Find magnetic susceptibility using partition function

In summary, the conversation discusses a magnetic system with n independent molecules per unit volume and four energy levels. The partition function and Helmholtz function are derived, and the magnetization and magnetic susceptibility are calculated. The correct expression for the free energy function is given, and the correct limit for the magnetic susceptibility is computed. The conversation also includes a mathematical error that is corrected.
  • #1
cozycoz

Homework Statement


A certain magnetic system contains n independent molecules per unit volume, each of which has four energy levels given by 0, ##Δ-gμ_B B##, ##\Delta##, ##\Delta +gμ_B B##. Write down the partition function, compute Helmholtz function and hence compute the magnetization ##M##. Hence show that the magnetic susceptibility ##χ## is given by [tex]χ = \lim_{B\rightarrow 0} {\frac{μ_0 M}{B}} = \frac{2 n μ_0 g^2 {μ_B}^2}{k_B T (3+e^{Δ/k_B T})} [/tex]

(Sorry I don't know how to type limit in mathematical form)
\lim_{n\rightarrow +\infty} {\frac{\sin(x)}{x}}

Homework Equations


##F=-k_B T \ln Z##, ##m=-(\frac{∂F}{∂B})##, ##m=MV##

The Attempt at a Solution


[/B]
If there are N=Vn molecules, the partition function is [tex] Z=(e^{0}+e^{-β(Δ-gμ_B)}+e^{-βΔ}+e^{-β(Δ+gμ_B)})^N [/tex].
Then Helmholtz free energy function F is
[tex] F=-k_B T[-βΔ+\ln (1+2\cosh {βgμ_B B})] [/tex]

Now I have to differentiate F by B, but I've noticed βΔ part would be gone because it doesn't have B at all.
My answer is
[tex] χ = \frac{2 n μ_0 g^2 {μ_B}^2}{3 k_B T}[/tex]
where βΔ part is omitted.

If you tell me what's wrong it would be lovely.
 
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  • #2
It is unclear how you got ## F ## from ## Z ##. It also isn't apparent how , given the ## F ## that you have presented, that a derivative w.r.t. ## B ## will not be a function of the magnetic field ## B ##, since ## cosh'(x)=sinh(x) ##. Also, in writing out ## Z ##, you left off the ## B ## dependence. You need to be more accurate in presenting it.
 
  • #3
Your expression for the free energy is incorrect. My best guess is that you forgot about the state with 0 energy when taking the natural log of ##Z##.

Let me show you how I did the problem:

Let ##\zeta## denote the one-particle partition function, i.e.,
$$\begin{eqnarray*}
\zeta &= 1 + e^{-\beta \Delta} (1 + e^{\beta g \mu_B B} + e^{-\beta g \mu_B B}) \\
&= 1 + e^{-\beta \Delta} (1 + 2 \cosh (\beta g \mu_B B))
\end{eqnarray*}$$
Then ##Z = \zeta^N##, so
$$\begin{eqnarray*}
m &= -\frac{\partial}{\partial B} (F) \\
&= \frac{\partial}{\partial B} (kT \ln Z) \\
&= \frac{kTN}{\zeta} \frac{\partial \zeta}{\partial B} \\
&= \frac{(kTN) (2\beta g \mu_B) e^{-\beta \Delta} \sinh(\beta g \mu_B B) }{ 1 + e^{-\beta \Delta} (1 + 2 \cosh (\beta g \mu_B B)) }
\end{eqnarray*}$$
We then have
$$ \chi = \lim_{B \to 0} \frac{\mu_0 m}{VB} = \lim_{B \to 0} \frac{\mu_0}{B} \frac{2ng \mu_B e^{-\beta \Delta} \sinh(\beta g \mu_B B) }{ 1 + e^{-\beta \Delta} (1 + 2 \cosh(\beta g \mu_B B)) } $$
Since we're computing a limit as ##B \to 0##, we need only expand the top and bottom of this fraction to order ##B##. Doing so yields the correct answer.
 
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  • #4
VKint said:
Your expression for the free energy is incorrect. My best guess is that you forgot about the state with 0 energy when taking the natural log of ##Z##.

Let me show you how I did the problem:

Let ##\zeta## denote the one-particle partition function, i.e.,
$$\begin{eqnarray*}
\zeta &= 1 + e^{-\beta \Delta} (1 + e^{\beta g \mu_B B} + e^{-\beta g \mu_B B}) \\
&= 1 + e^{-\beta \Delta} (1 + 2 \cosh (\beta g \mu_B B))
\end{eqnarray*}$$
Then ##Z = \zeta^N##, so
$$\begin{eqnarray*}
m &= -\frac{\partial}{\partial B} (F) \\
&= \frac{\partial}{\partial B} (kT \ln Z) \\
&= \frac{kTN}{\zeta} \frac{\partial \zeta}{\partial B} \\
&= \frac{(kTN) (2\beta g \mu_B) e^{-\beta \Delta} \sinh(\beta g \mu_B B) }{ 1 + e^{-\beta \Delta} (1 + 2 \cosh (\beta g \mu_B B)) }
\end{eqnarray*}$$
We then have
$$ \chi = \lim_{B \to 0} \frac{\mu_0 m}{VB} = \lim_{B \to 0} \frac{\mu_0}{B} \frac{2ng \mu_B e^{-\beta \Delta} \sinh(\beta g \mu_B B) }{ 1 + e^{-\beta \Delta} (1 + 2 \cosh(\beta g \mu_B B)) } $$
Since we're computing a limit as ##B \to 0##, we need only expand the top and bottom of this fraction to order ##B##. Doing so yields the correct answer.

Yes i forgot extra 1 taking log..
Thanks a lot
 

Related to Find magnetic susceptibility using partition function

1. What is magnetic susceptibility?

Magnetic susceptibility is a measure of the degree to which a material can be magnetized in the presence of an external magnetic field. It is a dimensionless quantity that indicates the extent to which a material can be magnetized compared to a vacuum.

2. What is the partition function method used for?

The partition function method is used to calculate the magnetic susceptibility of a material. It involves calculating the average energy of a system at different temperatures and magnetic field strengths, and then using this information to determine the magnetic susceptibility.

3. How does the partition function relate to magnetic susceptibility?

The partition function is a mathematical function that represents the statistical distribution of energy levels in a system. By calculating the partition function, we can determine the average energy of the system, which is directly related to the magnetic susceptibility of the material.

4. What factors affect the magnetic susceptibility of a material?

The magnetic susceptibility of a material is affected by several factors, including the type of material, its chemical composition, temperature, and the strength of the external magnetic field.

5. How does magnetic susceptibility impact the behavior of materials?

Magnetic susceptibility plays an important role in determining the magnetic properties of a material. Materials with high magnetic susceptibility are more easily magnetized and have stronger magnetic fields, while materials with low magnetic susceptibility are less affected by external magnetic fields.

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