Exercises — Cyclic coordinates — corresponding conservation law

Hum kya check karte hain: kya $x$ appear karta hai? kya $y$ appear karta hai? Na $x$ na $y$ $L$ mein dikhta hai (sirf unki velocities $\dot{x},\dot{y}$ hain). Toh $\partial L/\partial x=0$ aur $\partial L/\partial y=0$. Answer: ==both $x$ and $y$ are cyclic==. Flat plane "wahi dikhta hai" jahan bhi tum khade ho — yeh translational sameness exactly cyclic property hai.

Hum kya check karte hain: term $-\frac{1}{2}k{x}^2$ mein $x$ explicitly hai. Toh $\partial L/\partial x=-kx\ne 0$ (jab tak $x=0$ na ho). Answer: ==$x$ is NOT cyclic==. Spring ki potential energy change hoti hai jab tum move karte ho, toh ek genuine force $-kx$ hai. Koi missing coordinate nahi ⇒ yahan koi free conservation law nahi.

Hum kya check karte hain: $r$ appear karta hai ($r^2{\dot{\varphi }}^2$ ke andar aur $V(r)$ ke andar), toh $r$ cyclic nahi hai. Angle $\varphi$ sirf apni velocity $\dot{\varphi }$ ke through appear karta hai — coordinate $\varphi$ khud absent hai. Answer: ==$\varphi$ is cyclic, $r$ is not==. Dekho Central Force Motion.

Step (WHAT): $L$ ko $\dot{x}$ ke saath differentiate karo, $\dot{y}$ ko freeze karte hue. Sirf $\frac{1}{2}m{\dot{x}}^2$ mein $\dot{x}$ hai; uska derivative hai $m\dot{x}$. $p_x=\frac{\partial L}{\partial \dot{x}}=m\dot{x}.$ Kyun conserved hai: $x$ cyclic (L1.1) ⇒ ${\dot{p}}_x=0$ ⇒ $p_x=m\dot{x}=\text{const}$. Yeh ordinary linear momentum hai. Translational symmetry ⇒ momentum conservation.

Step: sirf term $\frac{1}{2}m{r}^2{\dot{\varphi }}^2$ mein $\dot{\varphi }$ hai. Differentiate karo: $\frac{1}{2}m{r}^2\cdot 2\dot{\varphi }=m{r}^2\dot{\varphi }$. $p_{\varphi }=\frac{\partial L}{\partial \dot{\varphi }}=m{r}^2\dot{\varphi }=\text{const}.$ Identify karo: yeh angular momentum about the axis perpendicular to the plane of motion hai — hum ise $J_z$ likhte hain (letter $J$ isliye choose kiya taaki $L$ se clash na ho, jo is poore page pe Lagrangian ko mean karta hai). $J_z$ measure karta hai "particle origin ke baare mein kitna rotation-oomph" rakhta hai: mass $\times$ radius${}^2$ $\times$ angular rate. Iska constant rehna Kepler's second law hai (equal areas in equal times).

$x$: $\partial L/\partial x=0$ ⇒ cyclic ⇒ $p_x=m\dot{x}=\text{const}$ (horizontal momentum frozen). $y$: $\partial L/\partial y=-mg\ne 0$ ⇒ not cyclic. Euler–Lagrange deta hai ${\dot{p}}_y=\partial L/\partial y=-mg$, toh $p_y=m\dot{y}$ steadily decreases — gravity vertically momentum pump kar raha hai bahar. Answer: ==horizontal $p_x$ conserved; vertical $p_y$ changes at rate $-mg$.==

Tool: $p_{\varphi }=m{r}^2\dot{\varphi }=\ell$ dono radii pe same number hai. $m=1$ ke saath: $r_1^2{\dot{\varphi }}_1=r_2^2{\dot{\varphi }}_2\Rightarrow (2{)}^2(3)=(6{)}^2{\dot{\varphi }}_2\Rightarrow 12=36{\dot{\varphi }}_2.$ $\boxed{{\dot{\varphi }}_2=\frac{12}{36}=\frac{1}{3}\approx \mathrm{0.333.}}$ Yeh kaisa dikhta hai (figure s01): per second swept shaded area $r_1$ aur $r_2$ dono pe same hai; bade radius pe planet crawl karta hai, chhote radius pe woh ghoom jaata hai.

Step: $p_{\varphi }=m{r}^2\dot{\varphi }$ se hum solve karte hain $\dot{\varphi }=\frac{p_{\varphi }}{m{r}^2}$. Square karo aur $mr$ se multiply karo: $mr{\dot{\varphi }}^2=mr{\left(\frac{p_{\varphi }}{m{r}^2}\right)}^2=\frac{p_{\varphi }^2}{m{r}^3}.$ Kyun useful hai: radial equation ban jaata hai $m\ddot{r}=\frac{p_{\varphi }^2}{m{r}^3}-\frac{dV}{dr}$ — ek equation sirf $r$ mein, kyunki cyclic coordinate eliminate ho gaya. Yeh exactly Central Force Motion ka "effective potential" reduction hai aur Hamiltonian Mechanics mein Routhian reduction ki spirit hai.

Cyclic check: $\varphi$ $L$ mein appear nahi karta (sirf $\dot{\varphi }$ karta hai, aur ${\sin }^2\theta$ depend karta hai $\theta$ pe, $\varphi$ pe nahi). Toh $\partial L/\partial \varphi =0$ ⇒ cyclic. Momentum: $p_{\varphi }=\partial L/\partial \dot{\varphi }=m{R}^2{\sin }^2\theta \dot{\varphi }=\text{const}$. $\theta =90^{\circ }$ pe: $\sin 90^{\circ }=1$, toh ${\sin }^2\theta =1$. $m{R}^2=1$ aur $p_{\varphi }=\frac{1}{2}$ ke saath: $\frac{1}{2}=(1)(1)\dot{\varphi }\Rightarrow \boxed{\dot{\varphi }=\frac{1}{2}=\mathrm{0.5.}}$ Pole ke paas ($\theta \to 0$), ${\sin }^2\theta \to 0$, toh $\dot{\varphi }$ blow up karna padega $p_{\varphi }$ fixed rakhne ke liye — bead upar ke paas furiously spin karta hai. Woh degenerate limit agले level ki geometric warning hai.

$L$ banao: kinetic energy $T$ (motion ki energy) dono beads ka sum hai, $T=\frac{1}{2}{m}_1{\dot{x}}_1^2+\frac{1}{2}{m}_2{\dot{x}}_2^2$; spring ki potential energy sirf stretch $x_2-x_1$ pe depend karti hai: $L=T-V=\frac{1}{2}{m}_1{\dot{x}}_1^2+\frac{1}{2}{m}_2{\dot{x}}_2^2-\frac{1}{2}k((x_2-x_1)-{\ell }_0{)}^2.$ Problem dhundho: na $x_1$ na $x_2$ akele cyclic hai, kyunki dono spring term mein appear karte hain. Clever variables: lo centre of mass $X=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$ aur separation $s=x_2-x_1$. Phir $L=\frac{1}{2}M{\dot{X}}^2+\frac{1}{2}\mu {\dot{s}}^2-\frac{1}{2}k(s-{\ell }_0{)}^2,M=m_1+m_2,\mu =\frac{m_1{m}_2}{M}.$ Padhho: $X$ ab $L$ mein absent hai ⇒ cyclic ⇒ $p_X=M\dot{X}=\text{const}=\text{total linear momentum}.$ Internal coordinate $s$ spring force carry karta hai; poora two-body system constant total momentum ke saath drift karta hai. Yeh analytical mechanics ka chehra hai "internal forces cannot move the centre of mass" ka.

$L$ banao: $L=T-V=\frac{1}{2}m({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2)-mgz.$ Cyclic check, teeno ke liye:

• $\partial L/\partial x=0$ ⇒ $x$ cyclic ⇒ $p_x=m\dot{x}=\text{const}$.
• $\partial L/\partial y=0$ ⇒ $y$ cyclic ⇒ $p_y=m\dot{y}=\text{const}$.
• $\partial L/\partial z=-mg\ne 0$ ⇒ $z$ not cyclic; ${\dot{p}}_z=-mg$, toh $p_z=m\dot{z}$ steadily falls. Answer: ==do conserved momenta $p_x,p_y$ (horizontal plane), vertical mein koi nahi.== Symmetry sirf gravity ke arrow ke along break hoti hai.

Pehle, woh object jo humein chahiye — energy function. Energy function define karo (Jacobi integral bhi kehte hain, ya momenta mein rewrite karne pe Hamiltonian): $h\equiv \dot{x}\frac{\partial L}{\partial \dot{x}}-L=\dot{x}{p}_x-L.$ Words mein: har velocity ko uske conjugate momentum se multiply karo, inhe add karo, aur $L$ subtract karo. $L=\frac{1}{2}m{\dot{x}}^2-V(x)$ ke liye yeh deta hai $h=\dot{x}(m\dot{x})-(\frac{1}{2}m{\dot{x}}^2-V)=\frac{1}{2}m{\dot{x}}^2+V(x)$ — total energy $T+V$. Key identity ($h$ kyun matter karta hai): $h$ ko motion ke saath differentiate karte hue aur Euler–Lagrange use karte hue, koi pata karta hai: $\frac{dh}{dt}=-\frac{\partial L}{\partial t}.$ Toh agar $L$ mein koi explicit time nahi ($\partial L/\partial t=0$), toh $dh/dt=0$ aur $h$ conserved hai. Yeh statement — ki time-translation symmetry energy function $h$ ko conserve karti hai — energy-function / Beltrami identity theorem hai. (Ek dimension mein yeh exactly Beltrami's identity hai; generally yeh energy-function argument hai jo Hamiltonian Mechanics mein jaata hai.) Student ke claim mein kya sahi hai: haan, ek conservation law follow karta hai "$L$ time pe depend nahi karta" se, aur conserved quantity is energy $h$ hai. Kya galat hai: $t$ ek generalized coordinate nahi hai — koi velocity $\dot{t}$ nahi hai aur isliye koi conjugate momentum $p_t\equiv \partial L/\partial \dot{t}$ nahi hai. Cyclic-coordinate theorem $t$ pe bilkul apply nahi hota; energy conservation alag energy-function/Beltrami theorem se aata hai upar, jo ek missing spatial coordinate ki bajaye time-translation symmetry se driven hai. Dekho Conservation of Energy. Yeh exactly parent note Cyclic coordinates — corresponding conservation law mein "Mistake C" hai.

Conservation intact hai: $p_{\varphi }=m{R}^2{\sin }^2\theta \dot{\varphi }$ saari valid motion ke liye exactly constant hai; $\dot{\varphi }$ ka divergence exactly wahi hai jo product fixed rakhta hai jab ${\sin }^2\theta \to 0$ hota hai. $\varphi$ ke cyclic hone ke baare mein kuch nahi tootta. Physical resolution — pole walled off hai. $\varphi$-motion mein stored kinetic energy dekho: $T_{\varphi }=\frac{1}{2}m{R}^2{\sin }^2\theta {\dot{\varphi }}^2=\frac{1}{2}\frac{p_{\varphi }^2}{m{R}^2{\sin }^2\theta }.$ Jab $\theta \to 0$ yeh without bound grow karta hai (yeh $1/{\sin }^2\theta$ ki tarah behave karta hai). Ek bead ke paas sirf finite total energy hai, toh woh nonzero $p_{\varphi }$ carry karte hue actually pole reach nahi kar sakta — $1/{\sin }^2\theta$ term ek infinite energy wall (ek "effective potential barrier") ki tarah act karta hai $\theta =0$ ke paas. Trajectory isliye kisi minimum ${\theta }_{\min }>0$ pe turn around karti hai jo energy balance se set hoti hai. Pole kab reach kiya ja sakta hai? Sirf agar $p_{\varphi }=0$ ho (pure meridian, "straight over the top" motion): tab $\dot{\varphi }=0$ poore time, koi barrier nahi, aur bead $\theta =0$ se guzar sakta hai. Toh divergence conservation law ka khud ko protect karna hai, fail nahi — yeh exactly woh trajectories forbid karta hai jo finite energy violate karengi.

(a) Translation symmetry ⇒ linear momentum. Agar coordinate $x\to x+a$ shift karna (har constant $a$ ke liye) $L$ unchanged chodh deta hai, toh $x$ cyclic hai aur Noether's Theorem ka conserved current exactly conjugate momentum $p_x=m\dot{x}$ hai — ordinary linear momentum. (b) Rotation symmetry ⇒ angular momentum. Agar angle $\varphi \to \varphi +\alpha$ rotate karna (har constant $\alpha$ ke liye) $L$ unchanged chodh deta hai, toh $\varphi$ cyclic hai aur Noether ka conserved current $p_{\varphi }=m{r}^2\dot{\varphi }$ hai — L2.2 ka angular momentum $J_z$. Noether strictly more general kyun hai: cyclic-coordinate rule sirf tab fire hota hai jab ek single coordinate literally $L$ se disappear ho. Noether's theorem kisi bhi continuous symmetry ko cover karta hai, jinmein woh bhi shamil hain jahan koi single $q_k$ vanish nahi hota lekin coordinates ka ek combination preserved hota hai — jaise rotations Cartesian coordinates mein expressed (jahan $x$ aur $y$ dono appear karte hain phir bhi mixture symmetric hai), Lorentz boosts, aur gauge transformations. Cyclic coordinates Noether ke far larger landscape ka easy, directly-visible corner hain.

• L1: kya tum absent coordinate ek nazar mein dhundh sakte ho? ::: Bare $q_k$ ko $L$ mein missing dhundho; ${\dot{q}}_k$ ko ignore karo.
• L2: kya tum $p_k=\partial L/\partial {\dot{q}}_k$ likh aur naam de sakte ho? ::: Haan — $x$ ke liye linear, $\varphi$ ke liye angular $m{r}^2\dot{\varphi }$.
• L3: kya tum conservation use karke ek coordinate eliminate kar sakte ho? ::: $\dot{\varphi }=p_{\varphi }/(m{r}^2)$ ke liye solve karo aur substitute karo.
• L4: kya tum variables invent karke ek cyclic coordinate create kar sakte ho? ::: Two-body systems ke liye centre-of-mass coordinate.
• L5: kya tum momentum theorem ko energy theorem se alag kar sakte ho? ::: Space symmetry (cyclic coord) vs time symmetry (energy function $h$) — alag theorems.