Hamiltonian — definition H = Σpᵢq̇ᵢ − L: Question bank

2.1.11 · D5 · Physics › Analytical Mechanics › Hamiltonian — definition H = Σpᵢq̇ᵢ − L

True or false — justify

Ek valid Hamiltonian ke final form mein phir bhi ek velocity

\overset{q}{˙}_{i}

ho sakti hai.

False.

H

(q, p, t)

ka function define kiya jaata hai; koi bhi bacha hua

\overset{q}{˙}

matlab hai ki tumne inversion

p_{i} = \partial L / \partial \overset{q}{˙}_{i} \Rightarrow \overset{q}{˙}_{i} (q, p)

skip kar di, isliye object ill-defined hai.

Hamiltonian hamesha total energy

T + V

ke barabar hoti hai.

False. Barabar hone ke liye

T

\overset{q}{˙}

mein homogeneous quadratic hona zaroori hai (koi explicit-time ya linear-velocity terms nahi) aur

V

velocity-independent honi chahiye; ek rotating wire par bead yeh condition tod deta hai, aur

H \neq = E

milta hai.

Agar

H

energy ke barabar hai, toh

H

automatically conserved hai.

False. "

H = E

T

aur

V

ki form ke baare mein hai; conservation ka matlab hai

\partial H / \partial t = 0

(yaani

\partial L / \partial t = 0

). Ek time-dependent driving potential

H = E

hone par bhi usse conserved nahi bana sakta.

Agar

H

conserved hai, toh use total energy ke barabar hona chahiye.

False. Rotating-wire bead ke liye

\partial L / \partial t = 0

hai isliye

H

conserved hai, lekin

H = E - m r^{2} ω^{2} \neq = E

. Conservation aur "

H = E

" logically independent hain.

Hamilton's equations Euler–Lagrange ki tarah second-order differential equations hain.

False. Ye

2 n

first-order ODEs hain (

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

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

), jo

n

second-order equations ko replace karti hain — dekho Hamilton's Canonical Equations.

Conjugate momentum

p_{i}

hamesha

m \overset{q}{˙}_{i}

(mass times velocity) ke barabar hota hai.

False.

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

; sirf ek simple

\frac{1}{2} m \overset{q}{˙}^{2}

term ke liye hi yeh

m \overset{q}{˙}

reduce hota hai. Pendulum ke liye yeh

m ℓ^{2} \dot{θ}

(angular momentum) hai, aur magnetic forces ke saath toh isme ek vector-potential term bhi shamil ho jaata hai.

Legendre transform original curve

L

ke baare mein information kho deta hai.

False. Convex

L

ke liye transform invertible hai:

L

aur

H

usi physics ko encode karte hain, ek velocity-as-argument se parametrised, doosra slope (momentum) se — dekho Legendre Transform.

H

ka conserved hona matlab hai ki har individual

p_{i}

conserved hai.

False.

H

conservation ek global statement hai; koi bhi

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

tabhi zero hoga jab

H

us particular

q_{i}

par depend nahi karta (ek cyclic coordinate), jo ek alag condition hai (dekho Conservation Laws & Noether's Theorem).

Spot the error

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

, isliye

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

Error yeh hai ki

\overset{x}{˙}

H

mein chod diya gaya. Tumhe likhna chahiye

H = p \overset{x}{˙} - L

aur phir

\overset{x}{˙} = p / m

substitute karna chahiye, jisse milta hai

H = \frac{p ^{2}}{2 m} + V

— jo

(x, p)

ka function hai.

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

, aur kyunki

H = p \overset{q}{˙} - L

hai, main bas

- L

mein

p \overset{q}{˙}

add karke ruk sakta hoon."

Tum ek step pehle ruk gaye: sum mein abhi bhi

\overset{q}{˙}

hai. Defining step elimination hai —

p (\overset{q}{˙})

ko invert karo taaki

\overset{q}{˙} (p)

mile aur har jagah substitute karo.

"Rotating bead ke liye,

H = T + V

kyunki

V = 0

hai isliye

H = T

Galat:

V = 0

hone par bhi,

H = \frac{p _{r}^{2}}{2 m} - \frac{1}{2} m r^{2} ω^{2}

hai, jabki

E = T = \frac{p _{r}^{2}}{2 m} + \frac{1}{2} m r^{2} ω^{2}

. Rotational term

H

mein sign flip kar leta hai kyunki constraint time-dependent hai.

"Kyunki

\sum_{i} p_{i} \overset{q}{˙}_{i} = 2 T

hai, hame hamesha

H = 2 T - L = T + V

milta hai."

Step

\sum p_{i} \overset{q}{˙}_{i} = 2 T

Euler's theorem use karta hai, jiske liye zaroori hai ki

T

\overset{q}{˙}

mein homogeneous quadratic ho. Agar

T

mein velocity-independent pieces hain (moving constraints), toh Euler's theorem alag multiple deta hai aur

H \neq = T + V

d H

mein

d \overset{q}{˙}

hai, isliye

H

\overset{q}{˙}

par depend karta hai."

Poora Legendre derivation dikhata hai ki

d H = \sum \overset{q}{˙}_{i} d p_{i} - \sum \overset{p}{˙}_{i} d q_{i} - \frac{\partial L}{\partial t} d t

—

p d \overset{q}{˙}

aur

\overset{q}{˙} d p

terms recombine ho jaati hain isliye koi $d \overset{q}{˙}$ survive nahi karta. Yahi vanishing proof hai ki

H = H (q, p, t)

H

define karne ke liye Euler–Lagrange equation ki zaroorat nahi hai."

H

define karne ke liye sirf

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

chahiye. Lekin

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

identify karne ke liye (jo Hamilton's equations padhne mein use hota hai) Euler–Lagrange invoke karna padta hai; uske bina tumhe

H

milta hai lekin clean canonical equations nahi.

Edge cases

Ek free particle (

V = 0

) ke liye, kya

H

phir bhi meaningful aur conserved hai?

Haan:

H = p^{2} /2 m = T = E

, aur kyunki

L

mein koi explicit time ya coordinate dependence nahi hai,

H

conserved hai aur har

p_{i}

bhi conserved hai (sab coordinates cyclic hain).

Kya hota hai Legendre transform ka agar

L

ek velocity mein linear ho (toh

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

Map

p (\overset{q}{˙})

invertible nahi hai (ek degenerate/constrained system);

\overset{q}{˙}

ko solve nahi kiya ja sakta aur standard Hamiltonian construction fail ho jaati hai, jiske liye constraint methods ki zaroorat padti hai.

Agar ek coordinate

q_{k}

cyclic hai (

H

se absent hai), toh uske momentum ka edge behaviour kya hai?

Tab

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

, isliye

p_{k}

ek conserved constant hai — us direction mein ek limiting "trivial" motion (link to Conservation Laws & Noether's Theorem).

Ek equilibrium point par jahan

\partial H / \partial q = 0

aur

p = 0

hai, Hamilton's equations kya predict karte hain?

Dono

\overset{q}{˙} = \partial H / \partial p = 0

aur

\overset{p}{˙} = - \partial H / \partial q = 0

, isliye system hamesha ke liye still baitha rehta hai — Phase Space mein ek fixed point.

Agar

\partial L / \partial t \neq = 0

lekin

H

ittifaaqan

E

ke barabar ho, toh kya energy conserved hai?

Zaroor nahi:

H = E

form ke baare mein hai, lekin

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

matlab hai ki dono

H

aur isliye

E

time mein change karte hain — ek explicitly driven system.

Zero kinetic energy region mein (particle momentarily at rest,

p = 0

) ek system ke liye

H

kya hai?

H

wahaan evaluate kiye gaye

- L

tak reduce ho jaata hai; ek natural system ke liye jisme

H = T + V

ho, yeh sirf

V

hai (saari energy potential) — motion ka ek valid, non-degenerate point.

Kya main, bina dekhe, do independent conditions " $H = E$ " versus " $H$ conserved" state kar sakta hoon? ::: $H = E$ ke liye $T$ ka $\overset{q}{˙}$ mein homogeneous-quadratic hona aur $V$ ka velocity-independent hona zaroori hai; $H$ conserved ke liye $\partial L / \partial t = 0$ chahiye. Dono ek doosre ko imply nahi karte.

Question bank — Hamiltonian — definition H = Σpᵢq̇ᵢ − L

True or false — justify

Spot the error

Why questions

Edge cases

Recall check