Einstein notation and differential operators.

Question

So I am dealing with the differential operator $\mathbf{D} = \mathbf{r} \times \boldsymbol{\nabla}$ where $r = x_i \mathbf{e}_i$. We then introduce $(\mathbf{D}f)_i$, which can be expressed as $\epsilon_{ijk}x_j\frac{\partial f}{\partial x_k}$. How would one express $\mathbf{\nabla}\cdot(\mathbf{D}f)_i$ in index notation? I think it includes using the Kronecker delta.

Answer 1

Let $\mathbf{A}=A_i\mathbf{e}_i$. Since

$$\nabla\cdot\mathbf{A}=\frac{\partial A_i}{\partial x_i}$$, we can write $$\nabla\cdot\left(\mathbf{D}f\right)=\frac{\partial}{\partial x_i}\left(\epsilon_{ijk}x_j\frac{\partial f}{\partial x_k}\right).$$ Using the product rule, we obtain $$\nabla\cdot\left(\mathbf{D}f\right)=\epsilon_{ijk}\delta_{ji}\frac{\partial f}{\partial x_k}+\epsilon_{ijk}x_j\frac{\partial^2 f}{\partial x_i\partial x_k},$$ because $\frac{\partial x_j}{\partial x_i}=\delta_{ji}$.

Since $\epsilon_{ijk}\delta_{ji}=\epsilon_{iik}=0$, we arrive at $$\nabla\cdot\left(\mathbf{D}f\right)=\epsilon_{ijk}x_j\frac{\partial^2 f}{\partial x_i\partial x_k}.$$ Note that $\epsilon_{ijk}=-\epsilon_{kji}$ and $\frac{\partial^2 f}{\partial x_i\partial x_k}=\frac{\partial^2 f}{\partial x_k\partial x_i}$. Therefore, $$\epsilon_{ijk}\frac{\partial^2 f}{\partial x_i\partial x_k}=-\epsilon_{kji}\frac{\partial^2 f}{\partial x_k\partial x_i}=-\epsilon_{ijk}\frac{\partial^2 f}{\partial x_i\partial x_k}\\ \Longrightarrow\epsilon_{ijk}\frac{\partial^2 f}{\partial x_i\partial x_k}=0.$$ Finally, we get $$\boxed{\nabla\cdot\left(\mathbf{D}f\right)=0}.$$