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 $T$?"
??? question "How does $C$ predicted by the Einstein model behave at low $T$?"
...
@@ -104,9 +104,7 @@ $$
...
@@ -104,9 +104,7 @@ $$
$$
$$
Here $v$ is the *sound velocity*.
Here $v$ is the *sound velocity*.
Now instead of $3N$ oscillators with the same frequency we have many oscillators with different frequencies $\omega(k) = v|\mathbf{k}|$.
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:
This makes the total energy equal to the sum over the energies of all the oscillators:
Turning back to our problem of computing heat capacity, we get:
Continuing with our calculation of the total energy, we get:
$$
$$
\begin{align}
\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),\\
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 @@ $$
...
@@ -179,9 +177,9 @@ $$
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.
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
With this definition, our integral becomes
$$
$$
...
@@ -194,20 +192,20 @@ $$
...
@@ -194,20 +192,20 @@ $$
We can trace back all the factors in the density of states to their origin:
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
* $(L/2\pi)^3$ 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 $dθ$
* $4π$ 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
* $ω^2$ is due to the area of this sphere being proportional to its squared radius.
* $3$ is the number of possible polarizations in 3D
* $3$ is the number of possible polarizations in 3D (two transversal, one longitudinal).
* $v^{-3}$ is due to $ω = v|k|$.
* $v^{-3}$ is due to $ω = v|k|$.
## Low $T$
## 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:
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.
Therefore we conclude that $C=\frac{ {\rm d}E}{ {\rm d}T}\propto T^3$.
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.
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 $N_\textrm{modes} \sim (k_B T L/\hbar v)^3$.
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.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.
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 the number of modes by the contribution of each mode we obtain $C\propto k_B (k_B T L/\hbar v)^3$.
4. Multiplying $N_\textrm{modes}$ by the contribution of each mode, we obtain $C\propto k_B (k_B T L/\hbar v)^3$.
