# Scalar and vector potentials

Show that $\nabla \times (\nabla U)=0$ and $\nabla(\nabla \times \overrightarrow{A})=0$  where $U$ is a scalar potential and $\overrightarrow{A}$ is a vector potential.

Showing that $\nabla \times (\nabla U) = 0$

$\nabla U =(dU/dx)*\overrightarrow {i} +(dU/dy)*\overrightarrow {j} + (dU/dz)*\overrightarrow {k}$

$\nabla \times (\nabla U)=\begin{vmatrix} i & j & k\\ d/dx & d/dy & d/dz\\ dU/dx & dU/dy & dU/dz \end{vmatrix}=$

$=\left( \frac {d^2U}{dydz} -\frac {d^2U}{dzdy} \right )i+\left( \frac {d^2U}{dzdx} -\frac {d^2U}{dxdz} \right )j+\left( \frac {d^2U}{dxdy} -\frac {d^2U}{dydx} \right )k=0$

Therefore $\nabla \times (\nabla U)=0$

Showing that $\nabla(\nabla \times \overrightarrow{A})=0$

$\nabla \times \overrightarrow{A}=\begin{vmatrix} i & j & k\\ d/dx & d/dy & d/dz\\ A_x & A_y & A_z \end{vmatrix}=\left( \frac {dA_z}{dy} -\frac {dA_y}{dz} \right )i+\left( \frac {dA_x}{dz} -\frac {dA_z}{dx} \right )j+\left( \frac {dA_y}{dx} -\frac {dA_x}{dy} \right )k$

$\nabla (\nabla \times \overrightarrow{A})=\frac {d}{dx}\left( \frac {dA_z}{dy} -\frac {dA_y}{dz} \right )+\frac {d}{dy}\left( \frac {dA_x}{dz} -\frac {dA_z}{dx} \right )+\frac {d}{dz}\left( \frac {dA_y}{dx} -\frac {dA_x}{dy} \right )=$

$=\left (\frac {d^2A_z}{dxdy}-\frac {d^2A_z}{dydx} \right )+\left(… \right )+\left(… \right )=0$

Therefore $\nabla (\nabla \times \overrightarrow{A})=0$