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Commit f27269a7 authored by Bowy La Riviere's avatar Bowy La Riviere
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Updates degeneracy error and latex error

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......@@ -87,7 +87,7 @@ $$
2.
$$
E = \frac{4}{5} (5.2)^{\frac{5/2}} 10^{10} eV
E = \frac{4}{5} (5.2)^{\frac{5}{2}} 10^{10} eV
$$
3.
......@@ -170,12 +170,12 @@ ax.xaxis.set_label_coords(1.0, 0.49)
2.The DOS for the positive energies is given by
$$
g(\varepsilon) = (2_s + 2_v) 2 \pi \left(\frac{L}{2 \pi}\right)^2 \frac{\varepsilon}{c^2},
g(\varepsilon) = (2_s \dot 2_v) 2 \pi \left(\frac{L}{2 \pi}\right)^2 \frac{\varepsilon}{c^2},
$$
where $2_s$ is the spin degeneracy and $2_v$ is the valley degeneracy.
If we account for the negative energies as well, we obtain
$$
g(\varepsilon) = (2_s + 2_v) 2 \pi \left(\frac{L}{2 \pi}\right)^2 \frac{|\varepsilon|}{c^2}.
g(\varepsilon) = (2_s \dot 2_v) 2 \pi \left(\frac{L}{2 \pi}\right)^2 \frac{|\varepsilon|}{c^2}.
$$
3.$g(\varepsilon)$ vs $\varepsilon$ is a linear plot. Here, the region marked by $-k_B T$ is a triangle whose area gives the number of electrons that can be excited:
......
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