Principle of least action — Hamilton's principle derivation: Question bank

2.1.19 · D5 · Physics › Analytical Mechanics › Principle of least action — Hamilton's principle derivation

True or false — justify karo

Action hamesha true path par minimize hota hai.

False —

δ S = 0

sirf ek stationary point maangta hai; lambe time ke baad (conjugate/focal point ke baad) true path ek saddle bhi ho sakta hai. Iska sahi naam hai "principle of stationary action."

Lagrangian total energy

T + V

ke barabar hota hai.

False — Lagrangian

L = T - V

hota hai. Sirf

T - V

ko Euler–Lagrange mein daalne par

F = ma

milta hai; conserved quantity

T + V

baad mein Hamiltonian ke roop mein aata hai.

Action

S

ek aisa number hai jo motion ke ek single instant par depend karta hai.

False —

S

ek functional hai: yeh poora path

q (t)

over

[t_{1}, t_{2}]

leta hai aur ek number return karta hai, isliye yeh poore route par depend karta hai, na ki ek instant par.

Hamilton's principle aur Newton's second law ek particle ke liye alag predictions dete hain.

False — ek 1D particle ke liye

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

ka Euler–Lagrange equation exactly

m \overset{x}{¨} = - d V / d x

deta hai, isliye ye equivalent hain, competing nahi.

Variation

q_{ε} = q + ε η

ke dauran, wiggle

η

endpoints par nonzero ho sakta hai.

False — endpoints fixed data hote hain, isliye

η (t_{1}) = η (t_{2}) = 0

; yahi integration by parts mein boundary term ko khatam karta hai.

Jab hum

L (q, \overset{q}{˙}, t)

ke partial derivatives lete hain, hum

q

aur

\overset{q}{˙}

ko independent variables maante hain.

True — partial derivative ke "slot" ke andar ye independent hote hain;

\overset{η}{˙} = \frac{d}{d t} η

ka link sirf variation ke dauran wapas aata hai aur yahi integration by parts exploit karta hai.

Action ke units Planck's constant ke units ke barabar hote hain.

True —

S

ke units energy

\times

time

= J\cdots

hain, jo

ℏ

se match karta hai; yeh coincidence path integral ka bridge hai jahan

S /ℏ

ek phase hai.

Euler–Lagrange equation path par har instant par sach hoti hai.

True — Fundamental Lemma bracketed expression ko har

t

ke liye zero hone par majboor karta hai, isliye chahe

S

ek global quantity hai, uski stationarity condition time mein local hai.

Error dhundo

δ S = 0

, aur kyunki

S

ek minimum hai,

\frac{d ^{2} S}{d ε ^{2}} > 0

hamesha."

First-order condition

δ S = 0

second-order sign ke baare mein kuch nahi kehta;

\frac{d ^{2} S}{d ε ^{2}}

positive (minimum), negative (maximum) ya zero (saddle) ho sakta hai, isliye yeh claim over-assert kar raha hai.

\int \frac{\partial L}{\partial q ˙} \overset{η}{˙} d t

ko parts se integrate karne par

- \int \frac{d}{d t} (\frac{\partial L}{\partial q ˙}) η d t

milta hai aur kuch nahi."

Boundary term

[\frac{\partial L}{\partial q ˙} η]_{t_{1}}^{t_{2}}

yeh bataye bina drop kar diya gaya ki kyun; yeh sirf isliye zero hota hai kyunki

η

dono fixed endpoints par zero hai.

"Kyunki

η

arbitrary hai,

\int (bracket) η d t = 0

ka matlab hai bracket ka integral zero hai."

Galat conclusion —

η

ki arbitrariness bracket ko khud pointwise zero hone par majboor karti hai, na sirf uske integral ko (warna ek localized

η

nonzero integral expose kar deta).

"Ek free particle ke liye

L = \frac{1}{2} m \overset{x}{˙}^{2}

, isliye

\frac{\partial L}{\partial x} = \frac{1}{2} m \overset{x}{˙}^{2}

Yeh ek slot-confusion error hai:

\partial L / \partial x

mein

\overset{x}{˙}

ko fixed treat kiya jaata hai, aur

L

mein koi explicit

x

nahi hai, isliye

\partial L / \partial x = 0

; likhne wale ne galat variable differentiate kiya.

"Euler–Lagrange hai

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

, isliye momentum

\frac{\partial L}{\partial q ˙}

hamesha conserved hai."

Sirf tab jab

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

ho (coordinate cyclic ho),

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

hota hai; generally momentum

\partial L / \partial q

rate se change karta hai — yeh Noether disguise mein hai.

"Falling body ke liye

L = \frac{1}{2} m \overset{y}{˙}^{2} - m g y

, isliye

\frac{\partial L}{\partial y} = + m g

Sign slip —

- m g y

ka derivative

- m g

hai, jo

m \overset{y}{¨} = - m g

deta hai, yaani neeche ki taraf acceleration; galat sign body ko upar bhej deta.

"Hum differentiate karne ke baad

ε

drop kar sakte hain aur true path paane ke liye

ε = 1

set kar sakte hain."

Nahi — true path

ε = 0

hai; hum

S (ε)

differentiate karte hain aur derivative ko $ε = 0$ par evaluate karte hain, kyunki true path woh reference hai jis se hum wiggle karte hain.

Why questions

Hum doosre term ko parts se integrate kyun karte hain

\overset{η}{˙}

ko waise rehne dene ki bajaye?

\overset{η}{˙}

η

mein convert karne ke liye taaki poore integrand se ek common factor

η

bahar nikala ja sake, jisse Fundamental Lemma kaam kar sake.

Endpoints ko fixed kyun rakhna zaroori hai?

Ye diye gaye start aur end data hain, aur inhe fix karne se

η

wahan zero ho jaata hai isliye boundary term khatam ho jaata hai, jo clean Euler–Lagrange equation deta hai; free endpoints ki jagah extra natural boundary conditions generate hote.

L = T - V

kyun,

T + V

kyun nahi?

Kyunki

T - V

ko Euler–Lagrange mein daalne par

m \overset{x}{¨} = F

reproduce hota hai, jo correct dynamics hai;

T + V

galat equations deta — yeh conserved energy hai, action ka integrand nahi.

Fundamental Lemma of calculus of variations ki zaroorat hi kyun hai?

Kyunki yeh "weighted integral har wiggle ke liye zero hai" ko upgrade karke "weight (bracket) identically zero hai" bana deta hai, jo actual equation of motion hai.

Hamilton's principle ko koi force vectors ya coordinate axes kyun nahi chahiye?

Kyunki

L = T - V

ek scalar hai aur action ek scalar integral hai, isliye poora formulation coordinate-free hai — tum koi bhi generalized coordinate

q

chunte ho aur machinery phir bhi kaam karti hai.

Action ka

ℏ

ke jaisi units mein hona physically meaningful kyun hai?

Path-integral view mein har path ek phase

e^{i S /ℏ}

contribute karta hai; classical stationary-action paths woh hain jahan neighbouring phases coherently add hote hain, isliye

S

ℏ

ki units mein hona hi woh phase ko dimensionless banata hai.

S

ko stationary banana optics mein Fermat's principle jaisa kyun "lagta" hai?

Fermat optical path time ko stationary banata hai jabki Hamilton action ko stationary banata hai; dono variational "laziness" laws hain, aur wohi integration-by-parts machinery unki governing equations deti hai.

Edge cases

Jab

L

explicitly

q

par depend nahi karta (ek cyclic coordinate) to Euler–Lagrange equation kya hoti hai?

Tab

\partial L / \partial q = 0

, isliye

\frac{d}{d t} (\partial L / \partial \overset{q}{˙}) = 0

aur conjugate momentum

\partial L / \partial \overset{q}{˙}

conserved hota hai — ek symmetry directly ek conservation law deti hai.

Ek free particle ke liye, stationary action

A

B

tak fixed-time routes mein se kya select karta hai?

Constant velocity (straight-line, uniform motion), kyunki koi bhi speeding-up ya slowing-down

\overset{x}{˙}^{2}

ka time-average badhata hai aur isliye

\int \frac{1}{2} m \overset{x}{˙}^{2} d t

badhata hai.

Kya hota hai derivation ke saath agar

η (t)

kahin nonzero ho lekin bracket sirf ek isolated point par nonzero ho?

Tum ek aisi

η

choose kar sakte ho jo exactly wahan bump ho jahan bracket nonzero hai, jisse integral nonzero ho jaata hai —

δ S = 0

ke contradict karta hai; isliye bracket ek bhi point par nonzero nahi ho sakta.

Jab potential

V

har jagah zero ho to

δ S = 0

kya reduce ho jaata hai?

Equation of motion

\frac{d}{d t} (m \overset{q}{˙}) = 0

ban jaati hai, yaani momentum constant — Newton's first law degenerate, force-free case ke roop mein emerge hoti hai.

Agar motion conjugate point ke baad bahut lambe time tak hoti hai, to kya true path abhi bhi

S

ka minimum hai?

Zaroor nahi — yeh stationary rehta hai, lekin saddle ban sakta hai, aur yahi reason hai ki "least action" general principle ke liye ek misleading naam hai.

ε

kya role play karta hai, aur

ε \to 0

limit mein kya hota hai?

ε

measure karta hai ki trial path true path se kitni door jaata hai; stationarity yeh statement hai ki

S

ka first-order-in- $ε$ change

ε \to 0

par zero ho jaata hai, isliye sirf linear term matter karta hai.

Kya do alag paths dono same endpoints ke beech stationary ho sakte hain?

Haan — focal/conjugate point ke baad multiple stationary paths (jaise multiple ray trajectories) same endpoints ko connect kar sakte hain, har ek Euler–Lagrange satisfy karta hai lekin is baat mein alag hain ki kya ye

S

minimize karte hain.

Recall One-line self-test

Agar tum bina dekhe bata sako ki boundary term kyun khatam hoti hai, $L = T - V$ kyun hai, aur stationary $\neq =$ minimum kyun hai, to yeh topic tumhara ho gaya. ::: Endpoints fixed $\Rightarrow η = 0$ wahan; $T - V$ $F = ma$ reproduce karta hai; $δ S = 0$ sirf ek first-order condition hai isliye saddles aur maxima bhi admit karta hai.