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

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

Relevant expressions:

$¯ ¯ ¯ E = N \sum i = 1 p_{i} E_{i}$ $barE = sum_(i=1)^(N) p_iE_i$ , the ensemble average energy

$¯ ¯¯¯ ¯ E^{2} = N \sum i = 1 p_{i} E_{i}^{2}$ $bar(E^2) = sum_(i=1)^(N) p_iE_i^2$ , the ensemble average squared energy

${¯ ¯ ¯ E}^{2} = {(N \sum i = 1 p_{i} E_{i})}_{i}^{2}$ $barE^2 = (sum_(i=1)^(N) p_iE_i)^2$ , the square of the ensemble average energy

$p_{i} = \frac{e_{i}^{- β E_{i}}}{Q}$ $p_i = (e^(-betaE_i))/Q$ , the distribution law for the canonical ensemble

$Q = N \sum i = 1 e_{i}^{- β E_{i}}$ $Q = sum_(i=1)^(N) e^(-betaE_i)$ , the canonical partition function

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Answer 1 · 2017-04-17T03:23:47

My professor answered back and said that there was. It was supposed to be ${(\frac{\partial^{2} ln Q}{\partial β^{2}})}_{V}$ $((del^2lnQ)/(delbeta^2))_V$ . Here is the solution...

$¯ ¯¯¯ ¯ E^{2} - {¯ ¯ ¯ E}^{2}$ $bar(E^2) - barE^2$

$= N \sum i = 1 p_{i} E_{i}^{2} - {(N \sum i = 1 p_{i} E_{i})}_{i}^{2}$ $= sum_(i=1)^(N)p_iE_i^2 - (sum_(i=1)^(N)p_iE_i)^2$

$= \frac{1}{Q} N \sum i = 1 (E_{i}^{2} e_{i}^{- β E_{i}}) - {(\frac{1}{Q} N \sum i = 1 E_{i} e_{i}^{- β E_{i}})}_{i}^{2}$ $= 1/Qsum_(i=1)^(N)(E_i^2e^(-betaE_i)) - (1/Qsum_(i=1)^(N)E_ie^(-betaE_i))^2$

Working backwards from the chain rule, we get:

$= \frac{1}{Q} {(\frac{\partial^{2} Q}{\partial β^{2}})}_{V} - {(\frac{1}{Q} {(\frac{\partial Q}{\partial β})}_{V})}_{V}^{2}$ $= 1/Q((del^2Q)/(delbeta^2))_V - (1/Q((delQ)/(delbeta))_V)^2$

$= \frac{1}{Q} {(\frac{\partial^{2} Q}{\partial β^{2}})}_{V} - \frac{1}{Q^{2}} {(\frac{\partial Q}{\partial β})}_{V}^{2}$ $= 1/Q((del^2Q)/(delbeta^2))_V - 1/Q^2((delQ)/(delbeta))_V^2$

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

$\frac{\partial}{\partial β} [\frac{1}{Q} {(\frac{\partial Q}{\partial β})}_{V}]$ $(del)/(delbeta)[1/Q((delQ)/(delbeta))_V]$

$= \frac{1}{Q} {(\frac{\partial^{2} Q}{\partial β^{2}})}_{V} + {(\frac{\partial Q}{\partial β})}_{V} (- \frac{1}{Q^{2}}) {(\frac{\partial Q}{\partial β})}_{V}$ $= 1/Q((del^2Q)/(delbeta^2))_V + ((delQ)/(delbeta))_V(-1/Q^2)((delQ)/(delbeta))_V$

$= \frac{1}{Q} {(\frac{\partial^{2} Q}{\partial β^{2}})}_{V} - \frac{1}{Q^{2}} {(\frac{\partial Q}{\partial β})}_{V}^{2}$ $= 1/Q((del^2Q)/(delbeta^2))_V -1/Q^2((delQ)/(delbeta))_V^2$

Hence, what we really have is:

$¯ ¯¯¯ ¯ E^{2} - {¯ ¯ ¯ E}^{2} = \frac{1}{Q} {(\frac{\partial^{2} Q}{\partial β^{2}})}_{V} - \frac{1}{Q^{2}} {(\frac{\partial Q}{\partial β})}_{V}^{2}$ $color(blue)(bar(E^2) - barE^2) = 1/Q((del^2Q)/(delbeta^2))_V - 1/Q^2((delQ)/(delbeta))_V^2$

$= \frac{\partial}{\partial β} [\frac{1}{Q} {(\frac{\partial Q}{\partial β})}_{V}]$ $= (del)/(delbeta)[1/Q((delQ)/(delbeta))_V]$

$= \frac{\partial}{\partial β} [{(\frac{\partial ln Q}{\partial β})}_{V}]$ $= (del)/(delbeta)[((dellnQ)/(delbeta))_V]$

$= {(\frac{\partial^{2} ln Q}{\partial β^{2}})}_{V}$ $= color(blue)(((del^2lnQ)/(delbeta^2))_V)$

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