 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/ii/paper-1/15c/a/solution

 Top articles  Latest articles New article in topic

Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 1 / 15C / a / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-14

Vary

q_{a}

and

p_{a}

independently, fixing the endpoint positions. Integration by parts gives

δ S = [p_{a} δ q_{a}]_{t_{1}}^{t_{2}} + \int_{t_{1}}^{t_{2}} [(\overset{q}{˙}_{a} - H_{p_{a}}) δ p_{a} + (- \overset{p}{˙}_{a} - H_{q_{a}}) δ q_{a}] d t .

The boundary term vanishes, so arbitrary interior variations imply

\overset{q}{˙}_{a} = H_{p_{a}}, \overset{p}{˙}_{a} = - H_{q_{a}} .

In the Lagrangian principle only

q

is varied independently, and

δ \overset{q}{˙} = d (δ q) / d t

; here positions and momenta are independent paths in phase space. The extremality condition is stationarity of the action, not necessarily a minimum. Along Hamiltonian motion,

\frac{d H}{d t} = H_{q_{a}} \overset{q}{˙}_{a} + H_{p_{a}} \overset{p}{˙}_{a} + H_{t} = H_{t},

so an explicitly time-independent Hamiltonian is conserved.

 Read the full article

 New to topics? Read the docs here!