Problem-52,page-165 of international 4th ed of Griffiths Introduction to electrodynamics. Show that the quadrupole term in the multipole expansion can be written: $$V_{quad}(r) = \frac{1}{4\pi \epsilon_o r^3} \sum_{i,j=1}^33 \hat{r}_i \hat{r}_j Q_{ij} \tag{1}$$ Where $$Q_{ij} = \frac12 \int \left[ 3r_i'r_j' -(r')^2 \delta_{ij}\right] \rho(r') d \tau' \tag{2}$$ And $\delta_{ij}$ is the kronecker delta
Here is the quadrupole term in the equation of multipole expansion:
$$ V_{quad} = \frac{k}{r^3} \int (r')^2 ( \frac{ 3 \cos^2 \theta-1}{2}) \rho dV' \tag{3}$$
Where, $r$ is the distance from centre of expansion to the point we evaluate potential at, $r'$ is the distance from centre of expansion to a point of our charge distribution,$\theta$ is the angle between these two vectors and $\rho$ is our charge density.
I'm having quite a hard time understanding the expression in (1), exactly what does $\hat{r}_i * \hat{r}_j$ mean? And, how exactly do we convert the integral in (3) into indicial notation
Ok, I got some help from a friend and I figured out what to do. Here is the deal, consider this integral: $$ \int (r' \cos \theta)^2 \rho dV \tag{4} $$
Now, now let the components of the vector for $r'$ (*) have components $r_i'$ and let $\hat{r}_i$ be the components of the $r$ vector normalized in cartesian, then:
$$ (r' \cos \theta)^2 = (\sum_i r_i' \hat{r}_i)^2 = \sum_i r_i' \hat{r}_i \cdot \sum_j r_j' \hat{r}_j= \sum_{i,j=0}^3 r_i' r_j' \hat{r}_i \hat{r}_j \tag{5}$$
Note that the third equality is an analogy to the continuous trick we do when evaluating the integral $\int e^{-x^2} dx$ by multiply the integral by itself and turning to polar.
Now, next we have:
$$ \int (r')^2 \rho dV \tag{6}$$
Here we will write the number 1 in possibly the most convoluted way possible:
$$ 1= \sum_i \hat{r}_i \hat{r}_i = \sum_i \hat{r}_i \hat{r}_j \delta_i^j \tag{7}$$
This turns out integral to:
$$ \sum_i \hat{r}_i \hat{r}_j \delta_i^j \int (r')^2 dV \tag{8}$$ Put everything back together in expression (3) , and we will have solved the question. Plug 8,6,5,4 into original equation