Find magnetic susceptibility using partition function

  • Thread starter cozycoz
  • Start date
  • #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.
 

Answers and Replies

  • #2
Charles Link
Homework Helper
Insights Author
Gold Member
4,740
2,089
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
139
12
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.
 
  • Like
Likes Charles Link
  • #4
cozycoz
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 Threads on Find magnetic susceptibility using partition function

Replies
1
Views
1K
Replies
3
Views
2K
  • Last Post
Replies
2
Views
5K
Replies
13
Views
5K
Replies
1
Views
2K
  • Last Post
Replies
2
Views
1K
Replies
2
Views
6K
Replies
0
Views
1K
  • Last Post
Replies
2
Views
3K
Replies
0
Views
5K
Top