Visual walkthrough — Generalized momenta and generalized forces

2.1.7 · D2 · Physics › Analytical Mechanics › Generalized momenta and generalized forces

Neeche sab kuch ek hi running example use karta hai: ek bead jo ek curved track par freely move kar sakta hai, jiski position hum ek number $q$ se label karenge. Hum literally sirf ek wire par ek dot se shuru karte hain.

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Step 1 — Ek dot, ek number

PICTURE.

Ek number kyun, do kyun nahi. Wire ek curve hai. Ek curve 1-dimensional hoti hai. $(x,y)$ ko us par force karne ka matlab hai ek leftover equation ($y=f(x)$) hamesha saath rakhna. $q$ choose karna us equation ko pehle hi phek deta hai, physics karne se pehle. Yahi poora reason hai ki Lagrangian Mechanics exist karta hai.

Real Cartesian position ek us ek number ka function hai: $\vec{r}=\vec{r}(q,t).$

• $\vec{r}$ — origin se bead tak ka sachha position arrow (Newton yahi use karta hai).
• $q$ — humara chosen dial.
• $t$ — allowed hai agar wire khud move ho rahi ho (yeh hume baad mein chahiye hoga).

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Step 2 — Dial ko wiggle karo: velocity ek same picture se rate hai

$\vec{v}=\frac{d\vec{r}}{dt}=\frac{\partial \vec{r}}{\partial q}\dot{q}+\frac{\partial \vec{r}}{\partial t}.$

Har symbol padho:

• $\vec{v}$ — bead ka sachha velocity arrow.
• $\frac{\partial \vec{r}}{\partial q}$ — ek unit dial-turn par bead real space mein kitna move karta hai, time frozen rakhte hue. Yeh ek vector tangent to the wire hai. Ise "sensitivity arrow" kaho.
• $\dot{q}$ — hum dial kitni tezi se ghuma rahe hain.
• $\frac{\partial \vec{r}}{\partial t}$ — extra motion agar wire khud drag ho rahi ho (fixed wire ke liye zero).

PICTURE.

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Step 3 — Bead ko push karo: real work, wire ke saath measure kiya gaya

Dot product kyun, sirf $F\delta r$ kyun nahi. Wire ke across point karne wala force bead par koi work nahi karta jo sirf wire ke along move kar sakta hai. Dot product $\vec{F}\cdot \delta \vec{r}$ automatically force ka sirf along-wire part rakhta hai. Yeh exactly woh sawaal hai "is push ka kitna hissa actually motion mein help karta hai?" — aur dot product woh tool hai jo jawab deta hai.

$\delta W=\vec{F}\cdot \delta \vec{r}.$

PICTURE.

• $\delta W$ — work (energy) ka tiny bit deliver hua.
• $\vec{F}$ — real applied force arrow.
• $\delta \vec{r}$ — tangent ke along tiny allowed step.
• $\vec{F}\cdot \delta \vec{r}=|\vec{F}||\delta \vec{r}|\cos \alpha$ — sirf $\vec{F}$ ka step ke along component count karta hai; $\alpha$ unke beech ka angle hai.

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Step 4 — Nudge ko dial ki terms mein rewrite karo

Ab Step 3 ke work mein substitute karo: $\delta W=\vec{F}\cdot \frac{\partial \vec{r}}{\partial q}\delta q=\underset{\text{ise }Q\text{ kaho}}{\underbrace{\left(\vec{F}\cdot \frac{\partial \vec{r}}{\partial q}\right)}}\delta q.$

PICTURE.

Term by term:

• $\delta q$ — tiny dial-turn (ek hi free cheez jo bachi hai).
• $\frac{\partial \vec{r}}{\partial q}$ — sensitivity arrow (Step 2): dial-turns ko real steps mein convert karta hai.
• $\vec{F}\cdot \frac{\partial \vec{r}}{\partial q}$ — woh bundle jo $\delta q$ ko multiply karta hai work dene ke liye. Hum ise $Q$ naam dete hain.

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Step 5 — $Q$ ek torque kyun ban jaata hai jab $q$ ek angle ho

• $q$ ek length (metres): $Q$ ki units hain J/m $=$ newtons → ek ordinary force.
• $q$ ek angle (radians, dimensionless): $Q$ ki units hain J/rad $=$ N·m → ek torque.

Yeh coincidence kyun nahi hai. Ek angle ke liye, sensitivity arrow $\partial \vec{r}/\partial q$ ki length $\approx r$ hoti hai (ek small angle ek arc $r\delta \theta$ sweep karta hai). Toh $Q_{\theta }=\vec{F}\cdot \frac{\partial \vec{r}}{\partial \theta }=(\text{tangential force})\times r,$ jo exactly schoolbook torque hai $=$ force $\times$ lever arm. Formula ne torque ko kisi special effort ke bina rediscover kar liya.

PICTURE.

Recall

Agar $q$ ek angle hai, toh generalized force $Q_{\theta }$ ek ::: torque hai, N·m (joules per radian) units ke saath. Iski units plain force se alag hain kyunki ::: weight $\partial \vec{r}/\partial q$ ki units length per radian hain, yaani ek lever arm.

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Step 6 — Momentum side: kinetic energy wahi tangent arrow se milti hai

Ek bead ki kinetic energy hai $T=\frac{1}{2}m\vec{v}\cdot \vec{v}$. $\dot{q}$ ke w.r.t. differentiate karo aur Step 2 ki cancel-the-dots identity use karo: $\frac{\partial T}{\partial \dot{q}}=m\vec{v}\cdot \frac{\partial \vec{v}}{\partial \dot{q}}=m\vec{v}\cdot \frac{\partial \vec{r}}{\partial q}.$ Agar $V$ mein koi $\dot{q}$ nahi (usual case), toh $p=\partial L/\partial \dot{q}=\partial T/\partial \dot{q}$.

Term by term:

• $m\vec{v}$ — sachha Newtonian momentum arrow.
• $\partial \vec{r}/\partial q$ — tangent sensitivity arrow (phir se!).
• $p=(m\vec{v})\cdot (\text{tangent arrow})$ — dial ki direction par true momentum ka projection.

PICTURE.

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Step 7 — Momentum ke liye angle case, aur cyclic-coordinate payoff

PICTURE.

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Ek-picture summary

Upar sab kuch wahi tangent arrow $\partial \vec{r}/\partial q$ ke do projections hain:

• Force ko us par project karo → generalized force $Q$ (angles ke liye ek torque).
• Momentum ko us par project karo → generalized momentum $p$ (angles ke liye angular momentum).
• Euler–Lagrange unhe stitch karta hai: $\dot{p}=Q$.

Recall Feynman retelling — poora walk plain words mein

Ek ball ko ek bent wire par rakho. Kyunki woh sirf ek taraf move kar sakta hai, ek dial $q$ batata hai woh kahaan hai. Dial ghao aur ball ek certain direction mein roll karta hai — woh direction (dial ke ek turn par) hamaara ek magic arrow hai. Ab do kaam karo. Pehle, apna push us arrow par shine karo: aapke force ka jo bhi fraction arrow ke saath align karta hai wahi akela part kaam karta hai, aur hum us shadow ko generalized force $Q$ karte hain. Doosra, ball ka momentum usi arrow par shine karo: woh shadow generalized momentum $p$ hai. Agar dial ek angle hota hai, toh dono shadows lever-arm length $r$ se badhte hain: push-shadow ek twist (torque) ban jaata hai aur momentum-shadow ek spin (angular momentum) — koi nayi physics nahi, bas arrow lamba ho gaya. Aakhir mein, energy bookkeeping (Euler–Lagrange) guarantee karta hai ki twist twist ke rate ke barabar hai jab spin change hota hai, $\dot{p}=Q$ — Newton ka law nayi clothes mein. Aur agar energy dial ki value ki parwah hi nahi karti, toh spin kabhi nahi badal sakta: isliye angular momentum conserved hota hai.

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