 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1/9/i/solution

q \mapsto q + ϵη

with

η (t_{0}) = η (t_{1}) = 0

δ S = \int_{t_{0}}^{t_{1}} (L_{q} η + L_{\overset{q}{˙}} \overset{η}{˙}) d t = [L_{\overset{q}{˙}} η]_{t_{0}}^{t_{1}} + \int_{t_{0}}^{t_{1}} (L_{q} - \frac{d}{d t} L_{\overset{q}{˙}}) η d t .

\frac{d}{d t} \frac{\partial L}{\partial q ˙} - \frac{\partial L}{\partial q} = 0.

Along such a trajectory,

\frac{d}{d t} (\overset{q}{˙} L_{\overset{q}{˙}} - L) = \overset{q}{˙} (\frac{d}{d t} L_{\overset{q}{˙}} - L_{q}) - L_{t} = - L_{t} .

L

has no explicit time dependence, the first integral is therefore

\overset{q}{˙} \frac{\partial L}{\partial q ˙} - L = constant .

Its negative is the equivalent Beltrami form. For the usual time-independent mechanical Lagrangian

L = T - V

with quadratic velocity dependence, this conserved quantity is the total mechanical energy

T + V

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