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Topic: Question about Helmholtz Free Energy and Partition Function  (Read 888 times)

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Offline riboswitch

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Question about Helmholtz Free Energy and Partition Function
« on: January 22, 2019, 10:51:58 AM »
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....

Offline mjc123

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Re: Question about Helmholtz Free Energy and Partition Function
« Reply #1 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.

Offline riboswitch

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Re: Question about Helmholtz Free Energy and Partition Function
« Reply #2 on: January 26, 2019, 05:23:51 PM »
dA/dT = dA/dβ*dβ/dT
dβ/dT = -1/kT2 = -β/T
I think you've missed out a minus sign.

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?

Offline mjc123

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Re: Question about Helmholtz Free Energy and Partition Function
« Reply #3 on: January 28, 2019, 07:26:43 AM »
Try expressing d(lnΣ)/dβ as (1/Σ)*dΣ/dβ

Offline riboswitch

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Re: Question about Helmholtz Free Energy and Partition Function
« Reply #4 on: January 30, 2019, 10:58:09 AM »
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.

Offline mjc123

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Re: Question about Helmholtz Free Energy and Partition Function
« Reply #5 on: January 30, 2019, 12:22:17 PM »
Quote
So now I can write the following, knowing that Q is also a function of β:
δ/δβ ln(Q(β))=δln(Q)/δβ δQ/δβ =1/Q δQ/δβ 
The middle expression ought to be dlnQ/dQ*dQ/dβ
Otherwise correct

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