Hamilton's equations of motion: Question bank — Hinglish deep dive

2.1.12 · D5 · Physics › Analytical Mechanics › Hamilton's equations of motion

Shuru karne se pehle, teen words kamane zaroori hain taaki koi line undefined symbol use na kare:

True or false — justify karo

H

hamesha total energy

T + V

hoti hai.

False.

H = T + V

tabhi hota hai jab kinetic energy velocities mein quadratic ho aur constraints scleronomic (time-independent) hon; rotating frame ya cranked wire yeh tod deta hai.

Agar

H

mein koi explicit time dependence nahi hai, toh

H

conserved hai.

True.

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

har trajectory ke saath chalti hai, toh

\partial H / \partial t = 0

H

ko constant force karta hai — chahe

H

energy ho ya na ho.

Agar

H

conserved hai, toh

H

zaroor total energy ke barabar hogi.

False. Conservation aur "

H = E

" alag-alag facts hain; ek rheonomic system mein conserved

H

ho sakta hai jo

E

se alag ho, ya

H = E

jo conserved na ho.

Hamilton's equations

2 n

first-order ODEs hain, jo

n

second-order waalon ko replace karti hain.

True. Har second-order Euler–Lagrange equation do first-order canonical equations mein split hoti hai, count double hota hai lekin order kam hota hai — yahi price hai symmetric

(q, p)

description ke liye.

\overset{p}{˙}_{i} = - \partial H / \partial q_{i}

mein minus sign ek sign convention hai jise hum flip kar sakte hain.

False. Yeh

d H

ko term-by-term match karne se forced hai; isko flip karna energy conservation tod dega aur phase-space flow ki area-preserving property khatam kar dega.

q

ke conjugate momentum

p

hamesha mass times velocity hoti hai.

False. Sirf simple

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

ke liye; magnetic term

q A \cdot \dot{r}

jaisi velocity-coupling ke saath,

p = m \dot{r} + q A

Tum

H

mein

\overset{q}{˙}

chod sakte ho.

False. Legendre transform ka poora point yahi hai ki har

\overset{q}{˙}_{i}

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

se eliminate karna zaroori hai; bacha hua

\overset{q}{˙}

matlab

H

abhi

(q, p, t)

ka function nahi bana.

Agar coordinate

q_{j}

cyclic hai (

H

mein absent hai), toh uska conjugate momentum

p_{j}

conserved hai.

True.

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

jab

q_{j}

appear na kare, toh

p_{j}

motion ka constant hai.

Do alag Lagrangians kabhi same Hamiltonian nahi de sakte.

False. Jo Lagrangians ek total time derivative

d F / d t

se differ karte hain unke equations of motion same rehte hain aur same physics de sakte hain; map

L \to H

us sense mein injective nahi hai.

Hamiltonian formalism require karta hai ki Lagrangian velocities ke liye invertible ho.

True. Tumhe

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

\overset{q}{˙}_{i}

ke liye solve karna hoga; agar yeh impossible hai (singular/degenerate

L

) toh standard Legendre transform fail ho jaata hai.

Error dhundho

H = \frac{1}{2} m \overset{x}{˙}^{2} + V

, toh bas plug in karo aur kaam khatam."

Error hai

\overset{x}{˙}

H

mein chod dena. Tumhe

\overset{x}{˙} = p / m

substitute karna hoga taaki

H = p^{2} /2 m + V

mile;

\overset{x}{˙}

ke saath

H

likhna Lagrangian aur Hamilton pictures ko mix kar deta hai.

\overset{q}{˙} = + \partial H / \partial p

ke saath symmetry se

\overset{p}{˙} = + \partial H / \partial q

bhi hoga."

Plus galat hai.

d H

mein

- \overset{p}{˙} d q

+ (\partial H / \partial q) d q

se match karne par

\overset{p}{˙} = - \partial H / \partial q

forced hai; asymmetry essential hai, koi symmetry restore nahi karni.

"Magnetic Lagrangian ke liye maine

p = m \dot{r}

set kiya aur aage badha."

Galat momentum. Yahan

p = \partial L / \partial \dot{r} = m \dot{r} + q A

; canonical momentum ki jagah kinetic momentum use karne se galat equations aate hain.

\frac{d H}{d t} = \frac{\partial H}{\partial q} \overset{q}{˙} + \frac{\partial H}{\partial p} \overset{p}{˙}

, aur yahi poora time derivative hai."

Yeh explicit

\partial H / \partial t

term chod deta hai.

H (q, p, t)

ke liye chain rule mein teen pieces hain; pehle do canonical equations se cancel ho jaate hain, sirf

\partial H / \partial t

bachta hai.

"Kyunki

\partial H / \partial t = - \partial L / \partial t

hai, aur energy conserved hai, toh

L

bhi time-independent hogi."

Do cheezein confuse ho rahi hain.

H

conserved ho sakta hai (

H

mein explicit

t

nahi) jabki

L

mein explicit

t

ho — agar transform aisa arrange ho — lekin generally

\partial H / \partial t = - \partial L / \partial t

hota hai; "energy conserved isliye

L

time-independent" ki reasoning yeh skip karti hai ki

L

ki time-dependence directly

H

ki time-dependence produce karti hai.

"Phase space mein trajectory khud ko cross kar sakti hai kyunki yeh sirf ek curve hai."

Time-independent

H

ke liye flow deterministic hai: har point ka ek unique tangent hota hai, toh trajectories cross nahi kar sakti — crossing ka matlab hoga ek state se do futures.

"SHO orbit

(x, p)

mein circle hai."

Yeh generally ellipse hai.

H = p^{2} /2 m + \frac{1}{2} m ω^{2} x^{2} = E

mein

x

aur

p

axes par alag scales hain; variables rescale karne ke baad hi circle banta hai.

Why questions

\overset{q}{˙}

p

se replace kyun karein, Lagrangian ki second-order equations kyun na rakhen?

First-order equations state ko phase space mein ek single point banate hain jisme ek unique flow line hoti hai, aur Poisson brackets aur Liouville's theorem ke peeche chhupa symmetric

(q, p)

structure expose hota hai.

Jab hum

d H

lete hain toh

d \overset{q}{˙}_{i}

term kyun vanish ho jaata hai?

Kyunki

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

definition se hai, toh term

p_{i} d \overset{q}{˙}_{i}

exactly

\frac{\partial L}{\partial q ˙ _{i}} d \overset{q}{˙}_{i}

ko cancel karta hai — yahi cancellation Legendre transform ka poora purpose hai.

d H / d t

compute karte waqt do chain-rule terms kyun cancel ho jaate hain?

\overset{q}{˙} = \partial H / \partial p

aur

\overset{p}{˙} = - \partial H / \partial q

substitute karne par

\frac{\partial H}{\partial q} \frac{\partial H}{\partial p} - \frac{\partial H}{\partial p} \frac{\partial H}{\partial q} = 0

milta hai; minus sign precisely yahi karta hai ki woh annihilate ho jayein.

Minus sign ko "area-preserving" kyun kehte hain?

Antisymmetric pairing

(\overset{q}{˙}, \overset{p}{˙}) = (+ \partial_{p} H, - \partial_{q} H)

phase-space flow ko divergence-free banata hai, toh system evolve hone par volumes (2D mein areas) conserved rehte hain (Liouville).

Canonical equations likhne se pehle

\overset{q}{˙}

(q, p)

ke through express karna kyun zaroori hai?

Warna

\partial H / \partial p

ambiguous hai — tum

H

p

ke saath differentiate nahi kar sakte agar

H

abhi bhi secretly

\overset{q}{˙}

par depend karta ho, kyunki

\overset{q}{˙}

aur

p

independent nahi hain.

Example A mein Newton's second law dobara kyun aata hai?

\overset{x}{˙} = p / m

aur

\overset{p}{˙} = - V^{'} (x)

combine karne par

m \overset{x}{¨} = - V^{'} (x)

milta hai; Hamilton's structure Newton ke equivalent hai, sirf doubled first-order form mein likha gaya hai.

Edge cases

Agar

L

kisi velocity

\overset{q}{˙}_{j}

se independent ho (koi

\overset{q}{˙}_{j}

term nahi), toh formalism ka kya hoga?

Tab

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

ek constraint hai, invertible relation nahi — Legendre transform singular ho jaata hai aur constrained (Dirac) treatment chahiye.

Free particle (

V = 0

) ke liye phase-space trajectory kaisi dikhti hai?

\overset{p}{˙} = 0

toh

p

constant hai, aur

\overset{x}{˙} = p / m

constant hai; trajectory

(x, p)

mein ek horizontal line hai jo uniform speed se traverse hoti hai.

SHO ka phase-space orbit energy

E \to 0

limit mein kya hoga?

Ellipse simat kar single fixed point

(x, p) = (0, 0)

ban jaata hai — particle rest par potential minimum mein, zero enclosed area ka degenerate orbit.

Agar

\partial H / \partial t \neq = 0

ho lekin trajectory momentarily ek aisi jagah ho jahan

\partial H / \partial t = 0

ho, toh kya

H

conserved hai?

Nahi. Conservation ke liye

\partial H / \partial t = 0

poori trajectory par zaroori hai; ek instantaneous zero

H

ko pehle ya baad mein change hone se nahi rokta.

Uniformly rotating frame mein jahan bead ek spinning wire par slide kare,

H

kya hoga?

H

conserved hai (agar setup mein explicit

t

na ho) lekin

H \neq = E

; yeh energy minus ek term

\propto ω^{2}

hai — classic case jahan "energy =

H

" intuition fail hota hai.

Cyclic coordinate ke liye reduced system kaisa dikhta hai?

p_{j}

ek fixed parameter (constant of motion) ban jaata hai, toh effective dynamics fewer dimensions mein live karti hain — yahi woh reduction ka seed hai jo Hamilton-Jacobi Theory aur Noether's Theorem mein use hoti hai.

Recall Aage badhne se pehle self-check

Kya tum, bina dekhe, woh ek reason bata sakte ho jis se $H \neq = E$ generally hota hai? ::: Kinetic energy velocities mein quadratic nahi hai aur/ya constraints explicitly time par depend karti hain (rheonomic). Kya tum bata sakte ho minus sign kyun flip nahi ho sakta? ::: Yeh $d H$ mein $- \overset{p}{˙} d q$ ko $+ \partial_{q} H d q$ se match karne se forced hai; isko flip karna $d H / d t$ cancellation aur isliye energy conservation tod deta hai.

Yeh bhi dekho: Legendre Transform, Phase Space and Liouville's Theorem, Poisson Brackets, Canonical Transformations.