Canonical transformations — generating functions: Worked examples

2.1.16 · D3 · Physics › Analytical Mechanics › Canonical transformations — generating functions

Scenario matrix

Har generating-function problem in cells mein se kisi ek mein fit hoti hai. Hamara kaam neeche har ek ko cover karna hai.

Cell	Kya cheez ise alag banati hai	Covered by
A. Direct forward	given $F$ , differentiate karke $(Q, P)$ padh lo	Ex 1
B. Implicit inversion	relations mein $q, Q$ aapas mein uljhe hain; solve karna padega	Ex 2
C. Sign / branch care	square roots aur inverse trig → quadrant + sign ambiguity	Ex 3
D. Degenerate input	koi variable zero hai; origin $q = p = 0$ ( $P = 0$ ) oscillator action–angle map ki ek sachchi coordinate singularity hai	Ex 4
E. Time-dependent $F$	$\partial F / \partial t \neq = 0$ , isliye $K \neq = H$	Ex 5
F. Test-only (no $F$ given)	Poisson brackets se prove karo ki map canonical hai	Ex 6
G. Word problem	physical oscillator, real numbers aur units ke saath	Ex 7
H. Exam twist	ek "trap": sahi type chuno, ya non-canonical map pakdo	Ex 8

Hum A→H chalte hain. Teen figures geometry carry karte hain: Figure 1 = phase-space picture (Ex 3), Figure 2 = sign-quadrant schematic (Ex 3), Figure 3 = non-canonical bracket plot (Ex 8).

Example 1 — Cell A: seedha padh lo

Worked example

$F_{2} = q P + \frac{1}{2} P^{2}$ (single degree of freedom) Transformation $(q, p) \to (Q, P)$ nikalo.

Forecast: $q P$ wala part identity hai (parent se). Extra $\frac{1}{2} P^{2}$ ka $Q$ ko $P$ mein linear kuch nudge karna chahiye. Guess karo $Q = q + P$ , $p = P$ .

Steps.

Yeh Type-2 generator hai, isliye $p = \partial F_{2} / \partial q$ aur $Q = \partial F_{2} / \partial P$ use karo. Yeh step kyun? $F_{2}$ old $q$ aur new $P$ par depend karta hai — yahi pairing Type 2 ki definition hai.
$p = \frac{\partial}{\partial q} (q P + \frac{1}{2} P^{2}) = P$ . Kyun? sirf $q P$ term mein $q$ hai; uska $q$ -derivative $P$ pull karta hai.
$Q = \frac{\partial}{\partial P} (q P + \frac{1}{2} P^{2}) = q + P$ . Kyun? $P$ mein har term differentiate karo: $q P \to q$ , $\frac{1}{2} P^{2} \to P$ .
New-in-old express karne ke liye invert karo: step 2 kehta hai $P = p$ ; use step 3 mein substitute karo: $Q = q + p$ . Yeh step kyun? steps 2–3 answer partly new variable $P$ ke terms mein dete hain; transformation new-in-old padhna chahiye, toh already-solved relation $P = p$ se $P$ eliminate karte hain.

Verify: Poisson bracket compute karo ${Q, P} = \frac{\partial Q}{\partial q} \frac{\partial P}{\partial p} - \frac{\partial Q}{\partial p} \frac{\partial P}{\partial q} = (1) (1) - (1) (0) = 1$ . ✓ Canonical. Forecast sahi tha.

Example 2 — Cell B: uljhe hue relations, solve karna padega

Worked example

$F_{1} = \frac{q ^{2}}{2 Q}$ $(Q, P)$ ko $(q, p)$ ke terms mein nikalo.

Forecast: $F_{1}$ mein $q$ aur $Q$ nonlinearly mix hain, toh $p$ aur $P$ dono ratios ke roop mein milenge; $Q (q, p)$ nikalne ke liye algebraically untangle karna hoga. Expect karo $Q \propto q / p$ .

Steps.

Type-1 rules: $p = \partial F_{1} / \partial q$ , $P = - \partial F_{1} / \partial Q$ . Kyun? $F_{1} (q, Q)$ = old $q$ , new $Q$ .
$p = \frac{\partial}{\partial q} \frac{q ^{2}}{2 Q} = \frac{q}{Q}$ . Kyun? $Q$ fixed hai; $\frac{d}{d q} q^{2} = 2 q$ .
$P = - \frac{\partial}{\partial Q} \frac{q ^{2}}{2 Q} = - (- \frac{q ^{2}}{2 Q ^{2}}) = \frac{q ^{2}}{2 Q ^{2}}$ . Kyun? $\frac{d}{d Q} Q^{- 1} = - Q^{- 2}$ , aur leading minus ise positive kar deta hai.
Untangle karo. Step 2 se, $Q = q / p$ . Step 3 mein substitute karo: $P = \frac{q ^{2}}{2 ( q / p ) ^{2}} = \frac{q ^{2} p ^{2}}{2 q ^{2}} = \frac{p ^{2}}{2}$ . Yeh step kyun? problem new variables ko old ones ke functions ke roop mein maangti hai — step 2/3 ne unhe leftover $Q$ ke terms mein diya, jise hum eliminate karte hain.

Toh $Q = \frac{q}{p}, P = \frac{p ^{2}}{2} .$

Verify: ${Q, P} = \frac{\partial Q}{\partial q} \frac{\partial P}{\partial p} - \frac{\partial Q}{\partial p} \frac{\partial P}{\partial q}$ . Compute karo: $\frac{\partial Q}{\partial q} = 1/ p, \frac{\partial P}{\partial p} = p, \frac{\partial Q}{\partial p} = - q / p^{2}, \frac{\partial P}{\partial q} = 0$ . Toh ${Q, P} = (1/ p) (p) - (- q / p^{2}) (0) = 1$ . ✓

Example 3 — Cell C: signs, branches aur charon quadrants

Yahan geometry matter karti hai, isliye hum phase plane draw karte hain. Phase point $(q, p)$ hai; origin se uski doori aur uska angle dono meaning carry karte hain. Figure 1 phase plane dikhata hai jisme har quadrant mein ek worked point hai.

Figure — Canonical transformations — generating functions

Worked example Oscillator generator, carefully kiya gaya:

$F_{1} = \frac{1}{2} mω q^{2} cot Q$ with $m = ω = 1$ Parent note mein humne $q = 2 P sin Q$ , $p = 2 P cos Q$ (with $m = ω = 1$ ) nikala tha. Phase point $(q, p) = (- 1, - 1)$ diya gaya, $(Q, P)$ nikalo — aur quadrant sahi karo.

Forecast: $q, p$ dono negative hain toh phase plane ke third quadrant mein hain (Figure 1 dekho). Radius $2$ hai. Toh expect karo $P = 1$ aur $Q$ ek angle jo third quadrant mein hai, $22 5^{\circ}$ ke paas.

Steps.

Angle kill karne ke liye square-and-add karo: $q^{2} + p^{2} = 2 P (sin^{2} Q + cos^{2} Q) = 2 P$ . Yeh step kyun? $sin^{2} + cos^{2} = 1$ se $P$ isolate hota hai bina kisi sign ambiguity ke — squares hamesha $\geq 0$ hote hain.
$P = \frac{q ^{2} + p ^{2}}{2} = \frac{1 + 1}{2} = 1$ . Kyun? bas plug in karo; $P$ = (radius $^{2}$ )/2.
Angle ke liye divide karo: $\frac{q}{p} = \frac{sin Q}{cos Q} = tan Q$ , toh naively $Q = arctan (q / p) = arctan (1) = 4 5^{\circ}$ . Caution kyun? $tan$ har $18 0^{\circ}$ pe repeat karta hai, toh $arctan$ akele nahi bata sakta $4 5^{\circ}$ vs $22 5^{\circ}$ .
Branch fix karo $q$ aur $p$ ke signs se (Figure 1 mein orange point dekho): $q < 0$ matlab $sin Q < 0$ , $p < 0$ matlab $cos Q < 0$ — dono negative ⟹ third quadrant ⟹ $18 0^{\circ}$ add karo: $Q = 4 5^{\circ} + 18 0^{\circ} = 22 5^{\circ} = \frac{5 π}{4}$ . Yeh step kyun? yahi hai arctan-quadrant fix: ratio sign information kho deta hai, individual signs use restore karte hain.

Charon cases (Figure 1 mein char coloured points; quadrant logic Figure 2 mein schematise kiya gaya hai), $r = q^{2} + p^{2}$ ke saath:

$(q, p)$	quadrant	$sin Q$	$cos Q$	$Q$	shift kyun?
$(+, +)$	I	$+$	$+$	$arctan (q / p)$	ratio $> 0$ , $arctan$ already $(0, 9 0^{\circ})$ mein land karta hai
$(+, -)$	II	$+$	$-$	$arctan (q / p) + 18 0^{\circ}$	$q / p < 0$ toh $arctan (q / p) \in (- 9 0^{\circ}, 0)$ — negative, $18 0^{\circ}$ off; add karo
$(-, -)$	III	$-$	$-$	$arctan (q / p) + 18 0^{\circ}$	$q / p > 0$ toh $arctan$ $(0, 9 0^{\circ})$ deta hai par sahi angle $18 0^{\circ}$ aage hai; add karo
$(-, +)$	IV	$-$	$+$	$arctan (q / p) + 36 0^{\circ}$ (ya $-	...

Boundary / axis cases (jahan ratio $q / p$ ill-posed hai):

point	kaun sa 0 hai	$tan Q = q / p$ ?	sahi $Q$	reason
$q = 0, p > 0$	$sin Q = 0$	$0/ p = 0$ , theek hai	$Q = 0^{\circ}$	$+ p$ axis par; $cos Q > 0$
$q = 0, p < 0$	$sin Q = 0$	$0/ p = 0$ , theek hai	$Q = 18 0^{\circ}$	$- p$ axis par; $cos Q < 0$
$q > 0, p = 0$	$cos Q = 0$	$q /0 = \infty$ undefined	$Q = 9 0^{\circ}$	division fail; $sin Q > 0$ se padho
$q < 0, p = 0$	$cos Q = 0$	$q /0 = \infty$ undefined	$Q = 27 0^{\circ}$	division fail; $sin Q < 0$ se padho
$q = 0, p = 0$	dono	$0/0$ indeterminate	$Q$ undefined	origin: $P = 0$ , angle meaningless (Ex 4 dekho)

Axis cases kyun matter karte hain: step 4 ka branch-fix procedure $p$ se divide karta hai; $p = 0$ axis par woh division blow up karta hai, toh ratio pe trust karne ki jagah surviving sign of $sin Q$ (yaani $q$ ka sign) se angle seedha padhna hoga.

Verify: reconstruct karo — $q = 2 \cdot 1 sin 22 5^{\circ} = 2 \cdot (- \frac{1}{2}) = - 1$ ✓, $p = 2 cos 22 5^{\circ} = - 1$ ✓.

Figure 2 un sign rules ko charon quadrants aur charon axis cases ka ek schematic map banata hai.

Example 4 — Cell D: degenerate / zero input

Worked example Oscillator generator mein

$Q = 0$ par kya hota hai? $F_{1} = \frac{1}{2} mω q^{2} cot Q$ use karke, woh point examine karo jahan $Q \to 0$ .

Forecast: $cot Q = cos Q / sin Q \to \infty$ jab $Q \to 0$ . Kuch blow up karta hai. Par physical answer ( $q, p$ ) finite rehna chahiye — yeh ek coordinate artefact hai, jaise Earth ke map par North Pole.

Steps.

Yaad karo $q = 2 P / (mω) sin Q$ . Jab $Q \to 0$ , $sin Q \to 0$ , toh $q \to 0$ . Kyun? sine factor position ko zero force karta hai $P$ se independent.
Aur $p = 2 P mω cos Q \to 2 P mω$ (maximum). Kyun? $cos 0 = 1$ ; particle origin par sabse tez chal raha hai — exactly oscillator $q = 0$ se guzarta hua.
Par generator khud: $F_{1} \propto q^{2} cot Q$ . $q = 2 P / (mω) sin Q$ rakhne par, $q^{2} cot Q = \frac{2 P}{mω} sin^{2} Q \cdot \frac{cos Q}{sin Q} = \frac{2 P}{mω} sin Q cos Q \to 0$ . Yeh step kyun? apparent blow-up cancel ho jaata hai: $sin^{2} Q$ single $1/ sin Q$ ko beat karta hai. Transformation $Q = 0$ par perfectly regular hai; sirf formula ki naive form singular dikhti hai.

Verify (limit): $Q \to 0 lim q^{2} cot Q$ with $q = 2 P sin Q$ ( $m = ω = 1$ ) $= lim_{Q \to 0} 2 P sin Q cos Q = 0$ . ✓ Koi true singularity nahi.

Jahan yeh actually break hota hai: Jacobian $\partial (Q, P) / \partial (q, p)$ nonzero hona chahiye taaki map locally invertible ho. $q = 0, p = 0$ (origin, $P = 0$ ) par angle $Q$ undefined hai — woh single point genuine coordinate singularity hai, exactly jaise pole par longitude.

Example 5 — Cell E: time-dependent generator, isliye $K \neq = H$

Worked example Uniformly accelerating frame,

$F_{2} = (q - \frac{1}{2} a t^{2}) P + ma t q$ se Free particle ka $H = \frac{p ^{2}}{2 m}$ hai. Upar diya gaya $F_{2}$ use karo (constant acceleration $a$ ke saath, aur $t$ woh independent time coordinate hai jisme har variable evolve karta hai) aur new Hamiltonian $K$ nikalo. Yahan $Q$ acceleration $a$ se accelerate karte frame mein position hai.

Forecast: kyunki $F_{2}$ explicitly $t$ par depend karta hai, parent ka rule $K = H + \partial F / \partial t$ matlab hai $K \neq = H$ . Accelerating frame mein fictitious force term aana chahiye (kuch $Q$ mein linear).

Steps.

Type-2 relations: $p = \partial F_{2} / \partial q$ , $Q = \partial F_{2} / \partial P$ , $K = H + \partial F_{2} / \partial t$ . Kyun? $F_{2} (q, P, t)$ ; last term isliye aata hai kyunki $F_{2}$ time coordinate $t$ par explicitly depend karta hai.
$p = \frac{\partial F _{2}}{\partial q} = P + ma t$ . Kyun? $\frac{\partial}{\partial q} [(q - \frac{1}{2} a t^{2}) P] = P$ aur $\frac{\partial}{\partial q} [ma tq] = ma t$ . Toh $P = p - ma t$ (momentum moving frame ke relative measure kiya gaya).
$Q = \frac{\partial F _{2}}{\partial P} = q - \frac{1}{2} a t^{2}$ . Kyun? sirf pehla bracket $P$ carry karta hai. Toh $Q$ position minus frame ka displacement $\frac{1}{2} a t^{2}$ hai — exactly "accelerating frame mein position."
$\frac{\partial F _{2}}{\partial t} = (- a t) P + ma q = - a tP + ma q$ . Kyun? explicit $t$ 's differentiate karo ( $q, P$ ko partial ki demand ke anusaar fixed rakhte hue): $\frac{\partial}{\partial t} (- \frac{1}{2} a t^{2}) P = - a tP$ , aur $\frac{\partial}{\partial t} (ma tq) = ma q$ .
$K = H + \partial_{t} F_{2} = \frac{p ^{2}}{2 m} - a tP + ma q$ assemble karo. Ab new variables mein rewrite karo ( $p = P + ma t$ , $q = Q + \frac{1}{2} a t^{2}$ ): $K = \frac{( P + ma t ) ^{2}}{2 m} - a tP + ma (Q + \frac{1}{2} a t^{2}) .$ $\frac{( P + ma t ) ^{2}}{2 m} = \frac{P ^{2}}{2 m} + a tP + \frac{m a ^{2} t ^{2}}{2}$ expand karo. $+ a tP$ , $- a tP$ cancel ho jaate hain: $K = \frac{P ^{2}}{2 m} + ma Q + m a^{2} t^{2} (\frac{1}{2} m a^{2} t^{2} + \frac{1}{2} m a^{2} t^{2}) .$ Yeh step kyun? $K$ ko apne variables mein express karna zaroori hai taaki dynamics padh sakein; leftover pure-time term $m a^{2} t^{2}$ koi force nahi add karta (uske $Q, P$ derivatives zero hain) aur drop kiya ja sakta hai.
Effective Hamiltonian: $K = \frac{P ^{2}}{2 m} + ma Q$ . Term $ma Q$ ek linear potential = fictitious force $- ma$ hai — exactly wahi jo accelerating car mein feel hota hai. $m a^{2} t^{2}$ drop karna legitimate kyun hai: sirf $t$ par dependent term function $\int m a^{2} t^{2} d t$ ka total time derivative hai, aur Hamiltonian mein total time derivative add karna ek canonical gauge change hai — woh action mein sirf endpoint term ke roop mein aata hai, isliye $\dot{Q} = \partial K / \partial P$ aur $\dot{P} = - \partial K / \partial Q$ bilkul unchanged rehte hain (uske $Q$ - aur $P$ -derivatives zero hain). Koi dynamics nahi khoti.

Verify: new vars mein Hamilton: $\dot{Q} = \partial K / \partial P = P / m$ , $\dot{P} = - \partial K / \partial Q = - ma$ . Toh frame mein particle $- a$ se accelerate karta hai — correct pseudo-force. ✓ (Extreme case ke liye Hamilton–Jacobi equation dekho jo $K \to 0$ drive karta hai.)

Example 6 — Cell F: koi generator nahi diya, Poisson brackets se canonicity prove karo

Worked example Kya

$Q = ln (\frac{s i n p}{q}), P = q cot p$ canonical hai? Koi $F$ supply nahi kiya gaya. Seedha test karo.

Forecast: yeh engineered lage hain taaki work out ho — ${Q, P} = 1$ pe bet karo.

Domain pehle. $Q = ln (sin p / q)$ real number ho, iske liye log ka argument positive chahiye: $\frac{sin p}{q} > 0$ . Natural patch pe restrict karo: $q > 0$ aur $0 < p < π$ (taaki $sin p > 0$ ); phir $P$ mein $cot p$ , $p = 0, π$ (jahan $sin p = 0$ aur dono $Q$ aur $P$ blow up) ke siwa finite hai. Yeh kyun matter karta hai: map sirf us open region par valid canonical transformation hai jahan uske formulas smooth aur invertible hain — log aur $cot$ dono woh region carve out karte hain. Boundary $p \to 0^{+}$ ya $p \to π^{-}$ par, $Q \to + \infty$ : ek edge hai, chart ka part nahi.

Steps.

Canonicity ka ek hi algebraic test hai ${Q, P} = 1$ (one degree of freedom). Yeh step kyun? fundamental Poisson bracket ${Q, P} = 1$ preserve karna Hamilton's equations preserve karne ke barabar hai — Poisson brackets dekho.
Likho $Q = ln sin p - ln q$ (valid kyunki dono factors hamare patch par positive hain). Partials: $\frac{\partial Q}{\partial q} = - \frac{1}{q}, \frac{\partial Q}{\partial p} = cot p$ . Kyun? har log par chain rule.
$P = q cot p$ : $\frac{\partial P}{\partial q} = cot p, \frac{\partial P}{\partial p} = - q csc^{2} p$ . Kyun? $\frac{d}{d p} cot p = - csc^{2} p$ .
${Q, P} = \frac{\partial Q}{\partial q} \frac{\partial P}{\partial p} - \frac{\partial Q}{\partial p} \frac{\partial P}{\partial q} = (- \frac{1}{q}) (- q csc^{2} p) - (cot p) (cot p) = csc^{2} p - cot^{2} p = 1.$ Last equality kyun? identity $csc^{2} p - cot^{2} p = 1$ (Pythagoras ko $sin^{2}$ se divide karo).

Verify: ${Q, P} = 1$ poore patch $q > 0, 0 < p < π$ par. ✓ Wahan canonical hai.

Example 7 — Cell G: numbers aur units ke saath real-world word problem

Worked example Spring par mass, generator se solve kiya gaya

$m = 2 kg$ ka block, $k = 8 N/m$ ki spring par, $q_{0} = 0.10 m$ se rest se shuru hota hai. Oscillator generator use karke action–angle form mein amplitude nikalo aur period nikalo.

Forecast: angular frequency $ω = k / m = 2 rad/s$ , toh period $\approx 3.14 s$ . Shuru mein sab potential energy hai: $E = \frac{1}{2} k q_{0}^{2}$ .

Steps.

$ω = k / m = 8/2 = 2 rad/s$ . Kyun? harmonic Hamiltonian $H = \frac{p ^{2}}{2 m} + \frac{1}{2} m ω^{2} q^{2}$ mein $m ω^{2} = k$ chahiye, toh $ω = k / m$ .
Total energy: $q_{0}$ par rest mein, $E = \frac{1}{2} k q_{0}^{2} = \frac{1}{2} (8) (0.10)^{2} = 0.04 J$ . Kyun? initially $p = 0$ , toh $H$ = sirf spring potential energy.
Parent se, $K = ω P$ aur $K = E$ (energy conserved, $F$ time-independent toh $K = H = E$ ). Isliye action $P = E / ω = 0.04/2 = 0.02 J\cdots$ . Yeh step kyun? $Q$ cyclic hai ⟹ $P$ conserved constant hai jo $E / ω$ ke barabar hai.
Angle evolve hota hai: $Q = ω t + ϕ$ ( $t$ = independent time coordinate). Period $T = 2 π / ω = 2 π /2 = π \approx 3.1416 s$ . Kyun? $Q$ har cycle mein constant rate $\dot{Q} = ω$ se $2 π$ badhta hai.
Amplitude check via $q = 2 P / (mω) sin Q$ : max $q_{m a x} = 2 (0.02) / (2 \cdot 2) = 0.01 = 0.10 m = q_{0}$ . Kyun? turning point par rest se shuru kiya, toh $q_{0}$ hai hi amplitude.

Verify (units aur numbers): $[P] = J\cdots$ (action, sahi dimension); $ω = 2 rad/s$ , $E = 0.04 J$ , $P = 0.02 J\cdots$ , $T = π s \approx 3.14 s$ , amplitude $= 0.10 m$ . ✓ Dekho Action–angle variables.

Example 8 — Cell H: exam twist (non-canonical map pakdo)

Worked example Trap: kya

$Q = q^{2}, P = p$ canonical hai? Nahi toh, momentum fix karo. Forecast: doubling-type maps usually bracket break karte hain. Bet hai yeh canonical nahi hai, aur repair $P$ ko $1/ (2 q)$ se scale karti hai.

Steps.

${Q, P}$ test karo. $\frac{\partial Q}{\partial q} = 2 q, \frac{\partial Q}{\partial p} = 0, \frac{\partial P}{\partial q} = 0, \frac{\partial P}{\partial p} = 1$ . Kyun? yahi $Q = q^{2}$ aur $P = p$ ke nonzero partial derivatives hain.
${Q, P} = (2 q) (1) - (0) (0) = 2 q \neq = 1$ . Yeh kyun matter karta hai? canonical map exactly $1$ deni chahiye; $2 q$ , $q = \frac{1}{2}$ ke siwa har jagah fail karta hai. Toh $Q = q^{2}, P = p$ canonical nahi hai.
Repair — momentum fix karo. $Q = q^{2}$ rakho par $P = f (q) p$ with ${Q, P} = 1$ dhundho. Phir ${q^{2}, f (q) p} = \frac{\partial ( q ^{2} )}{\partial q} \frac{\partial ( f ( q ) p )}{\partial p} = (2 q) f (q) = 1 \Rightarrow f (q) = \frac{1}{2 q}$ . Yeh step kyun? hum woh multiplier solve karte hain jo bracket ko $1$ force kare; sirf $p$ ki $q$ -dependent scaling offending factor $2 q$ cancel kar sakti hai.
Toh canonical version hai $Q = q^{2}, P = \frac{p}{2 q}$ . Yeh statement mein promised fix kyun hai: coordinate change $Q = q^{2}$ rehta hai, aur sirf momentum $1/ (2 q)$ se rescale hota hai taaki ${Q, P} = 1$ restore ho.

Verify: ${q^{2}, \frac{p}{2 q}} = \frac{\partial ( q ^{2} )}{\partial q} \frac{\partial}{\partial p} (\frac{p}{2 q}) - 0 = (2 q) (\frac{1}{2 q}) = 1$ . ✓

Generator se sanity check: yeh exactly Type-2 $F_{2} = q^{2} P$ hai. Check karo: $Q = \partial_{P} F_{2} = q^{2}$ ✓ aur $p = \partial_{q} F_{2} = 2 q P \Rightarrow P = p / (2 q)$ ✓. Consistent. Figure 3 ${Q, P} = 2 q$ ko required value $1$ ke against plot karta hai, $q = \frac{1}{2}$ par single accidental crossing dikhata hai.

Active recall

Recall Khud se test karo (answers chhupaao)

Ex 3 mein, $arctan (q / p)$ akele $Q$ kyun nahi de sakta, aur $p = 0$ axis par kya hota hai?
Ex 4 mein, kya $Q = 0$ par $F_{1}$ blow up karna matlab hai transformation wahan singular hai? Ek sachchi singularity kahan hai?
Ex 5 mein, $K$ ka kaun sa term fictitious force hai, aur pure-time term kyun drop kar sakte hain?
Ex 6 mein, kaun sa domain $Q = ln (sin p / q)$ well-defined banata hai?
Ex 8 mein, $Q = q^{2}$ repair karne wala multiplier kya hai?

Kis point par

Q = q^{2}, P = p

bracket accidentally satisfy karta hai?

Sirf

q = \frac{1}{2}

par, jahan

{Q, P} = 2 q = 1

m = 2, k = 8

block ka action?

P = E / ω = 0.04/2 = 0.02 J\cdots

Us block ka period?

T = 2 π / ω = π \approx 3.14 s

Ex 5 mein effective Hamiltonian?

K = \frac{P ^{2}}{2 m} + ma Q

, jo pseudo-force

- ma

deta hai.

Pure-time term

K

se kyun drop kar sakte hain?

Woh total time derivative hai — ek canonical gauge change — isliye uske

Q, P

derivatives zero hain aur dynamics unchanged rehti hai.

p = 0

axis par, arctan branch-fix kyun fail karta hai?

Woh

p

se divide karta hai; iske bajaye

Q

seedha

sin Q

ke sign se (yaani

q

se) padho:

q > 0 \Rightarrow 9 0^{\circ}

q < 0 \Rightarrow 27 0^{\circ}

Oscillator action–angle map ki ek sachchi coordinate singularity kahan hai?

Origin

q = p = 0

(

P = 0

) par, jahan angle

Q

undefined hai.

Domain jo Ex 6 ka map well-defined banata hai?

q > 0

aur

0 < p < π

(taaki

sin p / q > 0

log ke liye aur

cot p

finite ho).

Connections

Poisson brackets — ${Q, P} = 1$ test jo Ex 6 aur Ex 8 mein use kiya gaya.
Hamilton's equations — woh form jo har example preserve karta hai.
Action–angle variables — jahan Ex 3 aur Ex 7 le jaate hain.
Hamilton–Jacobi equation — ultimate time-dependent generator (Ex 5 ka cousin).
← parent topic par wapas