Visual walkthrough — Kinetic energy in generalized coordinates

2.1.3 · D2 · Physics › Analytical Mechanics › Generalized coordinates mein Kinetic Energy

Step 1 — Ek point jo track par tika hua hai: position map

Symbols ko wahan padhein jahan woh baithe hain:

• $\mathbf{r}$ ::: origin se particle tak ka arrow, room mein uski asli position.
• $q$ ::: woh ek number jo hum change kar sakte hain — track par position.
• $t$ ::: ghadi. Yeh tabhi aata hai jab koi track ko physically hila raha ho. Yeh soch apne paas rakhein; yeh Steps 5–7 ka hero hai.

MAP KYUN? Kyunki energy room-coordinates ($x,y$) mein janam leti hai lekin Lagrangian mechanics ise track-coordinates ($q$) mein chahti hai. Ek map donon ke beech translation dictionary hai. Figure dekho: jaise $q$ slide karta hai, $\mathbf{r}$ ki tip printed curve par chalne lagti hai.

Step 2 — Tip kitni tez chalti hai? (chain rule, drawn)

Ab actual velocity. Tip do independent reasons se chalti hai:

$\dot{\mathbf{r}}=\underset{\text{tangent arrow}}{\underbrace{\frac{\partial \mathbf{r}}{\partial q}}}\underset{\text{kitni tez }q\text{ change hota hai}}{\underbrace{\dot{q}}}+\underset{\text{track khicha ja raha hai}}{\underbrace{\frac{\partial \mathbf{r}}{\partial t}}}.$

Term by term:

• $\dot{\mathbf{r}}$ ::: room mein sachchi velocity arrow (upar dot matlab "rate per second").
• $\frac{\partial \mathbf{r}}{\partial q}\dot{q}$ ::: (tangent direction) $\times$ (coordinate ki speed). Yeh hai track ke saath slide karna.
• $\frac{\partial \mathbf{r}}{\partial t}$ ::: tip ka tabhi bhi chalna jab $q$ frozen ho, kyunki track khud carry kiya ja raha hai. Zero hai jab tak constraint na chale.

Figure donon contributions ko do alag arrows ki tarah dikhata hai, phir unka sum sachchi velocity ke roop mein.

Step 3 — Energy ko speed-squared chahiye: dot product aata hai

Toh total kinetic energy, har particle $a$ (har mass $m_a$) par summed, hai $T={\sum }_a\frac{1}{2}{m}_a{\dot{\mathbf{r}}}_a\cdot {\dot{\mathbf{r}}}_a.$

• ${\sum }_a$ ::: saare particles par add karo.
• $\frac{1}{2}{m}_a$ ::: jaana-pehchana "half mass".
• ${\dot{\mathbf{r}}}_a\cdot {\dot{\mathbf{r}}}_a$ ::: particle $a$ ka speed-squared.

DOT PRODUCT KYUN, NA KI SIRF NUMBERS MULTIPLY KARNA? Kyunki velocity ek arrow hai, number nahi. Sirf dot product do arrows ko ek single scalar "speed-squared" mein badalta hai, $x$ aur $y$ (aur $z$) par automatically summing karte hue. Figure ek velocity arrow aur uska apna length-squared shaded square ke roop mein dikhata hai.

Step 4 — Do brackets multiply karo: teen families of terms appear hoti hain

Step-2 velocity ko dot product mein substitute karo. Har particle ke liye: ${\dot{\mathbf{r}}}_a\cdot {\dot{\mathbf{r}}}_a=\left({\sum }_i\frac{\partial {\mathbf{r}}_a}{\partial q_i}{\dot{q}}_i+\frac{\partial {\mathbf{r}}_a}{\partial t}\right)\cdot \left({\sum }_j\frac{\partial {\mathbf{r}}_a}{\partial q_j}{\dot{q}}_j+\frac{\partial {\mathbf{r}}_a}{\partial t}\right).$

Ab hum kai coordinates $q_1,\dots ,q_n$ allow karte hain — isliye ek bracket index $i$ use karta hai aur mirror bracket index $j$: woh independently run karte hain, toh har $i$ har $j$ se milta hai.

Figure mein multiplication grid ke chaar cells hain; do off-diagonal cells identical hain (isliye $T_1$ mein $\times 2$ hota hai jo baad mein $\frac{1}{2}$ se cancel ho jaata hai). Yeh decomposition ka poora raaz hai — pure algebra of $(A+B{)}^2$.

Step 5 — Teen pieces padhna

Velocity ke powers ke hisaab se collect karne par headline milti hai: $\boxed{T=T_2+T_1+T_0}$

$T_1={\sum }_i{a}_i{\dot{q}}_i,a_i={\sum }_a{m}_a\frac{\partial {\mathbf{r}}_a}{\partial q_i}\cdot \frac{\partial {\mathbf{r}}_a}{\partial t}$ $T_0=\frac{1}{2}{\sum }_a{m}_a{\left(\frac{\partial {\mathbf{r}}_a}{\partial t}\right)}^2.$

• $a_i$ ::: tangent arrow aur drag arrow ke beech overlap. Nonzero tabhi jab track chale.
• $T_0$ ::: half-mass times drag-speed-squared. Woh energy jo particle ke paas hogi track par bilkul bhi na chal raha ho tab bhi.

Figure teen terms ko teen coloured bars ki tarah stack karta hai jinki heights "power of $\dot{q}$" = 2, 1, 0 hain.

Step 6 — Degenerate case: track koi nahi hilata (scleronomic)

Yeh 95% textbooks wala case hai. Kyunki $T$ ab $\dot{q}$'s mein degree 2 ka homogeneous hai, Euler's theorem deta hai ${\sum }_i{\dot{q}}_i\partial T/\partial {\dot{q}}_i=2T$ — woh identity jo baad mein $H=T+V$ banati hai energy function mein aur momentum–energy machinery of Euler-Lagrange equations ko power deti hai.

Worked check — plane polar ($x=r\cos \theta ,y=r\sin \theta$), coordinates $(r,\theta )$, no time: $\frac{\partial \mathbf{r}}{\partial r}=(\cos \theta ,\sin \theta ),\frac{\partial \mathbf{r}}{\partial \theta }=(-r\sin \theta ,r\cos \theta ).$

• $M_{rr}=m({\cos }^2\theta +{\sin }^2\theta )=m$.
• $M_{\theta \theta }=m{r}^2({\sin }^2\theta +{\cos }^2\theta )=m{r}^2$.
• $M_{r\theta }=m(\cos \theta )(-r\sin \theta )+m(\sin \theta )(r\cos \theta )=0$ — do tangent arrows perpendicular hain, toh koi overlap nahi. $\boxed{T=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\theta }}^2)}.$ Figure radial aur angular tangent arrows ko right angles par draw karta hai — woh right angle hi $M_{r\theta }$ ka zero hona hai.

Step 7 — Moving-track case: jahan $T_1$ aur $T_0$ actually appear hote hain

Ek line mein lesson: $T_1$ nonzero hota hai exactly tab jab tangent arrow aur drag arrow overlap karte hain ($a_i=$ unka dot product). Perpendicular ⇒ $T_1=0$ (bead case). Parallel ⇒ $T_1$ maximal (sliding-frame case). Figure donon geometries ko side by side contrast karta hai.

Ek-picture summary

Upar sab kuch ek identity hai — $(A+B)\cdot (A+B)$ — sort ki gayi ki kitne velocity factors bachte hain. Final figure poora flowchart hai: position map → chain-rule velocity → dot-product square → teen sorted piles, "freeze the clock" switch ke saath jo unme se do ko maar deta hai.

Active recall