Exercise 4

The following coordinate transformations are given: \[ x \rightarrow x + y \delta \] \[ y \rightarrow y - x \delta \]

The Lagrangian is \[ L = \frac{m}{2} (\dot{x}^2 + \dot{y}^2) - V (x^2 + y^2) \]

and \(V\) is a function of \(x^2 + y^2\): \[ x^2 + y^2 \rightarrow (x+y\delta)^2 + (y-x\delta)^2\] \[ = x^2 + y^2 + y^2 \delta^2 + x^2 \delta^2\]

Since \(\delta\) is an infinitesimal small number, squares and powers of higer order of \(\delta\) are neglected. Only the linear part up to 1st order is taken into account. With these assumptions, the value of \((x^2 + y^2)\) is invariant under the coordinate transformation given above.

The same holds true for the kinetic energy: \[ \dot{x} \rightarrow \dot{x} + \dot{y} \delta \] \[ \dot{y} \rightarrow \dot{y} - \dot{x} \delta \]

Here, too, the same calculation as above shows that the term \(\dot{x}^2 + \dot{y}^2\) is invariant under the given coordinate transformation. This means that the Lagrangian \(L\) is also invariant and there is a rotational symmetry.