Improve knowledge base

1 files changed, 23 insertions, 18 deletions

diff --git a/source/know/concept/bose-einstein-distribution/index.md b/source/know/concept/bose-einstein-distribution/index.md
index e420d7c..5640e69 100644
--- a/source/know/concept/bose-einstein-distribution/index.md
+++ b/source/know/concept/bose-einstein-distribution/index.md
@@ -11,21 +11,22 @@ layout: "concept"
 
 **Bose-Einstein statistics** describe how bosons,
 which do not obey the [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/),
-will distribute themselves across the available states
+distribute themselves across the available states
 in a system at equilibrium.
 
 Consider a single-particle state $$s$$,
 which can contain any number of bosons.
 Since the occupation number $$N$$ is variable,
-we turn to the [grand canonical ensemble](/know/concept/grand-canonical-ensemble/),
-whose grand partition function $$\mathcal{Z}$$ is as follows,
+we use the [grand canonical ensemble](/know/concept/grand-canonical-ensemble/),
+whose grand partition function $$\mathcal{Z}$$ is as shown below,
 where $$\varepsilon$$ is the energy per particle,
-and $$\mu$$ is the chemical potential:
+and $$\mu$$ is the chemical potential.
+We evaluate the sum in $$\mathcal{Z}$$ as a geometric series:
 
 $$\begin{aligned}
     \mathcal{Z}
-    = \sum_{N = 0}^\infty \Big( \exp(- \beta (\varepsilon - \mu)) \Big)^{N}
-    = \frac{1}{1 - \exp(- \beta (\varepsilon - \mu))}
+    = \sum_{N = 0}^\infty \Big( e^{-\beta (\varepsilon - \mu)} \Big)^{N}
+    = \frac{1}{1 - e^{-\beta (\varepsilon - \mu)}}
 \end{aligned}$$
 
 The corresponding [thermodynamic potential](/know/concept/thermodynamic-potential/)
@@ -34,41 +35,45 @@ is the Landau potential $$\Omega$$, given by:
 $$\begin{aligned}
     \Omega
     = - k T \ln{\mathcal{Z}}
-    = k T \ln\!\Big( 1 - \exp(- \beta (\varepsilon - \mu)) \Big)
+    = k T \ln\!\big( 1 - e^{-\beta (\varepsilon - \mu)} \big)
 \end{aligned}$$
 
-The average number of particles $$\Expval{N}$$
-is found by taking a derivative of $$\Omega$$:
+The average number of particles $$\expval{N}$$ in $$s$$
+is then found by taking a derivative of $$\Omega$$:
 
 $$\begin{aligned}
-    \Expval{N}
+    \expval{N}
     = - \pdv{\Omega}{\mu}
     = k T \pdv{\ln{\mathcal{Z}}}{\mu}
-    = \frac{\exp(- \beta (\varepsilon - \mu))}{1 - \exp(- \beta (\varepsilon - \mu))}
+    = \frac{e^{-\beta (\varepsilon - \mu)}}{1 - e^{-\beta (\varepsilon - \mu)}}
 \end{aligned}$$
 
-By multitplying both the numerator and the denominator by $$\exp(\beta(\varepsilon \!-\! \mu))$$,
+By multiplying both the numerator and the denominator by $$e^{\beta(\varepsilon \!-\! \mu)}$$,
 we arrive at the standard form of the **Bose-Einstein distribution** $$f_B$$:
 
 $$\begin{aligned}
     \boxed{
-        \Expval{N}
+        \expval{N}
         = f_B(\varepsilon)
-        = \frac{1}{\exp(\beta (\varepsilon - \mu)) - 1}
+        = \frac{1}{e^{\beta (\varepsilon - \mu)} - 1}
     }
 \end{aligned}$$
 
-This tells the expected occupation number $$\Expval{N}$$ of state $$s$$,
+This gives the expected occupation number $$\expval{N}$$
+of state $$s$$ with energy $$\varepsilon$$,
 given a temperature $$T$$ and chemical potential $$\mu$$.
-The corresponding variance $$\sigma^2$$ of $$N$$ is found to be:
+
+{% comment %}
+The corresponding variance $$\sigma^2 \equiv \expval{N^2} - \expval{N}^2$$ is found to be:
 
 $$\begin{aligned}
     \boxed{
         \sigma^2
-        = k T \pdv{\Expval{N}}{\mu}
-        = \Expval{N} \big(1 + \Expval{N}\big)
+        = k T \pdv{\expval{N}}{\mu}
+        = \expval{N} \big(1 + \expval{N}\!\big)
     }
 \end{aligned}$$
+{% endcomment %}