Show that the divergence of the curl of a vector is zero. (asked in MU June 2025 KT; also Dec 2024)

Answer: \(\nabla\cdot(\nabla\times\vec A)=0\) for any vector field with continuous second derivatives, because the mixed partial derivatives cancel in pairs.

Let \(\vec A=A_x\hat i+A_y\hat j+A_z\hat k\). Then

\[\nabla\times\vec A=\left(\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}\right)\hat i+\left(\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}\right)\hat j+\left(\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}\right)\hat k\]

Taking the divergence:

\[\nabla\cdot(\nabla\times\vec A)=\frac{\partial^2A_z}{\partial x\partial y}-\frac{\partial^2A_y}{\partial x\partial z}+\frac{\partial^2A_x}{\partial y\partial z}-\frac{\partial^2A_z}{\partial y\partial x}+\frac{\partial^2A_y}{\partial z\partial x}-\frac{\partial^2A_x}{\partial z\partial y}\]

Since the order of differentiation does not matter for smooth functions (\(\frac{\partial^2A_z}{\partial x\partial y}=\frac{\partial^2A_z}{\partial y\partial x}\), etc.), the terms cancel in pairs:

\[\nabla\cdot(\nabla\times\vec A)=0\]

Physical meaning: the curl of a field is solenoidal; for example \(\nabla\cdot\vec B=0\) is consistent with \(\vec B=\nabla\times\vec A\).