Support

Trouble Report

For immediate assistance browse through our support center. You can find answers to many questions in just a few minutes.

If still experiencing problems, send us a report.

required
Why the math question?

Online Simulation

And More

Top Tags

  1. radiation
  2. heat transfer
  3. carbon nanotubes
  4. microscale heat conduction
  5. molecular dynamics
  6. phonons
  7. heat conduction
  8. note
  9. boiling
  10. thermal radiation
  11. radiativeproperties
  12. index of refraction
  13. thermal transport
  14. nanotechnology
  15. optical thickness
  16. acetylene
  17. equation
  18. boiling flow regime
  19. thin films
  20. transport
  21. flow visualization
  22. infrared
  23. microchannels
  24. thermal circuit
  25. two-phase flow

Other

Notes on the quantum of thermal conductance

Table of Contents

Abstract

On this topic page, we derive the quantum of thermal conductance, and along the way, concepts related to phonon density of states and spatial dimensionality are discussed and applied. The end result is a simple expression for how much heat a given acoustic phonon branch, or polarization, can carry between two isothermal reservoirs divided by the temperature difference between the reservoirs.


Energy of a Collection of Phonons

The total energy U of a set of phonons in a solid at equilibrium can be expressed as the sum over all allowable wavevectors K and polarizations p as (see Ashcroft and Mermin and Kittel).

U = \sum_\textbf{K} \sum_p
\left( \langle n_{\textbf{K},p} \rangle + \frac{1}{2}\right) \hbar \omega_{\textbf{K},p} (1)

where \langle n_{\textbf{K},p} \rangle represents number of occupied normal modes (or number of phonons) given by the Planck (or Bose-Einstein) distribution function, and each pair (K, p) represents a so-called normal mode (i.e., a phonon) corresponding to a given wavevector (wavelength) and polarization. The Planck distribution is given by

\langle n_{\textbf{K},p} \rangle = \left[\exp\left(\frac{\hbar \omega_{\textbf{K},p}}{k_B T}\right) - 1 \right]^{-1} (2)

Often, we are more concerned with the energy density (u), rather than the total energy (U), and this new quantity can be calculated simply by dividing Eq. 1 by the system’s ‘volume’. However, we emphasize that the present concept of ‘volume’ depends on the dimensionality of the system, and this point often raises some confusion. Here, let us think of ‘volume’ as length for a 1-D (quantum wire) system, as area for a 2-D (quantum well) system, and as the ordinary volume for a bulk solid.

Our interest is to study transport processes, and as such, the double summation in Eq. 1 is cumbersome. In particular, we’d like to turn the summation over \textbf{K}-space (\emph{i.e.}, over all active phonon wavelengths and directions) into an integral. The summation over polarization branches (e.g., longitudinal-acoustic, transverse-optical, etc.) normally must remain (or else fall victim to approximation). Therefore, we introduce the concept of density of states. This concept is used primarily to convert summations or integrals over K-space to integrals over frequency. Kittel’s textbook (chapter 5) does a particularly nice job of explaining this concept. Here, we will focus on 1-D systems.

In brief, each allowable normal mode in K-space is separated by a distance 2π / L, where L is a characteristic macroscopic length of the medium (in this case a quantum wire) containing the phonons. One subtlety of the foregoing development and its description in some textbooks is that K can take both positive and negative values, and therefore, if we define K as the absolute magnitude of K (i.e., K = |K|), then there is one allowable wavevector for each increment of π / L in K-space (which, by definition, is strictly positive). In other words, the number of allowed phonon states in a 1-D region from 0 to K (which, recall, is the absolute magnitude of K) is

N = \frac{K}{\pi/L} (3)

Once this number of allowed modes or phonons is known, the density of such ‘states’ can be expressed as

D_{1D}(K)= \frac{dN}{dK} = \frac{L}{\pi} (4)

More often, the density of states is described with respect to the phonon frequency instead of wavevector. This transformation is made quite readily using the chain rule and the definition of phonon group velocity

D_{1D}(\omega) = \frac{dN}{d\omega} = \frac{dN}{dK} \frac{dK}{d\omega} = \frac{L}{v_g\pi} (5)

where vg = dω / dK is the phonon group velocity and does, in general, depend on frequency (or wavevector).

Once we have established the density of states, we can convert the K-space summation of Eq. (1) to an integral in energy density form as

u =  \frac{U}{V} =  \frac{1}{V} \sum_{\textbf{K}} \sum_p \left( \langle n_{\textbf{K},p} \rangle + \frac{1}{2}\right) \hbar \omega_{\textbf{K},p}
= \frac{1}{V} \sum_p \int d\textbf{K} D(K) \left( \langle n_{\textbf{K},p} \rangle + \frac{1}{2}\right) \hbar \omega_{\textbf{K},p}
= \frac{1}{V} \sum_p \int_{0}^{\infty}d\omega D(\omega) \left[ n(\omega,T) + \frac{1}{2} \right] \hbar \omega (6)

where we have used n(\omega, T) = \langle n_{\textbf{K},p} \rangle to emphasize this function’s dependence on frequency and temperature (see Eq. 2). We have also dropped the K and p subscripts on ω, recognizing that the frequency of a phonon depends implicitly on its wavevector and polarization. We re-state that the ‘volume’ V can be a length, area, or regular volume, depending on the dimensionality of the system. The foregoing equations actually apply to any dimensionality. For the the 1D systems of interest here, the energy density (i.e., energy per unit length) is

u =  \frac{U}{L} = \frac{1}{L} \sum_p \int_{0}^{\infty}d\omega D_{1D}(\omega) \left[ n(\omega,T) + \frac{1}{2}\right] \hbar \omega
= \sum_p \int_{0}^{\infty}d\omega \frac{1}{v_{g} \pi} \left[ n(\omega,T) + \frac{1}{2}\right] \hbar \omega (7)

The foregoing equations give the phonon energy per unit length in a quantum wire.

Energy Flux

The energy density derived above can be carried by phonons that move through the quantum wire with group velocity vg. Noting that, at a given location on the wire, half of the phonons move leftward while the other half move rightward, we can write the leftward energy flux as

J_{Q,R\rightarrow L} = \frac{1}{2} \sum_p \int_{0}^{\infty}d\omega \frac{1}{v_{g} \pi} v_g \left[ n(\omega,T) + \frac{1}{2}\right] \hbar \omega
= \frac{1}{2 \pi} \sum_p \int_{0}^{\infty}d\omega \left[ n(\omega,T) + \frac{1}{2}\right] \hbar \omega (8)

A similar expression would be used for the rightward heat flux. We note here that while heat flux normally has units of energy flow rate per unit area (e.g., W/m2), the units of JQ,RL above are Watts because the K-space is one-dimensional for the quantum wire. For a general 3D system, the units of u would be energy per volume (e.g., W/m3), and the energy flux would have the usual units of energy flow rate per area.

Schematic of a quantum wire with a device section between two thermal reservoirs.

Our primary purpose is to study the \emph{net} flow of heat through a 1D wire whose ends are held at constant temperatures, as shown in the foregoing figure. Rego and Kirczenow formulated the net flow of heat from right to left as

J_{Q,net} = \frac{1}{2 \pi} \sum_p \int_{0}^{\infty}d\omega \left[ n(\omega, T_R) - n(\omega, T_L) \right] \hbar \omega \Xi_p (\omega (9)

where Ξp(ω) is the probability of transmission of a phonon with frequency ω and polarization p.

Thermal Conductance

Thermal conductance h is defined as the ratio of heat flux to temperature difference. In the present context, it becomes the ratio of $J_{Q,net}$ in Eq.\ \ref{eq:1D_qnet} to the temperature difference ΔT = TRTL. For the simple case of only acoustic phonons (i.e., those with zero frequency at zero wavevector), Rego and Kirczenow showed that for perfect transmission (i.e., Ξp(ω) = 1) the integrals can be evaluated analytically, and the thermal conductance can be expressed as

h_{1D} = \frac{1}{2 \pi} \sum_{p=1}^{n_p} \int_{0}^{\infty}d\omega \left[ \frac{n(\omega, T_R) - n(\omega, T_L)}{\Delta T} \right] \hbar \omega

= \frac{k_{B}^{2} \pi}{6 \hbar} \left( \frac{T_R + T_L}{2} \right) N_p (10)

where Np is simply the number of active phonon polarizations. Therefore, if we assume a small temperature difference between the reservoirs and express their arithmetic average with the symbol T, the following equation defines the quantum of thermal conductance

h_q = \frac{k_B^2 T \pi}{6 \hbar} \left( = T \times 9.462 \times 10^{-13} \frac{W}{K^2} \right) (11)

This important result provides an upper limit on the amount of heat that can flow between two reservoirs connected by a quantum wire by means of a given acoustic phonon polarization. Notably, this upper limit exists even though we have assumed perfect transmission (i.e., Ξp(ω) = 1). In order to increase the thermal conductance, one would need to add more polarizations (Rego and Kirczenow derive results for higher-energy, non-acoustic phonon modes as well) or increase the wire’s cross-sectional area such that the effective \textbf{K}-space becomes two-dimensional and more modes are available to traverse from one side to the other. For our immediate purposes, the purely 1D representation suffices and provides a convenient model system for the study of lattice dynamics and phonon transport.

Last modified on 20 Dec, 2008