L3 · Chapter 6

Interface thermal resistance: why the temperature jumps where two materials meet

Sandbox Physics · 2026-10-30

When heat flows across the boundary between two solids, the temperature drops abruptly there, even if the contact is perfect. The reason is that phonons can be reflected at the interface: not every phonon on one side can enter the other. The jump divided by the heat flux is the interface thermal resistance, and its inverse the interface conductance. This chapter estimates how many phonons cross with two classic models, computes the conductance between silicon and a soft, heavy film, and points out a problem with the textbook formula that is easy to miss: it predicts a jump even when both sides are the same material.

The temperature jump at an interface

In steady state a heat flux $q$ crosses the interface, and the temperatures on the two sides, extrapolated to the interface, differ by $\Delta T$ . The Wikipedia article on interfacial thermal resistance defines their ratio as the interface resistance; its inverse is the interface conductance $G$ . The article also says most phonon-mediated interfaces have conductances between 80 and 300 MW/(m² K).

An interface resistance acts like an extra slab of material. Converted into a thickness of the same material it is the Kapitza length, $\ell_K=\kappa/G$ . Devices much larger than this can ignore the interface; at comparable sizes the interface becomes the main resistance.

\Delta T=\frac{q}{G},\qquad \ell_K=\frac{\kappa}{G}

The interface resistance is

1/G

, in m² K/W.

A steady heat flux

q

crosses two layers. Inside each layer the temperature falls in a straight line; at the boundary it drops by

\Delta T=q/G

. Extending the first layer’s line until it covers the jump gives the Kapitza length

\ell_K=\kappa/G

, the thickness of material with the same resistance as the interface. Original schematic, not to scale.

Source: Wikipedia: Interfacial thermal resistance · Ioffe Institute

Acoustic mismatch: a perfectly flat interface

The acoustic mismatch model assumes a geometrically perfect interface that phonons cross elastically as waves in a continuum, with no scattering. Waves refract at the interface like light, following the law of refraction, and the transmitted fraction is set by the acoustic impedances $Z=\rho v$ on the two sides: the closer the impedances, the more gets through.

Going from a fast medium into a slow one, waves bend towards the normal and can always enter. Going from slow to fast, waves beyond the critical angle $\arcsin(v_1/v_2)$ are totally reflected. The Lab’s packet tally samples every angle of incidence and should match the formula below within its error bars.

\frac{\sin\theta_2}{v_2}=\frac{\sin\theta_1}{v_1},\qquad \alpha_{12}(\mu_1)=\frac{4Z_1Z_2\,\mu_1\mu_2}{\left(Z_1\mu_2+Z_2\mu_1\right)^{2}}

Here

\mu=\cos\theta

, the cosine of the angle from the interface normal. This is the scalar plane-wave result; real solids also convert between longitudinal and transverse waves, which this model leaves out.

Source: Wikipedia: Interfacial thermal resistance

Diffuse mismatch: an interface that forgets

The diffuse mismatch model goes to the other extreme. The interface is fully rough, and a phonon reaching it scatters at random and forgets which side it came from. The chance of entering either side then depends only on how many states each side offers, not on the angle of incidence, and detailed balance fixes the number: in equilibrium the energy flows in the two directions must be equal.

For two gray Debye solids, the density of states times the speed at a given frequency is proportional to $1/v^2$ , which gives the formula below. Wikipedia says the elastic assumption makes acoustic-mismatch transmission lower, so the acoustic model typically gives an upper limit for the resistance and the diffuse model a lower limit. That is a rule of thumb, not a theorem: for the soft, heavy film in this chapter’s Lab the gray acoustic model gives 95.2 MW/(m² K), a little more conductance than the diffuse model’s 90.9.

Song and Chen tested the forgetting assumption with atomistic calculations and found that a single disordered interface does not make a phonon lose its memory completely, which limits where the diffuse mismatch model applies. Both models are textbook limits, not predictions for any real interface.

\alpha_{12}^{\mathrm{DMM}}=\frac{v_2^{-2}}{v_1^{-2}+v_2^{-2}}

Both sides count only phonons below both cutoff frequencies: silicon phonons above the film’s cutoff cannot enter the film in either model.

Source: Wikipedia: Interfacial thermal resistance · Q. Song and G. Chen

From transmission to conductance

Average the transmission over the incoming flux to get $\Gamma=2\int_0^1\alpha(\mu)\,\mu\,d\mu$ . The interface conductance is the ballistic bound of Chapter 2 times this average, with the heat capacity counting only the phonons that can enter the other side.

With transmission 1 the formula becomes $C_1v_1/4$ , called the radiation limit. It is the same number as the ballistic bound between two black walls in Chapter 2.

G=\frac{v_1}{4}\,\Gamma_{12}\,C_1(\omega<\omega_{D,2}),\qquad \Gamma_{12}=2\int_0^1\alpha_{12}(\mu)\,\mu\,d\mu

Detailed balance makes the conductance computed from side 2 equal to the one from side 1; the Lab computes both and checks one against the other.

Source: Wikipedia: Interfacial thermal resistance · Ioffe Institute

A jump between identical solids? Correcting the textbook formula

Make both sides the same material (the “Matched solids” preset). The transmission is 1 everywhere, the formula gives $G=C_1v_1/4$ , and so there is still a jump. But there is no interface at all, so there should be no jump.

The question is what “temperature” means. The textbook formula compares the temperatures of the phonons heading into the interface: those moving right on the left side and those moving left on the right side. When heat flows, the phonon distribution in a diffusive region is itself lopsided, and phonons heading downstream are a little hotter than the local equilibrium. The first-order distribution of Chapter 2 shows that phonons heading into the interface are $2q/(Cv)$ warmer than the local equilibrium temperature.

Subtracting this on both sides gives the interface resistance in terms of equilibrium temperatures (below), which is exactly zero for identical solids. This step is this course’s own derivation, based on the first-order approximation of Chapter 2, not taken from a reference. The extra wall resistance of 1 in the film of Chapter 3 is the same correction: the black-wall resistance $4/(Cv)$ minus $2/(Cv)$ on the film side, which for two walls adds up to 1 in units of the ballistic resistance. The exact solution gives about 1.06, so the first-order approximation is off by a few percent.

The Lab’s readouts still use the textbook formula, and its model notes say that identical solids keep the radiation-limit jump. That keeps the acoustic and diffuse mismatch values directly comparable with the literature.

R_{\mathrm{eq}}=\frac{1}{G}-\frac{2}{C_1v_1}-\frac{2}{C_2v_2}

For identical sides with transmission 1,

1/G=4/(Cv)

and the corrected

R_{\mathrm{eq}}=0

Source: Wikipedia: Interfacial thermal resistance · A. A. Maznev

Worked example: silicon and a soft, heavy film

Problem: side 1 is gray silicon (sound speed 6351 m/s, density 2.329 g/cm³, Debye heat capacity $1.603\times10^{6}\,\mathrm{J\,m^{-3}\,K^{-1}}$ at 300 K); side 2 is an assumed film with a sound speed of 2 km/s and a density of 11 g/cm³. With the diffuse mismatch model, find the conductance at 300 K, the Kapitza length, and the interface jump in a 1 µm stack with the interface in the middle and 10 K across it.

The Debye temperatures are 696 K and 219 K, and the impedances 14.79 and 22.00 MRayl. The diffuse transmission is $\alpha=2^{-2}/(6.351^{-2}+2^{-2})=0.910$ . The film’s cutoff is low, and only silicon phonons below it can enter; their heat capacity is just $62{,}900\,\mathrm{J\,m^{-3}\,K^{-1}}$ , about 3.9% of the total.

So $G=\tfrac14\times6351\times0.910\times62{,}900=90.9\,\mathrm{MW\,m^{-2}\,K^{-1}}$ , inside the typical range Wikipedia gives and far below the radiation limit of 2545 MW/(m² K). The gray silicon conductivity is 128 W/(m K), so the Kapitza length is $\ell_K=128/(90.9\times10^{6})=1.41\,\mu\mathrm m$ .

The stack carries $q=10\,\mathrm K/(R_1+1/G+R_2)=403\,\mathrm{MW/m^2}$ , and the interface jump is $q/G=4.44\,\mathrm K$ , 44.4% of the total drop. With the corrected definition, $R_{\mathrm{eq}}=1.031\times10^{-8}$ , about 6% below $1/G=1.101\times10^{-8}\,\mathrm{m^2\,K/W}$ , which does not change the conclusion.

Source: Wikipedia: Interfacial thermal resistance · Ioffe Institute · Q. Song and G. Chen

Try it in the Lab

Read the jump from the thermometers

Choose “Soft, heavy film: a clear jump”. In the temperature profile, see the two line fits miss each other at the white line, and read the fitted conductance over the model.
Switch to the acoustic mismatch model and compare the two conductances. Why does the acoustic model give the smaller resistance here?
Set the stack thickness to 10 µm and see what share of the total drop the interface keeps.

The critical angle and the cold interface

Choose “Stiff, light film: a critical angle” and find, in the crossing-chance chart, the angle where transmission falls to zero.
Choose “Matched solids: the radiation limit”, check that the transmission is 1 everywhere, and notice that the readouts still show a jump; explain it with this chapter’s correction.
Choose “Cold interface: the jump takes over”, read the Kapitza length and compare it with the stack thickness.

Exercises with explained answers

Two materials have acoustic impedances of 14.79 and 22.00 MRayl. What does the acoustic mismatch model give for transmission at normal incidence?
Show the answer
$4\times14.79\times22.00/(14.79+22.00)^2=0.962$ . The impedances differ by half, yet 96% of the energy gets through at normal incidence.
What is the acoustic-mismatch critical angle between silicon (6351 m/s) and a stiff film with a sound speed of 12 km/s?
Show the answer
$\arcsin(6.351/12)=32.0^\circ$ . Silicon phonons arriving at larger angles are all reflected, so the average transmission of this interface is only about 0.17.
For the interface in the worked example, thicken the stack to 10 µm with the interface still in the middle. What share of the total drop is the interface jump?
Show the answer
Both layers now have ten times the resistance, $q=10/(3.91\times10^{-8}+1.10\times10^{-8}+9.89\times10^{-8})=67.1\,\mathrm{MW/m^2}$ , and the jump is $q/G=0.739\,\mathrm K$ , or 7.39%. The thicker the stack compared with the Kapitza length, the less the interface matters.
For two identical sides with transmission 1, what interface resistance do the textbook and the corrected formulas give?
Show the answer
The textbook formula gives $1/G=4/(Cv)$ , about $3.9\times10^{-10}\,\mathrm{m^2\,K/W}$ for silicon; the corrected formula subtracts $2/(Cv)$ on each side and gives zero, as it should with no interface.

Sources and what we recomputed

Ioffe Institute, NSM Archive: Thermal properties of Silicon (Si) and Mechanical properties, elastic constants, lattice vibrations of Silicon (Si)
Both pages read in full (the mechanical page is at mechanic.html in the same folder). Values used at 300 K: specific heat 0.7 J/(g K), thermal conductivity 1.3 W/(cm K), thermal diffusivity 0.8 cm²/s, density 2.329 g/cm³, and the sound speeds 8.43 km/s (longitudinal) and 5.84 km/s (transverse) along the [100] direction. The course combines the two speeds into one Debye average; that average is our step, not a value from the pages.
A. A. Maznev, J. A. Johnson and K. A. Nelson, Onset of nondiffusive phonon transport in transient thermal grating decay, Physical Review B 84, 195206 (2011), arXiv:1108.3770v2
Read in full (arXiv v2). Introduction: the kinetic-theory mean free path of tens of nanometres, the gray-body approximation as an oversimplification, the slab between two black walls favoured by theorists, the grating period set by the crossing angle of the pump beams, and grating decay times of about 0.3 to 30 ns in silicon for periods of 1 to 10 µm. The abstract states that the reduction of the effective conductivity in silicon is significant at periods as large as 10 µm. The course quotes no equation numbers from this paper.
Wikipedia, Interfacial thermal resistance, revision 1325816272 (5 December 2025)
This revision read in full. Used here: the definition of the temperature jump, the acoustic and diffuse mismatch models and their assumptions, detailed balance, the statement that the acoustic model typically gives an upper limit and the diffuse model a lower limit for the resistance, and the range of 80 to 300 MW/(m² K) for most phonon-mediated interfaces. Wikipedia is a secondary source; the course uses it for definitions and typical ranges only.
Q. Song and G. Chen, Evaluation of diffuse mismatch model for phonon scattering at disordered interfaces, Physical Review B 104, 085310 (2021), arXiv:2106.04745v2
Partly read: the abstract, the introduction and the opening subsection of the theory section on the assumptions of the diffuse mismatch model and detailed balance. Used here: the model assumes a phonon forgets its origin after diffuse scattering, and the paper finds that a single disordered interface does not erase that memory, which limits the model. The atomistic Green’s-function results and later sections were not read and are not quoted.