Landau: Explaining the Definition of "Number of States with Energy

Click For Summary
SUMMARY

The discussion focuses on the definition of the "number of states with energy" as presented in Landau's "Statistical Physics Part 1." The equation for the number of states within an energy interval is expressed as \(\frac{\mathrm{d}\Gamma(E)}{\mathrm{d}E} \mathrm{d}E\), indicating that \(\Gamma(E)\) is treated as a continuous function. The energy probability distribution \(W(E) = \frac{\mathrm{d}\Gamma(E)}{\mathrm{d}E} w(E)\) differentiates between the probability density function \(w(E)\) and the overall probability distribution, clarifying that \(W(E)\) represents the probability of finding the system in a state with energy between \(E\) and \(E + \mathrm{d}E\).

PREREQUISITES
  • Understanding of statistical mechanics concepts
  • Familiarity with Landau's "Statistical Physics Part 1"
  • Knowledge of probability density functions (PDFs)
  • Basic principles of quantum mechanics
NEXT STEPS
  • Study the concept of continuous functions in statistical mechanics
  • Explore the relationship between probability distributions and density functions
  • Review the derivation of the density matrix in quantum mechanics
  • Investigate the implications of large ensembles in statistical physics
USEFUL FOR

Students of statistical mechanics, physicists studying quantum systems, and anyone seeking to deepen their understanding of probability distributions in thermodynamic contexts.

Stalafin
Messages
21
Reaction score
0
Question about Landau: Definition of "Number of states with energy" in an interval

Hey! I am currently reading Landau's Statistical Physics Part 1, and in Paragraph 7 ("Entropy") I am struggling with a definition.

Right before Equation (7.1) he gives the "required number of states with energy between E and E+\mathrm{d}E" as:
\frac{\mathrm{d}\Gamma(E)}{\mathrm{d}E} \mathrm{d}E

I don't understand this equation. Am I supposed to understand \Gamma(E) as a continuous function, and therefore \frac{\mathrm{d}\Gamma(E)}{\mathrm{d}E} as a derivative?

Furthermore, how is the energy probability distribution
W(E) = \frac{\mathrm{d}\Gamma(E)}{\mathrm{d}E} w(E)
different from w(E). Isn't w(E) kind of a probability distribution by itself?
 
Physics news on Phys.org


For your first question, yes. For the enormous ensembles considered by stat mech (typically 10^23 particles or larger), the number of discrete states approaches infinity so their distribution may be considered continuous.

For the second question, what is w(E)? I don't have access to the text right now...
 


Landau considers that part for a quantum mechanical system. w_n = w(E_n) is the distribution function for the system.

He gets there from the diagonal elements of the density matrix w_n = w_{nn} (since the statistical distributions must be stationary), which can be expressed as functions of the energy levels alone (assuming we have a system in a coordinate system, such that it is at rest and apart from the energy the other integrals of motion don't factor in) w_n=w_(E_n).
 


I think this is the usual distinction discussed in statistics between probability P and probability density function (PDF) p. Here W(E)dE is the probability of seeing the system in a state with energy between E and E+dE. It is given in terms of the product of the PDF w(E) and the number of states d\Gamma(E) in that interval.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 0 ·
Replies
0
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 0 ·
Replies
0
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K