 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-1/9/a/solution

p = \frac{\partial L}{\partial x ˙} = \frac{m x ˙}{1 - x ˙ ^{2}}, \frac{\partial L}{\partial x} = - V^{'} (x) .

\frac{d}{d t} (\frac{m x ˙}{1 - x ˙ ^{2}}) + V^{'} (x) = 0, \frac{m x ¨}{( 1 - x ˙ ^{2} ) ^{3/2}} = - V^{'} (x) .

Because

L

has no explicit time dependence, the energy form of the Beltrami identity gives a conserved quantity

E = \overset{x}{˙} \partial L / \partial \overset{x}{˙} - L

. Here

E = \frac{m x ˙ ^{2}}{1 - x ˙ ^{2}} + m 1 - \overset{x}{˙}^{2} + V (x) = \frac{m}{1 - x ˙ ^{2}} + V (x) .

For a direct check of conservation of energy, differentiate:

\frac{d E}{d t} = \overset{x}{˙} [\frac{m x ¨}{( 1 - x ˙ ^{2} ) ^{3/2}} + V^{'} (x)] = 0.

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