UPSC 2023 Maths Optional Paper 2 Q5d — Step-by-Step Solution

10 marks · Section B

Hamilton's equations · Mechanics & Fluid Dynamics · asked 10× in 13 yrs · Read the full method →

Question

A planet of mass $m$ is revolving around the sun of mass $M$ . The kinetic energy $T$ and the potential energy $V$ of the planet are given by $T=\tfrac12 m(\dot r^2+r^2\dot\theta^2)$ and $V=GMm\bigl(\tfrac{1}{2a}-\tfrac{1}{r}\bigr)$ , where $(r,\theta)$ are the polar coordinates of the planet at time $t$ , $G$ is the gravitational constant and $2a$ is the major axis of the ellipse (the path of the planet). Find the Hamiltonian and the Hamilton equations of the planet’s motion.

Technique

Standard Legendre transform from $L(q,\dot q)$ to $H(q,p)$ ; the system is natural so $H=T+V$ .

Solution

Generalised coordinates. $(q_1,q_2)=(r,\theta)$ .

Step 1 — Lagrangian and generalised momenta.

L=T-V=\tfrac12 m\bigl(\dot r^{2}+r^{2}\dot\theta^{2}\bigr)-GMm\!\left(\frac{1}{2a}-\frac{1}{r}\right).

Momenta:

p_r=\frac{\partial L}{\partial\dot r}=m\dot r,\qquad p_\theta=\frac{\partial L}{\partial\dot\theta}=mr^{2}\dot\theta.

Invert: $\dot r=\dfrac{p_r}{m},\quad\dot\theta=\dfrac{p_\theta}{mr^{2}}$ .

Step 2 — Hamiltonian.

For a holonomic, scleronomic system (constraints time-independent and kinetic energy quadratic in velocities), $H=T+V$ expressed in terms of $(q,p)$ :

T=\tfrac12 m\!\left(\dot r^{2}+r^{2}\dot\theta^{2}\right)=\frac{p_r^{2}}{2m}+\frac{p_\theta^{2}}{2mr^{2}},

V=GMm\!\left(\frac{1}{2a}-\frac{1}{r}\right).

\boxed{\;H(r,\theta,p_r,p_\theta)=\frac{p_r^{2}}{2m}+\frac{p_\theta^{2}}{2mr^{2}}+GMm\!\left(\frac{1}{2a}-\frac{1}{r}\right).\;}

Step 3 — Hamilton’s equations.

\dot r=\frac{\partial H}{\partial p_r}=\frac{p_r}{m},

\dot\theta=\frac{\partial H}{\partial p_\theta}=\frac{p_\theta}{mr^{2}},

\dot p_r=-\frac{\partial H}{\partial r}=\frac{p_\theta^{2}}{mr^{3}}-\frac{GMm}{r^{2}},

\dot p_\theta=-\frac{\partial H}{\partial\theta}=0.

(The last equation says angular momentum $p_\theta$ is conserved — Kepler’s second law in disguise.)

Answer

\boxed{\;\dot r=\frac{p_r}{m},\quad\dot\theta=\frac{p_\theta}{mr^{2}},\quad\dot p_r=\frac{p_\theta^{2}}{mr^{3}}-\frac{GMm}{r^{2}},\quad\dot p_\theta=0.\;}