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*Introduction to solid state Physics*3rd edition, page 87). For noble gases, this interaction could be a Van der Waals interaction, then for any two atoms the interaction potential is given by

##U=4 \epsilon \left [ \left(\frac{\sigma}{R} \right )^{12}-\left(\frac{\sigma}{R} \right )^{6} \right ]##.

So the total cohesive energy in the solid will be given by

##U_T=\frac{1}{2} \sum_{i,j} 4 \epsilon \left [ \left(\frac{\sigma}{R_{i,j}} \right )^{12}-\left(\frac{\sigma}{R_{i,j}} \right )^{6} \right ]##

##R_{i,j}=|\vec{r}_i-\vec{r}_j|##.

Now, this double sum can be converted into a single sum, and here is my doubt, which is how the reasoning is made to get that the sum over all i's gives a term of N, you can see this result in the book by Kittel, where besides it is taken a geometrical factor for the distance bewtween atoms, where appears the distance to first neighboors ##R##, so that:

##R_{i,j}=p_{i,j}R##.

But basically my doubt is on this step:

##U_T=\frac{1}{2} \sum_{i,j} 4 \epsilon \left [ \left(\frac{\sigma}{R_{i,j}} \right )^{12}-\left(\frac{\sigma}{R_{i,j}} \right )^{6} \right ]=\frac{N}{2} 4 \epsilon \sum_{j \neq i} \left [ \left(\frac{\sigma}{R_{i,j}} \right )^{12}-\left(\frac{\sigma}{R_{i,j}} \right )^{6} \right ]##

Which I interpret that states, for an arbitrary potential between atoms ##u_{i,j}## that:

##\sum_{i,j} u_{i,j}=N \sum_{j\neq i} u_{i,j}##.

That result is the one which I can't understand, I don't know how to get it. I think that it is a fundamental fact to demonstrate this result that ##u_{i,j}=u_{j,i}=u(|\vec{r}_i-\vec{r}_j|)##, but I still don't know how to show that the sum over one index gives N times the same thing.

Thanks in advance.