# Transforming from the non-traceless to the traceless quadrupole moment

Today I was working over the multipole expansion for electric potentials, and on my own, I got to the non-traceless definition of the quadrupole moment

$$Q_{\alpha \beta} = \int{\rho(\mathbf{r'})\cdot r'_{\alpha}r'_{\beta}\cdot d^3r'}$$

via a Taylor series expansion on $$\frac{1}{|\mathbf{r'} - \mathbf{r}|}$$.

My class notes reference this definition of the quadrupole, but they also mention the traceless one

$$\overline{Q}_{\alpha \beta} = \int{\rho(\mathbf{r'})\cdot (3r'_{\alpha}r'_{\beta}-r'^2\delta_{\alpha \beta})\cdot d^3r'}$$

and in another question on this site, it's explained that the traceless definition is preferable.

My class notes also explain that the integral for the quadrupole component of the potential is "symmetrical in $$\mathbf{r}$$ and $$\mathbf{r'}$$", and so, both definitions are equivalent.

My question is the following, does this symmetry in the integral have to do at all with the fact that point charge by point charge, one could swap $$\mathbf{r}$$ and $$\mathbf{r'}$$ and get the same potential, because it's only dependent on the distance between the point charge and the point where the potential is to be determined? If so, how could one rigurously change from one definition to the other? Would performing the changes of variables $$\mathbf{r} \rightarrow \mathbf{r'}$$ and $$\mathbf{r'} \rightarrow \mathbf{r}$$ suffice?