Solving Summations: Tips & Tricks for Homework

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Homework Statement



See the attachment, I am stuck as to how the summation sign [itex]\sum_{b\neq a}^{}[/itex] in (2.1.1) ends up as [itex]\sum_{ab}^{}[/itex] in the term with the red dot above (2.1.5).

Homework Equations


The Attempt at a Solution



As I understand you end up taking the product of two summations such that [itex]\sum_{a}^{}(\sum_{b\neq a}^{})=\sum_{ab}^{}[/itex], but I don't really understand the logic here.

just trying to understand, thanks in advance.
 

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yeh I got that but once you insert 2.1.1 i don't get how the summation in front of the red dot term is [itex]\sum_{ab}^{}[/itex] once you sub 2.1.1 you get [itex]\sum_{a}^{}(\sum_{b\neq a}^{})[/itex], I don't really understand how that works
 
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You are making a sum of sums. It may help to write the sums out for a small number of particles, let us say 3:
$$
\sum_{a} \sum_{b\neq a} F_{ab} = (F_{12} + F_ {13}) + [F_{21} + F_{23}] + \{F_{31} + F_{21}\}
$$
where the term in () is the term originating from the sum over ##b \neq 1## for ##a = 1##, [] for ##a = 2##, and {} for ##a = 3##. Now ##\sum_{ab}## is a bit of a bastard notation. If assuming that we by this mean ##\sum_{a=1}^3 \sum_{b=1}^3##, then we get some additional terms ##F_{11} + F_{22} + F_{33}##, but the particles do not exert forces onto themselves so these can be taken to be zero.
 
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Yeh I sort of came to a similar conclusion myself, the issue was that I didn't get why you can use the notation [itex]\sum_{ab}^{}[/itex] if you neglect particle self interactions, the notation in that case is not strictly true then? Wouldn't it be better to keep it in the form [itex]\sum_{a}^{}(\sum_{b\neq a}^{})[/itex], anyway thanks for the clarifications!
 
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