What is the divergence of $\frac{\vec{r} (\vec{r}.\vec{a})}{r^3}$?

Question

I'm trying to calculate the divergence of $\frac{\vec{r} (\vec{r}.\vec{a})}{r^3}$ and I've tried using 2 separate methods. First using expression for divergence in spherical polar coordinates (since the above field depends only on radial component) and the other by brute force using cartesian coordinates and Einstein summation convention but I'm getting different answers. I'm getting $\frac{\vec{r}.\vec{a}}{r^3}$ when solving in spherical polar coordinates and $ - \frac{\vec{r}.\vec{a}}{r^3}$ with Einstein summation convention. Can someone help me out here?

Answer 1

Assuming $\mathbf a$ is a constant vector, let $$\mathbf F = \frac{\mathbf r(\mathbf r \boldsymbol \cdot \mathbf a)}{r^3} = \frac{\mathbf r \boldsymbol \cdot \mathbf a}{r^2}\hat{\mathbf r}.$$ Using the formula for the divergence in spherical polar coordinates, \begin{align*} \boldsymbol \nabla \boldsymbol \cdot \mathbf F &= \frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 F_r \right) = \frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\mathbf r \boldsymbol \cdot \mathbf a}{r^2}\right)\\ &= \frac{1}{r^2} \frac{\partial}{\partial r}(\mathbf r \boldsymbol \cdot \mathbf a) \\ &= \frac{1}{r^2} \left( \frac{\partial \mathbf r}{\partial r} \boldsymbol \cdot \mathbf a + \mathbf r \boldsymbol \cdot \frac{\partial \mathbf a}{\partial r}\right) \\ &= \frac{1}{r^2} \left(\frac{\mathbf r}{r} \boldsymbol \cdot \mathbf a\right) \\ &= \frac{\mathbf r \boldsymbol \cdot \mathbf a}{r^3}. \end{align*} Now using the summation convention, note that $$\mathbf F = \frac{1}{r^3} x_j a_j x_i \mathbf e_i \implies F_i = \frac{x_i x_j a_j}{r^3},$$ where $(\mathbf e_1, \mathbf e_2, \mathbf e_3) = (\hat{\boldsymbol{\imath}}, \hat{\boldsymbol{\jmath}}, \hat{\boldsymbol k})$. Then, \begin{align*} \boldsymbol \nabla \boldsymbol \cdot \mathbf F &= \partial_i F_i \\ &= \frac{\partial}{\partial x_i} \left(\frac{x_i x_j a_j}{r^3}\right) \\ &= \frac{1}{r^6} \left(r^3 \frac{\partial}{\partial x_i}(x_i x_j a_j) - \frac{\partial r^3}{\partial x_i} x_i x_j a_i\right) \\ &= \frac{1}{r^6} \left(\color{red}{3}r^3 x_j a_j + r^3 x_i \delta_{ij} a_j - 3r^2 \frac{x_i}{r} x_i x_j a_j\right) \tag{$*$}\\ &= \frac{1}{r^3} \left(3x_j a_j + x_j a_j - 3 x_j a_j\right) \\ &= \frac{x_j a_j}{r^3} = \frac{\mathbf r \boldsymbol \cdot \mathbf a}{r^3}, \end{align*} as before. My guess is that the mistake was on line $(*)$, where the $\color{red} 3$ appears as $\frac{\partial x_i}{\partial x_i} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 3$.