Heat Capacity of Ferromagnets and Antiferromagnets

Question

I want to calculates (at least qualitatively) the heat capacities of a ferromagnet and an antiferromagnet, say on a cubic lattice so life is simple. I want to make sure my approach is legit and also ask how to deal with certain regimes.

Ferromagnets:

• Low Temperature $T\ll T_c$: I think in the regimes one uses the dispersion of magnons, $\omega_k = Ak^2$, recognizes these are bosonic quasi-particles and from there it's easy.
• $T\sim T_c$: Here Landau-Ginzburg tells us that phenomenologically $f=tm^2+um^4$, which leads to a discontinuous heat capacity at $T=T_c$.
• High temperature? This is a bit confusing for me. In reality in high temperature a ferromagnet becomes a paramagnet. Then each spin is independent of any other and therefore heat capacity is zero? What if $T>T_c$ but not $\gg T_c$?

Antiferromagnets:

• $T\ll T_c$ is similar to ferromagnets, just the magnons now have two linear branches (instead of one quadratic one)
• I have no idea what happens around $T\sim T_c$.
• Seems to me very high temperature should be the same as ferromagnetic case, but the intermediate temperatures, i.e. $T>T_c$ but not $\gg T_c$, are a complete mystery to me.

Answer 1

Your understanding of the heat capacities in the low temperature ferromagnetic (FM) and antiferromagnetic (AFM) phases are correct. At the critical point, things get tricky because there are critical fluctuations that one has to go beyond the mean-field description and use renormalization groups or scaling arguments to understand the singularity of heat capacity around the critical point. But in the high temperature phase, things get simple again. There is only one single phase in the high-temperature regime: the paramagnetic (PM) phase. In this phase, the spins are disordered and the coupling $J$ is irrelevant in the RG sense, so there are no real distinctions between ferromagnets and antiferromagnets in the PM phase. The heat capacity $C$ always decay with the temperature $T$ as

$$C\sim \frac{NJ^2}{k_B T^2}+\mathcal{O}[T^{-3}],$$

where $J$ is the (effective) spin-spin interaction strength which depends on the microscopic model details.

This form of the specific heat can be argued from general principles of thermodynamics. Let us start from the infinite temperature ($T\to\infty$) limit, where the spins are completely uncorrelated and random. Each individual spin will contribute a finite amount of entropy $s_0$. For example, each Ising spin will give $s_0=k_B \ln 2$ at $T\to\infty$. So the total entropy $S_0=Ns_0$ will be a constant independent of the temperature. As the spins are uncorrelated, the internal energy vanishes as $U_0=0$, so the free energy should be $F_0=U_0-TS_0=-TNs_0$, from which the heat capacity $C_0\equiv -T \partial_T^2 F_0=0$ is zero as expected. Now if we go away from the $T\to\infty$ limit, the free energy should receive some correction order by order from the high-temperature expansion (i.e. we expand the free energy $F(T)$ as a power series of $1/T$):

$$F=F_0+F_1 \left(\frac{J}{k_B T}\right)+F_2 \left(\frac{J}{k_B T}\right)^2+\cdots.$$

Obviously, $J/k_BT$ is the only dimensionless small parameter that controls the high-temperature expansion (because $J$ is the only energy scale in the system to cancle the dimension of $k_BT$). Because the spins are disordered in the PM phase, so $J<0$ (FM coupling) and $J>0$ (AFM coupling) should not make a difference (before the loop diagram appears in the high-temperature expansion). Therefore we know that the linear terms in $J$ must vanish in the free energy, i.e. $F_1=0$. By dimensional analysis, we can show that $F=F_0-aNJ^2/(k_BT)+\cdots$ where $a$ is a dimensionless constant that can be absorbed to the (re)definition of an effective coupling $J$. From this form of the free energy, the heat capacity of the lowest order of $1/T$ in the PM phase can be derived and the result is just as shown above.