Is there an error in this question: Show that #((del^2Q)/(delbeta^2))_V = bar(E^2) - barE^2# (the variance of the energy)? #beta# is only a function of temperature #T#, not a function of volume #V#. #E# is a function of only #V#.

Relevant expressions:

#barE = sum_(i=1)^(N) p_iE_i#, the ensemble average energy

#bar(E^2) = sum_(i=1)^(N) p_iE_i^2#, the ensemble average squared energy

#barE^2 = (sum_(i=1)^(N) p_iE_i)^2#, the square of the ensemble average energy

#p_i = (e^(-betaE_i))/Q#, the distribution law for the canonical ensemble

#Q = sum_(i=1)^(N) e^(-betaE_i)#, the canonical partition function

1 Answer
Truong-Son N.
Apr 17, 2017

My professor answered back and said that there was. It was supposed to be #((del^2lnQ)/(delbeta^2))_V#. Here is the solution...

#bar(E^2) - barE^2#

#= sum_(i=1)^(N)p_iE_i^2 - (sum_(i=1)^(N)p_iE_i)^2#

#= 1/Qsum_(i=1)^(N)(E_i^2e^(-betaE_i)) - (1/Qsum_(i=1)^(N)E_ie^(-betaE_i))^2#

Working backwards from the chain rule, we get:

#= 1/Q((del^2Q)/(delbeta^2))_V - (1/Q((delQ)/(delbeta))_V)^2#

#= 1/Q((del^2Q)/(delbeta^2))_V - 1/Q^2((delQ)/(delbeta))_V^2#

Working backwards from the product rule, it is difficult but we recognize that

#(del)/(delbeta)[1/Q((delQ)/(delbeta))_V]#

#= 1/Q((del^2Q)/(delbeta^2))_V + ((delQ)/(delbeta))_V(-1/Q^2)((delQ)/(delbeta))_V#

#= 1/Q((del^2Q)/(delbeta^2))_V -1/Q^2((delQ)/(delbeta))_V^2#

Hence, what we really have is:

#color(blue)(bar(E^2) - barE^2) = 1/Q((del^2Q)/(delbeta^2))_V - 1/Q^2((delQ)/(delbeta))_V^2#

#= (del)/(delbeta)[1/Q((delQ)/(delbeta))_V]#

#= (del)/(delbeta)[((dellnQ)/(delbeta))_V]#

#= color(blue)(((del^2lnQ)/(delbeta^2))_V)#

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