How would one go about proving the following result in $\mathbb R^3$ for the divergence of vector field $\vec F = F_i \hat e^i$ $$ \nabla \cdot {\mathbf F} = \frac{1}{h_1 h_2 h_3} \left[\frac \partial {\partial q^1} (F_1 h_2 h_3) + \frac \partial {\partial q^2} (F_2 h_3 h_1) + \frac {\partial }{\partial q^3} (F_3 h_1 h_2)\right]$$
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1 Answer
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Here is a heuristic argument.
The flux across an area element perpendicular to $\hat{e}^{1}$ is \begin{align*} F_{1}\left(h_{2}dq_{2}\right)\left(h_{3}dq_{3}\right) & =\left(F_{1}h_{2}h_{3}\right)dq_{2}dq_{3}. \end{align*} Thus, the differential of this flux, along $\hat{e}^{1}$-direction, is \begin{align*} \frac{\partial}{\partial q^{1}}\left(F_{1}h_{2}h_{3}\right)dq_{1}dq_{2}dq_{3}, \end{align*} and so the volume density is \begin{align*} & \frac{1}{\left(h_{1}dq_{1}\right)\left(h_{2}dq_{2}\right)\left(h_{3}dq_{3}\right)}\frac{\partial}{\partial q_{1}}\left(F_{1}h_{2}h_{3}\right)dq_{1}dq_{2}dq_{3}\\ = & \frac{1}{h_{1}h_{2}h_{3}}\frac{\partial}{\partial q^{1}}\left(F_{1}h_{2}h_{3}\right). \end{align*}
