1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Confused by sum

  1. Mar 23, 2009 #1
    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 [itex]i_1[/itex] in the following sum - where did the [itex]i_1[/itex] sum go to?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 24, 2009 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    The idea is that you are keeping [itex]i_1[/itex] 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
    [tex]N = \sum_{i_2 = B, R} \sum_{i_3 = B, R} \cdots \sum_{i_{10} = B, R} 1[/tex]
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Confused by sum
  1. Gibbs sum for particle (Replies: 3)

  2. A little confused (Replies: 2)

  3. Sum of torques (Replies: 1)

  4. Sum of Cosines (Replies: 6)

Loading...