 Solution

ID: past-exam-of-the-geometry-course-of-the-university-of-oxford/2009/a/3/ii/solution

Since

γ^{'} = r_{u} u^{'} + r_{v} v^{'}

, we have

γ^{'} \cdot r_{u} = E u^{'} + F v^{'}

. The normal-acceleration characterization of a geodesic gives

\frac{d}{d s} (E u^{'} + F v^{'}) = γ^{'} \cdot (r_{uu} u^{'} + r_{uv} v^{'}) .

Expanding and differentiating the metric coefficients yields

r_{u} \cdot r_{uu} = E_{u} /2

r_{v} \cdot r_{uu} + r_{u} \cdot r_{uv} = F_{u}

, and

r_{v} \cdot r_{uv} = G_{u} /2

. Therefore

\frac{d}{d s} (E u^{'} + F v^{'}) = \frac{1}{2} E_{u} (u^{'})^{2} + F_{u} u^{'} v^{'} + \frac{1}{2} G_{u} (v^{'})^{2} .

Applying the same calculation to

γ^{'} \cdot r_{v} = F u^{'} + G v^{'}

gives

\frac{d}{d s} (F u^{'} + G v^{'}) = \frac{1}{2} E_{v} (u^{'})^{2} + F_{v} u^{'} v^{'} + \frac{1}{2} G_{v} (v^{'})^{2} .

These are also the Euler-Lagrange equations of the quadratic metric energy.

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