= Fourth-power thermal boundary model
{title2=$q=\Sigma(T_h^4-T_c^4)$}
Choose a positive constant $\Sigma$ and the illustrative interface law
$$
q=\Sigma(T_h^4-T_c^4),\qquad
[\Sigma]=\mathrm{W\,m^{-2}\,K^{-4}},\qquad T_h\geq T_c>0.
$$
Here $T_h$ and $T_c$ are the hot and cold interface-side <temperatures>, not distant <thermometer> readings. This is a low-<temperature> <phonon> model under restrictive assumptions, not a universal fit to Kapitza's measured exponent or a material-property table. Its coefficient $\Sigma$ is not the <Stefan-Boltzmann constant>: the carriers in this model are <phonons>, rather than <photons> crossing a vacuum.
The fourth power has a simple counting explanation. In the three-dimensional low-<temperature> <Debye model>, a linear <phonon dispersion relation> gives a <density of states> proportional to $\omega^2$. A <phonon> has <energy> $\hbar\omega$ and <Bose-Einstein distribution> occupation $[\exp(\hbar\omega/k_BT)-1]^{-1}$. If propagation speeds and the angle-averaged transmission introduce no additional frequency scale over the occupied modes, the outgoing <energy> flux has the form
$$
J(T)\propto\int_0^\infty\frac{\omega^3\,d\omega}{e^{\hbar\omega/(k_BT)}-1}
=\left(\frac{k_BT}{\hbar}\right)^4\int_0^\infty\frac{x^3\,dx}{e^x-1}\propto T^4.
$$
Extending the upper integration limit assumes occupied modes lie well below the cutoff. Balancing the two outgoing fluxes, with coefficients satisfying zero net <heat flux> at equal <temperatures>, gives the chosen difference of fourth powers. Frequency-dependent transmission, different excitations or additional transport channels can change this conclusion.
The finite-drop area-specific <thermal resistance> follows by factoring a difference of powers:
$$
r_{K,\rm sec}=\frac{T_h-T_c}{q}
=\frac{1}{\Sigma(T_h+T_c)(T_h^2+T_c^2)}.
$$
Near <thermal equilibrium>, write $T_h=T+\delta T$ and $T_c=T$. Expansion gives $q=4\Sigma T^3\delta T+O((\delta T)^2)$, so the differential area-specific <Kapitza resistance> is
$$
R_K(T)=\lim_{\delta T\to0}\frac{\delta T}{q(T+\delta T,T)}
=\frac{1}{4\Sigma T^3}\quad[\mathrm{K\,m^2\,W^{-1}}],\qquad
\mathcal R_K(T)=\frac{R_K(T)}{A}\quad[\mathrm{K\,W^{-1}}].
$$
At a finite bias, varying only the hot side gives $(\partial q/\partial T_h)^{-1}=1/(4\Sigma T_h^3)$; this is generally different from $r_{K,\rm sec}$. Stating what is held fixed makes a differential resistance reproducible.
\Image[/kapitza-fourth-power-boundary-model.png]
{title=A chosen <fourth-power thermal boundary model> and its differential <Kapitza resistance>}
{id=kapitza-fourth-power-boundary-model}
{description=Original dimensionless calculations. The left curve fixes $T_c=T_0$ and compares exact <heat flux> with its equilibrium tangent. The right curve varies the common equilibrium <temperature> used to define $R_K$. Neither panel contains experimental measurements or fitted <helium> properties.}
{source=kapitza-fourth-power-boundary-model.py}
For an original normalized example, choose reference scales $T_0>0$ and $A_0>0$, and set
$$
\theta_h=\frac{T_h}{T_0}=1.2,\qquad \theta_c=\frac{T_c}{T_0}=1,\qquad
j=\frac{q}{\Sigma T_0^4}=1.2^4-1=1.0736.
$$
For contact <area> $A=2A_0$, the total <heat transfer> rate is $\dot Q=2.1472A_0\Sigma T_0^4$. The finite-drop resistance satisfies $\Sigma T_0^3r_{K,\rm sec}=0.2/1.0736=125/671\approx0.18629$. Using the equilibrium tangent at $T_0$ instead would predict $j=4(0.2)=0.8$, illustrating why a substantial <temperature> rise requires the nonlinear law. Doubling the common equilibrium <temperature> reduces $R_K$ by eight in this chosen model.
A series check can be done without guessing any material properties. Stipulate a hot-side slab with dimensionless areal bulk resistance $b_s=\Sigma T_0^3L_s/k_s=0.10$, and neglect the cold-side bulk drop. At the same $j=1.0736$, the hot reservoir must have $T_H/T_0=1.2+b_sj=1.30736$. The imposed reservoir drop splits into $0.10736$ in the bulk and $0.20$ at the interface. Applying the fourth-power law directly to $T_H$ would assign the bulk drop to the interface and overestimate the <heat flux>.
Finally, transfer <heat> at rate $Aq$ between maintained interface-side <temperatures> $T_h>T_c>0$. The hot side loses <entropy> at rate $Aq/T_h$, and the cold side gains it at rate $Aq/T_c$. With no interface <entropy> accumulation, the <entropy production> per unit <area> is
$$
\frac{\dot S_{\rm prod}}{A}=q\left(\frac1{T_c}-\frac1{T_h}\right)
=\Sigma\frac{(T_h-T_c)^2(T_h+T_c)(T_h^2+T_c^2)}{T_hT_c}>0.
$$
For the normalized example it is $\dot S_{\rm prod}/(A\Sigma T_0^3)=1.0736(1-1/1.2)=0.178933\ldots$. The same <energy> crosses both sides in a <steady state>, while total <entropy> increases, consistently with the <Second law of thermodynamics>. A finite <temperature> jump during <heat transfer> therefore expresses dissipation, not a failure of <energy> conservation.
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