Average Energy of Grand Canonical Ensemble

Click For Summary
SUMMARY

The discussion focuses on the derivation of the average energy (\(\bar{E}\)) in the context of the Grand Canonical Ensemble (GCE). The entropy formula is presented as \(S = KB \ln ZG + (\bar{E}/T) - μo\bar{N}/T\), leading to the Helmholtz function \(F = -TKB \ln ZG + μo\bar{N}\). A critical thermodynamic relationship is highlighted, \(\partial F/\partial T = -S\), which leads to an expression for \(\bar{E}\) that incorrectly includes the term \(μo\bar{N}\). The discussion concludes with the identification of this term as an error in the derivation.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically the Grand Canonical Ensemble.
  • Familiarity with statistical mechanics and the concept of partition functions.
  • Knowledge of Helmholtz free energy and its relation to entropy.
  • Basic proficiency in calculus, particularly partial derivatives.
NEXT STEPS
  • Study the derivation of the Grand Canonical Ensemble and its applications in statistical mechanics.
  • Explore the concept of partition functions in greater detail, focusing on \(Z_G\).
  • Learn about the implications of Helmholtz free energy in thermodynamic systems.
  • Investigate common pitfalls in thermodynamic derivations to avoid similar errors.
USEFUL FOR

This discussion is beneficial for physicists, particularly those specializing in statistical mechanics, as well as students and researchers looking to deepen their understanding of thermodynamic ensembles and their mathematical formulations.

phys_student1
Messages
104
Reaction score
0
Hello,

The entropy of the Grand Canonical Ensemble (GCE) is:

S = KB ln ZG + (\bar{E}/T) - μo\bar{N}/T

Helmholtz function is:

F = \bar{E} - TS = \bar{E} - TKB ln ZG - \bar{E} + μo\bar{N}
= -TKB ln ZG + μo\bar{N}

But

\partialF/\partialT = -S (From thermodynamics).

Then,

-TKB \partialln ZG/\partialT - KBln ZG = -kB ln ZG - \bar{E}/T + μo\bar{N}/T

This gives:

\bar{E} = kBT2 \partialln ZG/\partialT + μo\bar{N}

This is not the correct answer. The correct answer does not have the μo\bar{N} term, what's wrong ?
 
Science news on Phys.org
Up...
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K