For single quanta with an arbitrary dispersion relation $\omega(\vec{q})$, you can always write $E=\hbar \omega(\vec{q})$ and $\vec{p}=\hbar\vec{q}$. Energies might depend on momenta in some way, but de Broglie's relations hold for all single quanta, whether they're bosons like phonons or photons, fermions like electrons or holes, or your funny looking bosons. It really doesn't matter!
Back to your example:
We have a 2-dimensional* gas of bosons with a dispersion $\omega(\vec{k}) = A \sqrt{|\vec{k}|}$ at a fixed volume $V$ (well, surface area, but "volume" is more general). To find the heat capacity, we first calculate the density of states, which is where the dispersion relation comes in, then from that and Bose-Einstein statistics we calculate the total energy U:
$g(E) = \frac{V}{(2\pi)^2} \frac{d^{2}\vec{k}}{dE}$
Since our dispersion relation is isotropic (angularly symmetric) this simplifies to
$= \frac{V}{(2\pi)^2} 2\pi|\vec{k}| (\frac{dE}{d|\vec{k}|})^{-1}$
$\frac{dE}{d|\vec{k}|} = \frac{A\hbar}{2\sqrt{|\vec{k}|}}$, so
$g(E) = \frac{V}{(2\pi)^2} \frac{4\pi|\vec{k}|\sqrt{|\vec{k}|}}{A\hbar}$
For convenience, call $k = |\vec{k}|$, $E = \frac{k^{1/2}}{\hbar A}$ and this simplifies to
$g(E) = \frac{V}{\pi\hbar A} k^{3/2} = \frac{V(\hbar A)^2}{\pi} E^3$
Now, to calculate our total internal energy, $U$:
$U(T) = \int \frac{E}{exp(E/k_{B}T)-1} g(E)dE = \frac{V(\hbar A)^2}{\pi} \int \frac{E^4}{exp(E/k_{B}T)-1} dE$
$ = \int \frac{z^4}{e^z-1} dz \frac{V(\hbar A)^2}{\pi} (k_{B}T)^{5} $
$U(T) \approx V\frac{120\zeta(5)\hbar^{2}A^{2}k_{B}^5}{\pi} T^5$
Where I've substituted $z = E/k_{B}T$ to simplify the numerical integral, and taken the limits of the integral to be $0$ and $\infty$, which is a fair approximation at low temperatures. This is just a numerical constant, though, and you can usually ignore this when working it through by hand and just throw it into Maple or your favourite computer algebra system later. The important thing is that at low temperatures, the internal energy is proportional to the volume $V$ and $T^5$.
Said more simply, $U(T) \propto V T^5$.
To get the volumetric heat capacity $C_V(T)$ from the internal energy $U(T)$, we simply differentiate by $T$ and then divide by the volume $V$,
$C_{V} = \frac{1}{V}\frac{dU}{dT} \propto \frac{1}{V}\frac{d V T^5}{dT}$
$C_{V} \propto T^4$
Which was what we wanted!
*(I found this after completing the derivations for both 2D and 3D. You didn't specify which in your original question, but 2D is consistent with the $T^4$ dependence of $C_V$.)
First let me make two comments before answering the question.
The difference between metal and insulator rest in the existence of the itinerant electron Fermi surface or not. Ising (or Heisenberg) model is just an effective theory of local moments (localized electrons in the atoms), which contains no information of the itinerant electron, so there is no hope to start from an Ising model and explain the difference between metal and insulator.
The observation that "ferromagnets are mostly metals, while antiferromagnets are mostly insulators" is not quite true. There are ferromagnetic insulators like $\text{Fe}_3\text{O}_4$, which is one of the earliest ferromagnets discovered in human history. There are also examples of antiferromagnetic metals, from the historical ones like $\text{Cr}$ to the most recent ones like the parent compounds of iron-base superconductors (e.g. $\text{Ba}\text{Fe}_2\text{As}_2$).
The different magnets arise from the different magnetic exchange mechanisms in the material. In the following, some most famous exchange mechanisms in the solid are listed, but as the material can be complicated, so the list is far from complete.
- Ferromagnetic metal: itinerant exchange (RKKY interaction),
- Ferromagnetic insulator: double exchange,
- Antiferromagnetic metal: Fermi suface nesting and SDW instability,
- Antiferromagnetic insulator: superexchange.
In many transition metals (e.g. $\text{Fe}$), the exchange interaction between magnetic ions are mediated by the itinerant (conduction) electrons. The transition metal system contains both the itinerant electrons and the local moments (typically from $d$ orbitals). Local moments just sit on each atom while the itinerant electron travels between the atoms. When the itinerant electron meets the local moment, they mutually polarize each other towards the same orientation. So as the itinerant electrons travel between the atoms, the message of the magnetic orientation is brought from one local moment to another. So eventually all local moments tends to align in the same direction with the itinerant electron, and as the local moment orders, more itinerant electron will be polarized to the ordering orientation to reinforce the ordering. Therefore the ferromagnetism is developed in the metal by this collective behavior. This mechanism is known as the itinerant exchange or the Ruderman-Kittel-Kasuya-Yosida (RKKY) interaction. In the real space, the RKKY interaction between two local moments of the distant $r$ follows the oscillatory behavior
$$J_\text{RKKY}(r)\sim-\frac{\cos(k_F r)}{r^3}.$$
As in the dilute limit $k_F r\ll 1$ for many metals, ferromagnetism will dominate.
Antiferromagetic insulators are usually Mott insulators, in which the Fermi surface is gapped out by interaction, and there is no itinerant electrons. In this case, the magnetic correlation must be mediated by another mechanism, which is known as the superexchange. In the simplest Mott insulator (e.g. $\text{Mn}\text{O}$), each magnetic ion $\text{Mn}^{2+}$ would have a single unpaired electron in a $d$ orbital, which can hop between $\text{Mn}$ sites as bridged by the $\text{O}^{2-}$ ion in between. When the electron spins on $\text{Mn}$ are opposite aligned, it can hybridize over the Mn-O-Mn unit and gain kinetic energy.
However if the electron spins on $\text{Mn}$ are ferromagneitcally aligned, such hybridization will be forbidden by the Pauli exclusion principle. Therefore superexchange favors antiferromagnetism, and the effective exchange interaction is given by $J\sim t^2/U$ where $t$ is the effective hoping integral between $\text{Mn}$ sites, and $U$ is the on-site Coulomb repulsion.
Best Answer
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.