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Question about Helmholtz Free Energy and Partition Function
riboswitch:
I'm trying to decipher a step undertaken to demonstrate the formula that:
[tex]A = - \frac{ln(Q_{NVT})}{ \beta } = - k_{B}Tln(Q_{NVT})[/tex]
One of the steps undertaken is:
[tex]A = E - TS[/tex]
[tex]\frac{dA}{dT}_{N,V} = -S[/tex]
Since my professor's notes defined the Boltzmann factor as [tex] \beta = \frac{1}{k_{B}T} [/tex]
He then wrote that:
[tex]\frac{dA}{dT}_{N,V} = \frac{dA}{d\beta}_{N,V} \frac{ \beta }{T} = -S[/tex]
I just don't understand how he obtained that last one... I don't think he just substituted T with 1/βkB. As someone who lacks basic knowledge of mathematics, I have no idea how we arrived at that last part....
mjc123:
dA/dT = dA/dβ*dβ/dT
dβ/dT = -1/kT2 = -β/T
I think you've missed out a minus sign.
riboswitch:
--- Quote from: mjc123 on January 22, 2019, 12:58:26 PM ---dA/dT = dA/dβ*dβ/dT
dβ/dT = -1/kT2 = -β/T
I think you've missed out a minus sign.
--- End quote ---
Now it makes sense. Thanks. Can I ask another question? I have another question related to the derivation of the average energy of a canonical ensemble and eventually proving that the Helmholtz free energy is related to the partition function QNVT:
[tex]<E> = \frac{\sum_v E_{v}e^{- \beta E_{v} } }{\sum_v e^{- \beta E_{v} }} [/tex]
[tex] = - \big(\frac{dln\big(\sum_v e^{- \beta E_{v} }\big) }{d \beta}\big) [/tex]
The transition from the first formula to the second formula is still not understood by me. Why are we using the derivative all of the sudden here? Am I missing something?
mjc123:
Try expressing d(lnΣ)/dβ as (1/Σ)*dΣ/dβ
riboswitch:
Ok, I think I get it. First I have to derive the Q function (also known as the partition function):
[tex]Q_{NVT} = \sum_v e^{- \beta E_{v} } [/tex]
So, knowing that the derivative of e-ax is equal to -ae-ax, then I am finally able to derive the function Q:
[tex]\frac{ \delta Q}{ \delta \beta } = \frac{ \delta \big(\sum_v e^{- \beta E_{v} }\big) }{ \delta \beta }[/tex]
[tex]\frac{ \delta Q}{ \delta \beta } = - \sum_v E_{v}e^{- \beta E_{v} } [/tex]
Then going back to the average energy in the canonical ensemble:
[tex]<E> = \sum_v P_{v} E_{v} = \frac{1}{Q} \sum_v E_{v} e^{- \beta E_{v} }[/tex]
[tex]<E> \ = \ -\frac{1}{Q} \frac{ \delta Q}{ \delta \beta }[/tex]
The last equation can be rewritten in another way knowing the chain rule from my calculus class for derivative of a function inside a function:
[tex] \frac{ \delta }{ \delta x} f \big(g \big(x\big) \big) = f' \big(g \big(x\big) \big) g' \big(x\big) [/tex]
So now I can write the following, knowing that Q is also a function of β:
[tex] \frac{ \delta }{ \delta \beta} ln \big(Q \big( \beta \big) \big) = \frac{ \delta ln \big(Q\big) }{ \delta \beta } \frac{ \delta Q}{ \delta \beta } = \frac{1}{Q} \frac{ \delta Q}{ \delta \beta } [/tex]
So now I can finally define the average energy in the canonical ensemble as:
[tex]<E> \ = - \frac{ \delta ln \big(Q_{NVT} \big) }{ \delta \beta } [/tex]
Please correct me if there are mistakes in the procedures I have written.
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