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\exnumber{1815}
\heading{Cooling by demagnetization} 
\auname{Roy Apsel}


{\bf The problem:}

\Dn

Consider a solid with ${N}$ non-magnetic atoms and $N_{i}$
non-interacting magnetic impurities with spin ${s}$. There is a
weak spin-phonon interaction which allows energy transfer between
the impurities and the non-magnetic atoms.

\begin{itemize}
\item [(a)]
A magnetic field is applied to the system at a constant
temperature ${T}$. The field is strong enough to line up the spins
completely. What is the change in entropy of the system due to the
applied field? (neglect here the spin-phonon interaction).
\item [(b)]
Now the magnetic field is reduced to zero adiabatically. What is
the qualitative effect on the temperature of the solid? Why is the
spin-phonon interaction relevant?
\item [(c)]
Assume that the heat capacity
of the solid is ${C_{V}=3Nk_{B}}$ in the relevant temperature
range. What is the temperature change produced by the process (b)?
(assume the process is at constant volume).\\
\end{itemize}


\Dn\Dn


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{\bf The solution:}

\Dn

(a)
Neglecting the spin-phonon interaction term, the hamiltonian of the system is
\begin{equation}
	H  =  H_{s} + H_{p}
\end{equation}
Where $H_{s}$ is the contribution of the spins, and $H_{p}$ is the contribution of the phonons.\\

However, since the magnetic field does not couple with the phonons, then
the change in entropy of the entire system depends only on the spins
\begin{equation}
	H_{s} =   -B \sum_{k=1}^{N_{i}} s_{k} 
\end{equation}
Where B is an external magnetic field, and $s_{k}$ can take any value from $-s$ to $s$\\

Before the field is turned on, there is no magnetization, and the energy is $0$.
After the field is turned on, it takes the value $B_{i}$ which is assumed to be strong enough so that all of the spins are pointing in the same direction.
The change in energy is then
\begin{equation}
	\Delta E = -B_{i} N_{i} s
\end{equation}


The partition function of the spins is given by
\begin{equation}
	Z_{s} = \left(\sum_{s_{1}=-s}^{s} e^{\beta Bs_{1}}\right)^{N_{i}}
			= \left(\frac {e^{-\beta Bs} - e^{\beta B(s+1)}} {1-e^{\beta B}}\right)^{N_{i}}
				=\left( \frac {\sinh \left(\frac{\beta B}{2} (2s+1)\right)} {\sinh \frac {\beta B} {2}} \right)^{N_{i}}
\end{equation}

The entropy of the spins is given by
\begin{equation}
	S_{s} = -\left( \frac{\partial F}{\partial T} \right)_{B}
			= \frac{\partial}{\partial T} (T\log Z_{s})
				= \log Z_{s} + \beta E_{s}
\end{equation}


Since the field $B_{i}$ is strong enough to line up all the spins, the following condition is met $\beta B_{i}\gg1$\\
The change in entropy of the entire system is then given by\\
\begin{equation}
\begin{split}
	\Delta S 	&	= S_{s}(B=B_{i}) - S_{s}(B=0) \\
			&	= \log Z_{s}(B=B_{i}) - \log Z_{s}(B=0) + \beta \Delta E \\
			&	= N_{i} \log \left( \frac {\sinh \left(\frac{\beta B_{i}}{2} (2s+1)\right)} {\sinh \frac {\beta B_{i}} {2}} \right)   -N_{i}\log(2s+1)   -\beta N_{i} B_{i}s \\
			&	\approx -N_{i} \log(2s+1)
\end{split}
\end{equation}
This result could generally be exected, given eq. (3).\\

\Dn\Dn
%%%%%%%%%%%%
%\pagebreak


(b)
\begin{itemize}

\item \textbf{"What is the qualitative effect on the temperature of the solid?"} \\
In this process the field B is reduced from some maximal value $B_{i}$ to $0$ adiabatically, i.e. the total entropy remains constant ($\Delta S=0$).
The field B couples only to the spins (i.e. it does not interact with the phonons, neglecting the spin-phonon interaction term),
it is expected therefore that a change in the temperature of the system is determined by a change in entropy of the degrees of freedom of the spins (as the total entropy is fixed).
Since the entropy of the spins increases as B decreases, the temperature correspondingly decreases as well (Figure 1).
The system therefore is expected to cool down as B is switched off in an adiabatic process.

\item \textbf{"Why is the spin-phonon interaction relevant?"} \\
As the field is reduced from $B_{i}$ to $0$, the entropy of the degrees of freedom of the spins may increase (as the spins become less ordered),
therefore a decrease in entropy of the lattice vibrational modes allows for the overall entropy to remain fixed.
This process is made possible by the spin-phonon interaction.


\end{itemize}

\begin{figure}[ht!]
\centering
\includegraphics[height=60mm,width=90mm]{entropy_paramagnet.png}
\caption
{
	General behavior of the entropy $S_{s}$ with the temperature $T$, of a paramagent in a 2-state per single spin system,
	for different values of the magnetic field $h=\beta B$.
}
\end{figure}


%%%%%%%%%%%%
\pagebreak

(c)
Neglecting the spin-phonon interaction, the total entropy is
\begin{equation}
	S = S_{s} + S_{p}
\end{equation}

The heat capacity of the solid during this process is
\begin{equation}
	C_{V} = \left(\frac{\partial E_{p}}{\partial T}\right)_{V} = T\left(\frac{\partial S_{p}}{\partial T}\right)_{V} = 3N
	\qquad\qquad (k_{B}=1)
\end{equation}

Since the process is adiabatic, $\delta S=0$. Combining (7) \& (8) gives the following relation
\begin{equation}
	\frac{\delta T}{T} = -\frac{1}{3N} \delta S_{s}
\end{equation}

Given initial and final temperatures $T_{i}$ and $T_{f}$, integrating both sides
\begin{equation}
\begin{split}
	\log \left( \frac{T_{f}}{T_{i}} \right)    &		=     	  -\frac{1}{3N} \Bigg( S_{s} (T_{f},0) - S_{s} (T_{i},B_{i}) \Bigg) \\
								       &		=	  	  -\frac{1}{3N} \Bigg( \log Z_{s}(T_{f},0) - \log Z_{s}(T_{i},B_{i}) + \beta_{i} B_{i} N_{i}s \Bigg) \\
								       &		\approx	  -\frac{N_{i}}{3N}\log(2s+1)
\end{split}
\end{equation}

The final temperature of the process is then
\begin{equation}
	T_{f} \approx T_{i} \ e^{-\frac{N_{i}}{3N} \log(2s+1)}
\end{equation}
Which means that the temperature had dropped, $T_{f} < T_{i}$.\\
The temperature change produced by the process is then
\begin{equation}
	\Delta T = T_{f} - T_{i} \approx T_{i} \left[ e^{-\frac{N_{i}}{3N} \log(2s+1)} - 1 \right]
\end{equation}




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