{\partial r _{i}}{\partial t} δ t

se displace karta hoon."

Error: ek virtual displacement time ko freeze karta hai (yehi woh basis hai jis par

δ W

bana hai — definition box dekho), toh

\partial r_{i} / \partial t

term drop ho jata hai. Use rakhna real (

d

) aur virtual (

δ

) displacements ko mix karta hai aur

Q_{j}

ko corrupt karta hai.

"Kyunki

Q_{j} = \sum_{i} F_{i} \cdot \partial r_{i} / \partial q_{j}

forces ka dot product hai, iske units hamesha newtons hote hain."

Error:

\partial r_{i} / \partial q_{j}

ka weight (figure 2 mein tangent arrow ke roop mein draw kiya gaya) ka unit (length / unit of

q_{j}

) hota hai. Angular

q_{j}

ke liye yeh weight ek length hai, toh

Q_{j}

N·m (torque) mein aata hai, N mein nahi.

"Euler–Lagrange equation kehta hai

\overset{p}{˙}_{j} = Q_{j}

, toh main

Q_{j} = - \partial V / \partial q_{j}

compute karta hoon aur ho gaya —

\partial T / \partial q_{j}

ki zaroorat nahi."

Error: sahi statement hai

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

. Jab bhi

T

q_{j}

par depend kare (curvilinear coordinates, rotation),

\partial T / \partial q_{j}

term zaroori hai.

p_{j}

ka conservation chahiye ki system par bilkul koi force na ho."

Error: iske liye sirf yeh chahiye ki

L

, us ek coordinate

q_{j}

se independent ho. Forces aur potentials bahut bade ho sakte hain; jab tak woh

q_{j}

par depend na karein,

p_{j}

constant rehta hai — yeh Noether's Theorem symmetry statement hai.

"Magnetism jaisi velocity-dependent force ko potential se handle nahi kiya ja sakta, toh uska

Q_{j}

mein koi jagah nahi."

Error: magnetic (Lorentz) forces ek velocity-dependent generalized potential

U (q, \overset{q}{˙}, t)

ke through enter karte hain, aur generalized force recover hoti hai

Q_{j} = \frac{d}{d t} \frac{\partial U}{\partial q ˙ _{j}} - \frac{\partial U}{\partial q _{j}}

ke roop mein. Yeh framework mein fit hote hain — bas simple

- \partial V / \partial q_{j}

shortcut mein nahi.

Why questions

p_{j}

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

kyun define kiya gaya hai na ki

m \overset{q}{˙}_{j}

se?

Kyunki woh definition Euler–Lagrange ko

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

padhne deti hai — Newton ka "rate of change of momentum = force" kisi bhi coordinates mein. Yeh precisely usi law ko preserve karne ke liye choose kiya gaya hai.

Ek angular coordinate automatically apne conjugate momentum ke roop mein angular momentum kyun deta hai?

Kyunki

p_{θ} = \partial L / \partial \dot{θ}

aur rotational motion ke liye

T \propto r^{2} \dot{θ}^{2}

, toh

p_{θ} = m r^{2} \dot{θ}

— jo ki Angular momentum ki definition hi hai. Coordinate ki geometry, momentum ki identity dictate karti hai.

Hum sirf

x, y, z

ki jagah generalized coordinates ki taklif kyun uthate hain?

Kyunki constraints bake in ho jaate hain: ek pendulum ko do Cartesian coordinates

x^{2} + y^{2} = ℓ^{2}

se bandhne ki jagah ek

θ

chahiye. Fewer variables, solve karne ke liye koi constraint forces nahi.

Same formula

Q_{j} δ q_{j} = δ W

ek coordinate ke liye force aur doosre ke liye torque kyun deta hai?

Kyunki

Q_{j}

defined hai "virtual work per unit of

q_{j}

" ke roop mein. Agar

q_{j}

ek length hai toh tumhe N milte hain; agar yeh angle hai toh N·m milta hai.

Q_{j}

ke units auto-adjust hote hain taaki product

Q_{j} δ q_{j}

hamesha energy ho.

"Coordinate cyclic hai" — yeh ek momentum ke conserved hone ka sabse gehra reason kyun hai?

Kyunki ek cyclic coordinate

L

ki symmetry hai —

q_{j}

ko shift karo toh physics unchanged rehti hai. Noether's Theorem ise precise banata hai: har continuous symmetry ek conserved quantity deti hai.

\dot{θ}

baadh sakta hai even jab angular momentum

p_{θ}

conserved ho, kyun?

Kyunki

p_{θ} = m r^{2} \dot{θ}

(figure 3 dekho). Agar

r

shrink kare, toh product constant rakhne ke liye

\dot{θ}

badhna chahiye — figure skater apni baahein khinchti hai toh faster spin karta hai.

"Cancellation of dots" identity

\partial v_{i} / \partial \overset{q}{˙}_{j} = \partial r_{i} / \partial q_{j}

kyun hold karta hai?

Kyunki

v_{i} = \sum_{j} (\partial r_{i} / \partial q_{j}) \overset{q}{˙}_{j} + \partial r_{i} / \partial t

har

\overset{q}{˙}_{j}

mein linear hai jiska coefficient

\partial r_{i} / \partial q_{j}

hai;

\overset{q}{˙}_{j}

ke saath differentiate karne par sirf wahi coefficient bachta hai.

L

mein explicit time-dependence energy conservation kyun todti hai lekin cyclic-momentum conservation nahi?

Kyunki energy conservation follow karta hai

L

mein koi explicit $t$ na hone se (

\partial L / \partial t = 0

), jabki cyclic-momentum conservation follow karta hai

L

mein koi particular $q_{j}$ na hone se (

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

). Yeh independent symmetries hain; ek ko hatana doosre ko zarori nahi hatata.

Edge cases

Agar ek system ka zero potential hai (

V = 0

), kya phir bhi ek generalized force ho sakti hai?

Haan. Curvilinear ya rotating coordinates mein

\partial T / \partial q_{j}

nonzero ho sakti hai (centrifugal, Coriolis-like terms), jo

V = 0

ke bawajood force-like effects produce karte hain.

Ek coordinate ke liye

p_{j}

kya hoga jo

L

mein appear karta hai lekin jiski velocity

\overset{q}{˙}_{j}

nahi karti (ek non-dynamical coordinate)?

Toh

p_{j} = \partial L / \partial \overset{q}{˙}_{j} = 0

. Aisi coordinate ka koi momentum nahi hota; iska Euler–Lagrange equation ek constraint

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

ban jaata hai, equation of motion nahi.

Jis instant pendulum

θ = 0

(bottom) se guzarta hai, generalized force

Q_{θ}

kya hai?

Q_{θ} = - m g ℓ sin 0 = 0

. Gravity bottom par koi restoring torque nahi deti; bob momentarily

θ

direction mein unforced hota hai chahe woh fastest move kar raha ho.

Free particle ke liye Cartesian coordinates mein, kya generalized aur physical momentum coincide karte hain?

Haan — yeh degenerate case hai.

q = x

L = \frac{1}{2} m \overset{x}{˙}^{2}

, toh

p_{x} = m \overset{x}{˙}

exactly.

p_{j}

ki sari "strangeness" precisely tab gayab ho jaati hai jab coordinate ek seedha Cartesian length ho jisme koi velocity cross-terms na hon.

Agar applied forces sab

\partial r_{i} / \partial q_{j}

ke perpendicular hain toh

Q_{j}

ka kya hoga?

Q_{j} = \sum_{i} F_{i} \cdot \partial r_{i} / \partial q_{j} = 0

. Forces us coordinate ke displacement direction mein koi kaam nahi karte, toh woh

Q_{j}

mein kuch contribute nahi karte — e.g. wire ke normal constraint force ka contribution zero hota hai.

Polar coordinates mein ek particle ke liye

r \to 0

limit mein

p_{θ} = m r^{2} \dot{θ}

ka kya hota hai?

p_{θ} \to 0

agar

\dot{θ}

finite rahe. Lekin agar

p_{θ}

conserved aur nonzero hai, toh

\dot{θ} \to \infty

jab

r \to 0

— coordinate system origin par singular ho jaata hai (isi liye polar coordinates wahan fail karte hain).

Ek system ke liye jisme explicitly time-dependent constraint hai (e.g. ek bead ek wire par jise force kiya gaya hai

θ = ω t

se rotate karne ke liye), kya mechanical energy

T + V

conserved hai?

Nahi. Rotating-wire constraint driving ke through energy andar/bahar feed karta hai, toh

T + V

conserved nahi hai chahe

p_{r}

apni equation obey kare. Frozen-time virtual displacement exactly woh hai jo humein moving constraint ke bawajood

Q_{r}

ko cleanly define karne deta hai.

Ek charged particle magnetic field mein, kya kinetic momentum

m \overset{x}{˙}

conserved hota hai jab

x

cyclic ho?

Nahi — yeh canonical momentum

p_{x} = m \overset{x}{˙} + q A_{x}

hai jo conserved hota hai. Magnetic vector potential contribution

q A_{x}

wahi hai jo conserved quantity ko plain

m \overset{x}{˙}

se alag banata hai.

Recall Ek-line summary jo saath le jao

$p_{j} = \partial L / \partial \overset{q}{˙}_{j}$ aur $Q_{j} = \sum_{i} F_{i} \cdot \partial r_{i} / \partial q_{j}$ ; dono apni identity (linear↔angular, force↔torque) coordinate ke saath badal lete hain, aur $\overset{p}{˙}_{j} = \partial L / \partial q_{j}$ Newton's law disguise mein hai.