Question bank — Hamilton-Jacobi equation

False ki ye equivalent hain, true ki $K=0$ sirf ek option nahi hai: koi bhi constant $K$ phir bhi $\dot{Q},\dot{P}$ uske derivatives se deta hai, lekin $K=0$ isliye choose kiya jaata hai kyunki isse saari naye variables zero extra bookkeeping ke saath constant ho jaati hain.

False. $S$ true path par $L$ ka time integral hai ($S=\int Ldt$); $L$ integrand hai, $S$ accumulated action hai.

False. $H$ ka $p$ mein quadratic hona $\partial S/\partial q$ mein quadratic hona ban jaata hai, isliye $(\partial S/\partial q{)}^2$ jaisi terms aati hain — equation unknown $S$ mein genuinely nonlinear hai.

True. Woh $n$ constants ${\alpha }_i$ precisely naye constant momenta $P_i$ hain; kam se full canonical transformation generate nahi ho sakta, aur ek extra additive constant trivial hai kyunki sirf $S$ ke derivatives matter karte hain.

False. $W$ ke liye energy conservation chahiye ($H$ mein explicit $t$ na ho). Agar $H=H(q,p,t)$ genuinely time par depend karta hai, toh $-Et$ separate nahi kar sakte aur $W$ undefined hai.

False. Naye variables $Q,P$ constant hain, lekin physical position $q$ ko $Q=\partial S/\partial P$ ko $q(t)$ ke liye invert karke recover kiya jaata hai — particle move karta hai; sirf cleverly chosen labels still rehte hain.

False. Trajectories $S$ ke derivatives se aati hain ($p$ aur $Q$ ke through); $S$ mein constant add karne se koi derivative nahi badlta, hence koi physics nahi badlti.

True — lekin woh single equation ek first-order PDE hai, kaafi ordinary equations ko ek partial equation mein compress karta hai; difficulty compress hoti hai, delete nahi.

Error: type-2 relations hain $p=\partial S/\partial q$ aur $Q=\partial S/\partial P$. $q=\partial S/\partial p$ jaisi koi relation nahi hai; $S$ depend karta hai $q$ aur $P$ par, $p$ par nahi.

Error: $K=0$ deta hai $H=-\partial S/\partial t$, na ki $H=0$. Energy-like $H$ generally nonzero hota hai; yeh exactly $S$ ke time-slope se balance hota hai.

Error: $S$ actual (extremal) trajectory par action hai, kisi arbitrary path par nahi. True path se hatke, $\int Ldt$ Hamilton's principal function nahi hai.

Error: $E$ ek constant hai (conserved energy), separation constants ${\alpha }_i$ mein se ek. Yeh har $q$ par same number hai; equation us fixed $E$ ko given manke $\partial W/\partial q$ determine karta hai.

Error: $P=E$ choose karna ek convenient option hai, rule nahi. Koi bhi independent constant of integration ${\alpha }_i$ ko naya momentum label kiya ja sakta hai; $E$ sirf natural pick hai jab energy conserved ho.

Error: equation unhe hide karti hai, lekin uska complete integral $n$ constants carry karta hai, aur $P_i={\alpha }_i$ identify karna hi hai jo $Q_i=\partial S/\partial P_i$ ko solution produce karne deta hai. Inhe drop karo toh transformation hi nahi banta.

Error: $\partial W/\partial t=0$ (ye time-independent hai), lekin $S=W-Et$ mein $\partial S/\partial t=-E\ne 0$ hai. Woh $-E$ exactly wahi hai jo full HJ equation ko feed karta hai.

Type-2 $F_2=S(q,P,t)$ naye momenta $P$ ko independent variables treat karta hai, jo exactly wahi hai jo hum constant rakhna chahte hain; type-1 depend karta hai $Q$ par, jo awkward hai jab goal constant $P$ rakhna ho. Dekho Canonical Transformations.

Hamilton's equations dete hain $\dot{Q}=\partial K/\partial P$ aur $\dot{P}=-\partial K/\partial Q$. Agar $K$ identically zero hai, toh dono partials har jagah vanish karte hain, isliye $\dot{Q}=\dot{P}=0$ — naye variables frozen constants hain.

HJ action $S$ ke liye solve karta hai, jiska pehla slope in $q$ already momentum encode karta hai ($p=\partial S/\partial q$); Newton ka "second order" absorb ho jaata hai kyunki momentum ab ek separate variable nahi, unknown ka derivative hai.

Separation $S=W(q)-Et$ ke liye $\partial S/\partial t=-E$ ko constant hona chahiye. Ye tab kaam karta hai jab $H$ (hence $E$) conserved ho; explicitly time-dependent $H$ ek constant $E$ produce nahi karta jise peel off kar sakein. Dekho Separation of Variables.

Kyunki $S=\int Ldt$ action hai, aur true path par $dS/dt=\sum p_i{\dot{q}}_i-H=L$; HJ equation phir sirf yeh statement hai ki yahi action satisfy karta hai $H(q,\partial S/\partial q,t)+\partial S/\partial t=0$.

Jab motion periodic hota hai, naye constant momenta ko action variables (integrals $\oint pdq$) choose karne se $W$ aise angle variables generate karta hai jo time mein linearly advance karte hain — HJ woh machinery hai jo unhe produce karta hai.

Small wavelength ki limit mein, quantum wavefunction ki phase HJ equation follow karti hai jahan $S$ woh phase hai; classical mechanics quantum mechanics ki "geometric optics" hai, aur HJ uska wavefront equation hai.

Ek particular solution mein free constants nahi hote; sirf ek complete integral (saare $n$ independent ${\alpha }_i$ ke saath) ko un constants ke respect mein differentiate karke $Q_i$ mila sakte hain aur thereby trajectories ki poori family.

$S=\sqrt{2mE}x-Et\to 0$: zero energy matlab zero momentum ($p=\sqrt{2mE}\to 0$) aur particle still baitha hai, isliye (motionless) path par action vanish ho jaata hai.

$E>0$ ke liye nahi: $v=\sqrt{2E/m}$ finite hai. Exactly $E=0$ par particle at rest hai ($v=0$), ek legitimate degenerate case, singularity nahi.

$W^{\prime }=\sqrt{2m(E-\frac{1}{2}m{\omega }^2{x}^2)}\to 0$: wahan momentum vanish karta hai (particle instantaneously ruk jaata hai), aur $W$ mein integrand ka square-root edge integrable hai, koi true blow-up nahi.

Haan — full HJ equation $H(q,\partial S/\partial q,t)+\partial S/\partial t=0$ unchanged aur valid hai; sirf shortcut $S=W-Et$ aur characteristic function $W$ unavailable ho jaate hain.

Transformation locally kaam karta rehta hai, lekin motion ki frequencies match karti hain, jo ek degeneracy signal karti hai; action-angle language mein extra conserved quantities appear hoti hain aur orbits fewer independent periods mein close ho jaate hain.

Exactly ek — commonly $E$ (ya $\alpha =P$). Yeh naya constant momentum hai; paired naya coordinate $Q=\partial S/\partial E$ woh constant hai jo phase ya start time fix karta hai (e.g. free particle ke liye $-t_0$).

$Q$ ek frozen constant hai jo (minus) us time ke barabar hai jab particle origin se guzra tha; ise invert karne par, $x=\sqrt{2E/m}(t-t_0)$, uniform motion recover hota hai — constant encode karta hai kab, na ki kahan.