Chemical Potential for Bosonic Particles

Click For Summary
SUMMARY

The discussion focuses on the chemical potential m(T) for bosonic particles as described in Griffiths' Quantum Mechanics textbook. It establishes that m(T) monotonically increases as temperature T decreases, with N (number of particles) and V (volume) held constant. The derived equation for m(T) is m(T) = (2/V)[NkTln(2)+(2π²/3)N²/2V²/3T²/3]. The differentiation of m(T) with respect to T shows that dm/dT is positive, confirming that m(T) increases as T approaches zero, leading to the conclusion that m(T) must be negative at absolute zero.

PREREQUISITES
  • Understanding of Griffiths' Quantum Mechanics concepts
  • Familiarity with chemical potential in statistical mechanics
  • Basic knowledge of differentiation and calculus
  • Concepts of bosonic statistics and thermodynamic limits
NEXT STEPS
  • Study Griffiths' Quantum Mechanics, focusing on Chapter 5 regarding chemical potential
  • Learn about the implications of bosonic statistics on thermodynamic properties
  • Explore the relationship between temperature and energy in quantum systems
  • Investigate the mathematical techniques for differentiating thermodynamic equations
USEFUL FOR

This discussion is beneficial for physics students, particularly those studying quantum mechanics and statistical mechanics, as well as researchers working on thermodynamic properties of bosonic systems.

YAHA
Messages
121
Reaction score
0

Homework Statement


I am working Problem 5.29 *** (b) in Griffiths QM. We are asked to show that m(T) monotonically increases as T decreases, assuming N and V are constants. m(T) - is chemical potential.



Homework Equations



Too many to list, probably easier to look in the book if you have it.

The Attempt at a Solution



Honestly, I played with this for 2 hours. I also have the solutions manual. Even after looking there, the logic is completely incomprehensible. Specifically, he concludes that as T->0, E->0 and thus, m(T) must be negative. This step is not evident at all (as are many others in Griffiths books and solutions manuals).



 
Physics news on Phys.org
I would really appreciate some help. The solution goes something like this: m(T) = (2/V)[NkTln(2)+(2π2/3)N2/2V2/3T2/3] Differentiating with respect to T: dm/dT = -(2/VT)[N(2π2/3)N2V-2/3T-1/2]Using the fact that dE/dT > 0 we get: dm/dT > 0 This means that m(T) is an increasing function of T As T--> 0, E --> 0 and thus m(T) must be negative.
 

Similar threads

Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
2
Views
2K
  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
Replies
16
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
Replies
1
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K