Proof That Divergence of the Cross Product of a Constant Vector With Radial Vector is Zero
Suppose
\[\mathbf{\omega} =( \omega_1 , \omega_2 , \omega_3 ) \]
is a constant vector and \[\mathbf{r} =(x,y,z) \]
is the position vector of a particle Then \[\mathbf{\nabla} \cdot (\mathbf{\omega} \times \mathbf{r})=0 \]
Proof
\[ \begin{equation} \begin{aligned} \mathbf{\nabla} \cdot (\mathbf{\omega} \times \mathbf{r}) &= \mathbf{\nabla} \cdot (( \omega_1 , \omega_2 , \omega_3 ) \times (x,y,z)) \\ &= \mathbf{\nabla} \cdot (\omega_2 z - \omega_3 y, \omega_3 x- \omega_1 z, \omega_1 y - \omega_2 x) \\ &= \frac{\partial (\omega_2 z - \omega_3 y)}{\partial x} + \frac{\partial ( \omega_3 x- \omega_1 z)}{\partial y} + \frac{\partial (\omega_1 y - \omega_2 x)}{\partial z} =0 \end{aligned} \end{equation}\]
