15.3: Adiabatic Demagnetization
- Page ID
- 7307
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)We are now going to do the same argument for adiabatic demagnetization.
We are going to calculate an expression for \((∂T/∂B)_S\). The expression will be positive, since \(T\) and \(B\) increase together. We shall consider the entropy as a function of temperature and magnetic field, and, with the variables
we shall start with the cyclic relation
\[ \left(\frac{\partial S}{\partial T}\right)_{B}\left(\frac{\partial T}{\partial B}\right)_{S}\left(\frac{\partial B}{\partial S}\right)_{T}=-1. \label{15.3.1}\]
The middle term is the one we want. Let’s find expressions for the first and third partial derivatives in terms of things that we can measure.
In a reversible process \(dS = dQ/T\), and, in a constant magnetic field, \(dQ = C_BdT\). Here I am taking S to mean the entropy per unit volume, and CB is the heat capacity per unit volume (i.e. the heat required to raise the temperature of unit volume by one degree) in a constant magnetic field.
Thus we have \( \left(\frac{\partial S}{\partial T}\right)_{B}=\frac{C_{B}}{T}\).
The Maxwell relation corresponding to \( \left(\frac{\partial S}{\partial P}\right)_{T}=-\left(\frac{\partial V}{\partial T}\right)_{P}\) is \(\left(\frac{\partial S}{\partial B}\right)_{T}=\left(\frac{\partial M}{\partial T}\right)_{B}\). Thus Equation \ref{15.3.1} becomes
\[ \left(\frac{\partial T}{\partial B}\right)_{S}=-\frac{T}{C_{B}}\left(\frac{\partial M}{\partial T}\right)_{B}\].
Now for a paramagnetic material, the magnetization, for a given field, is proportional to B and it falls off inversely as the temperature (that’s the equation of state). That is, M = aB/T. and therefore \( \left(\frac{\partial M}{\partial T}\right)_{B}=-\frac{a B}{T^{2}}=-\frac{M}{T}\). Equation 15.3.2 therefore becomes
\[ \left(\frac{\partial T}{\partial B}\right)_{s}=\frac{M}{C_{B}}.\]
You should check the dimensions of this equation.
The cooling effect is particularly effective at low temperatures, when \(C_B\) is small.