Improve knowledge baseHEADmaster

1 files changed, 12 insertions, 21 deletions

diff --git a/source/know/concept/fermi-gas/index.md b/source/know/concept/fermi-gas/index.md
index 6d316cc..cd3d1a1 100644
--- a/source/know/concept/fermi-gas/index.md
+++ b/source/know/concept/fermi-gas/index.md
@@ -165,47 +165,38 @@ $$\begin{aligned}
     &= \frac{S}{6 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2} \varepsilon_F^{3/2}
 \end{aligned}$$
 
-The total population $$N = L^3 n$$ can therefore be expressed
-as a function of $$\varepsilon_F$$:
-
-$$\begin{aligned}
-    N(\varepsilon_F)
-    &= \frac{S L^3}{6 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2} \varepsilon_F^{3/2}
-\end{aligned}$$
-
-And from this we obtain a formula for the
-[density of states](/know/concept/density-of-states/)
-$$g$$ of a 3D Fermi gas:
+And from this we obtain the [density of states](/know/concept/density-of-states/)
+$$g$$ per unit volume of a 3D Fermi gas:
 
 $$\begin{aligned}
     \boxed{
         g(\varepsilon_F)
-        = \dv{N}{\varepsilon_F}
-        = \frac{S L^3}{4 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2} \varepsilon_F^{1/2}
+        = \dv{n}{\varepsilon_F}
+        = \frac{S}{4 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2} \varepsilon_F^{1/2}
     }
 \end{aligned}$$
 
 Now, $$\varepsilon_F$$ is the highest energy of a single fermion,
-but what about the total $$N$$-particle energy $$E$$?
+but what about the total $$N$$-particle energy $$E$$ of the cube?
 This is easy to calculate using the density of states:
 
 $$\begin{aligned}
-    E
+    \frac{E}{L^3}
     &= \int_0^{\varepsilon_F} \varepsilon \: g(\varepsilon) \dd{\varepsilon}
     \\
-    &= \frac{S L^3}{4 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2}
+    &= \frac{S}{4 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2}
     \int_0^{\varepsilon_F} \varepsilon^{3/2} \dd{\varepsilon}
     \\
-    &= \frac{3}{2} \frac{S L^3}{6 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2} \: \frac{2}{5} \varepsilon_F^{5/2}
+    &= \frac{3}{2} \frac{S}{6 \pi^2} \bigg( \frac{2 m}{\hbar^2} \bigg)^{3/2} \: \frac{2}{5} \varepsilon_F^{5/2}
 \end{aligned}$$
 
-Here, we recognize $$N(\varepsilon_F)$$ from earlier,
-leading to the following expression for the total $$E$$:
+Here, we recognize $$n(\varepsilon_F)$$ from earlier,
+leading to the following total $$E$$ per unit volume:
 
 $$\begin{aligned}
     \boxed{
-        E
-        = \frac{3}{5} N \varepsilon_F
+        \frac{E}{L^3}
+        = \frac{3}{5} n \varepsilon_F
     }
 \end{aligned}$$