DEBYE MODEL AND HEAT CAPACITY FOR TWO-ATOM BASIS

Detailed Derivation, Counting of Modes, and Optical Contributions

1. Counting of Modes for Two-Atom Basis

From Kittel (Ch. 4, Fig. 7), if there are $ p $ atoms per primitive cell:

$$ \text{Number of branches} = 3p $$

For $ p = 2 $:

\[ \text{Total branches} = 6 \]

They split as:

3 acoustic branches (1 LA + 2 TA)
3 optical branches (1 LO + 2 TO)

The total number of vibrational modes is $ 6N $, where $ N $ is the number of primitive cells. Since the number of atoms is $ 2N $:

\[ \text{Total degrees of freedom} = 3(2N) = 6N \]

So the counting is consistent.

2. Important Physical Fact (Debye Approximation)

The standard Debye model does not include optical modes.

Why?
Because the Debye model assumes linear dispersion $ \omega = v K $, which implies $ \omega \to 0 $ as $ K \to 0 $. However, optical branches have a non-zero frequency at $ K=0 $ ($ \omega(0) \neq 0 $).

Therefore, the Debye model is strictly an approximation for acoustic modes only. This distinction is critical for a two-atom basis.

3. Density of States in 3D (Derivation)

Apply periodic boundary conditions to a cube of side $ L $. The allowed $ K $-values are:

\[ K_x = \frac{2\pi}{L} n_x \]

The volume per allowed $ K $-point in reciprocal space is $ (2\pi)^3 / V $. The number of states in a spherical shell between $ K $ and $ K+dK $ is:

\[ \frac{V}{(2\pi)^3} 4\pi K^2 dK \]

Thus, the density of states in K-space is:

$$ D(K) dK = \frac{V K^2}{2\pi^2} dK $$

4. Convert to Frequency Space

Using the Debye linear assumption $ \omega = v K $:

\[ K = \frac{\omega}{v} \implies dK = \frac{d\omega}{v} \]

Substituting into $ D(K) $:

\[ D(\omega) d\omega = \frac{V}{2\pi^2} \left(\frac{\omega}{v}\right)^2 \frac{d\omega}{v} \] \[ D(\omega) = \frac{V \omega^2}{2\pi^2 v^3} \]

This is for one polarization. Multiplying by 3 acoustic polarizations:

$$ D(\omega) = \frac{3V \omega^2}{2\pi^2 v^3} $$

5. Debye Cutoff Frequency

We must enforce the correct mode counting. For the Debye model (acoustic only), we count $ 3N $ modes (where $ N $ is the number of primitive cells):

\[ \int_0^{\omega_D} D(\omega) \, d\omega = 3N \]

Substituting the density of states:

\[ \int_0^{\omega_D} \frac{3V \omega^2}{2\pi^2 v^3} d\omega = 3N \] \[ \frac{3V}{2\pi^2 v^3} \frac{\omega_D^3}{3} = 3N \] \[ \omega_D^3 = \frac{6\pi^2 v^3 N}{V} \]

Solving for $ \omega_D $:

$$ \omega_D = v \left( 6\pi^2 \frac{N}{V} \right)^{1/3} $$

6. Debye Temperature

Defined as:

$$ \Theta_D = \frac{\hbar \omega_D}{k_B} $$

Physical Meaning: It is the temperature at which $ k_B T \sim \hbar \omega_D $, meaning the highest frequency acoustic phonons become thermally excited.

7. Total Energy (Acoustic Only)

Using the Bose-Einstein distribution $ n(\omega) = \frac{1}{e^{\hbar\omega/k_B T} - 1} $, the total energy is:

\[ U = \int_0^{\omega_D} D(\omega) \hbar \omega n(\omega) d\omega \]

Substituting $ D(\omega) $:

\[ U = \frac{3V}{2\pi^2 v^3} \int_0^{\omega_D} \frac{\hbar \omega^3}{e^{\hbar\omega/k_B T} - 1} d\omega \]

8. Calculation of U

Let $ x = \frac{\hbar \omega}{k_B T} $. Then $ \omega = \frac{k_B T}{\hbar} x $ and the upper limit becomes $ x_D = \frac{\Theta_D}{T} $.

Substituting and using the definition of $ \omega_D $:

$$ U = 9N k_B T \left(\frac{T}{\Theta_D}\right)^3 \int_0^{x_D} \frac{x^3}{e^x - 1} dx $$

9. Heat Capacity ($ C_V $)

Differentiating $ U $ with respect to $ T $:

$$ C_V = 9N k_B \left(\frac{T}{\Theta_D}\right)^3 \int_0^{x_D} \frac{x^4 e^x}{(e^x - 1)^2} dx $$

10. Low Temperature Limit ($ T \ll \Theta_D $)

As $ T \to 0 $, the upper limit $ x_D \to \infty $. The definite integral evaluates to $ \frac{4\pi^4}{15} $.

$$ C_V = \frac{12\pi^4}{5} N k_B \left(\frac{T}{\Theta_D}\right)^3 $$

This is the famous Debye $ T^3 $ Law. It arises physically because at low temperatures, only long-wavelength acoustic modes are excited, and the phase space volume scales as $ \omega^2 $.

11. High Temperature Limit ($ T \gg \Theta_D $)

Here $ x $ is small, so $ e^x \approx 1+x $. The integral simplifies, leading to:

$$ C_V \to 3N k_B $$

This is the Dulong-Petit Law.

12. What Changes for Two-Atom Basis?

This is the crucial extension. For two atoms per cell, total modes = $ 6N $. The Debye model only accounts for the $ 3N $ acoustic modes.

Optical Contribution

Optical phonons have a flat dispersion near the zone center, so $ \omega \approx \omega_0 $. We can approximate the optical branch as Einstein oscillators:

\[ U_{opt} = 3N \frac{\hbar \omega_0}{e^{\hbar\omega_0/k_B T} - 1} \]

At Low T: The contribution is exponentially suppressed ($ \sim e^{-\hbar\omega_0/k_B T} $). Thus, the $ T^3 $ law still holds.

At High T: The optical modes contribute another $ 3N k_B $, making the total heat capacity $ 6N k_B $. The pure Debye model underestimates this.

13. Modified Debye-Einstein Model

A more accurate model for diatomic crystals is:

$$ C_V = C_{\text{acoustic}}^{\text{Debye}} + C_{\text{optical}}^{\text{Einstein}} $$

14. Physical Interpretation

Acoustic Velocity: Depends on the total mass $ M_1 + M_2 $. Heavier crystals have lower $ \Theta_D $.
Optical Frequency: Depends on the reduced mass. Optical energy scales are typically much higher than acoustic ones.
Why Debye Works: At low T, optical modes are frozen out, so only the acoustic modes (modeled well by Debye) matter. At high T, corrections are needed.

Deep Insight: The Debye model essentially replaces the complex Brillouin zone with a sphere in K-space. This assumption of isotropy and continuum elasticity is valid only for long-wavelength acoustic modes near $ K=0 $.

18nucleons

Search This Blog