2.1.16 · D2 · HinglishAnalytical Mechanics
Visual walkthrough — Canonical transformations — generating functions
2.1.16 · D2· Physics › Analytical Mechanics › Canonical transformations — generating functions
Hum assume karte hain ki tumhe almost kuch nahi pata. Har symbol ko earn karte hain.
Step 0 — Stage set karo: "phase space" kya hota hai?
Figure dekho. Horizontal axis hai, vertical axis hai. Ek dot = abhi ka poora state. Curve = uska future aur past.
Ek canonical transformation is plane ki ek relabelling hai: naye axes (nayi position) aur (naya momentum), is tarah chune gaye ki physics clean rahe.
Step 1 — Woh rule jo motion obey karta hai (Hamilton's equations)
mein dot ka matlab hai "time ke saath change ki rate" — woh quantity abhi kitni tezi se move kar rahi hai. Hamilton's equations kehte hain:
\underbrace{\dot p}_{\text{momentum kaise drift karta hai}}=-\underbrace{\frac{\partial H}{\partial q}}_{\text{energy ka slope }q\text{ direction mein}}.$$ > [!intuition] Ise ek flow ki tarah padho > $H$ ek *pahaadi* ki tarah hai $(q,p)$ plane ke upar. Hamilton's equations kehte hain ki state *neechay nahi jaati* — woh $H$ ki contour lines ke *saath saath* flow karti hai (slope ke perpendicular, us crucial minus sign ke saath jo usse circulate karata hai). Figure mein energy contours loops ki tarah hain aur flow arrows unke **around** chal rahe hain. ![[deepdives/dd-physics-2.1.16-d2-s02.png]] **Hamara goal:** naye coordinates $(Q,P)$ dhundna jismein yeh flow *boring simple* ho — ideally straight lines. Lekin hum sirf wahi relabellings use kar sakte hain jo flow rule ko bilkul same dikhne deti hain: $$\dot Q=\frac{\partial K}{\partial P},\qquad \dot P=-\frac{\partial K}{\partial Q},$$ kisi *naye* energy function $K(Q,P)$ ke liye. Yeh "form same rakhna" ki requirement hi ==canonical== word ka matlab hai. --- ## Step 2 — Rule kahan se aata hai: the action principle > [!definition] Action $S$ > Nature actual trajectory woh choose karti hai jo ek certain running total — ==action== — > $$S=\int\big(p\,\dot q-H\big)\,dt$$ > ko **stationary** banati hai: path ko thoda nudge karo (endpoints fixed rakhte hue) aur $S$ first order mein nahi badlata. Woh "no first-order change" $\delta S=0$ likha jaata hai. Integrand $p\dot q - H$ ko symbol by symbol decode karte hain: - $p\dot q$ — momentum times velocity, ek piece of "motion kaise accrue hoti hai". - $-H$ — energy subtract karo. - $\int\,dt$ — time par sab add karo. > [!intuition] Picture (ek stretched-elastic path) > Sochi ki $q$-vs-time graph par asli path ek tani hui elastic band hai. Koi bhi wiggle jo tum add karo *action ko lambi kar deti hai* — isliye bina wiggle ki, tani hui band real motion hai. Figure mein asli path coral mein hai aur ek wiggled competitor dashed lavender mein; sirf wiggle-free wala "stationary" hai. ![[deepdives/dd-physics-2.1.16-d2-s03.png]] Key fact yeh hai: **jo bhi integrand tumhe yeh stationary path deta hai, wahi Hamilton's equations bhi deta hai.** Isliye agar do integrands *same* stationary paths par laate hain, toh woh same physics describe karte hain. --- ## Step 3 — Ek move: do integrands, ek physics Hum chahte hain ki *naye* variables apne *khud ke* action principle obey karein, $$\delta\!\int\big(P\dot Q-K\big)\,dt=0,$$ **same** trajectories ke saath (same physics!). Do integrands kab identical stationary paths dete hain? > [!definition] Total time derivative > $\dfrac{dF}{dt}$ ek function $F$ ki ==motion ke saath change ki rate== hai. Iska integral sirf $F(\text{end})-F(\text{start})$ hai — yeh *sirf endpoints par* depend karta hai, path par nahi. > [!intuition] Yeh magic key kyun hai > Agar tum integrand mein $\dfrac{dF}{dt}$ add karo, toh extra bit integrate hokar $F_\text{end}-F_\text{start}$ deta hai. Lekin hum endpoints ko **fixed** rakhte hain wiggle karte waqt — isliye yeh extra bit ek *constant* hai, aur iska variation $\delta$ **zero** hai. Yeh change nahi kar sakta ki kaunsa path stationary hai! > > Figure mein: shaded endpoint disks pinned hain. Dashed detour aur solid path extra area enclose karte hain, lekin $dF/dt$ sirf pinned ends ki parwah karta hai — isliye woh *har* competing path mein same number contribute karta hai. ![[deepdives/dd-physics-2.1.16-d2-s04.png]] Isliye do integrands *at most* ek total time derivative se differ kar sakte hain: $$\underbrace{p\dot q - H}_{\text{purani duniya}}=\underbrace{P\dot Q - K}_{\text{nayi duniya}}+\underbrace{\frac{dF}{dt}}_{\text{harmless endpoint term}}.$$ > [!formula] Master relation (neeche sab kuch isi se nikalta hai) > $$\boxed{\;p\,\dot q - H \;=\; P\,\dot Q - K \;+\;\frac{dF}{dt}\;}$$ > $F$ ==generating function== hai. $F$ choose karo, aur poori transformation $(q,p)\to(Q,P)$ khud nikal aati hai. --- ## Step 4 — Rules squeeze karo: differentials match karo ($F_1$) Master relation ko $dt$ se multiply karo taaki dots clean differentials ban jayein ($\dot q\,dt=dq$): $$p\,dq - H\,dt = P\,dQ - K\,dt + dF.$$ Ab suppose karo ki $F$ **purani position $q$ aur nayi position $Q$** par depend karta hai: likho $F=F_1(q,Q,t)$. Tab chain rule se, $$dF_1=\underbrace{\frac{\partial F_1}{\partial q}}_{\text{F}_1\text{ kaise badlata hai }q\text{ ke saath}}\,dq+\underbrace{\frac{\partial F_1}{\partial Q}}_{\text{...}Q\text{ ke saath}}\,dQ+\underbrace{\frac{\partial F_1}{\partial t}}_{\text{...}t\text{ ke saath}}\,dt.$$ Substitute karo aur **same differential wale terms collect karo**: $$\Big(p-\tfrac{\partial F_1}{\partial q}\Big)dq \;=\; \Big(P+\tfrac{\partial F_1}{\partial Q}\Big)dQ \;+\;\Big(K-H+\tfrac{\partial F_1}{\partial t}\Big)dt.$$ > [!intuition] Coefficients alag alag kyun match hone chahiye (picture) > $q$, $Q$, aur $t$ **independent knobs** hain — ek ko turn karo baaki ko freeze karte hue. Figure mein teen sliders hain. Agar equation hold karni hai chahe tum sirf $dq$ slider ko kaise bhi jiggle karo ($dQ,dt$ frozen), tab sirf $dq$ ka coefficient akele zero hona chahiye. Har slider ke liye same baat. Teen independent directions ⟹ teen alag equations. ![[deepdives/dd-physics-2.1.16-d2-s05.png]] Har coefficient ko zero karne par: > [!formula] Type-1 relations (derive kiye gaye, yaad nahi kiye) > $$p=\frac{\partial F_1}{\partial q}\quad\text{(purana momentum)},\qquad > P=-\frac{\partial F_1}{\partial Q}\quad\text{(naya momentum, dhyan do }{-}),\qquad > K=H+\frac{\partial F_1}{\partial t}.$$ > [!mistake] Minus sign kahan se aaya? > Yeh koi convention **nahi** hai jo hat se nikala gaya ho: $dQ$ master relation ki **right-hand side** par baitha tha, isliye use left par laane se sign flip ho jaata hai. Woh variable ($Q$) jo *sirf $F$ ke andar* rehta hai — left par "$p$-partner" ki tarah nahi — minus inherit karta hai. Yahi mnemonic hai *"lonely-new gets the minus."* --- ## Step 5 — Ek variable ko uske partner se swap karo (Legendre → $F_2$) Type 1 nayi *position* $Q$ use karta hai. Agar hum nayi *momentum* $P$ control karna chahein toh? Hum sirf rename nahi kar sakte — humein $F$ ke andar $Q$ ko $P$ se **properly swap** karna hoga. Ek variable ko uske slope-partner se swap karne ka tool [[Legendre transformation]] hai. > [!definition] Legendre swap ek line mein > Product rule deta hai $P\,dQ = d(PQ) - Q\,dP$. Isliye $P\,dQ$ (jo "$dQ$ use karta hai") ban jaata hai $-Q\,dP$ (jo "$dP$ use karta hai") *plus ek total differential* $d(PQ)$ — aur total differentials harmless hain (Step 3!). $F_2(q,P,t)=F_1+PQ$ define karo. Picture woh trade dikhati hai: tangent-line construction jo ek curve ki *height at a point* ko uski *slope* mein badal deta hai, yani ek $Q$-description ko $P$-description mein convert karta hai bina information khoye. ![[deepdives/dd-physics-2.1.16-d2-s06.png]] Same "match the differentials" move chalate hue (ab independent knobs $q,P,t$ ke saath) milta hai: > [!formula] Type-2 relations > $$p=\frac{\partial F_2}{\partial q},\qquad Q=\frac{\partial F_2}{\partial P}\quad(\text{ab minus nahi!}),\qquad K=H+\frac{\partial F_2}{\partial t}.$$ > [!intuition] Minus yahan kyun gayab hua > Legendre swap ke baad, $P$ ek genuine *differential partner* $-Q\,dP$ ki tarah appear karta hai; $dP$ coefficients match karne se $Q=+\partial_P F_2$ milta hai. Sign arbitrary nahi hai — yeh bookkeeping hai ki har differential kis side se shuru hua tha. Doosri pair par same swap se $F_3(p,Q)$ aur $F_4(p,P)$ milte hain; har minus/plus traceable hai ki kaunsa differential Legendre-flip hua. --- ## Step 6 — Edge aur degenerate cases (reader ko kabhi stranded mat chhodo) > [!example] Case A — Time-independent $F$ > Agar $F$ mein explicit $t$ nahi hai, tab $\partial F/\partial t=0$, isliye $\boxed{K=H}$. **Naya energy purani energy ke barabar hai** — transformation sirf axes relabel karta hai bina hill ko reshape kiye. > [!example] Case B — "Do-nothing" generator $F_2=qP$ > Type-2 rules mein plug karo: > $p=\partial_q(qP)=P$ aur $Q=\partial_P(qP)=q$. Isliye $Q=q,\;P=p$ — ==identity==. Ek generator *trivial* ho sakta hai; $qP$ mein chhote tweaks infinitesimal flows produce karte hain. > [!example] Case C — Degenerate choice $F_1=qQ$ (swap) > $p=\partial_q(qQ)=Q\Rightarrow Q=p$, aur $P=-\partial_Q(qQ)=-q$. Coordinates aur momenta **swap** hote hain (ek sign ke saath): $Q=p,\;P=-q$. Yeh prove karta hai ki $q$ aur $p$ equal footing par hain — kuch bhi "intrinsically" position nahi hai. > [!mistake] Jab ek type *fail* kare: forbidden generators > Agar $F_1$ sirf $q$ aur $Q$ par depend karta hai lekin map $Q\leftrightarrow p$ invertible nahi hai (jaise $F_1$ linear ho isliye $p=\partial_q F_1$ ko $Q$ ke liye solve nahi kar sakte), toh Type 1 toot jaata hai — tumhe koi aisa type ($F_2$, $F_3$, $F_4$) use karna hoga jiske variables *independent* hain. Identity $Q=q$ literally **ek $F_1(q,Q)$ ke roop mein likha hi nahi ja sakta** isi reason se (tumhe $q$ aur $Q$ dependent chahiye honge). Isliye *char types exist karte hain* — koi ek akela har canonical map cover nahi kar sakta. --- ## Step 7 — Payoff ek picture mein: flow ko straighten karo > [!example] Harmonic oscillator, geometrically dekha gaya > $F_1=\tfrac12 m\omega q^2\cot Q$ ke saath parent note ne paaya $K=\omega P$. Dekho iska flow ke liye kya matlab hai: > - $\dot P=-\partial_Q K=0$ ⟹ $P$ **constant** hai (energy/$\omega$). > - $\dot Q=\partial_P K=\omega$ ⟹ $Q=\omega t+\phi$ ek **steady rate** par tick karta hai. > > Purane $(q,p)$ coordinates mein state ek ellipse ke around ghoomti hai (Step 2 ke loops). Naye $(Q,P)$ coordinates mein wahi motion ek point hai jo **horizontal line** par constant speed se slide karta hai. Figure dono side by side dikhata hai — curved flow *unbent* hokar straight ho jaata hai. Yahi [[Action–angle variables]] ka seed hai. ![[deepdives/dd-physics-2.1.16-d2-s07.png]] --- ## Ek picture mein summary ![[deepdives/dd-physics-2.1.16-d2-s08.png]] Poori derivation ek canvas par: **action principle** → do integrands sirf $dF/dt$ se differ kar sakte hain (pinned endpoints par harmless) → **master relation** → independent differentials match karo → **char type-rules** → ek smart $F$ **phase-space flow ko straighten karta hai**. ```mermaid graph TD A["Action principle: delta S = 0"] --> B["Two worlds, same physics"] B --> C["Integrands differ by dF over dt"] C --> D["Master relation p qdot minus H = P Qdot minus K plus dF/dt"] D --> E["Match dq dQ dt coefficients"] E --> F["Type 1 rules with minus on lonely new"] E --> G["Legendre swap gives Types 2 3 4"] F --> H["Pick clever F to straighten the flow"] G --> H ``` > [!recall]- Feynman retelling (poora walkthrough seedha seedha explain karo) > Kisi system ki motion ek ball ki tarah hai jo "kahan hai aur kaise move kar rahi hai" ke 2D map mein energy hills ke around flow karti hai. Main map ka grid re-draw karna chahta hoon taaki woh flow ek boring straight line ban jaaye. Lekin mujhe sirf wahi grids allowed hain jo flow-rule ko same dikhne dein — woh "canonical" hain. Trick yeh hai: nature ki motion woh hai jo ek running total (the action) ko tab rok deti hai jab tum path ko wiggle karo. Agar main ek aisa term add karo jo sirf path ke *start aur end* par depend karta ho — ek total time derivative $dF/dt$ — toh wiggle use notice nahi karta, kyunki ends pinned hain. Isliye purani recipe aur nayi recipe exactly $dF/dt$ se differ kar sakti hain aur *kuch nahi*. Yeh likhkar aur poochhkar "kaunsa piece har independent chhote change $dq$, $dQ$, $dt$ se multiply hota hai?" purane aur naye variables ko connect karne wale rules force out ho jaate hain. Akela function $F$ hi nayi grid ki recipe *hai*. $F$ achhe se choose karo — jaise oscillator ka $\cot$ generator — aur ek baar ki curly flow ek seedhi boring line ban jaati hai, jo exactly "problem solve karna" ka matlab hai. > [!recall]- Quick self-test > Hum integrand mein $dF/dt$ freely kyun add kar sakte hain? ::: Iska integral sirf fixed endpoints par depend karta hai, isliye iska variation zero ho jaata hai. > $q$, $Q$, $t$ ke coefficients alag alag kyun match hote hain? ::: Yeh independent knobs hain; equation har ek ke kisi bhi independent jiggle ke liye hold karni chahiye. > $P=-\partial_Q F_1$ mein minus kahan se aaya? ::: $dQ$ master relation ki right side par shuru hua tha; use left par laane se sign flip ho jaata hai. > Ek achha $F$ flow ke saath geometrically kya karta hai? ::: Yeh curved phase-space flow ko ek line par constant-speed motion mein straighten kar deta hai. ## Connections - [[Hamilton's equations]] — woh form jise hum preserve karte hain. - [[Legendre transformation]] — woh swap jo Type 1 ko Types 2–4 mein badalta hai. - [[Poisson brackets]] — algebraic canonicity test. - [[Hamilton–Jacobi equation]] — $K\to 0$ push karna time-dependent $F$ ke saath. - [[Action–angle variables]] — jahan Step 7 le jaata hai. - [[Symplectic geometry]] / [[Liouville's theorem]] — iske peeche ki area-preserving structure.