﻿ Ferromagnetism from the Heisenberg model in the mean-field approximation

## Ferromagnetism from the Heisenberg model in the mean-field approximation

In contrast to dia- and paramagnetism, ferromagnetic materials (e.g. Fe, Ni, Co) have a non-vanishing magnetization $\vec{M} \neq 0$ also at a vanishing external magnetic field $\vec{B}_a = 0$. In this section the magnetization of the isotropic Heisenberg model in the mean field approximation (german Molekularfeldnäherung or mittlere Feldnäherung) is calculated. The Heisenberg model is a very good model for describing the exchange interaction between spins and therefore ideal for modelling ferromagnetism. The Hamiltonian of the isotrope Heisenberg model is given by:

$$$H = - \sum_{i \: n.n. \: j}J'_{ij} \: \vec{S}_i \cdot \vec{S}_j - \mu \vec{B}_a \cdot \sum_i \vec{S}_i$$$

Here, the sum goes only over the nearest neighbor sites $j$ of lattice site $i$ ($i \: n.n. \: j$). This is because it is (reasonably) assumed that the spin on site $i$ only interacts directly with the neighboring spins but not with those further away. The number of nearest neighbor sites $z$ is called the coordination number. The exchange coupling constant between the spins on sites $i$ and $j$ is given by $J'_{ij}$ (the prime avoids confusion with the total angular momentum $J$). The term 'spins' is a bit misleading, since the projection in the $z$-direction of $\vec{S}_i$ and $\vec{S}_j$ can also be any number $m_J$ of total angular momentum $J = L + S$:

$$S^z_i = m_J = (-J,-J+1,\cdots , J-1, J)$$

In ferromagnetism, parallel spins decrease the energy and $J'<0$. For antiferromagnetism, antiparallel spins decrease the energy and $J'>0$. For paramagnetism $J'=0$. The Hamiltonian given above is referred to as isotropic Heisenberg Hamiltonian because in the full Heisenberg Hamiltonian $J$ couples the $(x,y,z)$ components of the spins differently and one has $J'^{xy}_{ij}$ and $J'^z_{ij}$ terms, but this is not considered here.

Without loss of generality, we let the external magnetic field point in the $z-$ direction: $\vec{B}_a = B_a \hat{z}$. The magnetic moment $\mu$ is $\mu = g_J \mu_B$ with $g_J$ being the Landé g factor and $\mu_B$ the Bohr magneton. A reasonable approximation is that in an isotropic crystal, the coupling $J'_{ij} = J'$ is the same for all sites. Then the Hamiltonian simplifies to,

$$$H = -J' \sum_{i \: n.n. \: j} \vec{S}_i \cdot \vec{S}_j - \mu B_a \sum_i S^z_i.$$$

Now mean field approximation is applied: The main idea is that the interaction of two neighboring spins $\vec{S}_i \cdot \vec{S}_j$ can be replaced by the interaction of a single spin of site $i$ with the mean spin value of its surrounding:

$$\vec{S}_i \cdot \vec{S}_j \rightarrow \vec{S}_i \cdot \langle \vec{S} \rangle .$$

Moreover, we will assume that all angular momenta, especially the mean field, point in the direction of the external magnetic field, i.e. in the $+z$ direction, such that the vectors are replaced according to:

$$\vec{S}_i \rightarrow S_i^z \equiv S_i = m_J^{(i)}$$ $$\langle S \rangle = \langle m_J \rangle$$

The mean field Hamiltonian can then be written as:

$$H_{MF} = -\sum_i S_i \left( z J' \langle S \rangle + \mu B_a \right).$$

The coordination number $z$ comes from the sum over nearest neighbors. The term in brackets doesn't depend on any lattice site and acts like a total magnetic field consisting of the external magnetic field, $B_a$ and the mean magnetic field $B_{MF}$:

$$H_{MF} \overset{\cdot}{=} -\mu \left(B_{MF} + B_a \right) \sum_i S_i.$$

The factor $\mu$ has been extracted to bring it in the original form. Now we have a new Hamiltonian with no coupling $J'$, but only a magnetic field $B_{total} = B_{MF} + B_a$. The mean magnetic field is given by:

$$B_{MF} = \frac{1}{\mu} \: z \: J' \langle S \rangle = \frac{zJ'}{\mu^2 n} \: M$$

Where in the last expression the average spin has been expressed in terms of the overall magnetization density:

$$M = \underbrace{\frac{N}{V}}_{n} \mu \langle S \rangle$$

and the spin density $n$ has been introduced. As defined on the paramagnetism page, the energies of the magnetic states are given by:

$$E_{m_J} = -m_J \mu B_{total}$$

The occupation probabilty of state $m_J$ is given by a Boltzmann distribution,

$$p_{m_J} = \frac{\exp\left(\frac{-E_{m_J}}{k_{B}T}\right)}{\sum\limits_{m_J = -J}^{m_J=J} \exp\left(\frac{-E_{m_J}}{k_{B}T}\right)}$$

The analysis is completely the same as for the general $J$ case in paramagnetism, just that the $B$ field is replaced by $B_{total}$.

The average value of $m_J$ is,

$$\langle m_J \rangle = \langle S \rangle = \frac{M}{n \mu} = \sum\limits_{m_J = -J}^{m_J=J}m_Jp_{m_J} = \frac{\sum\limits_{m_J = -J}^{m_J=J}m_J\exp\left(\frac{m_J \mu B_{total}}{k_{B}T}\right)}{\sum\limits_{m_J = -J}^{m_J=J} \exp\left(\frac{m_J \mu B_{total}}{k_{B}T}\right)}$$

The calculation of this expression can be found in the paramagnetism section, the result is:

$$\langle m_J \rangle = \langle S \rangle = \frac{M}{n \mu} = J B_J \left(J \frac{\mu B_{total}}{k_B T}\right)$$

where $B_J(x)$ is the Brillouin function,

$$B_J(x)= \frac{2J+1}{2J}\text{coth}\left(\frac{2J+1}{2J}x\right)-\frac{1}{2J}\text{coth}\left(\frac{x}{2J}\right)$$
 $\langle m_J \rangle = JB_J(J x)$ $\large x=\frac{\mu B_{total}}{k_BT}$

The magnetization density can then be expressed as,

$$$M=n \mu J \left(\frac{2J+1}{2J} \coth\left( \frac{2J+1}{2J} J \frac{\mu B_{total}}{k_BT}\right)-\frac{1}{2J} \coth\left(\frac{1}{2J} J \frac{\mu B_{total}}{k_B T}\right)\right) = n \mu J B_J\left( J \frac{\mu B_{total}}{k_BT} \right)$$$

$$M = n \mu J B_J\left(\frac{J \mu}{k_BT} \left( B_a + \frac{z J'}{\mu^2 n} M \right) \right)$$

Spontaneous magnetization density

To investigate the spontaneous magnetization of ferromagnetism, we set $B_a =0$ so that $B_{total} = B_{MF}$. Since $B_{MF}$ depends on $M$ as well, one obtains a self-consistent equation of state for $M$: The magnetization appears on both sides of the equation, but can not be solved analytically for $M$:

$$$M = n \mu J B_J\left( J \frac{\mu B_{MF}}{k_BT} \right)= n \mu J B_J\left( \frac{z J J'}{\mu n k_BT} M \right).$$$

$$M = M_s B_J\left(\frac{t_c}{T} \frac{M}{M_s}\right)$$

with $M_s=n \mu J$ and $t_c = \frac{z J^2 J'}{k_B}$. The significance of the two constants will be apparent soon, for now it is just a convenient parametrisation for plotting the magnetization as $m = M/M_s$ vs. $t=T/t_c$. Then one has

$$m = B_J\left(\frac{m}{t}\right)$$

Graphical solution

This can be solved graphically: The lhs ($m$) is drawn as a straight line and the right hand side (rhs, $B_J\left(\frac{m}{t}\right)$) as well for fixed $t$ and $J$. The solutions to the equation are the points where the two curves cross (apart from the trivial point $m=0$). The figure below shows the case for $J=\frac{3}{2}$, the critical value for a crossing is $t = 5/9$.

 $B_J\left(\frac{m}{t}\right)$ $m$

One can see from the graphical solutions, that only those lines have a non trivial crossing for which $t<\frac{1+J}{3 J}$, i.e. $T < \frac{1+J}{3 J} t_c$. This means that for temperatures above the critical temperature $T_c = \frac{z J' J(J+1)}{3 k_B}$ no spontaneous magnetization density can occur. This is the critical temperature for the phase transition from the ferromagnetic to the paramagnetic state.

Now, for a given $J$, one can solve the transcendental equation given above for various values of $t$ (e.g. by binary search) and thereby obtain the magnetization density as a function of temperature.

 $\large \frac{M}{M_s^{Ising}}$ $\large \frac{T}{T_c^{Ising}}$

Algebraic solution

Apart from numerical solutions, this equation can't be solved for $M$ in a closed form. However, in limiting cases analytic results for $M$ (valid only in a certain temperature regime) can be derived: The magnetization density depends on the ratio $\frac{\mu B_{MF}}{k_BT}$, so we can distinguish two limiting cases: i) Strong magnetic mean field = low temperature and ii) Weak magnetic mean field = high temperature. Strong magnetic field means that the magnetic energy $E_{magn} = \mu B$ is larger than the thermal energy $E_{therm} = k_B T$. For these two limiting cases, the equation for $M$ can be simplified, but we need the asymptotic behaviour of the Brillouin function:

$$$B_J (x) = \begin{cases} \frac{1+J}{3J} x - \frac{1+ 3J+4 J^2+ 2 J^3}{90 J^3}x^3 + \mathcal{O}(x^5) & x << 1, \text{ high temperature, low magnetization} \\ 1 & x >> 1, \text{ low temperature, high magnetization} \end{cases}, \quad x = \frac{z J J'}{\mu n k_B T} M$$$

The second limiting case can be interpreted right away: The Brillouin function approaches one for arbitrary values of $J$, so the total magnetization density is given by:

$$M = M_s = n \mu J$$

This result makes totally sense, because at very low temperatures (very high fields), all angular momenta are fully aligned in the $+z$ direction of the mean magnetic field. The resulting magnetization density is called the saturation magnetization $M_s$. This result is identical to the one found in the paramagnetism section.

The first limiting case (high temperatures, low fields) is more sophisticated: The expression for the Brillouin function is obtained by a simple (but tedious) Taylor expansion. The equation for $M$ is now of order 3, but it is of the form $M = \alpha M + \beta M^3$, so we can identify the so-called trivial solution to this equation, namely $M=0$. But this is not the only possible solution, because after cancelling one $M$ from the equation, one ends up with a quadratic equation for $M$:

$$1 = \frac{J(J+1)}{3} \frac{z J'}{k_B T} - \frac{J+ 3 J^2 +4 J^3+ 2 J^4}{90 n^2 \mu^2} \left(\frac{z J'}{k_B T}\right)^3 M^2$$

The solution to this equation gives the amount of the spontaneous magnetization density at zero external field. It is given by:

$$M = \pm \sqrt{30} n \mu k_B T \sqrt{\frac{zJ'J(J+1)-3 k_B T}{z^3 J'^3 \left(J+ 3 J^2+4 J^3 + 2 J^4\right)}}$$

The sign is arbitrary, because if no external field specialises one direction, the direction of the magnetization is also random and both solutions are equivalent. It has to be remembered that this is the solution for high temperatures (low magnetization), so we expect that at a certain critical temperature $T_c$ the magnetization should vanish again. The condition $M(T=T_c)=0$ can indeed be solved for $T_c$ and one obtains:

$$T_c = \frac{z J'}{3 k_B} J (J+1)$$

As one can see, we have derived $T_c$ a second time and obtained the same value as with the graphical solution. Is it really necessary to expand the Brillouin function up to $x^3$? In the lecture the expansion is only performed up to first order, but with $B_a \neq 0$. The problem is that up to first order $M \propto B_a$, so no spontaneous magnetization density ($M \neq 0$ and $B_a =0$) comes out. Therefore one has to expand the Brillouin function up to $x^3$.

For plotting purposes, the quantities $m=M/M_s$ and $t=T/T_c$ are introduced and the formula given above forms into:

$$m = \pm \sqrt{\frac{10}{3}} \frac{J+1}{\sqrt{1+2 J+2 J^2}} t \sqrt{1-t}$$

The formula is only valid for $M << M_s$, i.e. $m << 1$, because we made the Taylor expansion for small $M$. Above $T_c$, the magnetization density $M=0$. Sharply below $T_c$, the magnetization is small, so we have to evaluate the expression for $t < 1$. In the plot below (for $J=1/2$, i.e. the Ising model), one should keep in mind that this result is only valid for $M << M_s$ or $T \approx T_c$! At lower temperatures, the magnetization curve starts to bend down towards zero instead of becoming constant, which indicates where the approximation breaks down (compare to the correct plot above). Above $T_c$, the magnetization density is zero, as expected. A comparison between the correct curve and the one obtained by third order Taylor expansion is given below.

 $\large \frac{M}{M_s}$ $\large \frac{T}{T_c}$

Susceptibility

The magnetic susceptibility is defined as

$$\chi_m = \left. \frac{\partial M}{\partial H_a} \right\vert_{H_a=0}$$

In spin models, the relative permeability $\mu_r = 1$, because the spins are in vacuum. The susceptibility should be understood in the sense of a response function. In vacuum it holds that $H_a = \frac{1}{\mu_0} B_a$. Then we get:

$$\chi_m = \mu_0 \left. \frac{\partial M}{\partial B_a} \right\vert_{B_a=0}$$

First we calculate

$$\left. \frac{\partial}{\partial B_a} B_J\left(\frac{J \mu}{k_BT} \left( B_a + \frac{z J'}{\mu^2 n} M \right)\right) \right\vert_{B_a=0} =$$

$$\left. \frac{\partial B_J(x)}{\partial x} \right \vert_{x = \frac{z J J'}{n \mu k_B T} M} \left( \frac{\mu J}{k_B T} + \frac{z J J'}{n \mu k_B T} \underbrace{\left. \frac{\partial M}{\partial B_a} \right\vert_{B_a=0}}_{=\frac{1}{\mu_0} \chi_m} \right) =$$

$$\left( \frac{J+1}{3 J} - \frac{1+3 J +4J^2+2J^3}{30 J^3} \left( \frac{z J J'}{n \mu k_B T} \right)^2 M^2 + \mathcal{O}(x^4) \right) \left( \frac{\mu J}{k_B T} + \frac{z J J'}{n \mu_0\mu k_B T} \chi_m \right)$$

where in the last line a Taylor expansion for the derivative of the Brillouin function for very small $M$ was used (we consider only small magnetizations):

$$\left. \frac{\partial}{\partial x} B_J(x) \right\vert_{x= \frac{z J J'}{n \mu k_B T} M} = \left. \frac{J+1}{3 J} - \frac{1+ 3J+4J^2+2J^3}{30 J^3} x^2 + \mathcal{O}(x^4) \right\vert_{x=\frac{z J J'}{n \mu k_B T} M}$$

Then applying the derivative also on the left hand side (lhs, so on $M$), one gets:

$$\frac{1}{\mu_0} \chi_m = n \mu J \left(\frac{J+1}{3 J} - \frac{1+3 J +4J^2+2J^3}{30 J^3} \left( \frac{z J J'}{n \mu k_B T} \right)^2 M^2 \right) \left( \frac{\mu J}{k_B T} + \frac{z J J'}{n \mu_0\mu k_B T} \chi_m \right) \Rightarrow$$

To simplify the expression a little bit (albeit it still looks horrible) we use the dimensionless quantities $m = M/M_s$ and $t = T/T_c$. Careful calculus yields:

$$\chi_m = \frac{n \mu_0 \mu^2}{z J'} \frac{10 t^2 (J+1)^2-9 m^2 \left( 1+ 2J+2J^2 \right)}{10 t^2 (t-1) (J+1)^2 + 9 m^2 \left( 1+ 2J+2J^2 \right)}$$

Here once again one could ask whether it is really necessary to express the derivative of the Brillouin function up to $x^2$ and why the first (constant) term is not sufficient. Keeping only the zeroth order corresponds indeed to the $m=0$, i.e. the $T > T_c$ case and it is sufficient for getting out a Curie law. But the result is not valid for $T < T_c$ because it predicts a negative susceptibility, which can't be the case in ferromagnetism.

The Taylor expansion above restricts us to values $t \approx 1$, i.e. $T \approx T_c$. We already calculated $m$ in this limit, so we insert from above:

$$m^2 = \frac{10}{3} \frac{(J+1)^2}{1+2J+2J^2} t^2 (1-t) \Theta(1-t)$$

where the Heaviside function ensures that $m=0$ for $t>1$. Once more we have to do some math to arrive at:

$$\chi_m = \frac{n \mu_0 \mu^2}{z J'} \frac{1 + 3 (t-1) \Theta(t-1)}{(t-1)\left(1 -3 \Theta(1-t) \right)}$$

For $T>T_c$, the Theta function gives zero, so the result is (reinserting $t=T/T_c$ with the known expression for $T_c$):

$$\left. \chi_m \right\vert_{T>T_c} = \frac{n \mu_0 \mu^2}{3 k_B} \frac{J (J+1)}{T - \frac{z J' J (J+1)}{3 k_B}} = \frac{C_+}{T-T_c}$$

Coming from above $T_c$, we see that the susceptibility diverges at $T=T_c$, where $T_c$ has been found before. It diverges following a Curie-Weiss-law. We could have obtained this result also if we kept only the first term in the expansion of the derivative of the Brillouin function. However, the result can't be valid for $T < T_c$, because a negative diverging susceptibiliy makes no sense. To obtain also an expression for $\left. \chi_m \right\vert_{T < T_c}$, one has to keep the second term in the expansion.

Coming from below $T_c$, the Theta function gives one and we end up with:

$$\left. \chi_m \right\vert_{T < T_c} = \frac{n \mu_0 \mu^2}{2 z J'} \left(-3 + \frac{z J' J(J+1)}{z J' J(J+1)-3 k_BT} \right) = \underbrace{\frac{C_{+}}{2}}_{C_{-}} \left(-\frac{3}{T_c} + \frac{1}{T_c-T}\right)$$

 $\frac{\chi_m}{C_{+}}$ $T/T_c$

After all this work, we recover Curie-Weiss laws according to which the magnetic susceptibility diverges at $T=T_c= \frac{z J' J (J+1)}{3 k_B}$. The Curie constants are given by $C_{+} = \frac{n \mu_0 \mu^2 J (J+1)}{3 k_B}$ and $C_{-} = \frac{C_{+}}{2}$.

The constant $-\frac{3}{2T_c}$ is irrelevant if one is close to $T_c$ due to the divergent $\frac{1}{T_c-T}$ term. The Curie constants depend on whether $T_c$ is approached from above ($C_{+}$) or from below ($C_{-}$). The important thing is that from both sides the divergence is of type $\chi_m \propto \frac{1}{|T-T_c|}$.

Comparison with the Landau theory of second order phase transistions

Close to criticality, Lev Landau constructed a model for the free energy density $f(T,M)$ (free energy per volume/area/length) close to the critical temperature. It is given by:

$$f(T,M) = f_0(T) + \alpha_0 \left(T-T_c\right) M^2 + \frac{1}{2} \beta M^4 + M B_a$$

In this formula, $f_0(T)$ represents the temperature dependence of the free energy above the phase transition.The constant, real and positive parameters $\alpha_0$ and $\beta$ must be determined from thermodynamic quantities (derivatives of $f(T,M)$) and $M$ is the so-called order parameter. The order parameter is defined by its property of being zero above $T_c$ and non zero below $T_c$. In the case of magnetism, it is the magnetization density separating the ferromagnetic ($M \neq 0, T < T_c$) from the paramagnetic phase ($M = 0,T>T_c$). This model can be understood as a Taylor expansion in $M^2$ up to order $M^4$, which is sufficient because, close to the critical point, $M$ is small. The expansion is performed in $M^2$, not in $M$, because without an external magnetic field there must be no difference in whether $M>0$ or $M<0$. An odd power in $M$ is said to be symmetry breaking and would spoil this property. The only allowed odd-power term in the expansion is the $+M B_a$ term, since the external magnetic field breaks the symmetry. The following relations hold (for derivation see the corresponding web page):

magnetization: $$\left. \frac{d f}{d M} \right\vert_{B_a = 0} = 0 \longrightarrow M = \pm \sqrt{\frac{\alpha_0\left(T_c-T\right)}{\beta}}, \quad T < T_c$$

This expression can be compared with the one that we obtained before for the magnetization density close to $T_c$:

$$M = \pm \sqrt{30} n \mu k_B T \sqrt{\frac{zJ'J(J+1)-3 k_B T}{z^3 J'^3 \left(J+ 3 J^2+4 J^3 + 2 J^4\right)}}$$

$$M = \pm \sqrt{\frac{10}{3}} n \mu J (J+1) \frac{T}{T_c^{3/2}} \sqrt{\frac{T_c-T}{1+2J+2 J^2}}, \quad T_c = \frac{z J'}{3 k_B}J(J+1)$$

$$M \approx \pm \sqrt{\frac{10}{3}} n \mu J (J+1) \sqrt{\frac{T_c-T}{1+2J+2 J^2}}$$.

The last expression is obtained from the Taylor expansion around $T_c$:

$$\frac{T}{T_c^{3/2}} \sqrt{T_c-T} = \sqrt{\frac{T_c-T}{T_c}} + \mathcal{O}\left( \left( \frac{T_c-T}{T_c} \right)^\frac{3}{2} \right)$$

However, the approximation is necessary to obtain by direct comparison that:

$$\frac{\alpha_0}{\beta} = \frac{10 n^2 \mu^2 J^2 (J+1)^2}{3 \left( 1+ 2J +2J^2 \right)}$$

As a next step we calculate the susceptibility in the Landau theory. The result is the same as shown in the corresponding article, but here also the $T < T_c$ case is covered. This part should not be seen different, but complementary to the analysis done here. The starting point is the definition of the susceptibility:

$$\chi_m := \mu_0 \frac{\partial M}{\partial B_a} = \mu_0 \left( \frac{\partial B_a}{\partial M} \right)^{-1} = \mu_0 \left( \frac{\partial^2 f}{\partial M^2} \right)^{-1}$$

$$\chi_m = \frac{\mu_0}{2 \alpha_0 \left(T-T_c\right) + 6 \beta M^2}$$

For $T>T_c, M = 0$ and we arrive at the result known from the lecture:

$$\left. \chi_m \right\vert_{T > T_c} = \frac{\mu_0}{2 \alpha_0 \left( T-T_c \right)} = \frac{C_{+}}{T-T_c}$$

The second equal sign comes from the comparison with the previous section, where we also calculated the susceptibility. We found $C_{+} = \frac{n \mu_0 \mu^2 J (J+1)}{3 k_B}$. Now we can directly calculate $\alpha_0$ and from this $\beta$:

$$\alpha_0 = \frac{3}{2} \frac{k_B}{n \mu^2 J (J+1)} = \frac{\mu_0}{2 C_{+}}$$

$$\beta = \frac{9}{20} \frac{k_B}{n^3 \mu^4} \frac{1+2J+2J^2}{J^3 (J+1)^3} = \frac{\mu_0^3 \mu^3}{60 k_B^2} \frac{1+2J+2J^2}{C_{+}^3}$$

Below $T_c$, the magnetization $M^2 = \frac{\alpha_0}{\beta} \left(T_c - T \right)$ must be inserted in the formula given above to yield:

$$\left. \chi_m \right\vert_{T < T_c} = \frac{\mu_0}{2 \alpha_0 \left(T-T_c\right) + 6 \beta \frac{\alpha_0 \left(T_c-T \right)}{\beta}} = \frac{1}{2} \frac{\mu_0}{2 \alpha_0 \left( T_c-T \right)} = \frac{C_{-}}{T_c-T}.$$

As found before, the constant of the diverging susceptibility differs by a factor of two between the two regimes $T < T_c$ and $T> T_c$. The values of $\alpha_0$ and $\beta$, however, stay the same.

The $J = \frac{1}{2}$ case: Ising model

In the Ising model the spin has just two possible orientations: $J^z = \pm \frac{1}{2}$. The magnetization is given by

$$M = M_s \tanh\left(\frac{\mu B_{total}}{2 k_B T} \right) = M_s \tanh\left(\frac{\mu}{2 k_B T} \left(B_a + \frac{z J'}{n \mu^2}M \right) \right)$$

The Brillouin function for $J =\frac{1}{2}$ is simply the hyperbolic tangent. In this case, $M_s = \frac{n \mu}{2}$. The susceptibility follows a Curie-Weiss-law according to

$$\chi_m = \frac{C}{|T-T_c|}$$

with $T_c = \frac{z J'}{4 k_B}$ and $C_{+} = \frac{n \mu_0 \mu^2}{4 k_B}$. The parameters of the Landau theory for the Ising model are $\alpha_0 = \frac{2 k_B}{n \mu^2}$ and $\beta = \frac{8 k_B}{3 n^3 \mu^4}$.