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### Topic: Question about Helmholtz Free Energy and Partition Function  (Read 1215 times)

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#### 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:

$$A = - \frac{ln(Q_{NVT})}{ \beta } = - k_{B}Tln(Q_{NVT})$$

One of the steps undertaken is:

$$A = E - TS$$

$$\frac{dA}{dT}_{N,V} = -S$$

Since my professor's notes defined the Boltzmann factor as $$\beta = \frac{1}{k_{B}T}$$

He then wrote that:

$$\frac{dA}{dT}_{N,V} = \frac{dA}{d\beta}_{N,V} \frac{ \beta }{T} = -S$$

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

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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.

#### 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:

$$<E> = \frac{\sum_v E_{v}e^{- \beta E_{v} } }{\sum_v e^{- \beta E_{v} }}$$

$$= - \big(\frac{dln\big(\sum_v e^{- \beta E_{v} }\big) }{d \beta}\big)$$

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

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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

#### 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):

$$Q_{NVT} = \sum_v e^{- \beta E_{v} }$$

So, knowing that the derivative of e-ax is equal to -ae-ax, then I am finally able to derive the function Q:

$$\frac{ \delta Q}{ \delta \beta } = \frac{ \delta \big(\sum_v e^{- \beta E_{v} }\big) }{ \delta \beta }$$
$$\frac{ \delta Q}{ \delta \beta } = - \sum_v E_{v}e^{- \beta E_{v} }$$

Then going back to the average energy in the canonical ensemble:

$$<E> = \sum_v P_{v} E_{v} = \frac{1}{Q} \sum_v E_{v} e^{- \beta E_{v} }$$
$$<E> \ = \ -\frac{1}{Q} \frac{ \delta Q}{ \delta \beta }$$

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:

$$\frac{ \delta }{ \delta x} f \big(g \big(x\big) \big) = f' \big(g \big(x\big) \big) g' \big(x\big)$$

So now I can write the following, knowing that Q is also a function of β:

$$\frac{ \delta }{ \delta \beta} ln \big(Q \big( \beta \big) \big) = \frac{ \delta ln \big(Q\big) }{ \delta \beta } \frac{ \delta Q}{ \delta \beta } = \frac{1}{Q} \frac{ \delta Q}{ \delta \beta }$$

So now I can finally define the average energy in the canonical ensemble as:

$$<E> \ = - \frac{ \delta ln \big(Q_{NVT} \big) }{ \delta \beta }$$

Please correct me if there are mistakes in the procedures I have written.

#### 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