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\exnumber{5716}
\heading{Ferromagnetism for Cubic Crystal} 
\auname{Tal Gur and Omer Goldberg}


\Dn
{\bf{The Problem:}} \\
A cubic crystal which exhibits ferromagnetism at low temperature,
can be described near the critical temperature ${T_{c}}$ by an
expansion of a Gibbs free energy
\begin{equation}
G({\bf H},T)=G_{0} +\frac{1}{2} r{\bf M}^{2} + u{\bf M}^{4} +
v\sum_{i=1}^{3}M_{i}^{4} -   {\bf H}\cdot {\bf M}
\end{equation}
where
${{\bf H}=\left(H_{1},H_{2},H_{3}\right)}$ is the external field
and ${{\bf M}=\left(M_{1},M_{2},M_{3}\right)}$ is the total
magnetization; ${r=a\left(T -Tc\right)}$ and ${G_{0}, a, u}$ and
${v}$ are independent of ${\bf H}$ and ${T, a>0, u>0}$. The
constant ${v}$ is called the cubic anisotropy and can be either
positive or negative.

\begin{itemize}
\item [(a)]
At ${{\bf H}=0}$, find the possible solutions of ${\bf M}$ which
minimize ${G}$ and the corresponding values of
${G\left(0,T\right)}$ (these solutions are characterized by the
magnitude and direction of ${\bf M}$. Show that the region of
stability of ${G}$ is ${u+v>0}$ and determine the stable
equilibrium phases when ${T<T_{c}}$ for the cases (i) ${v>0}$,
(ii) ${-u<v<0}$.
\item [(b)]
Show that there is a second order phase transition at ${T=T_{c}}$,
and determine the critical indices $\alpha,\,\beta$ and $\gamma$
for this transition, i.e. ${C_{V,H=0} \sim |T- T_{c}|^{-\alpha}}$
for both ${T>T_{c}}$ and ${T<T_{c}}$, ${|{\bf M}|_{H=0} \sim (Tc-
T)^{\beta}}$ for ${ T<T_{c}}$ and ${\chi _{ij} =
\partial M_{i}/\partial H_{j}\sim ~\delta_{
ij} |T-T_{c}|^{-\gamma}}$ for ${T>T_{c}}$.\\
\end{itemize}

\Dn\Dn


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}

\Dn

(a) 
For ${\bf{H}}=0$ and ${\bf{M}}=\sqrt{M_1^2+M_2^2+M_3^2}$ one can find ${\bf{M}}$ which minimizes Gibbs free energy by applying a gradient in ${\bf{M}}$
\begin{equation}
\vec{\nabla}_{\bf{M}}G(0,T) =\sum_i \hat{i}\frac{\partial}{\partial M_i}G(0,T) = \sum_i \hat{i}(r M_i + 4u M_i {\bf{M}}^2+ 4v M_i^3) =  \sum_i \hat{i}M_i(r+ 4u {\bf{M}}^2+ 4v M_i^2) 
\end{equation}
Where $i=1,2,3$ and the corresponding directions. 
\newline
For stability, we demand that $G\rightarrow \infty$ as each $M_i\rightarrow \pm \infty$ .
Because each $M_i$ is independent and from symmetry considerations, we can check what happens to G by taking the limit in each direction separately.
It would be easier to see that for each $M_i$ , $\vec{\nabla}_M G$ will give an increasing monotonic function* as $M_i \rightarrow  \infty$.
\\
\begin{equation}
\lim_{M_i \to \infty}\vec{\nabla}_{\bf{M}}G(0,T) =\lim_{M_i \to  \infty}  \sum_i \hat{i}M_i(r+ 4u {\bf{M}}^2+ 4v M_i^2) 
\end{equation}
A polynomial function at infinity governed by the highest power of the polynomial function. therefore 
\begin{equation}
\lim_{M_i \to  \infty}\vec{\nabla}_{\bf{M}}G(0,T) \rightarrow \lim_{M_i \to  \infty}  4(u+v)M_i^3 > 0 
\end{equation}
The latter requires $u+v >0$, so this is the stability region of G.


\begin{tabbing}	\tiny * From symmetry decreasing monotonic function as each $M_i\rightarrow -\infty$  requires the same condition \end{tabbing}


In order to find the extrema points, we may require $\vec{\nabla}_{\bf{M}}G(0,T) =0$, meaning each direction must equal 0 separately. 
We can find solutions in different cases which are distinct by the number of non-zero components of the magnetization vector.
\\ 
The trivial solution is $M_i = 0$ for every $\hat{i}$ direction.
\newline
Next let's choose two of the magnetization components to be zero $M_i,M_j=0$ and one non-zero $M_k\neq 0$.
This will leave us with the equation
\begin{equation}
r+4(u+v)M_k^2=0 \rightarrow M_k^2= \frac{-r}{4(u+v)} 
\end{equation}
Where $i,j,k$ can be any permutation of $1,2,3$.
\\
Another option for solution is $M_i=0$ and $M_j,M_k\neq 0$ and we have 2 coupled equations
\begin{equation}\begin{split}
&r+4(u+v)M_j^2+4u M_k^2=0 \\
&r+4(u+v)M_k^2+4u M_j^2=0
\end{split}\end{equation}
or in matrices form
\begin{equation}\begin{split}
-\frac{r}{4}
\begin{pmatrix}
1\\1
\end{pmatrix}
=
\begin{pmatrix}
u+v & u\\u & u+v
\end{pmatrix}
\begin{pmatrix}
M_j^2\\M_k^2
\end{pmatrix}
\end{split}\end{equation}
Leading to the solution $M_j^2=M_k^2= \frac{-r}{4(2u+v)}$.
The last option is 3 non-zero components $M_i,M_j,M_k\neq 0$ leading to the matrix equation
\begin{equation}\begin{split}
-\frac{r}{4}
\begin{pmatrix}
1\\1\\1
\end{pmatrix}
=
\begin{pmatrix}
u+v & u & u\\u & u+v & u \\ u & u & u+v
\end{pmatrix}
\begin{pmatrix}
M_i^2\\M_j^2\\M_k^2
\end{pmatrix}
\end{split}\end{equation}
and its solution $M_{i,j,k}^2 = \frac{-r}{4(3u+v)}$.
\\
So we have the possible extrema points for $(M_i^2,M_j^2,M_k^2)$
\begin{equation}
(0,0,0) , (0,0,\frac{-r}{4(u+v)}), (0,\frac{-r}{4(2u+v)},\frac{-r}{4(2u+v)}), (\frac{-r}{4(3u+v)},\frac{-r}{4(3u+v)},\frac{-r}{4(3u+v)})
\end{equation}
in any permutation of 1,2,3.
\newline
Since this is the solution for the quadratic components of the extrema point of the magnetization, we can see that in the case of $r>0$ the last 3 solutions for the magnetization are non-physical, leaving the trivial solution the only physical solution.
\\ 
\\ 
Next, we want to check the stability of the solution (if it's a minimum), so we calculate the Hessian matrix
\begin{equation}\begin{split}
\begin{pmatrix}
r+4u{\bf{M}}^2+(12v+8u)M_x^2 & 8uM_xM_y  & 8uM_xM_z\\
8uM_yM_x & r+4u{\bf{M}}^2+(12v+8u)M_y^2 & 8uM_yM_z\\
8uM_zM_x & 8uM_zM_y & r+4u{\bf{M}}^2+(12v+8u)M_z^2\\
\end{pmatrix}
\end{split}\end{equation}
Due to the symmetry in this problem and since the solution found above is for quadratic components $M_i^2$, if $(M_1,M_2,M_3)$ is a solution then any permutation of $(\pm M_1,\pm M_2, \pm M_3)$ is also a solution.
\\
We can map all the degenerate solutions of $(M_i,M_j,M_k)$ to the corresponding eigenvalues of the Hessian
\begin{center}
    \begin{tabular}{ | l | l | l | p{5cm} |}
    \hline
    Magnetization Vector & Eigenvalues of the Hessian \\ \hline
   $ {\bf{M}^{(1)}} =\Big(\pm\sqrt{\frac{-r}{4(3u+v)}},\pm\sqrt{\frac{-r}{4(3u+v)}},\pm\sqrt{\frac{-r}{4(3u+v)}}\Big)$ & $-2r, -\frac{2rv}{3u+v},-\frac{2rv}{3u+v}$ \\ \hline
   ${\bf{M}^{(2)}} = \Big(\pm\sqrt{\frac{-r}{4(2u+v)}},\pm\sqrt{\frac{-r}{4(2u+v)}},0\Big)$ &  $-\frac{2rv}{2u+v}, \frac{rv}{2u+v},-2r $ \\ \hline
   ${\bf{M}^{(3)}} =\Big(\pm\sqrt{\frac{-r}{4(u+v)}},0,0\Big) $ & $\frac{rv}{u+v},\frac{rv}{u+v},-2r $\\ \hline
      ${\bf{M}^{(4)}} =\Big(0,0,0\Big)$ & $r,r,r$\\ \hline
    \end{tabular}
\end{center}

As we could predict, the trivial solution ${\bf{M}^{(4)}}$ is unstable point for $r<0$. We can also say that ${\bf{M}^{(2)}}$ is always unstable since its first two eigenvalues of the Hessian must have opposite signs.
\\
\\
Now let us determine the stable equilibrium phases for $T<T_c$ (meaning $r<0$) and 
\\
(i) $v>0$
\\
In this case the eigenvalues of the Hessian of ${\bf{M}^{(1)}}$ are all positive, and some of the eigenvalues of the Hessian of ${\bf{M}^{(3)}}$ are negative, meaning these are saddle points.
In this case, the corresponding value of the Gibbs free energy at the stable points is $G(0,T)=G_0 -\frac{9r^2}{16(3u+v)}$
\\
(ii) $-u<v<0$ 
\\
In this case the eigenvalues of the Hessian of  ${\bf{M}^{(3)}}$ are all positive, and some of the eigenvalues of the Hessian of  ${\bf{M}^{(1)}}$  are negative, meaning these are saddle points.
In this case, the corresponding values of the Gibbs free energy at the stable points is $G(0,T)=G_0 -\frac{3r^2}{16(u+v)}$

%%%%%%%%%%%%%%%
\Dn

(b) 
\\
Second order transition is defined by continuity of magnetization at stable equilibrium phase (which is the order parameter of the system) as a function of the temperature and a discontinuity in the derivative at $T=T_c$. 
\\
With that in mind we see that for $T>T_c  $, $M=0 $ and for $T<T_c$ ,$   M\propto \sqrt{-r}=\sqrt{a(T_c-T)}$.
\\
For $ T \rightarrow T_c $, $ \sqrt{-r}\to 0 $ which gives us the continuity at $ T_c $ for $M(T)$, while $ \frac{\partial M}{\partial T}(T) \propto {\frac{1}{\sqrt{-r}}}$ for  $T<T_c $ and $ \frac{\partial M}{\partial T}(T)=0 $ for $ T>T_c $ which gives us the discontinuity in the derivative  at $T=T_c $, and as expected, a second order phase transition.
\\
\\
We want to find the critical indices. 
The index $\alpha$ is defined by ${C_{V,H=0} \sim |T- T_{c}|^{-\alpha}}$ for both ${T>T_{c}}$ and ${T<T_{c}}$.
\begin{equation}
C_{V,H=0} = -T \frac{\partial^2 G(T,H=0)}{\partial T^2} = -T \frac{\partial}{\partial T} \big(\vec{\nabla}_{\bf{M}}G(T,H=0)  \frac{\partial M}{\partial T}\big)
\end{equation}
For ${T>T_{c}}$, $G(T,H=0) = G_0$ so the heat capacity is $C_{V,H=0} = 0$ and index $\alpha = 0$ in that case.\\
For ${T<T_{c}}$ , $|{\bf M}|\propto \sqrt{-r} \sim \sqrt{T_c-T}$ so $\frac{\partial M}{\partial T} \propto -\frac{1}{2\sqrt{T_c-T}}$.
\begin{equation}
C_{V,H=0} \sim -T \frac{\partial}{\partial T} \big(M_i(r+4u{\bf{M}}+4vM_i^2) \frac{\partial M}{\partial T}\big)
\end{equation}
and for $T$ around $T_c$ we can take only the first order in $M_i$
\begin{equation}
C_{V,H=0} \sim -T \frac{\partial}{\partial T} \big(r M_i  (-\frac{1}{2\sqrt{T_c-T}})\big)\sim -T \frac{\partial}{\partial T} (r)\sim T_c
\end{equation}
\\
which means $\alpha = 0$.
\\
The index $\beta$ is defined by ${|{\bf M}|_{H=0} \sim (T_c-T)^{\beta}}$ for ${ T<T_{c}}$. In both cases for stable phases for $r<0$ in the previous section the magnitude of the magnetization is
\begin{equation}
|{\bf M}|_{H=0} \propto \sqrt{-r}= \sqrt{T_c-T}=(T_c-T)^{\frac{1}{2}}
\end{equation} 
meaning $\beta = \frac{1}{2}$.
\\
\\
The index $\gamma$ is defined by ${\chi _{ij} =\partial M_{i}/\partial H_{j}\sim ~\delta_{ij} |T-T_{c}|^{-\gamma}}$ for ${T>T_{c}}$. In order to find the index, we need to calculate the gradient of the Gibbs free energy again for ${\bf{H}}\neq 0$ and find the magnetization that minimize it.
\begin{equation}
\vec{\nabla}_{\bf{M}}G(0,T) = \sum_i \hat{i}M_i(r+ 4u {\bf{M}}^2+ 4v M_i^2)-H_i=0
\end{equation} 
We get a vector equation, and for each direction
\begin{equation}
M_i(r+ 4u {\bf{M}}^2+ 4v M_i^2)=H_i
\end{equation} 
For ${T>T_{c}}$  and weak magnetic field  $\bf{H}$ we know that $M_i \to 0$, so we can take the first order of $M_i$ 
\begin{equation}
H_i \approx M_i r  \to M_i \approx \frac{1}{r} H_i
\end{equation} 
and the susceptibility
\begin{equation}
\chi _{ij} =\partial M_{i}/\partial H_{j}\approx \frac{1}{r} \frac{\partial H_i}{\partial H_j} \sim \delta_{ij} (T-T_c)^{-1}
\end{equation} 
meaning $\gamma =1$.





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