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$$(φ_j − φ_i )J_{ji} = (ψ_j − ψ_i )I_{ji} ⋯(1)$$

$$\sum_j φ_j\sum_i J_{ji} - \sum_i φ_i\sum_j J_{ji} = \sum_j ψ_j\sum_i I_{ji} - \sum_i ψ_i\sum_j I_{ji}⋯(2)$$

This is related to reciprocity theorem in circuits.

Just taking $\sum_j $ leaves $φ_jJ_{ji}$ look alike terms for instance.

Can someone explain how (1) becomes (2) ?

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    $\begingroup$ I don't understand the question. It seems that the answer is already in the title: You get $(2)$ from $(1)$ by summing over all $(i,j)$ pairs. Why do you say "Just taking $\sum_i$"? As you say yourself in the title, you need to sum over $i$ and $j$, not just $j$. $\endgroup$ – joriki Jan 8 '20 at 20:24
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First of all, they're finite sums, so $$\sum_i\sum_j=\sum_j\sum_i.$$ That is, we can always invert the order of summation.

The second thing to note is that $$\sum_i\sum_j a_ib_j =\sum_i a_i \sum_j b_j$$ To see this, note that the right-hand side is $$(a_1+\cdots+a_n)(b_1+\cdots b_m)$$ To evaluate this, we choose one of the $a$'a and one of the $b$s, and take their product. We add up the products over all choices of $a$ and $b$. That's exactly what the left hand side says.

So take $\sum_i\sum_j$ of both sides of the given equation; distribute the summation by linearlty; apply the two observations above to massage the equation into the desired form.

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