Prove the 3 definitions of entropy are equivalent (stat. mechanics)

Click For Summary
SUMMARY

The discussion focuses on proving the equivalence of three definitions of entropy in statistical mechanics: S(E,V) = kln(Γ(E)), S(E,V) = kln(ω(E)), and S(E,V) = kln(Σ(E)). The key variables include Γ(E), which represents the phase space volume, ω, the density of states, and Σ, the cumulative density of states. The relationship between these definitions is established through the integration of Hamiltonian dynamics and the properties of the energy variable E. The conclusion emphasizes that these definitions are equivalent up to an additive constant.

PREREQUISITES
  • Understanding of statistical mechanics concepts, particularly entropy.
  • Familiarity with Hamiltonian mechanics and the Hamiltonian function H.
  • Knowledge of phase space and integration in multiple dimensions.
  • Basic grasp of thermodynamic principles and Boltzmann's constant (k).
NEXT STEPS
  • Study the derivation of phase space volume Γ(E) in detail.
  • Explore the relationship between density of states ω and cumulative density of states Σ.
  • Learn about the implications of entropy in thermodynamic systems.
  • Investigate the role of additive constants in thermodynamic equations.
USEFUL FOR

This discussion is beneficial for students and researchers in physics, particularly those focusing on statistical mechanics, thermodynamics, and entropy-related studies.

Tosh5457
Messages
130
Reaction score
28

Homework Statement



<br /> S(E,V) = kln(\Gamma(E) )\\<br /> S(E,V) = kln(\omega(E) )\\<br /> S(E,V) = kln(\Sigma(E) )\\<br />

S entropy, k Boltzmann's constant. Prove these 3 are equivalent up to an additive constant.

Homework Equations



<br /> \Gamma(E) = \int_{E&lt;H&lt;E+\Delta}^{&#039;}dpdq\\<br /> \Gamma(E)=\omega\Delta \\<br /> \Delta &lt;&lt; E\\<br /> <br /> \Sigma(E) = \int_{H&lt;E}^{&#039;}dpdq\\<br /> \omega = \frac{\partial \Sigma}{\partial E}\\<br />

H is the system's Hamiltonian and E is an arbitrary energy. These are integrations over all the p and q's, I wrote them like that to abbreviate.

The Attempt at a Solution



Using the 1st definition I can get to the 2nd one, but I can't reach at sigma's definition.
<br /> <br /> kln(\Gamma(E)) = kln(\omega\Delta) = kln(\omega) + kln(\Delta)\\<br /> ln(\Delta) &lt;&lt; ln(\omega) =&gt; S = kln(\omega)\\<br />
 
Last edited:
Physics news on Phys.org
I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?
 
I would say ##\Gamma (E)=\Sigma (E+\Delta) -\Sigma (E)=\Delta \frac{\partial \Sigma}{\partial E}+...##
 

Similar threads

Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K