Exercises — Hamiltonian — definition H = Σpᵢq̇ᵢ − L

(b). Conjugate (generalized) momentum defined hai $L$ ke velocity ${\dot{q}}_i$ ke respect mein derivative ke roop mein, coordinate ke nahi.

• (a) sirf free particle ke liye sach hai jahan $L=\frac{1}{2}m{\dot{q}}^2$; yeh ek special case hai, definition nahi.
• (c) Euler–Lagrange equation se ${\dot{p}}_i$ hai, $p_i$ nahi.
• (d) $-{\dot{p}}_i$ hai (Hamilton's equation), $p_i$ nahi.

Nahi. $H$ sirf $(q,p,t)$ ka function hona chahiye. Is expression mein abhi bhi $\dot{x}$ hai. Tumhe $p=m\dot{x}\Rightarrow \dot{x}=p/m$ invert karke substitute karna hoga, jisse $H=\frac{p^2}{2m}+V(x)$ milega.

$2n$ first-order ODEs: ${\dot{q}}_i=\partial H/\partial p_i$ aur ${\dot{p}}_i=-\partial H/\partial q_i$. Yeh Lagrangian Mechanics ke $n$ second-order Euler–Lagrange equations ko replace karta hai.

Momentum: $p=\partial L/\partial \dot{x}=m\dot{x}$. Invert: $\dot{x}=p/m$. Assemble & substitute: $H=p\dot{x}-L=p\cdot \frac{p}{m}-\left(\frac{1}{2}m\frac{p^2}{m^2}-\frac{1}{2}k{x}^2\right)=\frac{p^2}{m}-\frac{p^2}{2m}+\frac{1}{2}k{x}^2.$ $\boxed{H=\frac{p^2}{2m}+\frac{1}{2}k{x}^2}=T+V.$

Momenta: $p_x=m\dot{x}$, $p_y=m\dot{y}$, isliye $\dot{x}=p_x/m$, $\dot{y}=p_y/m$. Assemble: $H=p_x\dot{x}+p_y\dot{y}-L=\frac{p_x^2+p_y^2}{2m}$. Hamilton's equations: $\dot{x}=\frac{\partial H}{\partial p_x}=\frac{p_x}{m},\dot{y}=\frac{p_y}{m},{\dot{p}}_x=-\frac{\partial H}{\partial x}=0,{\dot{p}}_y=0.$ Momenta constant hain → straight-line motion, free particle ke liye expected hai.

Momentum: $p=\partial L/\partial \dot{q}=m\dot{q}$. Invert: $\dot{q}=p/m$. (Ab hum isse har $\dot{q}$ hatane ke liye use karenge.) Assemble (substitute karne se pehle, $\dot{q}$ visible rakho): $H=p\dot{q}-L=p\dot{q}-\left(\frac{1}{2}m{\dot{q}}^2-mgq\right).$ Substitute $\dot{q}=p/m$ har jagah — pehla $\dot{q}$ ban jaata hai $p/m$, aur $\frac{1}{2}m{\dot{q}}^2$ ban jaata hai $\frac{1}{2}m(p/m{)}^2=p^2/(2m)$: $H=p\cdot \frac{p}{m}-\frac{1}{2}m{\left(\frac{p}{m}\right)}^2+mgq=\frac{p^2}{m}-\frac{p^2}{2m}+mgq=\frac{p^2}{2m}+mgq.$ Yeh substitution valid hai kyunki $\dot{q}=p/m$ identically hold karta hai jab hum $p$ define karte hain; koi leftover velocities nahi hain, isliye $H=H(q,p)$ as required. Yahan $T=p^2/2m$ ($\dot{q}$ mein quadratic hai) aur $V=mgq$ (velocity-independent hai), isliye $H=T+V=E$. ✓

$\dot{\theta }=\frac{\partial H}{\partial p_{\theta }}=\frac{p_{\theta }}{m{\ell }^2},{\dot{p}}_{\theta }=-\frac{\partial H}{\partial \theta }=-mg\ell \sin \theta .$ Pehle ko differentiate karo: $p_{\theta }=m{\ell }^2\dot{\theta }\Rightarrow {\dot{p}}_{\theta }=m{\ell }^2\ddot{\theta }$. Doosre ke barabar set karo: $m{\ell }^2\ddot{\theta }=-mg\ell \sin \theta .$ Yeh exactly Lagrangian equation of motion hai. ✓ Dono first-order Hamilton equations saath milke ek second-order Euler–Lagrange equation ke equivalent hain.

$H(q,p,t)$ ka total time derivative: $\frac{dH}{dt}={\sum }_i\frac{\partial H}{\partial q_i}{\dot{q}}_i+{\sum }_i\frac{\partial H}{\partial p_i}{\dot{p}}_i+\frac{\partial H}{\partial t}.$ Hamilton's equations ${\dot{q}}_i=\partial H/\partial p_i$ aur ${\dot{p}}_i=-\partial H/\partial q_i$ insert karo: $\frac{dH}{dt}={\sum }_i\frac{\partial H}{\partial q_i}\frac{\partial H}{\partial p_i}-{\sum }_i\frac{\partial H}{\partial p_i}\frac{\partial H}{\partial q_i}+\frac{\partial H}{\partial t}=\frac{\partial H}{\partial t}.$ Dono sums term by term cancel ho jaate hain. Isliye agar $\partial H/\partial t=0$ ho, to $dH/dt=0$ — $H$ conserved hai. Yeh Conservation Laws & Noether's Theorem ka Hamiltonian face hai.

Momentum: $p=\frac{\partial L}{\partial \dot{x}}=-m{c}^2\cdot \frac{-\dot{x}/c^2}{\sqrt{1-{\dot{x}}^2/c^2}}=\frac{m\dot{x}}{\sqrt{1-{\dot{x}}^2/c^2}}.$ Assemble (invert karne se pehle, structure dekhne ke liye): $H=p\dot{x}-L=\frac{m{\dot{x}}^2}{\sqrt{1-{\dot{x}}^2/c^2}}+m{c}^2\sqrt{1-{\dot{x}}^2/c^2}.$ Common denominator $\sqrt{1-{\dot{x}}^2/c^2}$ pe laao: $H=\frac{m{\dot{x}}^2+m{c}^2(1-{\dot{x}}^2/c^2)}{\sqrt{1-{\dot{x}}^2/c^2}}=\frac{m{\dot{x}}^2+m{c}^2-m{\dot{x}}^2}{\sqrt{1-{\dot{x}}^2/c^2}}=\frac{m{c}^2}{\sqrt{1-{\dot{x}}^2/c^2}}.$ $p$ ke terms mein: $p\sqrt{1-{\dot{x}}^2/c^2}=m\dot{x}$ se, algebra se ${\dot{x}}^2=\frac{p^2{c}^2}{p^2+m^2{c}^2}$ milta hai, isliye $\sqrt{1-{\dot{x}}^2/c^2}=\frac{mc}{\sqrt{p^2+m^2{c}^2}}$ aur $\boxed{H=\sqrt{p^2{c}^2+m^2{c}^4}}$ — famous relativistic energy. Isliye $L$ (velocity ka function) ko Legendre-transform karke $H$ (momentum ka function) banana itna powerful hai: momentum energy ka natural variable hai.

Momentum: $p=\partial L/\partial \dot{x}=m\dot{x}+eA$. Notice: momentum $m\dot{x}$ nahi hai! Iske saath extra $eA$ term hai. Invert: $\dot{x}=\frac{p-eA}{m}$. Assemble: $H=p\dot{x}-L=p\dot{x}-\frac{1}{2}m{\dot{x}}^2-eA\dot{x}+e\phi =(p-eA)\dot{x}-\frac{1}{2}m{\dot{x}}^2+e\phi .$ Lekin $(p-eA)=m\dot{x}$, isliye $(p-eA)\dot{x}=m{\dot{x}}^2$: $H=m{\dot{x}}^2-\frac{1}{2}m{\dot{x}}^2+e\phi =\frac{1}{2}m{\dot{x}}^2+e\phi =\frac{(p-eA{)}^2}{2m}+e\phi .$ Kya $H=T+V$ hai? Numerically $H=\frac{1}{2}m{\dot{x}}^2+e\phi =T+e\phi$. Kinetic energy sahi aayi, lekin canonical momentum $p=m\dot{x}+eA$, kinetic momentum $m\dot{x}$ se alag hai. Isliye $H$ ka energy value to hai, lekin $p$ mein express karne par yeh tell-tale $(p-eA{)}^2$ structure carry karta hai. Yahi electromagnetic Hamiltonians ka seed hai.

$\{x,H\}=\frac{\partial x}{\partial x}\frac{\partial H}{\partial p}-\frac{\partial x}{\partial p}\frac{\partial H}{\partial x}=1\cdot \frac{p}{m}-0=\frac{p}{m}=\dot{x}.✓$ $\{p,H\}=\frac{\partial p}{\partial x}\frac{\partial H}{\partial p}-\frac{\partial p}{\partial p}\frac{\partial H}{\partial x}=0-1\cdot kx=-kx=\dot{p}.✓$ Dono $\dot{f}=\{f,H\}$ se match karte hain, jo Phase Space mein equations of motion ka Poisson-bracket form hai.

(i) Momentum: $p_r=\partial L/\partial \dot{r}=m\dot{r}\Rightarrow \dot{r}=p_r/m$. Assemble karo aur $\dot{r}=p_r/m$ substitute karo: $H=p_r\dot{r}-L=\frac{p_r^2}{m}-\left(\frac{p_r^2}{2m}+\frac{1}{2}m{r}^2{\omega }^2\right)=\frac{p_r^2}{2m}-\frac{1}{2}m{r}^2{\omega }^2.$ (ii) Energy: $E=T+V=\frac{1}{2}m{\dot{r}}^2+\frac{1}{2}m{r}^2{\omega }^2=\frac{p_r^2}{2m}+\frac{1}{2}m{r}^2{\omega }^2$ ($\dot{r}=p_r/m$ use karte hue; yaad karo $V=0$). (iii) Gap: $H-E=\left(-\frac{1}{2}m{r}^2{\omega }^2\right)-\left(+\frac{1}{2}m{r}^2{\omega }^2\right)=-m{r}^2{\omega }^2\ne 0.$ Isliye $H=E-m{r}^2{\omega }^2$. Yeh exactly figure ke right panel mein vertical red gap hai. Kyun fail hota hai: constraint $\varphi =\omega t$ time-dependent hai, isliye $T$ $\dot{r}$ mein pure homogeneous quadratic nahi hai — iska ek $\dot{r}$-independent "centrifugal" piece $\frac{1}{2}m{r}^2{\omega }^2$ hai. Euler's theorem (jo $\sum p_i{\dot{q}}_i=2T$ deta) poore $T$ par laagu nahi hota. (iv) Conserved? $L$ mein koi explicit $t$ nahi hai ($\omega$ constant hai, $r$ variable hai), isliye $\partial L/\partial t=0\Rightarrow \partial H/\partial t=0\Rightarrow H$ conserved hai. Ek conserved quantity jo energy nahi hai.

Momentum: $p=\partial L/\partial \dot{x}=m\dot{x}+bx\Rightarrow \dot{x}=\frac{p-bx}{m}$. Assemble: $H=p\dot{x}-L=p\dot{x}-\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}m{\omega }^2{x}^2-bx\dot{x}=(p-bx)\dot{x}-\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}m{\omega }^2{x}^2.$ Kyunki $(p-bx)=m\dot{x}$ hai, $(p-bx)\dot{x}=m{\dot{x}}^2$ milta hai, isliye $H=m{\dot{x}}^2-\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}m{\omega }^2{x}^2=\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}m{\omega }^2{x}^2=\frac{(p-bx{)}^2}{2m}+\frac{1}{2}m{\omega }^2{x}^2.$ Conserved? $\partial L/\partial t=0$ (koi explicit $t$ nahi), isliye $H$ conserved hai. ✓ $H=T+V$? Value hai $\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}m{\omega }^2{x}^2=T+V$ — haan, numerically $H=E$ yahan. $bx\dot{x}$ term ek total time derivative hai ($\frac{d}{dt}(\frac{1}{2}b{x}^2)$), isliye yeh canonical momentum ko $bx$ se shift karta hai lekin energy nahi badalta. Ek subtle case: momentum disguised hai, phir bhi $H=E$ hold karta hai.

(i) Hessian: $\frac{\partial L}{\partial \dot{x}}=bx$, jo $\dot{x}$ contain nahi karta. Isliye $W=\frac{{\partial }^{2}L}{\partial {\dot{x}}^2}=0.$ Kyunki $\det W=0$ hai, recipe ki warning ka non-degeneracy condition fail ho jaata hai. Yeh ek singular (degenerate) Lagrangian hai. (ii) Kya problem aata hai: momentum hai $p=bx$. Notice karo ki yeh sirf $x$ par depend karta hai, $\dot{x}$ par nahi — isliye hum isse $\dot{x}$ ke liye solve nahi kar sakte. $\dot{x}$ ki har value ek hi momentum $bx$ deti hai. Recipe ka step 2 ($\dot{x}(x,p)$ paane ke liye $p$ invert karo) impossible hai. Iski jagah hum ek constraint paate hain: $p-bx=0$ identically hold karna chahiye. Velocity undetermined reh jaati hai. (iii) Hamiltonian: $H=p\dot{x}-L=p\dot{x}-bx\dot{x}+\frac{1}{2}k{x}^2=(p-bx)\dot{x}+\frac{1}{2}k{x}^2.$ Constraint surface $p=bx$ par, bracket $(p-bx)$ vanish ho jaata hai, isliye $H=\frac{1}{2}k{x}^2.$ Velocity $\dot{x}$ completely disappear ho gayi — lekin isliye nahi ki humne use substitute kiya; yeh isliye drop out hua kyunki iska coefficient $(p-bx)$ zero hone par forced hai. Yeh singular system ka fingerprint hai: $\dot{x}$ kabhi bhi $p$ ke terms mein expressible nahi hota, aur dynamics clean invertible $H(x,p)$ ki jagah constraints se govern hoti hai. (Aisi systems ko systematically handle karna Dirac's constraint theory hai.) Isko page ke har doosre problem se compare karo, jahan $W=m\ne 0$ tha aur inversion hamesha succeed ki.