Visual walkthrough — Derivation of Euler-Lagrange equations from D'Alembert's principle

2.1.5 · D2 · Physics › Analytical Mechanics › Derivation of Euler-Lagrange equations from D'Alembert's pri

---

Step 0 — Woh picture jise hum fix karne ki koshish kar rahe hain

Picture dekho. Bead se do arrows nikalte hain:

• burnt-orange arrow applied force ${\vec{F}}^{(a)}$ hai — yahan gravity. Chhota $(a)$ sirf "applied" matlab hai, yaani real physics jis ki hume parwah hai.
• teal arrow constraint force ${\vec{F}}^{(c)}$ hai — wire ka push. Yeh wire ke across point karta hai.

Green dashed arrow woh akela direction hai jisme bead ko move karne ki permission hai: wire ke saath. Uss allowed direction mein ek tiny imagined step ko $\delta \vec{r}$ bolo.

Dot product kyun, aur kuch kyun nahi? Kyunki hamari poori trick hai "kya wire ka push bead ko uske allowed step mein help karta hai?" Yeh exactly wahi sawaal hai "yeh do arrows kitne aligned hain?" — aur woh tool jo alignment ka jawab deta hai woh dot product hai. Perpendicular arrows $0$ dete hain; woh zero hi constraint force ko delete karega.

---

Step 1 — Constraint force ko project karke khatam karo

KYA. Newton kehta hai, har particle $k$ ke liye, $\underset{\text{applied}}{\underbrace{{\vec{F}}_k^{(a)}}}+\underset{\text{constraint}}{\underbrace{{\vec{F}}_k^{(c)}}}-\underset{\text{momentum ka rate}}{\underbrace{{\dot{\vec{p}}}_k}}=0.$ Yahan ${\vec{p}}_k=m_k{\dot{\vec{r}}}_k$ momentum hai (mass $\times$ velocity), aur ${\dot{\vec{p}}}_k$ par dot ka matlab hai "per second change ka rate". Toh ${\dot{\vec{p}}}_k=m_k{\ddot{\vec{r}}}_k=m_k{\vec{a}}_k$ — bas $m\vec{a}$, Newton ka jaana-pehchana form, ek taraf move kiya gaya.

KYUN. Hum poori line ko allowed step $\delta {\vec{r}}_k$ ke saath dot karte hain aur har particle par add up karte hain: ${\sum }_k({\vec{F}}_k^{(a)}+{\vec{F}}_k^{(c)}-{\dot{\vec{p}}}_k)\cdot \delta {\vec{r}}_k=0.$ Constraint force $\delta {\vec{r}}_k$ ke perpendicular hai (Step 0 mein teal $\perp$ green), isliye ${\vec{F}}_k^{(c)}\cdot \delta {\vec{r}}_k=0$. Yeh gayab ho jaata hai. Jo bachta hai woh D'Alembert's principle hai: ${\sum }_k({\vec{F}}_k^{(a)}-{\dot{\vec{p}}}_k)\cdot \delta {\vec{r}}_k=0.$

PICTURE. Green line par teal arrow ki shadow ki length zero hai — yahi "virtual work nahi karta" ka geometric matlab hai.

---

Step 2 — $x,y,z$ ki jagah smart coordinates $q_i$ lo

KYA. Bead ko $x,y,z$ se describe karne ki bajaye (jo wire ki equation maanni padte hain), hum ek aisa coordinate chunte hain jo pehle se wire par hi rehta hai: pendulum ke liye angle $\theta$; bead ke liye wire-along-distance $r$. Inhe generalized coordinates $q_1,\dots ,q_n$ bolo. Position tab ek function ban jaati hai ${\vec{r}}_k={\vec{r}}_k(q_1,\dots ,q_n,t)$. Dekho Constraints and generalized coordinates.

KYUN. $q$ is tarah choose karne par, $q$ ki har value automatically legal hai — constraint bake in ho jaati hai, aur hum ise kabhi dobara nahi likhte. Ek tiny allowed step ban jaata hai $\delta {\vec{r}}_k={\sum }_i\frac{\partial {\vec{r}}_k}{\partial q_i}\delta q_i.$ Ise term by term padho:

• $\partial {\vec{r}}_k/\partial q_i$ (curly $\partial$ = "partial derivative": $q_i$ ko thoda change karo, baaki saare $q$'s aur time ko freeze karo, poochho ki ${\vec{r}}_k$ kaise move karta hai) ek arrow hai jo wire ke along point karta hai — woh direction jisme knob $q_i$ particle ko push karta hai.
• $\delta q_i$ = hum knob $i$ ko kitna ghumaate hain.
• Sum ${\sum }_i$ saare knobs ke pushes ko add up karta hai.
• Koi $\partial {\vec{r}}_k/\partial t\delta t$ term nahi aata — kyunki time frozen hai, $\delta t=0$.

PICTURE. Har knob $q_i$ ek tangent arrow contribute karta hai; $\delta {\vec{r}}_k$ unka weighted sum hai. Knobs independent hain — aap ek ko bina doosron ko disturb kiye ghuma sakte ho. Woh word independent yaad rakho; yeh Step 7 ka punchline hai.

---

Step 3 — "Force part" aur "inertia part" mein split karo

KYA. D'Alembert ke do pieces hain. Step-2 expression ko dono mein substitute karo: $\underset{\text{force part}}{\underbrace{\sum _k{\vec{F}}_k^{(a)}\cdot \delta {\vec{r}}_k}}-\underset{\text{inertia part}}{\underbrace{\sum _k{\dot{\vec{p}}}_k\cdot \delta {\vec{r}}_k}}=0.$

KYUN (force part). $\delta {\vec{r}}_k={\sum }_i(\partial {\vec{r}}_k/\partial q_i)\delta q_i$ plug in karo aur $\delta q_i$ gather karo: ${\sum }_k{\vec{F}}_k^{(a)}\cdot \delta {\vec{r}}_k={\sum }_i\underset{Q_i}{\underbrace{(\sum _k{\vec{F}}_k^{(a)}\cdot \frac{\partial {\vec{r}}_k}{\partial q_i})}}\delta q_i.$ Bundle $Q_i$ generalized force hai — real forces, knob $i$ par project kiye gaye (phir ek dot product!). Dekho Generalized forces and potentials.

PICTURE. Orange force arrow har tangent arrow par shadow daalta hai; us shadow ki length hi $Q_i$ hai.

Inertia part, ${\sum }_k{\dot{\vec{p}}}_k\cdot (\partial {\vec{r}}_k/\partial q_i)$, mein ek acceleration ${\ddot{\vec{r}}}_k$ hai. Accelerations ugly lagte hain. Agla step ek trick hai unhe kisi aisi cheez mein badalne ki jो hume pasand hai: energy.

---

Step 4 — Product-rule trick (acceleration ko energy mein badlo)

KYA. Ek term $m_k{\ddot{\vec{r}}}_k\cdot \frac{\partial {\vec{r}}_k}{\partial q_i}$ par focus karo. Reverse product rule use karo:

= \frac{d}{dt}\!\Big(m_k\dot{\vec r}_k\cdot\frac{\partial\vec r_k}{\partial q_i}\Big) - m_k\dot{\vec r}_k\cdot\frac{d}{dt}\frac{\partial\vec r_k}{\partial q_i}.$$

YEH TOOL KYUN. Do cheezein ki derivative ke liye product rule, $(fg{)}^{\prime }=f^{\prime }g+f{g}^{\prime }$, rearrange hota hai $f^{\prime }g=(fg{)}^{\prime }-f{g}^{\prime }$ mein. Set karo $f={\dot{\vec{r}}}_k$ (toh $f^{\prime }={\ddot{\vec{r}}}_k$) aur $g=\partial {\vec{r}}_k/\partial q_i$. Yeh unwanted acceleration $\ddot{\vec{r}}$ ko do velocity-flavoured pieces mein launder karta hai, jo hum jaldi kinetic energy ke roop mein pehchaan lenge. Yahi kyun product rule aur koi identity nahi: yeh woh akela move hai jo "second derivative" ko "first derivative" quantities tak neeche laata hai.

PICTURE. $\ddot{\vec{r}}$ ko socho "velocity arrow mein change." Identity ise dobara draw karti hai (ek product ki total change) minus (tangent arrow ki change). Do chhote, tame arrows.

---

Step 5 — Do lemmas, drawn

Hum energy ko pehchaan sakein, usse pehle hume $q$ aur uski speed $\dot{q}$ ke baare mein do chhote facts chahiye.

PICTURE. Lemma 1 kehta hai: woh arrow jo aap "velocity speed-knob $i$ ke saath kaise change hoti hai" poochh kar paate ho, woh wahi arrow hai jo "position position-knob $i$ ke saath kaise change hoti hai." Lemma 2 kehta hai: chhote square ke do raaste — differentiate-then-time vs. time-then-differentiate — ek hi corner par pahunchtein hain.

---

Step 6 — Kinetic energy $T$ ko pehchano

KYA. Do lemmas ko Step 4 mein feed karo:

=\frac{d}{dt}\Big(m_k\dot{\vec r}_k\cdot\frac{\partial\dot{\vec r}_k}{\partial\dot q_i}\Big) - m_k\dot{\vec r}_k\cdot\frac{\partial\dot{\vec r}_k}{\partial q_i}.$$ Ab key sighting: kyunki $\dfrac{\partial}{\partial u}\big(\tfrac12 m_k\dot{\vec r}_k^{\,2}\big)=m_k\dot{\vec r}_k\cdot\dfrac{\partial\dot{\vec r}_k}{\partial u}$ (square par chain rule), har piece **kinetic energy** $\tfrac12 m_k\dot{\vec r}_k^{\,2}$ ki derivative hai. Saare particles par sum karo aur $T=\sum_k\tfrac12 m_k\dot{\vec r}_k^{\,2}$ likho: $$\sum_k m_k\ddot{\vec r}_k\cdot\frac{\partial\vec r_k}{\partial q_i} =\frac{d}{dt}\frac{\partial T}{\partial\dot q_i}-\frac{\partial T}{\partial q_i}.$$

KYUN. $T$ kinetic energy $q$'s mein likhi hai. Poora ugly inertia sum do clean energy-derivatives mein collapse ho gaya — ek ke aage $d/dt$ hai, ek bare hai.

PICTURE. Velocity arrow, apne khud ke rate-of-change ke saath dotted, literally $\frac{1}{2}m{v}^2$ energy bowl ki slope hai. Do terms woh do tarike hain jisme woh bowl tilt hoti hai jab hum $q_i$ aur ${\dot{q}}_i$ wiggle karte hain.

---

Step 7 — Assemble karo, aur independence use karo

KYA. Force part (Step 3) aur energy form (Step 6) ko D'Alembert mein wapas daalo: ${\sum }_i[\frac{d}{dt}\frac{\partial T}{\partial {\dot{q}}_i}-\frac{\partial T}{\partial q_i}-Q_i]\delta q_i=0.$

KYUN har bracket zero hai. Holonomic system ke liye knobs $\delta q_i$ independent hain — main akele knob 1 wiggle kar sakta hoon, baaki sab zero. Tab sum bas (bracket 1)$\times \delta q_1=0$ hai, forcing bracket 1 $=0$. Har knob ke liye repeat karo. Independent nudges ka weighted sum tab zero hota hai jab har weight zero ho: $\boxed{\frac{d}{dt}\frac{\partial T}{\partial {\dot{q}}_i}-\frac{\partial T}{\partial q_i}=Q_i}$

PICTURE. Independent axes: ek ke along push karna baaki ko untouched chhod deta hai, isliye total zero matlab har axis par alag se zero.

---

Step 8 — Potential fold in karo aur $L=T-V$ paao

KYA & KYUN. Agar applied forces ek potential $V$ se aate hain, toh $Q_i=-\partial V/\partial q_i$ (dekho Generalized forces and potentials). Kyunki $V$ mein koi $\dot{q}$ nahi hai, ise $T$ mein add karne se $\partial /\partial \dot{q}$ term par koi effect nahi padta. Lagrangian define karo $L=T-V$: $\frac{d}{dt}\frac{\partial L}{\partial {\dot{q}}_i}-\frac{\partial L}{\partial q_i}=0.■$ Yahi equation Hamilton's principle (least action) ek action minimize karke produce karta hai — ek summit tak do raaste. Iske symmetries Noether's theorem ko feed karte hain.

Humne Newton's laws of motion se shuroo kiya; hum constraint forces gayab hone aur sab kuch energy mein likha hone par khatam hue.

---

Ek-picture summary

Flow: Newton $\to$ allowed step ke saath dot karo (constraint force delete karo) $\to$ $q$-knobs mein change karo $\to$ force/inertia split karo $\to$ product-rule + do lemmas acceleration ko $T$ mein launder karte hain $\to$ independence sum ko split karti hai $\to$ $V$ fold in karo $\to$ Euler–Lagrange.

Recall Feynman retelling (plain words mein wapas bolo)

Newton ka law forces mein likha hai, lekin wire ka push ek nuisance hai jo hum jaante nahi. Trick: sirf woh directions mein bead ko nudge karo jo wire allow karti hai, aur poori law ka us nudge ke saath dot product lo. Wire ka push nudge ke sideways hai, isliye uski shadow zero hai — yeh gayab ho jaata hai. Phir, $x,y,z$ ki jagah jo wire ki equation maanni padti hai, bead ko ek knob $q$ se describe karo jo already wire par rehta hai, toh constraint bake in ho jaati hai aur kabhi dobara nahi likhte. Jo bachta hai use ek "real forces" part mein split karo (jo generalized force $Q_i$ ban jaata hai) aur ek "inertia" part jo ek ugly acceleration carry karta hai. Product rule ko ulta use karo us acceleration ko velocity terms mein demote karne ke liye, aur do chhote lemmas ("dots cancel karo", "derivatives ka order swap karo") unhe kinetic energy $T$ ki slopes ke roop mein pehchaanne dete hain. Ab sab kuch independent knobs ka sum hai times ek bracket; independent cheezein jinka sum zero ho matlab har bracket zero hai. Woh bracket hai $\frac{d}{dt}\partial T/\partial \dot{q}-\partial T/\partial q=Q_i$. Finally, agar forces ek potential $V$ se aate hain, toh use $L=T-V$ mein absorb karo aur clean Euler–Lagrange equation pao. Newton, annoying constraints algebraically sweep away ho jaane ke saath.

Recall Quick self-check

D'Alembert mein constraint force kyun gayab ho jaata hai? ::: Yeh virtual displacement ke perpendicular hai, isliye iska dot product (allowed direction par uski shadow) zero hai. Hum generalized coordinates $q_i$ kyun use karte hain? ::: Har $q_i$ already constraint respect karta hai, isliye constraint equation bake in ho jaati hai aur constraint forces kabhi wapas nahi aate. Product-rule trick kya achieve karti hai? ::: Yeh unwanted acceleration $\ddot{\vec{r}}$ ko first-derivative (velocity) terms mein convert karti hai jo kinetic energy mein assemble ho jaate hain. Hum har bracket ko zero kyun set kar sakte hain? ::: Kyunki holonomic system ke liye $\delta q_i$ independent hain, isliye unka weighted sum zero tab hi hota hai jab har weight zero ho. Constraints non-holonomic hone par kya toot jaata hai? ::: $\delta q_i$ ab independent nahi rehte; aap brackets split nahi kar sakte aur Lagrange multipliers use karne padte hain.