UPSC 2016 Maths Optional Paper 2 Q5c — Step-by-Step Solution

10 marks · Section B

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

Question

Consider a single free particle of mass $m$ , moving in space under no forces. If the particle starts from the origin at $t=0$ and reaches the position $(x,y,z)$ at time $\tau$ , find the Hamilton’s characteristic function $S$ as a function of $x,y,z,\tau$ .

Technique

Identify $S$ as Hamilton’s principal function = action along the actual free-particle path; straight-line constant-velocity motion gives $\dot x_i=x_i/\tau$ ; integrate $L$ over $[0,\tau]$ ; confirm via the Hamilton–Jacobi equation.

Solution

Setup. For a free particle the Lagrangian is $L=\tfrac12 m(\dot x^2+\dot y^2+\dot z^2)$ and the Hamiltonian $H=\dfrac{1}{2m}(p_x^2+p_y^2+p_z^2)$ . Hamilton’s principal function $S(x,y,z,\tau)$ is the action along the actual trajectory from $(0,0,0)$ at time $0$ to $(x,y,z)$ at time $\tau$ , regarded as a function of the endpoint and time. It satisfies the Hamilton–Jacobi equation

\frac{\partial S}{\partial \tau}+H\!\left(\nabla S\right)=0,\qquad \frac{\partial S}{\partial\tau}+\frac{1}{2m}\Big[(\partial_x S)^2+(\partial_y S)^2+(\partial_z S)^2\Big]=0.

Step 1 — Actual motion (no forces)

With no forces the particle moves in a straight line at constant velocity. Starting at the origin and arriving at $(x,y,z)$ in time $\tau$ :

\dot x=\frac{x}{\tau},\quad \dot y=\frac{y}{\tau},\quad \dot z=\frac{z}{\tau}\qquad(\text{constant along the path}).

Step 2 — Action integral

S=\int_0^\tau L\,dt=\int_0^\tau \tfrac12 m(\dot x^2+\dot y^2+\dot z^2)\,dt.

The integrand is constant in time, equal to $\tfrac12 m\dfrac{x^2+y^2+z^2}{\tau^2}$ . Multiplying by the duration $\tau$ :

S=\tfrac12 m\,\frac{x^2+y^2+z^2}{\tau^2}\cdot\tau=\frac{m(x^2+y^2+z^2)}{2\tau}.

Step 3 — Result

Answer

\boxed{\;S(x,y,z,\tau)=\frac{m\,(x^2+y^2+z^2)}{2\tau}.\;}