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  GR0177 #98
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Statistical Mechanics}Average Energy

The average energy is given by \sum_i E_i e^{-E_i/kT}/Z, where Z=\sum_i e^{-E_i/kT} is the partition function.

The above is just the stat mech generalization of the baby-averaging \bar{u}=\sum_i U_i/N, where N is the number of energies summed over. Z is like a generalized N.

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2008-07-26 17:15:06
No knowledge of what this function this represents is needed to solve this problem. Use units.

The units of the given expression are the units of Ei which as given is energy.

a) Has units of energy. (yes)
b) Unitless. (no)
c) Unity is unity this one can be ruled out if you are not a moron. (n0)
d) Probability is unitless. (no)
3) Entropy has units of Q/T or energy/temperature. (no)
2007-09-22 06:14:51

(For completeness sake and a deeper understanding) one might want to add that

(D) P(Ei) = exp(-Ei/kT)/Z (see how close!!!)


(E) S/k = Sum[(Exp(-Ei/kT))/Z)(Ei/kT+lnZ)]

(C) Unity is unity ;)

or something like that
Da Broglie
2011-04-08 17:33:44
yep pretty important, just to put the equations into LaTeX for ease of reading,

(D) P\left(E_i\right)=\frac{e^{\frac{-E_i}{kT}}}{Z}

(E) \frac{S}{k}=\Sigma[\frac{e^{\frac{-E_i}{kT}}}{Z}(\frac{E_i}{kT})+ln(Z)]

just to add here that average energy of a state is given by

E_{avg}=\frac{1}{Z}\frac{\partial Z}{\partial\beta} where \beta=\frac{1}{kT}

which equals the given expression in the problem

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