Lagrangian L = T − V: Question bank — Hinglish deep dive

2.1.4 · D5 · Physics › Analytical Mechanics › Lagrangian L = T − V

Shuru karne se pehle, symbols ka ek reminder taaki kuch bina samjhe use na ho:

$T$ = kinetic energy (woh "chalti" energy, $\frac{1}{2} m v^{2}$ ek point mass ke liye).
$V$ = potential energy (woh "stored" energy, jaise $m g y$ ya $\frac{1}{2} k x^{2}$ ).
$L = T - V$ = Lagrangian, har instant par ek scalar number.
$q$ = ek generalized coordinate: koi bhi independent number jo configuration fix karta hai (ek position $x$ , ya sirf ek angle $θ$ ).
$\overset{q}{˙}$ = uska time-rate-of-change (generalized velocity), "q-dot" padha jata hai.
$\overset{q}{¨}$ = doosra time-derivative, yaani acceleration: $\overset{q}{¨} = \frac{d ^{2} q}{d t ^{2}}$ (do dots = time ke saath do baar differentiate karo). Toh $\ddot{θ}$ angular acceleration hai, $\overset{x}{¨}$ ordinary acceleration hai.
$δ$ = variation symbol: $δ q$ matlab ek path mein tiny "wiggle" jo tum add karte ho (ek chhota trial change), aur $δ S$ us se action mein first-order change hai. "Stationary" matlab $δ S = 0$ : kisi bhi chhoti wiggle ke under koi first-order change nahi.
$S = \int_{t_{1}}^{t_{2}} L d t$ = action, ek poore path ke liye ek number.

Char pictures poore page ko anchor karti hain — shuru karne se pehle inhe ek nazar dekho taaki neeche ke words ke paas kuch point karne ko ho.

Picture 1 — action ek area hai. $L$ sirf ek number hai jo system ke chalte waqt badalta hai; $L$ ko time ke against plot karo aur us curve ke neeche ka area action $S$ hai. Do alag paths do alag curves dete hain, isliye do alag shaded areas.

Picture 2 — path ko vary karna. Real path (solid) ke aas paas wiggled "trial" paths (dashed) hain jo same do endpoints share karte hain. Nature woh path chunti hai jiska action tiny wiggles ke under nahi badalta, yaani $δ S = 0$ .

Picture 3 — stationary aur minimum same nahi hote. Paths ki family ke saath slide karo aur $S$ dekho: true path par slope flat hai, lekin flat valley bottom (minimum) ya saddle ho sakta hai. "Least action" ek nickname hai, law nahi.

Picture 4 — pendulum coordinate. Vertical se ek angle $θ$ bob ko completely fix kar deta hai, chahe woh 2D plane mein ho. Rod woh constraint hai jisne ek degree of freedom mita di.

True or false — justify

Kya

L = T + V

hai?

False.

L = T - V

hai; plus-sign wali object total energy

E = T + V

hai, jo alag role play karti hai. Sirf minus sign se equations of motion sahi force sign ke saath nikalti hain.

Action

S

ek single instant se attached number hai.

False. Picture 1 dekho:

S = \int L d t

do fixed events ke beech poore

L

-vs-

t

curve ke neeche ka area hai, toh

S

har candidate path ke liye ek number hai, har instant ke liye nahi.

Nature hamesha action ko minimize karti hai.

Generally false. Picture 3 yahi point dikhati hai: nature action ko stationary banati hai (

δ S = 0

), jo ek flat slope hai — saddle ya maximum ka bhi flat slope hota hai, toh "least action" ek historical naam hai, hamesha literally minimum nahi hota.

Euler–Lagrange equation poore system ke liye ek saath ek baar solve karni padti hai.

False. Yeh har generalized coordinate

q_{i}

ke liye independently hold karti hai: har degree of freedom ke liye ek equation, jo sirf us coordinate ko vary karke milti hai baaki ko fixed rakhte hue.

L (q, \overset{q}{˙}, t)

ke andar, partial derivatives lete waqt

q

aur

\overset{q}{˙}

ko independent treat kiya jata hai.

True. Jab

\partial L / \partial q

aur

\partial L / \partial \overset{q}{˙}

banate ho to doosre ko freeze karte ho; sirf

\frac{d}{d t} \frac{\partial L}{\partial q ˙}

banane ke baad chain rule unhe reconnect karta hai.

Lagrangian method mein bhi tension aur normal forces draw karne padte hain.

False. Picture 4 dekho: kyunki

θ

us circle ke saath chalti hai jis par bob move kar sakta hai, rod ka tension (jo rod ke along, motion ke perpendicular point karta hai) koi kaam nahi karta aur kabhi

L

mein nahi aata.

Kisi bhi smooth function

f (q, t)

ke liye, Lagrangians

L

aur

L^{'} = L + \frac{df ( q , t )}{d t}

same equations of motion dete hain.

True. Yahan

f (q, t)

coordinate aur time ka koi bhi differentiable function hai (ek free "gauge" choice, physical energy nahi). Path par uska integral

\int \frac{df}{d t} d t = f (q (t_{2}), t_{2}) - f (q (t_{1}), t_{1})

hai, jo endpoints se fixed hai, toh yeh

S

mein ek constant add karta hai aur nahi badal sakta ki

δ S = 0

kahan ho.

Agar

T

aur

V

dono zero hain toh Euler–Lagrange equation meaningless hai.

False lekin degenerate.

L = 0

0 = 0

trivially milta hai; simply koi dynamics nahi hai — ek free, forceless, potential-free "system" consistent hai, bas empty hai.

Magnetic field mein charged particle ke liye bhi Lagrangian hamesha

T - V

hota hai.

Roughly false as stated. Velocity-dependent (magnetic) forces ko generalized potential

U (q, \overset{q}{˙})

chahiye, aur

L = T - U

hota hai; plain "

T

minus position-only

V

" form conservative forces ka special case hai.

Jo Lagrangian ka kinetic part

\frac{1}{2} m \overset{q}{˙}^{2}

hai, usse hamesha

\overset{q}{¨}

ke liye solve kiya ja sakta hai.

Generally false. Agar

\overset{q}{˙}^{2}

ka coefficient ("mass matrix") zero ho jata hai toh Lagrangian singular hai:

\partial^{2} L / \partial \overset{q}{˙}^{2} = 0

aur tum

\overset{q}{¨}

ke liye invert nahi kar sakte. Aise singular Lagrangians (field theory / constraints mein common) ko constrained dynamics ki special machinery chahiye.

Spot the error

"Kyunki

E = T + V

aur

L = T - V

, toh

L = E - 2 V

, isliye

L

aur

V

jaanne se energy free mein milti hai."

Algebra theek hai lekin identity ek condition chhupa rahi hai:

E

conserved tabhi hoti hai jab

\partial L / \partial t = 0

ho (

L

mein explicit time nahi). Concretely, conserved quantity

H = \overset{q}{˙} \frac{\partial L}{\partial q ˙} - L

hai, aur

\frac{d H}{d t} = - \frac{\partial L}{\partial t}

; agar

\partial L / \partial t \neq = 0

toh

H

(isliye

E

) drift karta hai aur "free energy" claim empty hai.

"Maine

L

banaya, phir motion find karne ke liye

\frac{\partial L}{\partial q} = 0

set kiya."

Error: yeh equation of motion nahi hai. Tumhe poori

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

use karni hai; pehla term drop karne se poori dynamics throw away ho jati hai (yeh sirf equilibrium points deta).

"Pendulum ke liye maine string tension ko

V

mein daala taaki gravity aur tension dono account ho jayein."

Error: tension ek ideal constraint force hai jo koi kaam nahi karta; ise

V

mein nahi daalna chahiye. Sirf genuine potentials (yahan gravity)

V

mein belong karti hain.

θ

choose karna tension ko already remove kar deta hai.

"Action vary karte waqt, boundary term

[\partial L / \partial \overset{q}{˙} δ q]

(parts integrate karne ke baad bacha) vanish hota hai kyunki

\partial L / \partial \overset{q}{˙} = 0

hai."

Error: momentum zero nahi hai. Boundary term isliye khatam hota hai kyunki dono endpoints par

δ q = 0

hai (fixed-endpoint condition, exactly Picture 2 mein shared endpoints), momentum factor ke vanish hone ki wajah se nahi.

"Kyunki

\overset{q}{˙}

q

ka derivative hai, jab main

L

q

ke saath differentiate karta hoon toh mujhe yeh bhi account karna chahiye ki

\overset{q}{˙}

kaise badlata hai."

Error: partial derivatives ke dauran

q

aur

\overset{q}{˙}

independent slots hain. Unhe pehle se join karne par spurious extra terms nikalte hain; linkage baad mein

\frac{d}{d t}

ke through apply hoti hai.

"Maine

L = T + V

use kiya aur

m \overset{x}{¨} + F = 0

mila; yeh basically Newton jaisa hai, bas rearrange kiya."

Error: sign galat hai. Newton deta hai

m \overset{x}{¨} = F

, yaani

m \overset{x}{¨} - F = 0

T + V

version anti-physics predict karta hai (force galat taraf push karta hai), yahi exactly kyun

T - V

mandatory hai.

"Mere block par friction hai, toh main ek friction term

V

mein add kar dunga aur

L = T - V

use karunga."

Error: friction dissipative hai aur kisi bhi potential

V (q)

se nahi aa sakta (yeh velocity direction par depend karta hai aur energy drain karta hai). Yeh instead right-hand side par ek generalized force

Q_{nc}

ke roop mein enter karta hai ("nc" = non-conservative: koi bhi force, jaise friction ya drag, jo potential se derivable nahi):

\frac{d}{d t} \frac{\partial L}{\partial q ˙} - \frac{\partial L}{\partial q} = Q_{nc}

. Plain

L = T - V

sirf conservative aur velocity-dependent-potential forces cover karta hai.

"Main rolling-without-slipping constraint ko

V

mein likh kar model karunga."

Error: rolling-without-slipping aksar non-holonomic hota hai — velocities ke beech ek aisa relation jo coordinate relation mein integrate nahi ho sakta, toh ise coordinates ya

V

mein absorb nahi kiya ja sakta. Ise Lagrange multipliers se impose karna padta hai, unlike holonomic pendulum rod jo

θ

mein bake in ho sakti hai.

Why questions

Hum path vary karte waqt fixed endpoints kyun demand karte hain?

Kyunki (Picture 2) hum same start aur end events ke beech alag routes compare karte hain; endpoints ko move karne dena ek alag sawaal poochhega aur ek nonzero boundary term chhodega jo derivation kharab karta hai.

Action banane ke liye

L

ko time par integrate kyun karte hain instead of sirf ek moment mein

L

examine karne ke?

Ek instant poora path decide nahi kar sakta. Integrate karne se har candidate path ke liye ek single scalar milta hai (Picture 1 ka shaded area), jo exactly wahi hai jo tumhe paths compare karne aur stationary wala choose karne ke liye chahiye.

T - V

mein minus sign physically kyun matter karta hai?

Kyunki

\frac{\partial L}{\partial q} = - \frac{d V}{d q} = F

; minus "stored energy" ko correct force direction mein turn karta hai, toh Euler–Lagrange result

F = m \overset{q}{¨}

se match karta hai. Ise flip karo aur force sign flip ho jata hai.

Ek single angle

θ

ek pendulum ko completely describe kyun kar sakta hai jo 2D space mein rehta hai?

Rigid rod ek constraint hai jo ek degree of freedom remove karta hai; bob ek circle par stuck hai (Picture 4), toh ek coordinate (angle) kaafi hai — yeh generalized coordinates ka payoff hai.

Agar

L

mein ek coordinate appear nahi karta (sirf uska

\overset{q}{˙}

appear karta hai) toh yeh conservation law kyun signal karta hai?

Agar

\frac{\partial L}{\partial q} = 0

toh

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

, toh generalized momentum

p = \frac{\partial L}{\partial q ˙}

constant hai — dekho Generalized Momentum and Conservation.

L

mein explicit time dependence energy conservation kyun tod deta hai?

Ise symmetry ki tarah socho: agar

L

har instant par identical lagta hai (

L

mein explicit

t

nahi), toh poore experiment ko time mein aage slide karna kuch nahi badalta, aur woh time-translation symmetry hi energy conservation hai. Lekin agar koi actively ek parameter crank kar raha hai (maan lo

L

t

par depend karta hai ek moving support ke through), toh "kal wahi experiment" ek alag experiment hai — symmetry toot gayi. Quantitatively

\frac{d H}{d t} = - \frac{\partial L}{\partial t}

, toh energy exactly us rate se leakti hai jis rate par clock

L

mein likha gaya hai.

Lagrangian method 1D mein ek single particle ke liye Newton's Second Law mein kyun reduce hota hai?

L = \frac{1}{2} m \overset{x}{˙}^{2} - V (x)

ko Euler–Lagrange mein plug karne par exactly

m \overset{x}{¨} = - \frac{d V}{d x} = F

milta hai; Newton special case hai, aur isse match karna hi hai jaise

L = T - V

justify hua.

Hamiltonian Lagrangian ko replace karne ki bajaye usse kyun banaya jata hai?

H

L

ka Legendre transform hai,

\overset{q}{˙}

ko momentum

p

se swap karta hai; yeh same physics ko energy ko star quantity banake re-express karta hai, lekin underlying content abhi bhi

L

se aata hai.

Edge cases

Free particle (

V = 0

) ke liye Euler–Lagrange equation ka kya hota hai?

\frac{\partial L}{\partial q} = 0

, toh

\frac{d}{d t} (m \overset{q}{˙}) = 0

: momentum conserved hai aur particle constant speed par straight line mein chalta hai — stationary path literally straight wali hai.

Pendulum ke stable equilibrium

θ = 0

par motion kya hai?

\ddot{θ} = - \frac{g}{ℓ} sin θ

(yaad raho

\ddot{θ} = d^{2} θ / d t^{2}

) se

θ = 0

par

\ddot{θ} = 0

milta hai, toh wahan rakha gaya toh wahi baitha rehta hai; paas mein oscillate karta hai kyunki restoring term

sin θ

zero ki taraf push karta hai.

Pendulum ki inverted position

θ = π

ke baare mein kya?

sin π = 0

toh

\ddot{θ} = 0

: yeh bhi ek equilibrium hai, lekin unstable — koi bhi tiny displacement badhta hai kyunki restoring sign flip ho jata hai, toh real bob wahan nahi tikta.

Kya method tab bhi kaam karta hai jab

L

mein explicit time dependence ho, jaise driven oscillator mein?

Haan, Euler–Lagrange abhi bhi hold karta hai; lekin tab energy generally conserved nahi hoti kyunki

\frac{d H}{d t} = - \partial L / \partial t \neq = 0

— woh time-translation symmetry jo conservation guarantee karta, toot gayi hai.

Chhote angles ke liye, kya pendulum equation kuch simpler mein collapse hoti hai?

Haan:

sin θ \approx θ

\ddot{θ} \approx - \frac{g}{ℓ} θ

milta hai, simple harmonic motion

ω = g / ℓ

ke saath — ek limiting case jo SHM form recover karta hai.

Agar do candidate paths ka action value same ho toh kya?

Action ka stationary hona first-order change

δ S = 0

ke vanish hone ke baare mein hai, value ki uniqueness ke baare mein nahi; degenerate actions ho sakti hain (jaise symmetric paths) lekin har true path phir bhi Euler–Lagrange satisfy karta hai.

Agar kinetic term degenerate ho (singular Lagrangian) toh kya?

Tab

\overset{q}{¨}

ko Euler–Lagrange se solve nahi kiya ja sakta, aur ordinary equations of motion ki jagah hidden constraints appear hoti hain — standard

L = T - V

recipe silently ek non-degenerate (invertible) kinetic term assume karti hai.

Connections

Euler–Lagrange Equation — woh equation jis par "spot the error" ke har item ka sahara hai.
Principle of Least Action — "stationary, not minimum" clarify karta hai.
Generalized Coordinates and Constraints — kyun tension kabhi $V$ mein nahi aata; holonomic vs non-holonomic.
Generalized Momentum and Conservation — cyclic-coordinate items.
Hamiltonian H = T + V — energy counterpart; $H = \overset{q}{˙} \partial L / \partial \overset{q}{˙} - L$ .
Newton's Second Law — 1D limiting case.