Finding beta for the boltzman distribution.

  • #1
Hello! I'm trying to do a satisfactory derivation of the boltzmann distribution. By using lagrange multipliers I've come as far as to prove that

[tex]P(i) = \frac{1}{Z} e^{-\beta E(i)}[/tex]
where
[tex]Z = \sum_i e^{-\beta E(i)},[/tex]

but how does one actually establish that
[tex]\beta = 1/kT?[/tex]
 

Answers and Replies

  • #2
vanhees71
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Take a monatomic ideal gas and derive the mean energy,

[tex]U=-\frac{\partial \ln Z}{\partial \beta}[/tex]

and compare with the definition of the temperature,

[tex]U=\frac{3}{2} N k T.[/tex]
 
  • #3
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I think that Leonard Susskind does this derivation in one of his lectures on statistical mechanics that is available on youtube.

http://www.youtube.com/watch?v=H1Zbp6__uNw"
 
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  • #4
Take a monatomic ideal gas and derive the mean energy
Ah, yes that is certainly a way to go. But how could that result possibly be general? Doesn't the distribution apply to any combination of systems who shares a total energy E and a number of particles N?
 

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