Using the Fermi wavevector, we calculate _Fermi wavelength_ $\lambda_F\equiv 2\pi/k_F$ and observe that it is on the order of the atomic spacing for typical free electron densities in metals[^3].
## Extension to finite temperatures
We now extend our discussion to $T > 0$ by taking a close look at the Fermi-Dirac distribution
## Heat capacity
We now extend our discussion to $T > 0$ by taking a closer look at the Fermi-Dirac distribution
We observe that at finite temperature $T>0$, thermal excitations _smear out_ $n_{F}(\beta(\varepsilon-\mu))$.
Below we plot $g(\varepsilon) n_{F}(\beta(\varepsilon-\mu))$ at $T = 0$ (blue + orange shaded area) and $T > 0$ (orange shaded area) alongside the 3D solid density of states to further illustrate the smearing effect.
We observe that at finite temperature $T>0$, thermal excitations _smear out_ the sharp change in the number of occupied electrons near $\varepsilon_F$.
Below we compare the number of occupied electron states at each energy $g(\varepsilon) n_{F}(\beta(\varepsilon-\mu))$ at $T = 0$ (blue shaded area) with $T > 0$ (orange shaded area).
```python
E=np.linspace(0,2,500)
...
...
@@ -358,7 +359,7 @@ ax.set_ylabel(r"$g(ε)$")
ax.set_xlabel(r"$ε$")
draw_classic_axes(ax,xlabeloffset=.2)
```
In order to estimate the fraction of thermally excited electrons, we approximate the blue and orange areas by triangles.
In order to estimate the fraction of thermally excited electrons, we approximate difference between the blue and orange areas by triangles.
The validity of the approximation depends on the ratio between the thermal energy $E_{\mathrm{T}} = k_{B}T$ and the Fermi energy $\varepsilon_{F}$.
If the ratio is large, there is a lot of thermal smearing and the approximation breaks down because the area cannot be accurately approximated by triangles.
However, if the ratio is small, there is barely any thermal smearing, and $n_{F}(\beta(\varepsilon-\mu))$ is approximately a step function.
