15.3: Adiabatic Demagnetization

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 C_B 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.