Derivation of Euler-Lagrange equations from D'Alembert's principle: Question bank

2.1.5 · D5 · Physics › Analytical Mechanics › Derivation of Euler-Lagrange equations from D'Alembert's pri

Definition Symbols yaad karo (taaki yeh page akela khada rahe)

$r_{k}$ — particle $k$ ki position; $\dot{r}_{k}$ uski velocity, $\ddot{r}_{k}$ uska acceleration.
$p_{k} = m_{k} \dot{r}_{k}$ — particle $k$ ka momentum (mass times velocity); $\dot{p}_{k} = m_{k} \ddot{r}_{k}$ iske change ki rate hai.
$F_{k}^{(a)}$ — particle $k$ par applied force: woh real pushes aur pulls (gravity, springs, tumhara haath) jinki hum parwah karte hain.
$F_{k}^{(c)}$ — constraint force: jo wire/rail/rod supply karta hai constraint satisfied rakhne ke liye (wire jo bead ko sideways push karta hai). Hum ise generally jaante nahi aur jaanna bhi nahi chahte.
$q_{i}$ — generalized coordinates (minimal free variables, jaise pendulum ke liye $θ$ ); $\overset{q}{˙}_{i}$ unki time-rates.
$δ r$ — ek virtual displacement: ek kalpana ki gayi shift jisme time frozen hai, $δ t = 0$ .
$Q_{i}$ — coordinate $q_{i}$ ke liye generalized force: applied forces ko $q_{i}$ par project karna, $Q_{i} = \sum_{k} F_{k}^{(a)} \cdot \frac{\partial r _{k}}{\partial q _{i}}$ . Yeh potential nahi hai — potential $V$ ek alag scalar energy hai jisse conservative $Q_{i}$ ko derive kiya ja sakta hai as $Q_{i} = - \partial V / \partial q_{i}$ .
$T$ — kinetic energy, $V$ — potential energy, $L = T - V$ — the Lagrangian.

Neeche ki picture start karne se pehle pin karne layak hai: yeh dikhati hai kyun allowed direction par project karne se constraint force delete hoti hai, aur virtual displacement real displacement se kaise alag hai.

Figure — Derivation of Euler-Lagrange equations from D'Alembert's principle

Aur woh do engine-room identities jinhe traps refer karti hain — "cancellation of dots" aur " $d / d t$ ka $\partial / \partial q$ ke saath commute karna" — yahan summarize ki gayi hain taaki tumhe back flip na karna pade:

Recall Teen derivation steps jinhe traps quote karti hain (mini-recap)

Step 3 (product-rule-backwards): $m_{k} \ddot{r}_{k} \cdot \frac{\partial r _{k}}{\partial q _{i}} = \frac{d}{d t} (m_{k} \dot{r}_{k} \cdot \frac{\partial r _{k}}{\partial q _{i}}) - m_{k} \dot{r}_{k} \cdot \frac{d}{d t} \frac{\partial r _{k}}{\partial q _{i}}$ . Yeh unwanted acceleration ko velocity-type terms ke liye trade karta hai.
Lemma 1 (cancellation of dots): $\frac{\partial r ˙ _{k}}{\partial q ˙ _{i}} = \frac{\partial r _{k}}{\partial q _{i}}$ .
Lemma 2 (commuting derivatives): $\frac{d}{d t} \frac{\partial r _{k}}{\partial q _{i}} = \frac{\partial r ˙ _{k}}{\partial q _{i}}$ .

True or false — justify karo

True or false: Virtual displacement

δ r

wahi cheez hai jo real displacement

d r

hai jo particle actually time

d t

mein karta hai.

False.

δ r

time freeze karta hai (

δ t = 0

), isliye yeh sirf explore karta hai ki constraint abhi tumhe kahan hone deta hai;

d r

additionally constraint ki apni time-drift (

\frac{\partial r}{\partial t} d t

) bhi carry karta hai. Yeh tab coincide karte hain jab constraint time-independent (scleronomic) ho — woh extra drift term tab zero hoti hai.

True or false: D'Alembert's principle kehta hai ki applied forces akele zero virtual work karte hain.

False. Yeh kehta hai

\sum_{k} (F_{k}^{(a)} - \dot{p}_{k}) \cdot δ r_{k} = 0

. Iska true hone ka reason yeh hai ki constraint forces

F_{k}^{(c)}

zero virtual work karte hain (woh allowed motion ke perpendicular push karte hain), isliye jab hum

δ r

se dot karte hain toh woh Newton's equation se drop out ho jaate hain — applied forces aur inertia bachti hai, sirf applied forces nahi.

True or false:

δ r_{k} = \sum_{i} \frac{\partial r _{k}}{\partial q _{i}} δ q_{i}

mein hum safe rehne ke liye ek

\frac{\partial r _{k}}{\partial t} δ t

term add kar sakte hain.

False. Virtual displacement definition se

δ t = 0

set karta hai, isliye woh term exactly zero hai. Ise rakhna purpose defeat kar deta: precisely time-drift ki absence hi hai jo

\sum_{k} F_{k}^{(c)} \cdot δ r_{k} = 0

ko moving constraints ke liye bhi hold karwati hai.

True or false: Constraint forces system par kabhi real work nahi karte.

False. Ek moving constraint par (ek sliding wall, ek driven rotating wire) real work

F^{(c)} \cdot d r

nonzero ho sakta hai, kyunki

d r

mein

F^{(c)}

ke saath ek component hota hai jo wall ki apni motion se aata hai. Sirf virtual work

F^{(c)} \cdot δ r

vanish karta hai, kyunki

δ r

mein aisa koi time-drift component nahi hota.

True or false: Jab hum

\sum_{i} [\dots] δ q_{i} = 0

par pahunchte hain, toh har bracket vanish honi chahiye.

True, lekin sirf tab jab holonomic system ke liye

δ q_{i}

independent hain. Logic yeh hai: independent, arbitrary quantities ka ek weighted sum sabhi choices ke liye zero tab hi hota hai jab har weight (bracket) zero ho. Independence hatao (non-holonomic case) aur ek

δ q_{i}

doosre ko constrain karta hai, isliye brackets trade off kar sakte hain aur individually vanish karna zaroori nahi.

True or false: Euler–Lagrange equation

\frac{d}{d t} \frac{\partial L}{\partial q ˙ _{i}} - \frac{\partial L}{\partial q _{i}} = 0

Newton se pare ek naya physical law hai.

False. Derivation ka har step ek equality hai: yeh Newton ka

F = m a

hai jisme unknown constraint forces algebraically project out ki gayi hain aur bachne wali cheezein energies mein repackage ki gayi hain. Wahi physics, cleaner bookkeeping — koi naya postulate smuggle nahi kiya gaya.

True or false: Hum

V

L

mein fold kar sakte hain kyunki

\partial V / \partial \overset{q}{˙}_{i} = 0

hai.

True usual case

V = V (q, t)

mein: kyunki

V

mein koi

\overset{q}{˙}

-dependence nahi hai,

- V

T

mein add karne se

\partial L / \partial q_{i}

change hota hai (giving

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

) lekin

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

untouched rehta hai — exactly wahi jo chahiye. Velocity-dependent potential

U (q, \overset{q}{˙}, t)

ke liye folding abhi bhi kaam karta hai lekin tumhe generalized potential form

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

use karni padegi; concrete magnetic example ke liye neeche edge case dekho.

True or false:

\frac{\partial r ˙ _{k}}{\partial q ˙ _{i}} = \frac{\partial r _{k}}{\partial q _{i}}

("cancellation of dots") sirf symbolically dots cancel karna hai.

Spirit mein False — yeh ek genuine result hai. Kyunki

\dot{r}_{k} = \sum_{j} \frac{\partial r _{k}}{\partial q _{j}} \overset{q}{˙}_{j} + \frac{\partial r _{k}}{\partial t}

\overset{q}{˙}_{j}

mein linear hai with coefficients jo

q, t

par depend karte hain lekin

\overset{q}{˙}

par nahi,

\overset{q}{˙}_{i}

mein differentiate karne par single coefficient

\frac{\partial r _{k}}{\partial q _{i}}

select hota hai. Mnemonic cancellation jaisi lagti hai sirf isliye kyunki underlying reason actually wahi quantity return karta hai.

Error dhundho

"Newton deta hai

F^{(a)} + F^{(c)} - \dot{p} = 0

;

δ r

se dot karke aur sum karke, applied forces cancel ho jaati hain, constraint forces bachti hain." — galti kahan hai?

Ulta hai. Yeh

\sum F^{(c)} \cdot δ r = 0

hai (constraint forces allowed motion ke perpendicular hoti hain) jo cancel hoti hai, isliye constraint forces vanish ho jaati hain aur applied forces generalized force

Q_{i}

ke roop mein bachti hain.

"Pendulum ke liye main Cartesian

x, y

use karunga toh mere paas do clean equations hongi." — is choice mein kya galat hai?

Tum rod tension (ek constraint force) ko ek unknown ke roop mein reintroduce kar lete ho aur ek constraint

x^{2} + y^{2} = ℓ^{2}

(rod length

ℓ

) milti hai. Iske bajaye

q = θ

choose karne se woh constraint automatically bake ho jaati hai, isliye tension kabhi appear nahi hoti.

\frac{d}{d t} \frac{\partial r _{k}}{\partial q _{i}}

— position

q_{i}

mein move nahi karta, isliye yeh zero hai." — error dhundho.

Yeh zero nahi hai. Lemma 2 se (upar recap ki gayi),

\frac{d}{d t} \frac{\partial r _{k}}{\partial q _{i}} = \frac{\partial r ˙ _{k}}{\partial q _{i}}

, jo generally nonzero hai kyunki

\dot{r}_{k}

apne coefficients ke through

q

's par depend karta hai.

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

hamesha, isliye main hamesha

L = T - V

likh sakta hoon." — hidden assumption dhundho.

Iske liye applied forces ko conservative hona chahiye (

F^{(a)} = - \nabla V

). Friction, driving forces, ya magnetic forces plain scalar

V

se capture nahi hote; tab tum raw

Q_{i} = \sum_{k} F_{k}^{(a)} \cdot \frac{\partial r _{k}}{\partial q _{i}}

rakhte ho.

"Step 3 mein humne likha

m_{k} \ddot{r}_{k} \cdot \frac{\partial r _{k}}{\partial q _{i}} = \frac{d}{d t} (m_{k} \dot{r}_{k} \cdot \frac{\partial r _{k}}{\partial q _{i}})

." — kya drop hua?

Correction term

- m_{k} \dot{r}_{k} \cdot \frac{d}{d t} \frac{\partial r _{k}}{\partial q _{i}}

. Product rule backwards mein do pieces hote hain; doosra drop karna eventual

- \partial T / \partial q_{i}

term kho deta hai.

T = \frac{1}{2} m \overset{r}{˙}^{2}

rotating wire par bead ke liye (radial coordinate

r

, wire

ω

rate par spin kar rahi hai)." — kya missing hai?

Tangential contribution: bead wire ke saath sideways bhi sweep hoti hai, isliye

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

r^{2} ω^{2}

miss karne par

\partial T / \partial r

khatam ho jaata hai aur centrifugal effect kho jaata hai.

"Kyunki

δ q_{i}

infinitesimal hain, main inhe specific tiny numbers set kar sakta hoon solve karne ke liye." — yeh galti kyun hai?

δ q_{i}

arbitrary aur independent symbols hain; "har bracket = 0" ka conclusion us arbitrariness se aata hai (yeh sabhi choices ke liye hold karna chahiye), kisi single value se nahi. Values fix karne se kuch sabit nahi hota.

Why questions

Hum arbitrary directions ki bajaye virtual displacements par project kyun karte hain?

Kyunki virtual displacements wahan point karte hain jo constraints allow karte hain, aur constraint forces unke perpendicular hoti hain — isliye dot product

F^{(c)} \cdot δ r

zero hai, un forces ko delete karte hue jo hum jaante nahi (figure s01 dekho).

Step 3 mein seedha differentiate karne ki bajaye hum product-rule-backwards trick kyun use karte hain?

Unwanted acceleration

\ddot{r}

ko velocity terms ke time-derivatives se trade karne ke liye, jo cleanly

\frac{d}{d t} \frac{\partial T}{\partial q ˙ _{i}} - \frac{\partial T}{\partial q _{i}}

mein repackage ho jaate hain. Direct differentiation

\ddot{r}

ko constraint geometry ke saath stuck rehne deti hai aur usse energy banana impossible ho jaata hai.

\partial T / \partial q_{i}

term kyun matter karta hai — kya kinetic energy "sirf speed ke baare mein" nahi hai?

Jab mapping

r_{k} (q, t)

curve ya rotate karta hai,

T

coordinate

q

par hi depend karta hai (jaise rotating wire ke liye

r^{2} ω^{2}

term, radial coordinate

r

aur rotation rate

ω

ke saath), isliye

\partial T / \partial q \neq = 0

. Woh term automatically centrifugal, Coriolis-like, aur geometric forces produce karta hai — dekho Kinetic energy in generalized coordinates.

Hum conservative

V

L

mein fold kar sakte hain lekin friction ko nahi — kyun?

Ek conservative force ek scalar

V (q)

ka gradient hai, isliye

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

aur

\partial V / \partial \overset{q}{˙}_{i} = 0

. Friction kisi bhi scalar ka gradient nahi hai (yeh motion ki direction par depend karta hai), isliye koi

V

exist nahi karta jise absorb kiya ja sake.

δ q_{i}

ki independence specifically ek holonomic property kyun hai?

Holonomic constraints position-equations hain, isliye inhe solve away kiya ja sakta hai, har coordinate ko kisi bhi instant par freely variable chhod ke. Non-holonomic constraints velocities ko aapas mein baandhte hain aur integrate out nahi ho sakte, isliye ek

δ q_{i}

ko vary karne par doosre bhi force hote hain — independence kho jaati hai.

Hamilton's principle is force-based derivation ke same equation kyun deta hai?

Dono same content encode karte hain: action integral extremize karna mathematically EL bracket ke vanish hone ki demand ke equivalent hai, jo time mein integrate kiya gaya D'Alembert's principle hai. Do darwaze, ek kamra.

Hum

L

ko differentiate karte waqt

\overset{q}{˙}

aur

q

ko independent variables kyun treat karte hain?

L

ke partial derivatives ke andar isko abstract configuration-velocity space par ek function treat kiya jaata hai, jahan

q

aur

\overset{q}{˙}

alag slots hain. Relation

\overset{q}{˙} = d q / d t

sirf baad mein impose hota hai, outer total derivative

\frac{d}{d t}

se; beech mein inhe mix karna partials corrupt kar deta hai.

Edge cases

Edge case: ek free particle,

V = 0

q = x

. EL kya reduce karta hai aur yeh reassuring kyun hai?

\frac{d}{d t} (m \overset{x}{˙}) - 0 = 0 \Rightarrow m \overset{x}{¨} = 0

— Newton's first law. Jab koi constraints ya potentials na hon toh formalism ko plain Newton reproduce karna chahiye.

Edge case: constraints time-independent hain (scleronomic). Kya virtual aur real displacements ab coincide karte hain?

Haan first order tak, kyunki

\partial r / \partial t = 0

woh time-drift remove kar deta hai jo

δ r

d r

se alag karti thi. Virtual-real distinction sirf moving (rheonomic) constraints ke liye bites karta hai.

Edge case: ek non-holonomic constraint (jaise rolling without slipping). Kya tum phir bhi har EL bracket ko zero set kar sakte ho?

Nahi —

δ q_{i}

ab independent nahi hain, isliye terms separate karne se pehle tumhe Lagrange multipliers se constraints adjoin karni padegi.

Edge case:

L

mein kisi coordinate

q_{j}

par explicit dependence nahi hai (

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

). Kya bachta hai?

Tab

\frac{d}{d t} \frac{\partial L}{\partial q ˙ _{j}} = 0

, isliye

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

conserved hai — ek cyclic coordinate. Yeh Noether's theorem ka seed hai:

L

ki ek symmetry ek conserved quantity deti hai.

Edge case: electromagnetic field mein ek charged particle (charge

e

) — concrete velocity-dependent potential. "

V

fold karo" ki jagah kya aata hai, aur kya yeh sahi force reproduce karta hai?

Tum generalized potential

U = e ϕ - e A \cdot \dot{r}

use karte ho, jahan

ϕ

scalar potential hai aur

A

vector potential hai (dono position aur time ke functions). Yeh

\dot{r}

par depend karta hai, isliye

L = T - U = \frac{1}{2} m \dot{r}^{2} - e ϕ + e A \cdot \dot{r}

banao. Ise

\frac{d}{d t} \frac{\partial L}{\partial r ˙} - \frac{\partial L}{\partial r} = 0

mein feed karne par,

E = - \nabla ϕ - \partial_{t} A

aur

B = \nabla \times A

use karne ke baad, exactly Lorentz force

m \ddot{r} = e (E + \dot{r} \times B)

milti hai. Yeh "minimal coupling" hai, aur yeh standard example hai ki folding $\overset{q}{˙}$ -dependent forces ko generalized-potential form

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

ke zariye handle kar sakta hai.

Edge case: agar do chosen generalized coordinates secretly independent nahi hain (ek redundant choice). Kya hoga?

"Har bracket = 0" step fail ho jaata hai, kyunki

\sum_{i} [\dots] δ q_{i}

ka vanish hona ab har coefficient ko zero force nahi karta. Tumhe pehle redundancy eliminate karni padegi ya multipliers use karne padenge.

Edge case: potential explicitly time par depend karta hai,

V = V (q, t)

. Kya derivation phir bhi hold karta hai?

Haan.

V

ki time-dependence

\partial V / \partial \overset{q}{˙}_{i} = 0

ko intact rehne deti hai, isliye

L

mein folding valid hai; sirf conserved-energy conclusion (ek alag result) kho jaata hai.

Recall Ek-line self-test

Agar koi tumhe ek system de, toh kaun sa single question decide karta hai ki tum "har bracket = 0" likh sakte ho ya nahi? Kya $δ q_{i}$ independent hain? ::: Equivalently: kya constraints holonomic hain? Agar haan, separate karo; agar nahi, multipliers use karo.