Six equal point masses \(m\) are located at the points \(\pm a \boldsymbol{i}, \pm a \boldsymbol{j}\) and \(\pm a \boldsymbol{k}\). Show that the quadrupole term in the potential vanishes, and find the leading correction to the monopole term \(-6 \mathrm{Gm} / \mathrm{r} .\) (Note: This requires expansion of the potential up to terms of order \(a^{4} / r^{5}\).)

First, let's recall the expression for the quadrupole moment tensor: $$ Q_{ij}=m\sum_{\alpha}\left(3x_{\alpha i}x_{\alpha j}-\delta_{ij}r_{\alpha}^{2}\right), $$ where \(\alpha\) enumerates the masses, and \(x_{\alpha i}\) denotes the \(i\)-th component of the \(\alpha\)-th mass. For our given points, we can compute \(Q_{ij}\): $$ Q_{ij}=m\left[3(atot)\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)-\left(\begin{array}{ccc} a^{2} & 0 & 0 \\ 0 & a^{2} & 0 \\ 0 & 0 & a^{2} \end{array}\right)\right]. $$

Now, we'll find the trace of the quadrupole moment tensor. The trace is the sum of the diagonal elements, denoted as \(Q_{ii}\): $$ \mathrm{Tr}(Q_{ij})=Q_{11}+Q_{22}+Q_{33}=3(atot)m-a^{2}m=0. $$ Since the trace of the quadrupole moment tensor vanishes, the quadrupole term also vanishes.

We know that the monopole term of the potential is given by \(-6 \mathrm{Gm} / \mathrm{r}\). Now, we need to find the leading correction term up to order \(a^{4} / r^{5}\). We can do this by using the general expression of potential in the form of multipole expansion: $$ V(r) = -GM \left(\frac{1}{r} + \frac{Q_{ij} x_i x_j}{2 r^5} + \ldots\right). $$ Considering only terms that have an order of \(a^{4} / r^{5}\), the potential expansion becomes: $$ V(r) = -6 \frac{GM}{r} - \frac{1}{2}\frac{GQ_{ij}x_i x_j}{r^5} + \dots. $$ Since we showed in step 2 that the quadrupole term vanishes, the only term we need to consider is the monopole term: $$ V(r) = -6 \frac{GM}{r}. $$ Thus, the quadrupole term vanishes and the leading correction to the monopole term is \(-6 \mathrm{Gm} / \mathrm{r}\).