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I'am reading John Taylor's Classical Mechanics chapter 1 page 20 where he proves the principle of conservation of momentum which states "If the net external force $F^{ext}$ on an $N$-particle system is zero,the system's total momentum $P$ is constant.

I'am not sure if i'am having trouble understanding the usage of summation notation or the entire derivation (which includes the concept) itself.Anyways here it is.A five particle system labelled by $\alpha$ or $\beta = 1,2,...,5$. The particle $\alpha$ is subject to four internal forces,shown by small arrows and denoted $F_{\alpha \beta}$ and the net external force shown by the large arrow and denoted by $F^{ext}_{\alpha}$. enter image description here

Thus the net force on particle $\alpha$ is $$F_{\alpha} = \sum_{\beta \neq \alpha}F_{\alpha \beta}+F^{ext}_{\alpha}$$

According to Newton's second law,this is the same as the rate of change of $P_{\alpha}$:$$\dot{P}_{\alpha} = \sum_{\beta\neq\alpha}F_{\alpha\beta}+F^{ext}_{\alpha}$$

Then he considers the total momentum of the $N$-particle system,$$P=\sum_{\alpha}P_{\alpha}$$ Differentiating wrt time we get $$\dot{P}=\sum_{\alpha}\dot{P}_{\alpha}$$ or,substituting for $\dot{P}_{\alpha}$,$$\dot{P}=\sum_{\alpha}\sum_{\beta\neq\alpha}F_{\alpha\beta}+\sum_{\alpha}F^{ext}_{\alpha}$$ .The above expression is what is bothering me .I need someone to give me an intuition behind the double summation sign. How come it's $\dot{P}=\sum_{\alpha}\sum_{\beta\neq\alpha}F_{\alpha\beta}+\sum_{\alpha}F^{ext}_{\alpha}$ and not just $\dot{P}=\sum_{\alpha}\sum_{\beta\neq\alpha}F_{\alpha\beta}+F^{ext}_{\alpha}$. He eventually concludes that $$\dot{P}=\sum_{\alpha}F^{ext}_{\alpha}\equiv F^{ext}$$.

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The only thing you seem to be asking is why the second term in $\dot P$ was summed over $\alpha$. It was always summed over alpha, so the sum over alpha couldn't disappear. After all, $\dot P$ doesn't depend on any alpha index so it would be inconsistent if the right hand side did depend on it. It's very far to trace what's your actual problem. – Luboš Motl Aug 24 '12 at 9:23
My actual problem is the derivation itself. – alok Aug 24 '12 at 9:43
Let me say the proof in words. The forces on each of 5 particles are either internal, from the other four; or the external, from someone else. Each of the forces may violate the momentum conservation i.e. change total momentum. However, the internal forces don't because they change the momentum of each of the 2 interacting particles oppositely, via action and reaction. So the only remaining term is the other one, from the external forces, and if the sum of these external forces vanishes, the total change of the momentum is zero. – Luboš Motl Aug 24 '12 at 11:49
@alok I'm not sure what specific step is troubling you, but I'll take a guess. Do you see that $\sum_{\alpha}(\sum_{\beta\neq\alpha}F_{\alpha\beta}+F^{ext}_{\alpha}) = \sum_{\alpha}\sum_{\beta\neq\alpha}F_{\alpha\beta}+\sum_{\alpha}F^{ext}_{\alpha}‌​$? – mmc Aug 24 '12 at 12:13
@alok. If you now see it, you are welcome to write an answer to avoid creating an orphan post. – Qmechanic Aug 25 '12 at 18:42
up vote 1 down vote accepted

The answer is this:$$\sum_{\alpha}\sum_{\beta\neq\alpha}F_{\alpha\beta}=\sum_{\alpha}\sum_{\beta>\alpha}(F_{\alpha\beta}+F_{\beta\alpha})$$ We conclude that $$\dot{P}=\sum_{\alpha}F^{ext}_{\alpha}\equiv F^{ext}$$.

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The main point is that if you have two sequences of numbers $a_i$ and $b_i$, the summation of both of them "splits": $$\sum_i\left(a_i+b_i\right)=\sum_i a_i+\sum_i b_i.$$ This is the same as saying that $$(a_1+b_1)+(a_2+b_2)+\cdots=(a_1+a_2+\cdots)+(b_1+b_2+\cdots),$$ and simply expresses the associativity and commutativity of the sum.

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