 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2013/2/a/solution

q_{j}

is absent explicitly from the Lagrangian:

\partial L / \partial q_{j} = 0

, even though

\overset{q}{˙}_{j}

may occur. The Euler-Lagrange equation then gives

\frac{d}{d t} (\frac{\partial L}{\partial q ˙ _{j}}) = 0, p_{j} = \frac{\partial L}{\partial q ˙ _{j}} = constant .

Thus a continuous coordinate-translation symmetry gives a conserved conjugate momentum. The conserved quantity need not be ordinary linear momentum: an absent azimuthal angle can give conserved angular momentum. The derivative is taken holding the other coordinates and velocities fixed.

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