Generalized coordinates — choosing them, degrees of freedom: Visual walkthrough

2.1.2 · D2 · Physics › Analytical Mechanics › Generalized coordinates — choosing them, degrees of freedom

Yeh page Generalized coordinates — choosing them, degrees of freedom ka picture-by-picture companion hai. Agar koi word undefined lagta hai, toh woh hamari zimmedaari hai — hum use neeche build karte hain.

Step 1 — Ek free particle: use pin karne ke liye kitne numbers chahiye?

KYA. Ek single dot (ek "particle" — ek object itna chhota ki hum use ek point maante hain) ko empty space mein kahi rakho. Ek dost ko exactly batane ke liye ki woh kahan hai, tum kitne numbers bologe?

PEHLE YEH SAWAAL KYUN. Is chapter mein sab kuch cheezein locate karne ke liye zaruri numbers ki bookkeeping hai. Kuch bhi subtract karne se pehle, hume yeh jaanna hai ki bina kisi restriction ke system describe karne ki cost kya hai. Woh cost hamaara starting budget hai.

PICTURE. Figure dekho. Red dot tak corner se pahunchne ke liye tum ek certain amount rightward ( $x$ ), ek certain amount forward ( $y$ ), aur ek certain amount up ( $z$ ) chalte ho. Teen arrows, teen numbers. Do nahi, chaar nahi — space khud teen independent directions rakhti hai.

Figure — Generalized coordinates — choosing them, degrees of freedom

Toh ek free particle ko 3 numbers chahiye. Woh "3" us space ki dimension hai jisme hum rehte hain — ise yaad rakho, yeh formula mein 3 ban jaata hai.

Step 2 — Bahut saare particles: raw budget $3 N$

KYA. Ab $N$ particles bikher do (yahan $N$ ka matlab hai "jitne bhi tumhare paas hain" — ek whole number jaise 2, ya 5, ya ek million). Har ek ko abhi bhi apne 3 numbers chahiye, aur ek particle ka address doosre ko dhundhne mein kaam nahi aata.

KYUN. Independent objects ke independent addresses hote hain. Bulk mein kharidne par koi discount nahi milti jab particles unconstrained hoon — toh total cost sirf per-particle cost times number of particles hai.

PICTURE. Teen dots, har ek ke paas apna chhota $x, y, z$ tripod of arrows. Saare arrows gino: $3 + 3 + 3 = 9$ . Generally $3 \times N$ .

2D problem ke liye (sab kuch ek flat plane mein trapped hai, toh $z$ hamesha 0 hai aur kuch kaam nahi karta) raw budget $3 N$ ki jagah $2 N$ hai. Same idea, ek direction kam.

Step 3 — Ek constraint ek equation hai, aur woh tumse ek number le leta hai

KYA. Ek constraint ek aisi rule hai jo kuch configurations ko forbid karti hai. Jis tarah ki rules hum yahan handle karte hain — ek holonomic constraint — woh ek aisi rule hai jo tum ek single equation ke roop mein likh sakte ho jo positions (aur shayad time) ko relate kare: $f (r_{1}, \dots, r_{N}, t) = 0.$ Yahan $f$ bas "coordinates se bani koi formula" hai aur use $0$ set karna rule hai.

EK EQUATION EXACTLY EK NUMBER KYUN REMOVE KARTA HAI. Ek equation ek relationship hai: yeh tumhe ek variable ko doosron ke terms mein solve karne deta hai. Jab tum baaki sab jaante ho, toh woh aakhri ek ab free nahi raha — woh force ho gaya. Toh equation exactly ek unit of freedom kha jaata hai. Do nahi (ek equation do cheezein fix nahi kar sakti), zero nahi (ek genuine rule hamesha kuch fix karti hai).

PICTURE. Left: ek particle jo poora flat sheet roam karne ke liye free hai — 2 numbers $(x, y)$ . Right: hum rule impose karte hain " $y = x$ " (yahi hamaara $f = y - x = 0$ hai). Ab particle ek line par stuck hai. Tum abhi bhi $x$ freely choose kar sakte ho, lekin jaise hi tum karte ho, $y$ decide ho jaata hai. Ek equation → ek line → ek number bacha, do nahi.

Step 4 — $k$ independent rules stack karo: $k$ subtract karo

KYA. Agar $k$ aisi holonomic constraints hain, aur woh independent hain (koi rule secretly kisi doosre ki repeat ya consequence nahi hai), toh har ek independently ek number kill karta hai. Toh woh total $k$ numbers kill karte hain.

INDEPENDENCE KYUN MATTER KARTA HAI. Agar do rules ek hi cheez kehte hain, toh doosra tumhe kuch nahi deta — woh number jo woh remove karta, pehle se chala gaya tha. Sirf naya information freedom reduce karta hai. Letter $k$ genuinely-naye rules count karta hai.

PICTURE. Ek particle 3D space mein (3 numbers). Rule 1: "is tilted plane par raho" → use ek 2D sheet pe drop karta hai. Rule 2 (independent): "is doosre plane par raho" → do planes ek line mein cross karte hain, 1 number bacha. Har independent plane-rule ne ek dimension peel off ki: $3 \to 2 \to 1$ .

Step 5 — Pendulum: dekho "2" kaise "1" banta hai

KYA. Length $ℓ$ ke rigid rod par ek mass, vertical plane mein swing kar raha hai. 2D mein raw budget: $2 N = 2$ . Rod ki rigidity rule hai "pivot se distance hamesha $ℓ$ hai": $x^{2} + y^{2} = ℓ^{2}$ , yaani $k = 1$ . Toh $n = 2 - 1 = 1$ .

ANGLE CLEVER COORDINATE KYUN HAI. Formula batata hai ek number kaafi hai, lekin kaun sa number? Agar hum $x$ par insist karte hain, toh hume hamesha yaad rakhna padega $y = - ℓ^{2} - x^{2}$ — rule hume homework ki tarah follow karta hai. Iske bajaye angle $θ$ chuno. Tab $x = ℓ sin θ, y = - ℓ cos θ,$ aur check karo: $x^{2} + y^{2} = ℓ^{2} (sin^{2} θ + cos^{2} θ) = ℓ^{2}$ automatically, har $θ$ ke liye. Constraint "solve" nahi hua — woh gayab ho gaya, coordinate ke meaning mein bake ho gaya.

PICTURE. Rod ek circle sweep karta hai. Jo ek arrow matter karta hai woh growing angle $θ$ hai, straight-down se measure kiya gaya. Forbidden directions (rod ke along in/out) faded draw ki gayi hain — tum wahan ja nahi sakte, toh unka koi cost nahi.

Step 6 — Rigid body: count 6 par kyun rukta hai

KYA. Ek rigid body mein infinitely many particles hote hain, toh naive $3 N = \infty$ . Lekin rigidity ("har pair of points fixed distance rakhta hai") ek colossal pile of constraints hai jo count ko flat 6 par collapse kar deti hai, chahe body kitni bhi badi ho.

6 KYUN, 3 + 3 SPLIT KE ROOP MEIN. Body ka ek point freeze karo: woh space mein place karne ke liye 3 numbers leta hai (translation). Ab body sirf uss point ke baare mein spin kar sakti hai. Uski orientation ke liye 3 aur numbers chahiye — Euler angles $(ϕ, θ, ψ)$ . Total $3 + 3 = 6$ . Har extra particle ki position tab rigidity se forced ho jaati hai, kuch naya contribute nahi karta.

PICTURE. Ek chalk brick. Blue arrow-triple: iska corner kahan baitha hai (3 translation numbers). Pink curved arrows: teen independent tarike jisme yeh tumble kar sakta hai (3 rotation numbers). Chhe freedoms, full stop.

Step 7 — Degenerate cases: jahan picture bend hoti hai

KYA. Do "watch out" scenarios jahan naive $n = 3 N - k$ mislead karta hai.

Case A — invisible rotation (dumbbell). Do point atoms, ek bond-length rule: $3 N - k = 6 - 1 = 5$ . Lekin naively tum expect karte $3 + 3 = 6$ (translation + full orientation). Ek missing hai? Bond axis ke baare mein spin karne se koi point nahi hilda — ek line par do points aisi spin ke baad identical dikhte hain. Woh "rotation" real freedom nahi hai, toh orientation ko sirf 2 angles chahiye, 3 nahi. Result $3 + 2 = 5$ ✓.

Case B — rolling disk (non-holonomic). Bina slipping ke roll karne wala coin ek velocity rule obey karta hai ( $\dot{r}$ spin se tied) jo position-only equation $f (r, t) = 0$ mein squash nahi ho sakti. Toh yeh non-holonomic hai: yeh restrict karta hai kaise move karo, nahi kahan ho sakte ho. Reachable positions ka set bada rehta hai, aur position coordinates ke liye $n = 3 N - k$ simply apply nahi hota. Tum coin ko same spot par new direction face karte hue wapas bhi roll kar sakte ho.

PICTURE. Left: dumbbell jiski bond axis ke baare mein ek faded circular arrow "moves nothing" mark kiya hua hai. Right: ek rolling coin apne ground track ke saath; ek green arrow velocity rule dikhata hai, "can't integrate to a position rule" cross out kiya gaya.

Ek-picture summary

Upar sab kuch, ek board par: raw budget $3 N$ (ya rigid body ke liye $6$ ) upar baitha hai; har independent holonomic rule ek knife hai jo exactly ek number slice off karta hai; neeche jo bachta hai woh $n$ hai, degrees of freedom — good generalized coordinates ki count jo tum tab ek Lagrangian aur Euler–Lagrange equations mein feed karte ho.

Recall Feynman: saara walkthrough plain words mein

Raw shuru karo. Kaho "main har dot ko teen numbers se describe karunga — right, forward, up." Sab jodo: woh $3 N$ hai, tumhara wasteful starting pile. Ab problem ki rules padho. Har rule jo tum ek equation ke roop mein likh sako ("rod itna lamba hai", "bead wire par rehta hai") ek knife hai: woh tumhare pile se exactly ek number slice karta hai, kyunki rule obey karte waqt ek number ab tumhara choose karne ka nahi raha. $k$ baar slice karo $k$ real, non-repeating rules ke liye. Pile mein jo bacha woh $n$ hai — sach mein kitne tarike hain tumhara system wiggle kar sakta hai. Do warnings: (1) ek rigid body rules ka ek huge pile hai jo hamesha 6 chhod deta hai, aur ek do-atom stick ka ek "spin" koi point nahi hilda toh count nahi hota; (2) ek rolling coin ki rule kitni tez ke baare mein hai, nahi kahan ke baare mein, toh yeh alag animal hai — isse mat slice karo. Raw shuru karo, rules subtract karo, sneaky cases respect karo.

Recall

$N$ particles ke liye 3D mein raw budget? ::: $3 N$ numbers. Ek holonomic constraint exactly ek DOF kyun remove karta hai? ::: Yeh ek equation hai, toh yeh tumhe baaki ke terms mein ek variable solve karne deta hai — woh variable ab free nahi raha. Pendulum ke liye $θ$ kyun $x$ se better hai? ::: $x = ℓ sin θ, y = - ℓ cos θ$ automatically $x^{2} + y^{2} = ℓ^{2}$ satisfy karta hai, toh constraint gayab ho jaata hai. Do-atom dumbbell ke 5 DOF kyun hain, 6 kyun nahi? ::: Rotation about the bond axis koi point nahi hilda, toh orientation ko 2 angles chahiye 3 nahi: $3 + 2 = 5$ . Rolling disk ke liye $n = 3 N - k$ kyun use nahi ho sakta? ::: Uska rolling constraint non-holonomic hai (ek velocity rule jo position equation mein integrate nahi ho sakta), toh yeh cleanly ek position coordinate remove nahi karta.