Classical reason for good heat conductance in diamond?

Question

My teacher in my physics course attributed this effect to phonons, more here, but I am skeptical about this argument, it feels like he is overlooking the whole question -- what about with Silica that has similar structure to diamond with covalent bonds?

My friend attributed this to the small mass of atoms and large Young constant, considering the system to be made of springs -- $v_{sound}=\sqrt{\frac{Y}{\rho}}$. He suggested to think of heat as sound to use the formula.

Silica has less strong bonds to diamond so perhaps this is a reason why diamondis a good conductor of hear and silica is not. Air is good insulator so does it mean that carbon atoms are so closely packed and silicon atoms not? Is this the reason for the difference in conductance?

Answer 1

I think that there is no really good classical reason. Heat conductivity in a solid depends on bonding in that solid and bonding is inherently a quantum mechanical phenomenon.

For instance, recall that graphite is a good conductor of heat (and electricity) in two dimensions and a fairly good insulator (heatwise and electrically) in the third. This is due to the layered structure in graphite and that is due to the bonding in graphite.

Answer 2

I think the classical explanation for good heat-conductance in diamond is the spring-analogy. But it does not explain all elements such as metals where electron-gas in valence shell contribute to the good heat-conductance. In order to understand this deeper, you need to study the statistics outlined below. The phonon gas occurs between the bonds in diamond while in the electron gas with metals of blocks I/II.

Let's think about classical features such as heat capacity and heat-conductance. As said, they are different in nature depending on the location in periodic table.

I/II blocks in periodic table

Let's consider the first block and second block. The good heat-conductance in metals such as Li, Na and Mg is due to the electron-gas that forms fermi-balls according to Fermi-Dirac statistics. When you experience the heat-change, phonos are the interactive quantitative pieces of energy.

Only the electrons in the highlighted region contribute meaningfully to the specific heat capacity. The source of this picture is the page 166 here, a course booklet in 2062 Physics course in Aalto University.

In order to understand this, you must understand Fermi-Dirac statistics and Fourier Transforms by which you can get the location-frequency-spectrums (in Finnish paikkataajuus spektri): you have distribution of particles and then you use that distribution to deduce the frequency plot.

IV block elements such as Si and C

Now the heat conductance here with structures such as diamond or Silica cannot be due to electron-gas because there is no moving electrons in diamond or silica. The very good heat conductance in diamond is due to mechanical vibration and very strong bonds. Silica has much weaker bonds than carbon atoms with four covalent bonds. Young's modulus is analogical to the mechanical Hooke's constant. Einstein and Debyen explained this so that atoms act like massless springs (page 155 of the earlier lecture slide). Now you can calculate the internal energy with the help of Bose-Einstein statistics.

This plot shows the heat capacities as a function of temperature. The parameter $\theta_D$ is the Debye temperature, $R$ is the gas-constant and dots show real temperatures. The source is page 158 of the earlier lecture slides.

Now in Chemistry, people often use the Maxwell-Boltzman statistics. In Quantum mechanics, they say that Maxwell-Boltzman is like an approximation -- it only work well with very low-temperature or very high temperature -- otherwise it is a bad approximation even though useful in practice but not correct. The statistics Fermi-Dirac and Bose-Einstein are the more realistic description of nature. Depending on the spin-number of atom, you choose the statistics. If your spin is some integer (or zero with modulo 1), you have a boson. If your spin has half (or half with modulo 1), you have a fermion. Examples of fermions contain electron, neutron, proton -- and they all have spin of half.

Let's consider Rubidium 86 (atomic number 37 so 37 electrons, 37 protons and 49 neutrnos) to its isotope 87 (37 electrons, 37 protons and 50 neutrnos). Because 86's spin is odd and 87's spin is even, 86 is fermion while 87 is boson. Now use Boson-Einstein statistics with 87 and Fermi-Dirac with 86.

P.s. You could mathematically approach this problem this way:

1. calculate amount of particles $N$

2. derivate $N$ with respect to $E$ energy
3. find the point when the derivative $\frac{\partial N}{\partial E}$ is zero i.e. $\frac{\partial N}{\partial E}=0$.