Symmetry due to Newton's Third Law in the proof of Virial Theorem 
Source: An Introduction to Modern Astrophysics
This is from the proof of Virial Theorem. In the above picture's second equation, how the first term is zero due to symmetry from Newton's third Law? I'm more specifically concerned with $(r_{i}+r_{j})$ which are position vector of ith and jth particle. After multiplying by Fij to both of them, how it results to the condition in which we can use $F_{ij}=-F_{ji}$?
 A: I think the simplest way is to take into account that the sum
$$\frac{1}{2}\sum_i\sum_{j, j\ne i} {\bf F}_{ij}\cdot ({\bf r}_i + {\bf r}_j)
$$
is a sum over all the values of $i$ and $j$, with the only constraint that $i \neq j$.
With respect to the exchange of $i$ and $j$, ${\bf F}_{ij}$ changes its sign (antisymmetric), by Newton's third law, while $({\bf r}_i + {\bf r}_j)$ is symmetric. The resulting scalar product is then antisymmetric. Therefore, each $i,j$ term of the sum is exactly canceled by the term $j,i$.
A: Newton's second law states that the force the $i^{th}$ particle exerts on the $j^{th}$ is equal in magnitude, but opposite in direction to the force that the $j^{th}$ particle exerts back on to the $i^{th}$ particle. That is, $$\vec F_{ij}=-\vec F_{ji}$$
If we look at the sum, we see Newton's third law terms$^*$ multiplied by distance, so that $$\frac{1}{2}\sum_i\sum_{j, j\ne i} F_{ij}\cdot (r_i + r_j)=\frac{1}{2}\underbrace{\sum_i\sum_{j,j\ne i}(F_{ij}\cdot r_i+F_{ij}\cdot {r_j})}_{\text{*3rd law pairs}\ \ \large F_{ij}\cdot r_i=-\vec F_{ji}\cdot r_i}$$ that will continually cancel throughout the sum. Given that $i\ne j$ over the summation, we will get terms like $$F_{12}\cdot r_1 - F_{21}\cdot r_1 +F_{13}\cdot r_1 - F_{31}\cdot r_1 + \cdots +F_{21}\cdot r_2 - F_{12}\cdot r_2 + F_{23}\cdot r_2-F_{32}\cdot r_2  + \cdots=0$$
where we know that all the $F_{ij}=-F_{ji}$ by Newton's third law, and by the explicit symmetry of the sum, the entire  term will vanish. This is a simple way of saying that since $F_{ij}$ is antisymmetric, while $(r_i+r_j)$ is purely symmetric, the dot product in the sum will also be antisymmetric, which once again means all of the terms in the original sum will vanish.
