📂Mathematical Physics

The Divergence of Curl is Always Zero

Formula

The divergence of the curl of a vector function $\mathbf{A}$ is always $0$.

$$\nabla \cdot (\nabla \times \mathbf{A}) = 0$$

Proof

The curl of $\mathbf{A}$ is as follows.

$$\begin{align*} \nabla \times \mathbf{A} &= \begin{vmatrix} \hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}} \\ \displaystyle \frac{\partial}{\partial x} & \displaystyle \frac{\partial}{\partial y} & \displaystyle \frac{\partial}{\partial z} \\ A_{x} & A_{y} & A_{z} \end{vmatrix} \\ &= \hat{\mathbf{x}} \left( \frac{\partial A_{z}}{\partial y} - \frac{\partial A_{y}}{\partial z} \right) + \hat{\mathbf{y}} \left( \frac{\partial A_{x}}{\partial z} - \frac{ \partial A_{z}}{\partial x} \right) + \hat{\mathbf{z}} \left( \frac{\partial A_{y}}{\partial x}-\frac{\partial A_{x}}{\partial y} \right) \end{align*}$$

The divergence of any vector function $\mathbf{F}$ is as follows.

$$\nabla \cdot \mathbf{F} = \dfrac{\partial F_{x}}{\partial x} + \dfrac{\partial F_{y}}{\partial y} + \dfrac{\partial F_{z}}{\partial z}$$

Therefore, the following result is obtained.

$$\begin{align*} \nabla \cdot (\nabla \times \mathbf{A}) &= \frac{\partial}{\partial x} \left( \frac{\partial A_{z}}{\partial y} - \frac{\partial A_{y}}{\partial z} \right) + \frac{\partial}{\partial y} \left( \frac{\partial A_{x}}{\partial z} - \frac{ \partial A_{z}}{\partial x} \right) + \frac{\partial}{\partial z} \left( \frac{\partial A_{y}}{\partial x}-\frac{\partial A_{x}}{\partial y} \right) \\ &= \left( {\color{blue}\frac{\partial^2 A_{z}}{\partial x \partial y}} {\color{green}- \frac{\partial ^2 A_{y}}{\partial x \partial z} } \right) + \left( {\color{red}\frac{\partial^2 A_{x}}{\partial y \partial z}} {\color{blue}- \frac{ \partial^2 A_{z}}{\partial y \partial x} }\right) + \left( {\color{green}\frac{\partial^2 A_{y}}{\partial z \partial x} } {\color{red}-\frac{\partial^2 A_{x}}{\partial z \partial y}} \right) \\ &= 0 \end{align*}$$

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