1. Not finding help here? Sign up for a free 30min 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!

Entropy as log of omega (phase space volume)

  1. Jan 29, 2015 #1
    1. The problem statement, all variables and given/known data
    I've seen this problem appear in more than one textbook almost without any changes. It goes like this:
    Assume the entropy ##S## depends on the volume ##\bar{\Omega}## inside the energy shell: ##S(\bar{\Omega})=f(\bar{\Omega})##. Show that from the additivity of ##S## and the multiplicative character of ##\bar{\Omega}##, it follows that ##S=const \times log \bar{\Omega}##

    2. Relevant equations


    3. The attempt at a solution
    I've found a couple of solutions already (one in Pathria), that consists of considering two subsystems and calculating the derivatives of S in respect to the omegas, plus a bunch of assumptions. But from the problem statement I can't help but think there's got to be a simpler way. I'm trying to expand S as a sum of ##f(\Omega_i)##, expand each f as a power series, and then use the fact that omega is the product of the omegas of the subsystems. But that leads nowhere.

    Slightly off-topic: I'm taking a grad course on statistical mechanics but my previous knowledge on stat mech is very weak, so I'll probably be on these forums frequently throughout the semester. Is this the appropriate forum for stat mech homework questions?
     
  2. jcsd
  3. Jan 30, 2015 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    I don't see how power series would be easier than the described approach.

    Looking at two subsystems together leads to ##f(xy)=f(x)+f(y)## (where x,y are the Ω of two different systems). Calculating the derivative with respect to x gives ##yf'(xy)=f'(x)##, the derivative of this with respect to y leads to ##f'(xy)+xyf''(xy)=0## or (using z=xy) ##f'(z) = -zf''(z)##. This differential equation is solved by ##f'(z)=\frac{c}{z}## which leads to ##f(z)=c \log(z)##.

    It is.
     
  4. Jan 31, 2015 #3
    Thanks :) That makes perfect sense. I solved it in a similar way but made some unnecessary turns here and there and it made things look more complicated.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Entropy as log of omega (phase space volume)
  1. Phase space (Replies: 8)

Loading...