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\exnumber{1816}
\heading{Cooling by adiabatic demagnetization} 
\auname{Zion Hazan and Yair Yifrach}


{\bf The problem:}

\Dn
Consider a system of ${N}$ spins on a lattice at temperature
${T}$, each spin has a magnetic moment  . In presence of an
external magnetic field each spin has two energy levels, ${\mu H}$.


\begin{itemize}
\item[1)]
Evaluate the changes in energy $\delta E$ and in entropy $\delta
S$ as the magnetic field increases from ${0}$ to ${H}$. Derive the
magnetization $M(H)$ and show that
\[\delta E=T\delta S - \int_{0}^{H}M\left(H'\right)dH '.\]\
Interpret this result.


\item[2)]
What is the temperature change when ${H}$ is
reduced to zero in an adiabatic process. Explain how can this
operate as a cooling machine to reach ${T\approx 10^{-4} K}$.
(Note: below ${10^{-4} K}$ in realistic systems spin-electron or
spin-spin interactions reduce ${S\left(T, H=0\right)\rightarrow
0}$ as ${ T\rightarrow 0}$. This method is known as cooling by
adiabatic demagnetization).

\end {itemize}


\Dn\Dn


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{\bf The solution:}

\Dn
 

\begin{enumerate}
\item[1)]
Let's recall the definition of Helmholtz free energy:
\begin{equation}
F=E-TS
\end{equation}

Solve for $E$:
\begin{equation}
E=TS+F
\end{equation}

Where the energy, entropy and Helmholtz free energy are functions of ${T,H,N}$.\\
Let's differentiate Eq.(2), remembering that the temperature and the number of spin are held fix:
\begin{equation}
dE=TdS+\left( \frac{\partial F}{\partial H} \right)_{T,N} dH
\end{equation}

Let's consider the average magnetization $M$ as the generalized force, conjugate to the external magnetic field $H$:
\begin{equation}
M(T,H,N)=-\left( \frac{\partial F}{\partial H} \right)_{T,N}
\end{equation}

Plug in Eq.(4) to Eq.(3) and integrate over the field:
\begin{equation}
\delta E = T\delta S - \int ^H _0 M(T,H',N)dH'
\end{equation}

Where $\delta$E and $\delta$S are respectively the change of the energy and entropy during the change of the external magnetic field.

\Dn\Dn
%%%%%%%%%%%%%%%%%%%%%
\item[2)]

The Hamiltonian of the interaction of a spin with magnetic moment $\mu$ in a magnetic field H is:
\begin{equation}
\mathcal{H}=-\mu_iH
\end{equation}

Where ${\mu_i}$ is either ${+\mu}$ or ${-\mu}$\\\\
Let's calculate the partition function of a single spin:
\begin{equation}
Z_1=\sum_{\mu_i} \exp(\beta\mu_iH)=\exp(\beta\mu H)+\exp(-\beta\mu H)=2\cosh(\beta\mu H)
\end{equation}

Thus, since we neglect interaction between the spins and consider only the interaction with the external field, the partition function for $N$ spins will be:
\begin{equation}
Z=Z_1^N=\left( 2\cosh(\beta\mu H) \right)^N
\end{equation}

In an adiabatic process there is no change in entropy.

The entropy is derived by:
\begin{equation}
\begin{aligned}
S=-\frac{\partial F}{\partial T}=\frac{\partial}{\partial T}Tln(Z)=ln(Z)+T\frac{\partial}{\partial T}ln(Z)=
\\\\=N\left ( ln\left(2\cosh\left(\frac{\mu H}{T}\right) \right)-\frac{\mu H}{T}\tanh\left(\frac{\mu H}{T}\right) \right )
\end{aligned}
\end{equation}

Where we used ${F=-Tln(Z)}$.

Let's take the differential of Eq.(9):
\begin{equation}
0=dS(T,H,N)=\frac{\partial S}{\partial T} dT+\frac{\partial S}{\partial H} dH
\end{equation}
Where in the last equation we assume that $N$ is held fix.

After some algebra, utilizing Eq.(9) in Eq.(10), we arrive at:
\begin{equation}
\frac{dT}{T}=\frac{dH}{H}
\end{equation}

Let's integrate Eq.(11) over the process:
\begin{equation}
\int^{T_f}_{T_i}\frac{dT}{T}=\int^{H_f}_{H_i}\frac{dH}{H} \Rightarrow T_f=\frac{H_f}{H_i}T_i
\end{equation}

So, we conclude that:
\begin{equation}
\Delta T_{adiabatic}=\frac{\Delta H}{H_i}T_i
\end{equation}

We can see, from Eq.(12), that as ${H_f\rightarrow 0; T_f\rightarrow 0}$.\\

This can operate as a refrigerator, in analogy to the carnot cycle applied in gas compression.\\
The process is illustrated in the images below.\\

\begin{figure}[H]
%%\advance\leftskip-0.4cm

  \centering
\includegraphics[width=90mm]{MCE}
 \caption{taken from wikipedia}
\label{fig:method}
\end{figure}

\begin{figure}[H]
%%\advance\leftskip-0.4cm
 \centering
\includegraphics[width=90mm]{fig1}
\caption{taken from newenergyandfuel.com/}
\label{fig:method}
\end{figure}



In realistic systems, there also spin-spin and spin-nuclear interactions, so we cannot reduce the effective magnetic field to zero, so there is a limit on how much we can cool the specimen.\\ For further reading see \textit{Thermal Physics, Kittel, SE, page 346}.
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\end{enumerate}
















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