Exercise 2

We prove that the curl of a gradient is equal to zero: \[ \vec{\nabla} \times \left[ \vec{\nabla} V(x, y, z) \right] = \vec{0} \]

Written in components the equations is as follows: \[ \left( \begin{matrix} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{matrix}\right) \times \left( \begin{matrix} \frac{\partial V(x, y, z)}{\partial x} \\ \frac{\partial V(x, y, z)}{\partial y} \\ \frac{\partial V(x, y, z)}{\partial z} \end{matrix}\right) = \left( \begin{matrix} \frac{\partial}{\partial y} \frac{\partial V(x, y, z)}{\partial z} - \frac{\partial}{\partial z} \frac{\partial V(x, y, z)}{\partial y} \\ \frac{\partial}{\partial z} \frac{\partial V(x, y, z)}{\partial x} - \frac{\partial}{\partial x} \frac{\partial V(x, y, z)}{\partial z} \\ \frac{\partial}{\partial x} \frac{\partial V(x, y, z)}{\partial y} - \frac{\partial}{\partial y} \frac{\partial V(x, y, z)}{\partial x} \end{matrix}\right) = \left( \begin{matrix} 0 \\ 0 \\ 0 \end{matrix}\right) \] since the sequence of partial derivatives can be changed (Exercise 1, Interlude 3).

In classcial mechanics vector fields are called conservative, if they can be derived by taking the gradient of a scalar field. An example for a conservative field is a force field, which can be calculated as a gradient of the potential. Since the curl of such a field vanishes, this means the work done along any closed path is always equal to zero.