Visual walkthrough — Hamiltonian — definition H = Σpᵢq̇ᵢ − L

2.1.11 · D2 · Physics › Analytical Mechanics › Hamiltonian — definition H = Σpᵢq̇ᵢ − L

Hum ek idea build karte hain: variable "velocity" ko variable "momentum" se kaise swap karte hain. Wohi swap poora game hai, aur isse Legendre Transform kehte hain.

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Step 1 — Ek curve, aur usse describe karne ke do seedhe tarike

KYA. Physics ko ek minute ke liye bhool jao. Koi bhi curve lo jo upar ki taraf mude — ek bowl jaisi shape. Horizontal axis ko $x$ kaho aur curve ki height ko $f(x)$. "$f(x)$" ka matlab bas itna hai: "point $x$ ke upar curve ki height".

KYUN. Variables swap karne se pehle yeh poochna zaroori hai: ek curve ko describe karne ka matlab kya hota hai? Aam tarika hai points ki list $(x,f(x))$. Lekin ek doosra, utna hi complete tarika bhi hai — curve ko uski slopes se describe karo. Yahi doosra description hai jo $L$ ko $H$ mein badalta hai, isliye hum yahan se shuru karte hain.

PICTURE.

Blue bowl $f(x)$ hai. Amber dot ek chosen $x$ par baitha hai. Wahan bowl ko chhune wali cyan line tangent hai — woh seedhi line jo curve ko sirf us ek point par graze karti hai. Uski steepness slope hai, jise $p$ likhte hain:

$p=\underset{\text{tangent line ki slope }x\text{ par}}{\underbrace{f^{\prime }(x)}}.$

Yahan $f^{\prime }(x)$ (padho "$f$-prime of $x$") derivative hai — ek number jo yeh jawab deta hai: "agar main $x$ ko thoda sa right nudge karun, toh height kitni tezi se badhti hai?" Steep bowl bada $p$ deta hai; flat bottom $p=0$ deta hai.

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Step 2 — Curve ko height ki jagah slope se padhna

KYA. Jaise tum amber dot ko bowl ke saath left se right slide karte ho, tangent ki slope $p$ steadily change hoti hai. Ek bowl ke liye (ek convex curve — jo sirf ek hi taraf mude) har position $x$ ek alag slope $p$ deta hai. Koi do points ek slope share nahin karte.

KYUN. Yahi "koi do points ek slope share nahin karte" property woh key hai jo hume description ulti chalane deti hai: agar koi tumhe ek slope $p$ de, toh bowl par exactly ek hi point hai us slope ke saath. Toh slope $p$, points ke liye ek label ka kaam kar sakta hai — bilkul $x$ jaisi hi acchi. Isliye hum insist karte hain ki curve convex ho — warna do points ek slope share karte aur label ambiguous ho jaata.

PICTURE.

Do amber dots, do cyan tangents, do clearly alag slopes. Arrow yaad dilata hai: slope ek legitimate name-tag hai convex curve par position ke liye. Hum abhi name-tag $x$ ko name-tag $p$ se trade karne wale hain.

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Step 3 — Tangent line aur curve ke beech ka gap

KYA. $x$ par ek point lo jiska slope $p$ hai. Uski tangent line ko vertical axis tak wapas extend karo (yaani $x=0$ tak). Dekho woh kahan cross karta hai. Us crossing height ko $-g$ kaho (minus sign ek convention hai jo baad mein kaam aayega). Tab

$g=px-f(x).$

Yahan har symbol ko abhi yahin name karte hain:

• $px$ — slope times horizontal distance = seedha tangent line $0$ se $x$ tak kitni height gain karta hai.
• $f(x)$ — curve ki actual height $x$ par.
• $g=px-f$ — woh vertical gap jo tum tangent ke back-projection se axis tak giroge; equivalently tangent ke intercept ka negative.

KYUN. Yahi ek algebraic combination $px-f$ poora Legendre transform hai. Iske baad sab bookkeeping hai. Hum ise pehle geometrically banate hain taki formula kabhi rabbit-out-of-a-hat na lage.

PICTURE.

Cyan tangent line ko axis tak wapas follow karo. Amber bracket $g=px-f$ hai. Notice karo: jaise hum $x$ slide karte hain, dono $p$ aur $f$ change hote hain, toh $g$ bhi change hota hai — aur hum ab $g$ ko slope $p$ ka function sochenge, $x$ ka nahin.

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Step 4 — Prove karo ki swap ne sach mein $x$ hata diya (differential trick)

KYA. Hume check karna hai ki $g$ sach mein $p$ par depend karta hai aur secretly $x$ par nahin. Har cheez ka ek tiny change lo (ek differential, likha $d(\cdot )$ = "ka infinitesimal change"):

$dg=d(px)-df=\underset{\text{product rule on }px}{\underbrace{pdx+xdp}}-\underset{df=f^{\prime }dx=pdx}{\underbrace{pdx}}.$

Dono $pdx$ terms cancel ho jaate hain, bacha:

$\boxed{dg=xdp}.$

KYUN. Dekho kya hua: $dg$ mein sirf $dp$ hai — koi leftover $dx$ nahin. Yahi mathematical proof hai ki $g$ cleanly $p$ ka function hai. Aur $dp$ ka coefficient $x$ hai, toh $x=dg/dp$: transform apna hi inverse hai — slopes aur positions ne bas roles badal liye hain.

PICTURE.

Cartoon mein cancellation dikhti hai: ek green $+pdx$ arrow aur ek red $-pdx$ arrow annihilate karte hain, aur sirf amber $xdp$ bachta hai. Yahi exact cancellation neche physics derivation mein hoti hai — dhyan se dekho.

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Step 5 — Axes rename karo: yahi Hamiltonian hai

KYA. Ab dictionary badlo. Replace karo:

geometry word	physics word
horizontal $x$	velocity $\dot{q}$ ("$\dot{q}$" = coordinate $q$ ke change ki rate)
height $f(x)$	[[Lagrangian Mechanics
slope $p=f^{\prime }(x)$	momentum $p=\partial L/\partial \dot{q}$
transform $g=px-f$	Hamiltonian $H=p\dot{q}-L$

KYUN. Lagrangian $L$ velocity $\dot{q}$ ka bowl-shaped function hai (normal kinetic term $\frac{1}{2}m{\dot{q}}^2$ ke liye yeh literally ek parabola hai). Velocity direction mein uski slope hi momentum hai — yeh coincidence nahin hai, yeh definition hai $p=\partial L/\partial \dot{q}$ (yahan $\partial$ ka matlab hai "doosre variables fixed rakh ke slope"). Toh $\dot{q}$ direction mein $L$ ka Legendre transform exactly $H$ hai.

PICTURE.

Wohi bowl jaise Step 1 mein tha, relabelled. Horizontal axis ab $\dot{q}$ padhti hai, curve $L$ hai, amber gap $H=p\dot{q}-L$ hai. Kuch naya nahin hua — humne sirf axes ke words badle.

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Step 6 — Ek real particle ke liye karo (bowl → parabola)

KYA. Lo $L=\frac{1}{2}m{\dot{x}}^2-V(x)$. Velocity-bowl parabola $\frac{1}{2}m{\dot{x}}^2$ hai.

• Slope (momentum): $p=\partial L/\partial \dot{x}=m\dot{x}$.
• Invert: $\dot{x}=p/m$ — slope se position padho.
• Assemble: $H=p\dot{x}-L=\frac{p^2}{m}-(\frac{p^2}{2m}-V)=\frac{p^2}{2m}+V(x).$

KYUN. Yeh Step 3 ka $g=px-f$ hai ek actual parabola ke saath, toh tum dekh sakte ho ki number $\frac{p^2}{2m}$ ek parabola ke tangent ke intercept-gap ke roop mein aata hai. Isse kinetic-plus-potential energy milti hai — system ki canonical energy.

PICTURE.

Cyan parabola $\frac{1}{2}m{\dot{x}}^2$; velocity $\dot{x}=p/m$ par padha gaya amber gap $\frac{p^2}{2m}$ ke barabar hai. Tangent ki slope momentum $p=m\dot{x}$ hai. Boxed answer mein har quantity drawing mein visible hai.

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Step 7 — Degenerate case: flat curve swap tod deti hai

KYA. Kya hoga agar $L$ $\dot{q}$ mein bowl nahin ho? Maano $L$ velocity mein linear hai, jaise $L=c\dot{q}+(\text{stuff without }\dot{q})$. Tab slope constant $p=c$ hai — yeh $\dot{q}$ par bilkul depend nahin karta.

KYUN. Ab swap fail ho jaata hai: equation $p=c$ ko $\dot{q}$ ke liye invert nahin kiya ja sakta (kaafi saare $\dot{q}$ ek hi slope dete hain). Geometrically "bowl" ek seedhe ramp mein degenerate ho gaya hai — har point ek hi slope share karta hai, toh slope ab valid name-tag nahin raha (Step 2 se compare karo). Yeh singular Legendre case hai; yahi exactly woh jagah hai jahan parent note ki condition "$T$ must be quadratic in $\dot{q}$" aati hai. Yeh constrained/gauge systems ko flag karta hai jo zyada advanced machinery se handle hote hain.

PICTURE.

Left: healthy bowl, slopes fan out karte hain, unique labels. Right: seedha ramp — tangent line hi curve hai, sab points ke liye ek slope, inversion $\dot{q}(p)$ ka koi jawab nahin. Amber "no unique $\dot{q}$" mark tumhe warn karta hai.

Recall Swap par kab trust kar sakte ho?

Legendre transform invertible hai tab hi jab curve swapped variable mein strictly convex ho ::: tab hi jab $L$ $\dot{q}$ mein curved ho (flat/linear nahin), yaani ${\partial }^{2}L/\partial {\dot{q}}^2\ne 0$.

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Ek-picture summary

Blueprint saatoon steps compress karta hai: velocity-bowl $L(\dot{q})$, uski tangent jiska slope momentum $p$ hai, aur amber intercept-gap jo hi Hamiltonian $H=p\dot{q}-L$ hai. Picture padho, physics padho.

Recall Feynman retelling (ise zor se bolo)

"Lagrangian ek bowl-shaped hill hai jo velocity ke against draw ki gayi hai. Kisi bhi velocity par, us hill ki steepness ko hum momentum kehte hain. Ab yeh trick hai: hill par har spot ko uski velocity se label karne ki jagah, uski steepness se label karo — kyunki bowl par har spot ki apni unique steepness hoti hai. Relabelling karne ke liye, apne spot par tangent line khiincho aur ise axis tak slide karo; woh chota sa gap jo woh chhod jaati hai, $p\dot{q}-L$, Hamiltonian hai. Jab main ek tiny step leta hun, velocity-change term perfectly cancel ho jaata hai aur sirf momentum-change bachti hai — yahi cancellation proof hai ki Hamiltonian sach mein velocity bhool gaya aur sirf momentum yaad rakhta hai. Agar hill sach mein ek parabola hai, toh yeh gap $p^2/2m+V$ nikalta hai — wahi purani energy. Aur agar hill bowl ki jagah flat ramp hoti, toh har spot ek hi steepness share karta, relabelling impossible hoti, aur poori trick toot jaati — jo physicist ke liye constrained system ka warning sign hai."

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Related: Legendre Transform · Hamilton's Canonical Equations · Phase Space · Poisson Brackets · Conservation Laws & Noether's Theorem · back to the parent topic.