Exercises — Noether's theorem — symmetry ↔ conservation law

Kya dhundna hai: wo coordinate jiska symbol $L$ mein appear na kare — sirf uski velocity ho. $L$ scan karo: hum dekhte hain $r$, $\dot{r}$, $\dot{\varphi }$ — lekin koi bare $\varphi$ nahi. To $\varphi$ cyclic hai. Iska matlab kyun important hai: $\varphi$ ke liye Euler–Lagrange kuch yun likhta hai $\frac{d}{dt}\frac{\partial L}{\partial \dot{\varphi }}=\underset{=0}{\underbrace{\frac{\partial L}{\partial \varphi }}}\Rightarrow \frac{d}{dt}\left(m{r}^2\dot{\varphi }\right)=0.$ Conserved quantity conjugate momentum hai $p_{\varphi }=\frac{\partial L}{\partial \dot{\varphi }}=m{r}^2\dot{\varphi }=\text{const}.$ Ye bead ki angular momentum $L_z$ hai. Physical symmetry: poore system ko kisi bhi angle $\varphi \to \varphi +ϵ$ se rotate karne par $L$ unchanged rehta hai (table ki koi preferred direction nahi hai).

(a) Time-translation invariance ($L$ mein koi explicit $t$ nahi) → energy conserved. (b) Space-translation invariance → total linear momentum $\hat{n}\cdot \vec{P}$ conserved. (c) Rotation invariance → total angular momentum $\hat{n}\cdot \vec{L}$ conserved. Pattern ye hai: ek continuous knob mein "doesn't-matter" hona ek number ko freeze kar deta hai.

Step 1 — transformation velocities ke saath KIYA karta hai. Shifts ko time mein differentiate karo: $\delta \dot{x}=-\dot{y},\delta \dot{y}=\dot{x}$. Step 2 — invariance check (WHY $\delta L=0$). $\delta L=\frac{\partial L}{\partial \dot{x}}\delta \dot{x}+\frac{\partial L}{\partial \dot{y}}\delta \dot{y}=m\dot{x}(-\dot{y})+m\dot{y}(\dot{x})=0.✓$ Step 3 — charge banao. $Q=p_x\delta x+p_y\delta y=m\dot{x}(-y)+m\dot{y}(x)=m(x\dot{y}-y\dot{x})=L_z.$ To rotation ↔ $L_z$, bilkul waisa hi jैसा parent §5 ne promise kiya tha.

Neeche diya figure padho taaki geometrically samjho WHY $\delta L=0$ hai. Kala arrow vector $(x,y)$ hai; lal arrow dikhata hai rotation use kahan bhejti hai. Notice karo ki lal push $\delta \vec{r}=(-y,x)$ kale arrow ke perpendicular hai (unka dot product $x(-y)+y(x)=0$ hai). Ek vector ke right angles par push karna use bina stretch kiye rotate karta hai — isliye length freeze ho jaati hai, aur exactly isliye kinetic energy (jo us velocity vector ki length se bani hai) change nahi hoti. Ye perpendicular-push picture "rotation is a symmetry" ka geometric heart hai, aur jo conserved charge ye produce karta hai wo $L_z$ hai.

Angular momentum (cyclic $\varphi$ se, jaise Ex 1.1 mein): $p_{\varphi }=m{r}^2\dot{\varphi }=\text{const}$. Energy (koi explicit $t$ nahi): $E=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\varphi }}^2)-\frac{GMm}{r}$. Circular orbit par $\dot{r}=0$, aur forces balance karne se $mr{\dot{\varphi }}^2=\frac{GMm}{r^2}$, to ${\dot{\varphi }}^2=\frac{GM}{r^3}=\frac{1}{64}$, $\dot{\varphi }=\frac{1}{8}$.

• $p_{\varphi }=1\cdot 16\cdot \frac{1}{8}=2$.
• $E=\frac{1}{2}\cdot 1\cdot 16\cdot \frac{1}{64}-\frac{1}{4}=\frac{1}{8}-\frac{1}{4}=-\frac{1}{8}$. To $p_{\varphi }=2$ aur $E=-0.125$ (bound orbit, isliye $E<0$).

Pehle master identity $\frac{dh}{dt}=-\frac{\partial L}{\partial t}$ derive karo — ise sirf quote mat karo. $h=\dot{x}p-L$ se shuru karo jahan $p=\frac{\partial L}{\partial \dot{x}}$, aur time mein product rule aur chain rule use karke differentiate karo: $\frac{dh}{dt}=\ddot{x}p+\dot{x}\dot{p}-\frac{dL}{dt}.$ Ab $\frac{dL}{dt}$ ko chain rule se expand karo (yaad raho $L$ depend karta hai $x$, $\dot{x}$, aur explicitly $t$ par): $\frac{dL}{dt}=\frac{\partial L}{\partial x}\dot{x}+\frac{\partial L}{\partial \dot{x}}\ddot{x}+\frac{\partial L}{\partial t}=\frac{\partial L}{\partial x}\dot{x}+p\ddot{x}+\frac{\partial L}{\partial t}.$ Ise substitute karo: $\frac{dh}{dt}=\ddot{x}p+\dot{x}\dot{p}-\frac{\partial L}{\partial x}\dot{x}-p\ddot{x}-\frac{\partial L}{\partial t}=\dot{x}(\dot{p}-\frac{\partial L}{\partial x})-\frac{\partial L}{\partial t}.$ Bracket Euler–Lagrange se vanish ho jaata hai ($\dot{p}=\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}=\frac{\partial L}{\partial x}$). Jo bachta hai wo hai $\boxed{\frac{dh}{dt}=-\frac{\partial L}{\partial t}}$ To energy sirf $L$ ke explicit time dependence se change hoti hai — ye precise Noether statement hai "time-translation symmetry ⇒ energy conserved." Yahan apply karo. Sirf potential mein $t$ hai: $\frac{\partial L}{\partial t}=-\frac{1}{2}\dot{k}{x}^2=-\frac{1}{2}{k}_0\alpha x^2$, to $\frac{dh}{dt}=+\frac{1}{2}{k}_0\alpha x^2\ne 0.$ Energy rate $\frac{1}{2}{k}_0\alpha x^2$ se leak ho rahi hai: tum knob $k$ ghumaa ke system par kaam kar rahe ho. Symmetry broken → charge not conserved, bilkul waisa hi jaisa Noether ka converse warn karta hai.

Step 1 — velocity shift. $\delta \dot{x}=\frac{d}{dt}(t)=1$. Step 2 — $L$ vary karo. $\delta L=m\dot{x}\delta \dot{x}=m\dot{x}=\frac{d}{dt}(mx)$. To ye $F=mx$ ke saath total time derivative hai. Action abhi bhi invariant hai (boundary term), isliye Noether abhi bhi apply hota hai — lekin correction ke saath. Step 3 — corrected charge. $Q=p\delta x-F=(m\dot{x})(t)-mx=m\dot{x}t-mx.$ Check karo ki ye constant hai: $\frac{dQ}{dt}=m\dot{x}+m\ddot{x}t-m\dot{x}=m\ddot{x}t=0$ kyunki free particle mein $\ddot{x}=0$ hota hai. ✓ Physically $Q=-m{x}_0$ hai, initial position encode karta hai — Galilean boosts ka conserved charge. Ye wahi case hai jo parent ne flag kiya tha: $\delta L\ne 0$ lekin $\delta L=dF/dt$ abhi bhi ek law deta hai.

WHY $V$ translation rok nahi sakti: $V$ depend karta hai ${\vec{r}}_1-{\vec{r}}_2$ par; ${\vec{r}}_a\to {\vec{r}}_a+ϵ\hat{n}$ ke under difference unchanged rehta hai, to $\delta V=0$, aur kinetic terms sirf velocities dekhte hain. Isliye $\delta L=0$. Noether charge: $Q={\vec{p}}_1\cdot \hat{n}+{\vec{p}}_2\cdot \hat{n}=\hat{n}\cdot ({\vec{p}}_1+{\vec{p}}_2)=\hat{n}\cdot \vec{P}.$ $\hat{n}=\hat{x}$ lo: $P_x=m_1{\dot{x}}_1+m_2{\dot{x}}_2=2(1)+3(-1)=-1$ — ek fixed number, bhaley hi $p_{1x}=2$ aur $p_{2x}=-3$ force ke through change ho sakein. Moral: center-of-mass coordinate cyclic hai; relative coordinate saari interaction carry karta hai.

Step 1 — time shift ke liye corrected charge: $Q=p\delta x-F=(m\dot{x})(\dot{x})-L=m{\dot{x}}^2-(\frac{1}{2}m{\dot{x}}^2-V)=\frac{1}{2}m{\dot{x}}^2+V.$ Ye exactly $h=T+V=E$ hai. To "clock shift karo" ↔ energy, cleanly derived, assumed nahi. Step 2 — number: $E=\frac{1}{2}(2)(3^2)+5=9+5=14$.

Step 1 — differentiate karo, is baat ka use karke ki $\vec{L}$ constant hai. $\frac{d\vec{A}}{dt}=\dot{\vec{p}}\times \vec{L}-mk\frac{d\hat{r}}{dt}.$ Step 2 — plane-polar unit vectors set up karo. Motion ek plane mein hai; $\hat{r}$ (radial) aur $\hat{\varphi }$ (tangential) use karo, jahan $\hat{z}=\hat{r}\times \hat{\varphi }$. Unke time derivatives standard results hain $\frac{d\hat{r}}{dt}=\dot{\varphi }\hat{\varphi }$ aur $\frac{d\hat{\varphi }}{dt}=-\dot{\varphi }\hat{r}$. Angular momentum hai $\vec{L}=m\vec{r}\times \dot{\vec{r}}=m{r}^2\dot{\varphi }\hat{z}$. Step 3 — pehla term explicitly expand karo (koi black box nahi). $\dot{\vec{p}}=-\frac{k}{r^2}\hat{r}$ aur $\vec{L}=m{r}^2\dot{\varphi }\hat{z}$ insert karo: $\dot{\vec{p}}\times \vec{L}=(-\frac{k}{r^2}\hat{r})\times \left(m{r}^2\dot{\varphi }\hat{z}\right)=-mk\dot{\varphi }(\hat{r}\times \hat{z}).$ Ab right-handed relation $\hat{r}\times \hat{z}=-\hat{\varphi }$ use karo (kyunki $\hat{z}\times \hat{r}=+\hat{\varphi }$). Isliye $\dot{\vec{p}}\times \vec{L}=-mk\dot{\varphi }(-\hat{\varphi })=+mk\dot{\varphi }\hat{\varphi }.$ Step 4 — doosra term expand karo. $mk\frac{d\hat{r}}{dt}=mk\dot{\varphi }\hat{\varphi }.$ Step 5 — combine aur cancel karo. $\frac{d\vec{A}}{dt}=mk\dot{\varphi }\hat{\varphi }-mk\dot{\varphi }\hat{\varphi }=\vec{0}.✓$ Saari $r$-dependence cancel ho gayi kyunki dono terms ne same $\hat{\varphi }$-vector produce kiya — ye $1/r$ potential ka special algebra hai. Meaning: $\vec{A}$ orbit ke major axis ke along point karta hai length $mke$ (eccentricity $e$) ke saath; uski constancy isliye hai ki Kepler ellipses precess nahi karti. Associated symmetry 4 dimensions mein ek rotation group mein rehti hai — dekho Symmetry Groups & Lie Algebras.

Step 1 — har ingredient kaise scale karta hai. $x\to \lambda x$ aur $t\to {\lambda }^{2}t$ ke under, velocity ek ratio hai: $\dot{x}=\frac{dx}{dt}\to \frac{d(\lambda x)}{d({\lambda }^{2}t)}=\frac{\lambda dx}{{\lambda }^{2}dt}={\lambda }^{-1}\dot{x}\Rightarrow {\dot{x}}^2\to {\lambda }^{-2}{\dot{x}}^2.$ Time measure scale hota hai $dt\to {\lambda }^{2}dt$. Step 2 — action assemble karo. $S=\int \frac{1}{2}m{\dot{x}}^{2}dt\to \int \frac{1}{2}m\left({\lambda }^{-2}{\dot{x}}^2\right)\left({\lambda }^{2}dt\right)=\int \frac{1}{2}m{\dot{x}}^{2}dt=S.$ Velocity se ${\lambda }^{-2}$ factor aur measure se ${\lambda }^2$ factor exactly cancel ho jaate hain — action strictly invariant hai. Ye action ki genuine symmetry hai. Step 3 — concrete numerical check (step by step reconciled). $m=1,v_0=1,T=1$ lo original path $x(t)=t$ ke liye: $S={\int }_0^1\frac{1}{2}(1)(1{)}^{2}dt=\frac{1}{2}.$ Ab $\lambda =2$ ke saath scaling apply karo. Transformed path hai $x^{\prime }=\lambda x=2x$ jo scaled time $t^{\prime }={\lambda }^{2}t=4t$ par express ki gayi; $t$ eliminate karne par $x^{\prime }=2t=2\cdot \frac{t^{\prime }}{4}=\frac{t^{\prime }}{2}$ milta hai, yaani nayi velocity $\frac{d{x}^{\prime }}{d{t}^{\prime }}=\frac{1}{2}={\lambda }^{-1}{v}_0$ hai (Step 1 se match karta hai), aur naya interval $t^{\prime }\in [0,{\lambda }^{2}T]=[0,4]$ hai. Scaled action directly compute karo: $S^{\prime }={\int }_0^4\frac{1}{2}(1)(\frac{1}{2}{)}^{2}d{t}^{\prime }=\frac{1}{2}\cdot \frac{1}{4}\cdot 4=\frac{1}{2}.$ $S^{\prime }=\frac{1}{2}=S$ — shrunken velocity-squared ($\times \frac{1}{4}$) exactly four-times-longer time interval ($\times 4$) se compensate ho jaati hai. Invariance real numbers par demonstrate ki gayi, sirf symbols par nahi.

Wo coordinate jo $L$ se missing ho use ::: cyclic (ignorable) kehte hain, aur uska conjugate momentum conserved hota hai. Energy exactly tab conserved hoti hai jab $L$ ka ::: time $t$ par koi explicit dependence na ho. Symmetry $q_i\to q_i+ϵ\delta q_i$ ke saath $\delta L=dF/dt$ ke liye Noether charge hai ::: $Q={\sum }_i{p}_i\delta q_i-F$. Rotation invariance ::: angular momentum $\hat{n}\cdot \vec{L}$ conserve karti hai.