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\heading{Ex 5956: Mean field approximation  classical spin vector model } 
\auname{yair massury} 

{\bf The problem:}

\Dn
Apply the mean field approximation to the  classical spin vector model
\[{\cal H}=-\epsilon\sum_{\langle i,j\rangle}{\vec{\bf  s}}_{i}\cdot { \vec{\bf s}}_{j}-{\bf h}\cdot \sum_{i}{ \vec{\bf s}}_{i}\]
where ${\vec{\bf s}_{i}}$ is a unit vector and ${i,j}$ are neighboring
sites on a lattice with coordination number $c$. The lattice has
${N}$ sites and each site has ${c}$ neighbors.

\Dn

\begin{itemize}
\item [(a)]
Assume that $\; \vec{h}=(0,0,h)\;$, define a mean field ${{\bf h}_{eff}}$ ,and evaluate the partition
function ${Z}$ in terms of ${{\bf h}_{eff}}$.
\Dn

\item [(b)]
Define $\;  \theta_i \;$ as the inclination angle of $\;  {\bf{s_i}} \;$ with respect to $ \vec{h}\;\;$  
Assume that at equilibrium $\;\; <{\bf{s_i}}=(0,0,m)>\; $ where $\; m =<\cos{\theta}>$.
Find the equation for m and find the
transition temperature $\;{T_{c}}\;\;$for$\;\; h\rightarrow0^{+}\;$.
\Dn
\item [(c)]
Write an expression for the mean field energy of the system assuming that $\;\;m(T)  \;\;$ is known.
\Dn
\item[(d)] Identify exponents $\gamma,\beta$ as $T\rightarrow T_c$ for
the susceptibility
${\chi \sim (T-T_c)^{-\gamma}}$ above $T_c$ and for
$M\sim (T_c-T)^{\beta}$ below $T_c$.
\item[(e)] Show that there is a jump in $C_V$ at
$T_c$.
\Dn

\end{itemize}

\Dn
\Dn

{\bf The solution:}
\Dn
\begin{itemize}
\item [(a)]
we make a roughly Mean field approximation and got the Hamiltonian for the singel spin at site $ i$: 
\begin{equation}
 {\cal H}^{i}=-\epsilon c{\langle {\vec{\bf  s}}\rangle}{ \vec{\bf s}}_{i}-{  h}\cdot { \vec{\bf s}}_{i} \;\;\;\;\;\;\;\;
\end{equation}

lets define ${\langle {\vec{\bf s}}\rangle}=\frac{ {M}}{N}=\vec{m}\;\;$ and $\; { \vec{\bf s}}_{i}=\cos{\theta_{i} } $



\begin{equation} 
 {\cal H}^{i}=-(\epsilon c {m}+h)\cos{\theta_{i}} \nonumber\\
 so\;\;\; that\;\; \rightarrow \; \;\; {{{\bf {h}}_{eff}}}=\epsilon c m+h \nonumber\\
 {\cal H}^{i}=(-{{{\bf h}_{eff}}}\cos{\theta_{i}} ) 
\end{equation}
 

 \Dn
$\ast$Notice that the Mean field approximation transform the problem from multi particle problem to a single particle problem (without a direct interaction).so that ${  Z_{N}}={  Z_1}^{N}$ the spins are pins to their positions.

\begin{equation} { Z_{1}}=\frac{1}{4\pi}\int^{1}_{-1}e^{\beta{{\bf {h_{1}}}}}d(\cos{\theta_{i}})\int^{2\pi}_{0}d\varphi=\frac{1}{4\pi}\int^{1}_{-1}e^{\beta{\bf h}_{eff}\cos{\theta}}d(\cos{\theta})\int^{2\pi}_{0}d\varphi=\frac{\sinh(\beta{\bf h}_{eff})}{\beta{\bf h}_{eff}}
\end{equation}

\begin{equation} { Z_{N}}=({ Z_{1}})^{N}={\left(\frac{\sinh(\beta{\bf h}_{eff})}{\beta{\bf h}_{eff}} \right)}^{N}
\end{equation}


\Dn

\item [(b)]

we define ${m(T)=\langle \cos \theta_{i}\rangle}$

\begin{equation}
 m\equiv \langle \cos (\theta_{i})\rangle=\frac{1}{Z} \sum cos (\theta) e^{\beta{\bf h}_{eff}\cos{\theta}}=\frac{1}{Z} \frac{1}{{\bf h}_{eff}} \frac{\partial}{\partial \beta}Z  
 \end{equation} 
For singel spin:
\begin{equation}
m= \frac{1}{{{\bf h}_{eff}} }\frac{\partial}{\partial \beta} \ln {Z } =\frac{1}{{\bf h}_{eff}} \frac{\partial}{\partial \beta} \left[    \ln\left(  \sinh(\beta{\bf h}_{eff})\right)  -\ln \left( \beta{\bf h}_{eff}  \right)          \right] 
\end{equation}
\begin{equation}
m =\coth(\beta{\bf h}_{eff})-\frac{1}{{ \beta { \bf h}_{eff}} }\cong \left\{  h\rightarrow 0,\;\;T\approx T_{c} \Rightarrow m\rightarrow 0    \right\}  
\end{equation}

Second approach:
\begin{equation}
F=-K_{B}T  \ln Z_{N}=-\frac{N}{\beta}\left[-\beta \frac{1}{2} \epsilon c m^2 + \ln \left[ sinh(\beta \epsilon c m)- \ln (\beta \epsilon c m) \right]
\right] \end{equation}
\Dn
$\ast$  we want to choose M from extremum condition.Where $\; {\frac{M}{ N}=m} \;$
\Dn
\begin{equation}
\frac{\partial {F}}{\partial M}=\frac{1}{ N} \frac{\partial {F}}{\partial {m}} =\epsilon c m -\frac{1}{\beta}\frac{\cosh(\beta \epsilon c m) }{\sinh(\beta \epsilon c m) }\beta \epsilon c + 
\frac {1}{m\beta} 
 = \epsilon c m  -\epsilon c  \coth (\beta \epsilon c m)+\frac{1}{\beta  m}\;\;\Rightarrow \;0
\end{equation}

\begin{equation}
 m= \left[\coth (\beta \epsilon c m)-\frac{1}{\beta \epsilon c m}\right]
\end{equation}



\begin{eqnarray}
\coth{x}\cong \frac{1}{x}+\frac{x}{3}-\frac{x^3}{45} \;\;\;\;\;\;\;\;\;\;\;
\tanh{x}\cong {x}+-\frac{x^3}{3}\;\;\; \rightarrow \;\;\coth{x}-\frac{1}{x}\approx\tanh{x}
\nonumber \\
m \approx \tanh{ { \beta { \bf h}^{0}_{eff}} }\;\;\Rightarrow  \:\:  \tanh^{-1}(m)=\beta(\epsilon c m+h)
\end{eqnarray}
 For first order $\; \tanh^{-1}(m)\cong m\;$ and to find $T_c\;$ we take $\; h\rightarrow 0$.
\begin{equation} 
m=\beta \epsilon c m\;\;\;\;\;\rightarrow \epsilon c=K_{B}T_c
 \end{equation}
 
 
 \Dn
\item [(c)]
 
 
In the mean field approximation the spins are independent of each other, and therefore \\
$ \left\langle {\bf  s}_{i}{\bf  s}_{j}\right\rangle=\left\langle {\bf  s}_{i}\right\rangle\left\langle {\bf  s}_{j}\right\rangle   $. It follow that the energy is:

\begin{equation}
E=\left\langle {\cal H}\right\rangle=\left\langle -\epsilon\sum_{\langle i,\epsilon\rangle}{\bf \vec{s}}_{i}\cdot {\bf \vec{s}}_{j}-{\bf h}\cdot \sum_{i}{ \vec{\bf s}}_{i}\right\rangle 
\end{equation}
\begin{equation}
E=-\frac{1}{ 2}\epsilon c m^2 N-\epsilon c m Nh
\end{equation}
For $ h\rightarrow 0 \;\;$we get the known result for the energy $\;E=-\frac{1}{ 2}\epsilon c m^2 N\;\;$
 \Dn
 

\item [(d)]
From section b we get:
\begin{equation}
{m =\coth(\beta{\bf h}_{eff})-\frac{1}{{ \beta { \bf h}_{eff}} }  \;\;\;\;\;\;\;\;\;\; \coth{x}-\frac{1}{x} \approx \tanh{x}}
\end{equation}
so we get the known result for m:
\begin{equation}
{m =\tanh(\beta{\bf h}_{eff})  \Rightarrow \:\: \tanh^{-1}(m)=\beta{\bf h}_{eff}}\nonumber\\
\tanh^{-1}(m)\cong m+\frac{m^3}{3}
\end{equation}
we take $\;\;h\rightarrow 0 \;\; $
\begin{equation}
\epsilon{c}{m}=k_B {Tm}+k_B T\frac{m^3}{3} \nonumber \\
\frac{3 (T_c-T)}{ T}m=m^3
\end{equation}
\begin{equation}
m=\sqrt{3 (\frac{T_c}{T}-1)}\Rightarrow \;\;\;\;\;  \beta=\frac{1}{2}
\end{equation}

\begin{eqnarray}
h=T tanh^{-1}(m)-\epsilon{c}{m} \Rightarrow (For \;\; T>T_c \;\;\;\;\; m\rightarrow 0)\;\;\; h\cong T(m+\frac{m^3}{3}) -T_c m 
\nonumber \\ 
\chi=\frac{\partial m}{\partial h} |_{h\rightarrow 0}=\frac{1}{\frac{\partial h}{\partial m}}=\frac{1}{(T-T_c)+m^2}\cong{(T-T_c)}^{-1} \;\; \Rightarrow \;\;\; \gamma=1
\end{eqnarray}
 \Dn
\item [(e)]

\begin{equation}  
U=-\frac{\partial\ln Z } {\partial \beta} =-\frac{\partial T } {\partial \beta} \frac{\partial\ln Z } {\partial T} =K_{B}T^2\frac{\partial\ln Z } {\partial T}
\end{equation}

\begin{equation}  
C_{v}= \frac{\partial U } {\partial T} =2K_{B} T \frac{\partial\ln Z } {\partial T} +K_{B}T^2\frac{\partial^2 \ln Z } {\partial T^2}
\end{equation}

\begin{equation}   
-\frac{\partial^2  F}{\partial T^2}=\frac{\partial }{\partial T^2}(KT \ln{Z})=\frac{\partial }{\partial T}(K \ln{Z}+KT\frac{\partial \ln{Z} }{\partial T})=2K\frac{\ln{Z} }{\partial T} +KT\frac{\partial^2\ln{Z} }{\partial T^2}
\end{equation}
\begin{equation}  
\Rightarrow \;\; C_{v}= -T\frac{\partial^2  F}{\partial T^2}=-T\frac{\partial }{\partial T}( \frac{\partial F }{\partial m}\cdot \frac{\partial m}{\partial T}  )
\end{equation}
\Dn
\Dn
 From section b we got: $\;\; m \approx \sqrt{\frac{T_{c}}{T}-1} \;\;$ to get: $ \;\; \frac{\partial m}{\partial T} \approx  \frac{-T_{c}/T^2}{2 \sqrt{\frac{T_{c}}{T}-1}} \;\;$
 \begin{equation}
F=-K_{B}T  \ln Z_{N}=-\frac{N}{\beta}\left[-\beta \frac{1}{2} \epsilon c m^2 + \ln \left[ sinh(\beta \epsilon c m)- \ln (\beta \epsilon c m) \right]
\right] \end{equation}
\begin{eqnarray}
\frac{\partial {F}}{\partial {m}} =\frac{N}{\beta}({\beta}\epsilon c m  -\coth(\beta \epsilon c m) \beta \epsilon c + \frac {1}{m} )\;\;\;\;\;\;\;\;\;\;\;\;\;\;  [\epsilon c\approx  {T_c}] \nonumber \\
 \cong N(\epsilon c m  -\epsilon c(\frac{1}{ {\beta}\epsilon c m }-\frac{{\beta}\epsilon c m}{3}\ )+\frac{1}{\beta  m})\cong N\epsilon c m(1-\frac{T_c}{T})
\end{eqnarray}
\begin{equation}
C_{v}=-T\frac{\partial }{\partial T}( \frac{\partial F }{\partial M}\cdot \frac{\partial M}{\partial T}  )=-T\frac{\partial }{\partial T}\left( N T_c m(1-\frac{T_c}{T})   \right)\left( \frac{-T_{c}/T^2}{2 \sqrt{\frac{T_{c}}{T}-1}}  \right)
\end{equation}

\begin{equation}
C_{v}\cong T\frac{\partial }{\partial T}\left( N \frac{T^2_{c}}{T^2} (1-\frac{T_c}{T})   \right)
\end{equation}
We search $C_v$ close to $T_c$ \; so $\;\;(T-T_c)<<1 \;$ and $\; \frac{T^2_{c}}{T^2}\cong 1 $
\begin{equation}
C_{v}\cong T_cN\frac{\partial }{\partial T}\left(  \frac{T-T_c}{T_c}   \right)=NT_c
\end{equation}
We can see the jump in $C_v$ for $T>T_c\;\;\;\;\;(m=0)$ and as a result $\;\;C_v=0$ ,for $T<T_c$ \\
$C_v=const$
\\
\Dn
Additional approach:  we consider the  mean field energy $\;E=-\frac{1}{ 2}\epsilon c m^2 N\;\;$:
\begin{equation}  
C_{v}= \frac{\partial U } {\partial T} =\frac{\partial } {\partial T}(-\frac{1}{2} \epsilon c m^2N)
\end{equation}
\begin{equation}  
C_{v}= -\frac{1}{2} NT_c  \frac{\partial m^2 } {\partial T} =-\frac{1}{2}NT_c\frac{\partial  } {\partial T}\left(\frac{T_c}{T}-1 \right)\cong -\frac{1}{2}NT_c\frac{\partial  } {\partial T}\left({T_c}-{T}\right)\cong NT_c
\end{equation}

\end{itemize}


\end{document} 
