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

2.1.5 · Physics › Analytical Mechanics

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KYA derive kar rahe hain hum

Hum ise scratch se derive karenge, kabhi assume nahi karenge.

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Starting ingredients

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Derivation KAISE chalti hai (step by step)

Hamare paas $N$ particles hain, positions ${\vec{r}}_k={\vec{r}}_k(q_1,\dots ,q_n,t)$.

Step 1 — Virtual displacements ko generalized coordinates mein express karo

$\delta {\vec{r}}_k={\sum }_i\frac{\partial {\vec{r}}_k}{\partial q_i}\delta q_i.$ Yeh step kyun? Kyunki $\delta t=0$ hai, explicit time dependence contribute nahi karta. $\delta q_i$ ek holonomic system ke liye independent hote hain — wahi independence poora fayda hai.

Step 2 — D'Alembert ko "force part" aur "inertia part" mein split karo

${\sum }_k{\vec{F}}_k^{(a)}\cdot \delta {\vec{r}}_k-{\sum }_k{\dot{\vec{p}}}_k\cdot \delta {\vec{r}}_k=0.$

Force part → generalized forces. Step 1 substitute karo: ${\sum }_k{\vec{F}}_k^{(a)}\cdot \delta {\vec{r}}_k={\sum }_i\underset{Q_i}{\underbrace{\left(\sum _k{\vec{F}}_k^{(a)}\cdot \frac{\partial {\vec{r}}_k}{\partial q_i}\right)}}\delta q_i.$ $Q_i$ generalized force hai. Kyun? Yeh saari real forces collect karta hai, $q_i$ direction par projected.

Step 3 — Inertia part ko ek key identity se rewrite karo

Yahan algebraic core hai. ${\dot{\vec{p}}}_k=m_k{\ddot{\vec{r}}}_k$ ke saath: ${\sum }_k{m}_k{\ddot{\vec{r}}}_k\cdot \frac{\partial {\vec{r}}_k}{\partial q_i}.$ Product-rule trick use karo (ulta chalao):

= \frac{d}{dt}\!\left(m_k\dot{\vec r}_k\cdot\frac{\partial\vec r_k}{\partial q_i}\right) - m_k\dot{\vec r}_k\cdot\frac{d}{dt}\frac{\partial\vec r_k}{\partial q_i}.$$ **Yeh step kyun?** Yeh $\ddot{\vec r}$ (ek acceleration jo hum nahi chahte) ko velocity-type quantities ke time-derivatives mein convert karta hai, jinhe hum kinetic energy mein pack kar sakte hain. Ab humein do **lemmas** chahiye. > [!formula] Lemma 1 — "cancellation of dots" > $$\frac{\partial \dot{\vec r}_k}{\partial \dot q_i} = \frac{\partial \vec r_k}{\partial q_i}.$$ > **Derivation:** $\dot{\vec r}_k = \sum_j \dfrac{\partial\vec r_k}{\partial q_j}\dot q_j + \dfrac{\partial\vec r_k}{\partial t}$. Coefficients $q,t$ par depend karte hain lekin $\dot q_i$ par **nahi**, isliye $\dot q_i$ ke w.r.t. differentiate karne par exactly $\partial\vec r_k/\partial q_i$ milta hai. > [!formula] Lemma 2 — "commuting d/dt and ∂/∂q" > $$\frac{d}{dt}\frac{\partial \vec r_k}{\partial q_i} = \frac{\partial \dot{\vec r}_k}{\partial q_i}.$$ > **Derivation:** $\dfrac{d}{dt}\dfrac{\partial\vec r_k}{\partial q_i} = \sum_j \dfrac{\partial^2\vec r_k}{\partial q_j\partial q_i}\dot q_j + \dfrac{\partial^2\vec r_k}{\partial t\,\partial q_i}$, jo exactly $\dfrac{\partial}{\partial q_i}\Big(\sum_j\frac{\partial\vec r_k}{\partial q_j}\dot q_j + \frac{\partial\vec r_k}{\partial t}\Big) = \dfrac{\partial\dot{\vec r}_k}{\partial q_i}$ hai (mixed partials commute karte hain). ### Step 4 — Lemmas insert karo Lemma 1 pehle term mein aur Lemma 2 doosre mein use karke: $$m_k\ddot{\vec r}_k\cdot\frac{\partial\vec r_k}{\partial q_i} = \frac{d}{dt}\!\left(m_k\dot{\vec r}_k\cdot\frac{\partial\dot{\vec r}_k}{\partial\dot q_i}\right) - m_k\dot{\vec r}_k\cdot\frac{\partial\dot{\vec r}_k}{\partial q_i}.$$ Ab pehchano ki $m_k\dot{\vec r}_k\cdot\dfrac{\partial\dot{\vec r}_k}{\partial(\cdot)} = \dfrac{\partial}{\partial(\cdot)}\left(\tfrac12 m_k \dot{\vec r}_k^2\right)$. $k$ par sum karke aur $T = \sum_k \tfrac12 m_k\dot{\vec r}_k^2$ define karke: $$\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}.$$ ### Step 5 — Assemble karo D'Alembert (Step 2) ban jaata hai $$\sum_i\left[\frac{d}{dt}\frac{\partial T}{\partial \dot q_i} - \frac{\partial T}{\partial q_i} - Q_i\right]\delta q_i = 0.$$ Kyunki $\delta q_i$ **independent** hain, har bracket zero hona chahiye: $$\boxed{\;\frac{d}{dt}\frac{\partial T}{\partial \dot q_i} - \frac{\partial T}{\partial q_i} = Q_i\;}$$ **"Har ek zero kyun hona chahiye"?** Independent quantities ka linear combination tabhi zero hoga jab har coefficient zero ho. Yeh independence holonomic constraints *dwara di* hoti hai. ### Step 6 — Potential laao Agar applied forces conservative hain, $\vec F_k^{(a)} = -\nabla_k V$, to $$Q_i = \sum_k \vec F_k^{(a)}\cdot\frac{\partial\vec r_k}{\partial q_i} = -\frac{\partial V}{\partial q_i}.$$ Kyunki $V$ usually sirf $q$ par depend karta hai (na ki $\dot q$ par), $\partial V/\partial\dot q_i = 0$, isliye hum ise fold kar sakte hain. $L = T - V$ define karo: $$\frac{d}{dt}\frac{\partial L}{\partial \dot q_i} - \frac{\partial L}{\partial q_i} = 0.\qquad\blacksquare$$ --- ## Worked examples > [!example] (1) 1D mein free particle, $q = x$ > $T = \tfrac12 m\dot x^2$, $V=0$. *Yeh kyun choose kiya?* Sabse simple sanity check. > $\partial L/\partial\dot x = m\dot x$; $\frac{d}{dt}(m\dot x) = m\ddot x$; $\partial L/\partial x = 0$. EL deta hai $m\ddot x = 0$ — Newton's first law. ✓ > [!example] (2) Simple pendulum, $q=\theta$ > Position: $x = \ell\sin\theta,\ y = -\ell\cos\theta$. *Yeh coordinates kyun?* $\theta$ automatically rigid-rod constraint satisfy karta hai, isliye tension kabhi appear nahi karta. > $T = \tfrac12 m\ell^2\dot\theta^2$, $V = -mg\ell\cos\theta$. To $L = \tfrac12 m\ell^2\dot\theta^2 + mg\ell\cos\theta$. > $\frac{\partial L}{\partial\dot\theta} = m\ell^2\dot\theta \Rightarrow \frac{d}{dt}(\cdot) = m\ell^2\ddot\theta$. $\frac{\partial L}{\partial\theta} = -mg\ell\sin\theta$. > EL: $m\ell^2\ddot\theta + mg\ell\sin\theta = 0 \Rightarrow \ddot\theta = -\frac{g}{\ell}\sin\theta$. ✓ (Tension free mein chali gayi!) > [!example] (3) Frictionless rotating wire par bead (forced) > Wire radial line ke saath $\omega$ par rotate kar rahi hai, bead coordinate $r$. *Kyun?* Dikhata hai ki $\partial T/\partial q \ne 0$ matter karta hai. > $T = \tfrac12 m(\dot r^2 + r^2\omega^2)$, $V=0$. > $\frac{\partial T}{\partial\dot r}=m\dot r\Rightarrow\frac{d}{dt}=m\ddot r$; $\frac{\partial T}{\partial r}=m r\omega^2$. > EL: $m\ddot r - m r\omega^2 = 0 \Rightarrow \ddot r = \omega^2 r$. Centrifugal term $\partial T/\partial r$ se nikla. ✓ --- ## Common mistakes (steel-manned) > [!mistake] "Constraint forces koi work nahi karte, to main unhe real motion mein bhi ignore kar sakta hoon." > **Kyun sahi lagta hai:** *virtual* displacement mein wo zero work karte hain, aur yeh sach hai. **Trap:** ek *moving* constraint ke liye, real work $\vec F^{(c)}\cdot d\vec r$ nonzero ho sakta hai (jaise moving wall). **Fix:** D'Alembert **virtual** ($\delta t=0$) displacements use karta hai precisely isliye ki $\sum\vec F^{(c)}\cdot\delta\vec r=0$ time-dependent constraints ke liye bhi hold kare. Time hamesha freeze karo. > [!mistake] "$\delta q_i$ cancel ho jaate hain, to main sum drop karke ek bracket = 0 hamesha likh sakta hoon." > **Kyun sahi lagta hai:** holonomic systems ke liye kaam karta hai. **Trap:** iske liye $\delta q_i$ ka **independent** hona zaroori hai. Non-holonomic constraints ke saath wo *nahi* hote, aur Lagrange multipliers chahiye. **Fix:** har bracket zero set karne se pehle independence check karo. > [!mistake] "Main $L$ differentiate kar sakta hoon $q$ aur $\dot q$ ko related treat karke (kyunki $\dot q = dq/dt$)." > **Kyun sahi lagta hai:** wo linked lagte hain. **Trap:** $\partial/\partial q_i$ aur $\partial/\partial\dot q_i$ mein tum $q_i$ aur $\dot q_i$ ko **independent variables** treat karte ho; link sirf $d/dt$ se reimpose hota hai. **Fix:** pehle partials (independent), phir total time derivative. --- > [!recall]- Feynman: 12-saal ke bacche ko explain karo > Socho ek bead curved wire par slide kar raha hai. Wire hamesha bead ko *sideways* push karti hai (taaki wo wire par rahe), kabhi us direction mein nahi jisme bead move kar sakta hai. To agar tum sirf un directions mein dekho jinmein bead *slide* karne ki permission hai, to wire ka push disappear ho jaata hai — wahan uska koi kaam nahi hota. D'Alembert kehta hai: sirf allowed directions mein dekho, "push minus mass×acceleration = 0" likho, aur annoying wire-force gayab ho jaati hai. Phir hum sab kuch **energy** ke zariye cleverly rewrite karte hain (motion energy minus stored energy), aur ek neat equation nikalti hai jo bead ko batati hai kaise move karna hai — bina kabhi jaane ki wire kitna hard push karta hai. > [!mnemonic] EL machine yaad rakho > **"Dot the Velocity, Minus the Position"** → $\frac{d}{dt}\partial_{\dot q}L - \partial_q L = 0$. > Aur derivation route ke liye: **D**'Alembert → **V**irtual work → **G**eneralized coords → **L**emmas (dots cancel / dots commute) → **K**inetic energy → **E**uler-Lagrange. *"Dare Very Good Lads Keep Equations."* --- ## #flashcards/physics Virtual displacement kya hota hai? ::: Ek imagined, instantaneous ($\delta t=0$) infinitesimal position change jo us instant par constraints ke saath consistent ho. D'Alembert's principle state karo. ::: $\sum_k(\vec F_k^{(a)} - \dot{\vec p}_k)\cdot\delta\vec r_k = 0$, sirf applied (non-constraint) forces par summing. D'Alembert's principle se constraint forces kyun drop ho jaate hain? ::: Wo allowed (virtual) displacements ke perpendicular hote hain, isliye $\sum_k\vec F_k^{(c)}\cdot\delta\vec r_k = 0$. Lemma 1 (cancellation of dots) state karo. ::: $\partial\dot{\vec r}_k/\partial\dot q_i = \partial\vec r_k/\partial q_i$. Lemma 2 (commuting derivatives) state karo. ::: $\frac{d}{dt}\partial\vec r_k/\partial q_i = \partial\dot{\vec r}_k/\partial q_i$. Generalized force $Q_i$ kya hota hai? ::: $Q_i = \sum_k\vec F_k^{(a)}\cdot\partial\vec r_k/\partial q_i$; conservative forces ke liye $-\partial V/\partial q_i$ ke barabar hota hai. $\delta q_i$ ki kaun si property har bracket ko zero set karne deti hai? ::: Unki independence, jo holonomic constraints dwara guaranteed hoti hai. $V$ ko $L$ mein absorb kyun kiya ja sakta hai? ::: Kyunki $V=V(q,t)$ mein $\partial V/\partial\dot q_i = 0$ hota hai, isliye $-V$ add karne se $\partial/\partial\dot q$ term nahi badlta lekin $-\partial V/\partial q = Q_i$ add ho jaata hai. Inertia term kis energy expression ke barabar hota hai? ::: $\sum_k m_k\ddot{\vec r}_k\cdot\partial\vec r_k/\partial q_i = \frac{d}{dt}\partial T/\partial\dot q_i - \partial T/\partial q_i$. EL equation ko $T$ aur $Q_i$ form mein likho. ::: $\frac{d}{dt}\partial T/\partial\dot q_i - \partial T/\partial q_i = Q_i$. --- ## Connections - [[D'Alembert's principle]] - [[Constraints and generalized coordinates]] - [[Hamilton's principle (least action)]] — same EL equation tak pahunchne ka alternative route - [[Generalized forces and potentials]] - [[Lagrange multipliers for non-holonomic constraints]] - [[Newton's laws of motion]] - [[Kinetic energy in generalized coordinates]] - [[Noether's theorem]] — $L$ ki symmetries → conserved quantities ## 🖼️ Concept Map ```mermaid flowchart TD N[Newton F=ma] CF[Constraint forces unknown] VD[Virtual displacement, time frozen] ZW[Constraint forces do zero work] DA[D'Alembert principle] GC[Generalized coordinates q_i] S1[Express delta r_k in q_i] QF[Generalized force Q_i] INE[Inertia part rewrite] KID[Key identity via product rule] EL[Euler-Lagrange equation] LAG[Lagrangian L = T - V] N -->|has nuisance| CF VD -->|allowed by constraints| ZW N -->|dotted with delta r| DA ZW -->|removes F_c| DA DA -->|rewrite in| GC GC -->|gives| S1 S1 -->|force part yields| QF S1 -->|inertia part yields| INE INE -->|uses| KID KID -->|combined with| QF QF -->|assemble| EL KID -->|assemble| EL LAG -->|substituted into| EL ```