# Confused by sum

1. Mar 23, 2009

### latentcorpse

At the top of page 26 here
http://www.ph.ed.ac.uk/~pmonthou/Statistical-Mechanics/documents/SM7.pdf [Broken]

when we talk about "summing out" the states of all the other particles, why are we not summing over $i_1$ in the following sum - where did the $i_1$ sum go to?

Last edited by a moderator: May 4, 2017
2. Mar 24, 2009

### CompuChip

The idea is that you are keeping $i_1$ unspecified, but you are summing over the state of all the other particles.

Maybe the following analogy will help you. Suppose that I have 10 vases with blue and red balls, and I draw a ball from each of them. In principle, I have to specify a "state" of the "system" by saying for each vase, whether I draw a red or a blue ball, for example: (BBRBRRBRRB). Now suppose that I am interested in the first one only, and I want to calculate how many systems there are in which I have a red or a blue ball in vase 1.
Let Z be the total number of possible configurations (in this case, clearly 2^10). To find out how many arrangements there are in which the first ball drawn is blue, I can simply sum over all configurations of numbers 2 -- 10, while the first one is fixed. So I get
(BRRRRRRRRRR), (BRRRRRRRRB), (BRRRRRRRBR), (BRRBRBRRBRB), etc.
and I need to count all of them.
So what I will get is
$$N = \sum_{i_2 = B, R} \sum_{i_3 = B, R} \cdots \sum_{i_{10} = B, R} 1$$
and the fraction (relative amount) of configurations with a blue one in the first vase is N / Z (which in this case, of course, will simply give 1/2).

The example you posted is very similar, only there every configuration isn't equivalent, but you have to weigh it by some Boltzmann factor.