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Commit b654f398 authored by T. van der Sar's avatar T. van der Sar
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Update 2_debye_model.md - polish

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......@@ -31,10 +31,10 @@ _(based on chapter 2.2 of the book)_
Before the start of this lecture, you should be able to:
- Describe how atoms are modeled in the Einstein model
- Derive the low-temperature behaviour of the Einstein heat capacity per atom
- Recall how atoms are modeled in the Einstein model
- Describe a procedure for calculating the heat capacity of a solid
- Describe how the frequency of a sound wave depends on the wavenumber $k=2\pi/\lambda$, where $\lambda$ is the wavelength.
- Formulate a volume integral in spherical coordinates
- Formulate volume integrals using spherical coordinates
!!! summary "Learning goals"
......@@ -71,7 +71,7 @@ T_E = fit[0][0]
fig, ax = pyplot.subplots()
ax.scatter(T, c)
ax.plot(temp, c_einstein(temp, T_E), label=f'Einstein model, $T_E={T_E:.5}K$')
ax.text(T_E, 2, r'$T=T_E$', ha='left', color='r');
ax.text(T_E+3, 1.5, r'$T=T_E$', ha='left', color='r');
ax.plot([T_E, T_E], [0, 3], 'r--')
ax.set_ylim(bottom=0, top=3)
ax.set_xlim(0, 215)
......@@ -79,7 +79,7 @@ ax.set_xlabel('$T(K)$')
ax.set_ylabel(r'$C/k_B$');
```
The einstein model works reasonably well, but it predicts a too small heat capacity at low $T$
The Einstein model works reasonably well, but it predicts a too small heat capacity at low $T$.
??? question "How does $C$ predicted by the Einstein model behave at low $T$?"
......@@ -104,9 +104,7 @@ $$
$$
Here $v$ is the *sound velocity*.
Now instead of $3N$ oscillators with the same frequency we have many oscillators with different frequencies $\omega(k) = v|\mathbf{k}|$.
This makes the total energy equal to the sum over the energies of all the oscillators:
Now instead of $3N$ oscillators with the same frequency we have many oscillators with different frequencies $\omega(k) = v|\mathbf{k}|$. The total energy is given by the sum over the energies of all the oscillators:
$$
E=\sum_\mathbf{k} \left(\frac{1}{2}\hbar\omega(\mathbf{k})+\frac{\hbar\omega(\mathbf{k})}{ {\rm e}^{\hbar\omega(\mathbf{k})/{k_{\rm B}T}}-1}\right)
......@@ -140,8 +138,8 @@ $$
(of course similar periodicity applies to $y$ and $z$ coordinates)
Periodicity means that not all the points in $k$-space are allowed.
Instead only waves with each component $k_x, k_y, k_z$ of the $\mathbf{k}$-vector belonging to a set
$$k=…, \frac{-4\pi}{L}, \frac{-2\pi}{L}, 0, \frac{2\pi}{L}, \frac{4\pi}{L}, …$$
Instead only waves for which each component $k_x, k_y, k_z$ of the $\mathbf{k}$-vector belongs to the set
$$k_{x,y,z}=…, \frac{-4\pi}{L}, \frac{-2\pi}{L}, 0, \frac{2\pi}{L}, \frac{4\pi}{L}, …$$
satisfy the periodic boundary conditions.
In 3D the allowed $k$-vectors form a regular grid:
......@@ -152,13 +150,13 @@ There is therefore exactly one allowed ${\bf k}$ per volume $\left(\frac{2\pi}{L
When we consider larger and larger box sizes $L→∞$, the volume per allowed mode becomes smaller and smaller, and eventually we obtain an integral:
$$
\sum_\mathbf{k} f(\mathbf{k}) ≈ \frac{L^3}{(2\pi)^3}\iiint\limits_{-∞}^{∞}dk_x dk_y dk_z f(\mathbf{k})
\sum_\mathbf{k} \rightarrow \frac{L^3}{(2\pi)^3}\iiint\limits_{-∞}^{∞}dk_x dk_y dk_z )
$$
## Density of states
Turning back to our problem of computing heat capacity, we get:
Continuing with our calculation of the total energy, we get:
$$
\begin{align}
E &= \frac{L^3}{(2\pi)^3}\iiint\limits_{-∞}^{∞}dk_x dk_y dk_z × 3×\left(\frac{1}{2}\hbar\omega(\mathbf{k})+\frac{\hbar\omega(\mathbf{k})}{ {\rm e}^{\hbar\omega(\mathbf{k})/{k_{\rm B}T}}-1}\right),\\
......@@ -179,9 +177,9 @@ $$
In the last expression everything inside the brackets is about Bose-Einstein statistics, while all the prefactors together are specific to the problem we are studying.
We can emphasize this further by introducing a new concept, _density of states_, $g(\omega)$.
We can emphasize this further by introducing a new concept, _density of states_, $g(\omega)$, which is a central concept in this course.
> Density of states $g(ω)$ is the number of available normal modes per infinitesimal interval $δω$.
> The density of states $g(ω)$ is the number of available normal modes per infinitesimal interval $δω$.
With this definition, our integral becomes
$$
......@@ -194,20 +192,20 @@ $$
We can trace back all the factors in the density of states to their origin:
* $(L/2\pi)^3$ is the number of possible waves per unit volume in the reciprocal space
* $4π$ is the area of a unit sphere, the result of integration over $d \varphi$ and $dθ$
* $ω^2$ is due to the area of this sphere being proportional to its squared radius
* $3$ is the number of possible polarizations in 3D
* $(L/2\pi)^3$ is the number of possible waves per unit volume in the reciprocal space.
* $4π$ is the area of a unit sphere, the result of integration over $d \varphi$ and $dθ$.
* $ω^2$ is due to the area of this sphere being proportional to its squared radius.
* $3$ is the number of possible polarizations in 3D (two transversal, one longitudinal).
* $v^{-3}$ is due to $ω = v|k|$.
## Low $T$
In general, ${\rm d}k/{\rm d}\omega$ can be difficult to calculate; we will see more of this later. But going back to the Debye model for now, and using $g(\omega)=V\omega^2/2\pi^2v^3$. The total energy then becomes:
Using $g(\omega)=V\omega^2/2\pi^2v^3$, the total energy becomes:
$$ E=E_{\rm Z}+\frac{3V}{2\pi^2 v_{\rm s}^3}\int\limits_0^\infty\left(\frac{\hbar\omega}{ {\rm e}^{\hbar\omega/k_{\rm B}T}-1}\right)\omega^2{\rm d}\omega$$
Here, the factor 3 comes from the fact that every wave has three polarizations (two transversal, one longitudinal). The term $E_{\rm Z}$ goes to infinity through integration. This is no problem, as it doesn't count towards the heat capacity.
The term $E_{\rm Z}$ goes to infinity through integration. This is no problem, as it doesn't count towards the heat capacity.
Substitute $x\equiv\frac{\hbar\omega}{k_{\rm B}T}$:
$$
......@@ -218,12 +216,12 @@ $$
Therefore we conclude that $C=\frac{ {\rm d}E}{ {\rm d}T}\propto T^3$.
Can we understand this without any calculation terms? Turns out we can!
Can we understand this without calculating any terms? Turns out we can!
1. At temperature $T$ only modes with $\hbar \omega \lesssim k_B T$ get thermally excited.
2. These phonons have wave vectors $|k| \lesssim k_B T /\hbar v$, and therefore their total number is proportional to the volume of a sphere with the same radius times the density of modes in $k$-space. This gives us $N_\textrm{modes} \sim (k_B T L/\hbar v)^3$.
3. Each of these modes is a harmonic oscillator at a reasonably high temperature. Therefore, similarly to the Einstein model, it contributes $\sim k_B$ to the heat capacity.
4. Multiplying the number of modes by the contribution of each mode we obtain $C\propto k_B (k_B T L/\hbar v)^3$.
2. These modes have wave vectors $|k| \lesssim k_B T /\hbar v$. Therefore their total number is proportional to the volume of a sphere with radius $|k|$ multiplied by the density of modes in $k$-space. This gives us $N_\textrm{modes} \sim (k_B T L/\hbar v)^3$.
3. As these modes are thermally excited, they behave like classical harmonic oscillators and contribute $\sim k_B$ to the heat capacity each (similar to the Einstein model).
4. Multiplying $N_\textrm{modes}$ by the contribution of each mode, we obtain $C\propto k_B (k_B T L/\hbar v)^3$.
## Debye's interpolation for medium $T$
......
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