Total momentum of a $N$-particle system, what is going on?

Question

This is a multiparticle system proposed in the taylor's classical mechanics textbook (pg 20-21) :

The derivation of the rate of change of total momentum is given as follows:

$$ \begin{aligned} \text{net force on particle} \ \alpha = F_{\alpha} &= \sum_{\beta \neq \alpha} F_{\alpha\beta} + F_{\alpha}^{ext} \\ \dot{p_\alpha} &= \sum_{\beta \neq \alpha} F_{\alpha\beta} + F_{\alpha}^{ext} \\ \text{total momentum of the system} = P &= \sum_{\alpha} p_\alpha \\ \dot{P} &= \sum_{\alpha} \dot{p_\alpha} \\ &= \sum_\alpha \sum_{\beta \neq \alpha} F_{\alpha\beta} + \sum_{\alpha}F_\alpha^{ext} \end{aligned} $$

The double sum here contains $N(N-1)$ terms in all. Each term $F_{\alpha\beta}$ in this sum can be paired with a second term $F_{\beta \alpha}$, so that

$$ \sum_{\alpha} \sum_{\beta \neq \alpha} F_{\alpha\beta} = \sum_\alpha \sum_{\beta > \alpha} (F_{\alpha\beta} + F_{\beta \alpha}) $$

Then it mentions that by third law the sum of action-reaction forces is zero (which makes sense), therefore the double sum is zero, we conclude that

$$ \dot{P} = \sum_{\alpha} F_{\alpha}^{ext} = F^{ext} $$

The things that doesn't make sense to me are:

1. Why do we only consider the momentum of $\alpha$ in the total momentum of the system? (or are there assumptions being made?)
2. What does the $N(N-1)$ part means? (the whole quoted sentence) How does it lead to the equation at its next line(the double summation)?

Answer 1

1. Note sure what you're asking here. $\alpha$ is a dummy summation index. We first figure out what is the momemtum of the first particle, what is the momentum of the second particle.... what is the momentum of the $N^{th}$ particle, and add it all up, to get the total momentum.

2. The fact that there are $N(N-1)$ terms is unimportant. Anyway, the way that number comes about is that if we assume $N$ particles, then there are $N$ possible values for the integer $\alpha$, and for each $\alpha$, there are $N-1$ possible values for the integer $\beta$, so a total of $N(N-1)$. But, like I said, this is just an unimportant counting remark. The way to "simplify" the double sum is to split up the condition $\beta\neq \alpha$ into $\alpha<\beta$ and $\beta<\alpha$, and then to relabel indices. Explicitly, \begin{align} \sum_{\substack{1\leq \alpha,\beta\leq N\\ \alpha\neq \beta}}F_{\alpha\beta}&= \sum_{\substack{1\leq \alpha,\beta\leq N\\ \alpha< \beta}}F_{\alpha\beta} + \sum_{\substack{1\leq \alpha,\beta\leq N\\ \beta< \alpha}}F_{\alpha\beta}\tag{i} \end{align} Now, in the second term, we can relabel indices, for example \begin{align} \sum_{\substack{1\leq \alpha,\beta\leq N\\ \beta<\alpha}}F_{\alpha\beta} =\sum_{\substack{1\leq \rho,\sigma\leq N\\ \sigma<\rho}}F_{\rho\sigma} =\sum_{\substack{1\leq @,\sharp\leq N\\ \sharp< @}}F_{@\sharp} =\sum_{\substack{1\leq *,\ddot{\smile}\leq N\\ \ddot{\smile}< *}}F_{*\ddot{\smile}} &=\sum_{\substack{1\leq \beta,\alpha\leq N\\ \alpha< \beta}}F_{\beta\alpha}\\ &=\sum_{\substack{1\leq \alpha,\beta\leq N\\ \alpha< \beta}}F_{\beta\alpha}\tag{ii} \end{align} I hope the first three equal signs are obvious: the choice of symbols for the indices DOES NOT MATTER. In the fourth equal sign, because the choice of symbols doesn't matter I have deiced to use $\beta,\alpha$ instead. Finally, in the last equal sign on the next line, I simply used the fact that the condition $1\leq \beta,\alpha\leq N$ (which in full means $1\leq \beta \leq N$ AND $1\leq \alpha\leq N$) is symmetric, and thus is the same as saying $1\leq \alpha,\beta\leq N$. Hence, in equation (i), I can use (ii) to get \begin{align} \sum_{\substack{1\leq \alpha,\beta\leq N\\ \alpha\neq \beta}}F_{\alpha\beta}&= \sum_{\substack{1\leq \alpha,\beta\leq N\\ \alpha< \beta}}F_{\alpha\beta} + \sum_{\substack{1\leq \alpha,\beta\leq N\\ \alpha< \beta}}F_{\beta\alpha}\\ &= \sum_{\substack{1\leq \alpha,\beta\leq N\\ \alpha< \beta}}(F_{\alpha\beta}+F_{\beta\alpha}), \end{align} which is precisely the same result you got. In essence, we're just relabeling the indices.

If this sounds too abstract, look at the following square array: \begin{align} \begin{pmatrix} F_{11}& F_{12}& \cdots & F_{1N}\\ F_{21}& F_{22}& \cdots & F_{2N}\\ \vdots & \vdots & \ddots & \vdots\\ F_{N1}& F_{N2} & \cdots & F_{NN} \end{pmatrix} \end{align} Here, $\sum\limits_{\substack{1\leq \alpha,\beta\leq N\\ \alpha\neq \beta}}F_{\alpha\beta}$ means we're adding up all the terms which are NOT on the main diagonal (i.e top left to bottom right). So, there are obviously two triangles, the upper triangle consists of those whose first index is smaller than the second index, i.e $\sum\limits_{\substack{1\leq \alpha,\beta\leq N\\ \alpha< \beta}}F_{\alpha\beta}$. The lower triangle consists of those whose first index is larger than the second index, i.e $\sum\limits_{\substack{1\leq \alpha,\beta\leq N\\ \alpha< \beta}}F_{\beta\alpha}$.

One final remark: the $N(N-1)$ can also be seen from the array because there are a total of $N^2$ terms, but we're disregarding those on the main diagonal, which is $N$ of them, so we're left with $N^2-N=N(N-1)$ terms.