1 files changed, 16 insertions, 11 deletions

diff --git a/source/know/concept/clausius-mossotti-relation/index.md b/source/know/concept/clausius-mossotti-relation/index.md
index 03bdcac..61332db 100644
--- a/source/know/concept/clausius-mossotti-relation/index.md
+++ b/source/know/concept/clausius-mossotti-relation/index.md
@@ -18,19 +18,21 @@ $$\begin{aligned}
 \end{aligned}$$
 
 If there are $$N$$ such bodies per unit volume,
-the polarization density $$\vb{P} = \varepsilon_0 \chi_e \vb{E}$$
-with $$\vb{P} = N \vb{p}$$ suggests that $$\chi_e = N \alpha$$.
-However, this is an underestimation:
+the macroscopic polarization density $$\vb{P} = \varepsilon_0 \chi_e \vb{E}$$
+with $$\vb{P} = N \vb{p}$$ may suggest that $$\chi_e = N \alpha$$.
+However, this turns out to be an underestimation:
 each body's induced dipole creates its own electric field,
 weakening the field felt by its neighbors.
-We need to include this somehow,
-but $$\alpha$$ is defined for a single dipole in a vacuum.
+To calculate $$\chi_e$$ from $$\alpha$$, we need to include this effect,
+but $$\alpha$$ is defined only for a single dipole in a vacuum.
 
-Let $$\vb{E}_\mathrm{int}$$ be the uniform internal field excluding the dipoles' contributions,
-and $$\vb{E}(\vb{r})$$ the net field including them.
+Let $$\vb{E}_\mathrm{int}$$ be the uniform internal field
+excluding the dipoles' contributions,
+and $$\vb{E}(\vb{r})$$ be the net field including them.
 Assume that the dipoles $$\vb{p}_i$$ are arranged
 in a regular crystal lattice at sites $$\vb{R}_i$$.
-Then $$\vb{E}(\vb{r})$$ is the sum of $$\vb{E}_\mathrm{int}$$ and all the dipoles' fields:
+Then $$\vb{E}(\vb{r})$$ is the sum of $$\vb{E}_\mathrm{int}$$
+and all the dipoles' fields:
 
 $$\begin{aligned}
     \vb{E}(\vb{r})
@@ -45,6 +47,7 @@ $$\begin{aligned}
     = - \frac{1}{4 \pi \varepsilon_0} \nabla \bigg( \frac{\vu{r} \cdot \vb{p}_i}{|\vb{r}|^2} \bigg)
 \end{aligned}$$
 
+
 {% include proof/start.html id="proof-dipole" -%}
 The atoms or molecules $$\vb{p}_i$$ need not be perfect dipoles,
 as long as they approximate one when viewed from a distance
@@ -84,6 +87,7 @@ $$\begin{aligned}
 Then the corresponding electric field $$\vb{E}_i$$ is given by $$- \nabla V_i$$ as is well known.
 {% include proof/end.html id="proof-dipole" -%}
 
+
 The dipole $$\vb{p}_0$$ at $$\vb{r} = 0$$
 feels a net local field $$\vb{E}_\mathrm{loc}$$, given below.
 The crystal's symmetry ensures that all its neighbors' fields cancel out:
@@ -101,9 +105,10 @@ $$\begin{aligned}
 Even if there is no regular lattice, this result still holds well enough,
 as long as the dipoles are uniformly distributed over a large volume.
 
-So what was the point of including $$\vb{E}_i(\vb{r})$$ in the first place?
-Well, keep in mind that the sum over neighbors is nonzero for $$\vb{r} \neq \vb{R}_i$$,
-which *does* affect the macroscopic field $$\vb{E}$$, defined as:
+So... if all the neighbors' fields cancel out at $$\vb{r} \in \vb{R}_i$$,
+then what was the point of including $$\vb{E}_i(\vb{r})$$ in the first place?
+Well, those contributions do *not* cancel out for $$\vb{r} \not{\!\!\in} \: \vb{R}_i$$,
+and this fact *does* affect the average macroscopic field $$\vb{E}$$, defined as:
 
 $$\begin{aligned}
     \vb{E}