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diff --git a/source/know/concept/clausius-mossotti-relation/index.md b/source/know/concept/clausius-mossotti-relation/index.md
new file mode 100644
index 0000000..a0f4916
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+++ b/source/know/concept/clausius-mossotti-relation/index.md
@@ -0,0 +1,264 @@
+---
+title: "Clausius-Mossotti relation"
+sort_title: "Clausius-Mossotti relation"
+date: 2024-04-14
+categories:
+- Physics
+- Electromagnetism
+layout: "concept"
+---
+
+The **polarizability** $$\alpha$$ of a small dielectric body (e.g. an atom or molecule)
+is defined to relate the [electric field](/know/concept/electric-field/) $$\vb{E}$$
+applied to that body to the resulting dipole moment $$\vb{p}$$:
+
+$$\begin{aligned}
+    \vb{p}
+    = \alpha \varepsilon_0 \vb{E}
+\end{aligned}$$
+
+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:
+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.
+
+Let $$\vb{E}_\mathrm{int}$$ be the uniform internal field excluding the dipoles' contributions,
+and $$\vb{E}(\vb{r})$$ 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:
+
+$$\begin{aligned}
+    \vb{E}(\vb{r})
+    = \vb{E}_\mathrm{int} + \sum_{i} \vb{E}_i(\vb{r} - \vb{R}_i)
+\end{aligned}$$
+
+Where the individual contribution $$\vb{E}_i(\vb{r})$$
+of each dipole $$\vb{p}_i$$ is as follows:
+
+$$\begin{aligned}
+    \vb{E}_i(\vb{r})
+    = - \frac{1}{4 \pi \varepsilon_0} \nabla \bigg( \frac{\vu{r} \cdot \vb{p}_i}{|\vb{r}|^2} \bigg)
+\end{aligned}$$
+
+{% 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
+much smaller than the crystal's lattice constant.
+Clearly, in a multipole expansion
+of the true charge distribution $$\rho_i(\vb{r})$$'s electric potential $$V_i(\vb{r})$$,
+the dipole term will be dominant in that case, given by:
+
+$$\begin{aligned}
+    V_i(\vb{r})
+    \approx \frac{1}{4 \pi \varepsilon_0} \frac{1}{|\vb{r}|^2} \int \rho_i(\vb{r}') \: |\vb{r}'| \cos{\theta} \dd{\vb{r}'}
+\end{aligned}$$
+
+Where $$\theta$$ is the angle between $$\vb{r}$$ and $$\vb{r}'$$,
+so this can be rewritten as a dot product
+with the unit vector $$\vu{r}$$, normalized from $$\vb{r}$$:
+
+$$\begin{aligned}
+    V_i(\vb{r})
+    = \frac{1}{4 \pi \varepsilon_0} \frac{1}{|\vb{r}|^2} \: \vu{r} \cdot \!\!\int \vb{r}' \rho_i(\vb{r}') \dd{\vb{r}'}
+\end{aligned}$$
+
+The integral is a more general definition of the dipole moment $$\vb{p}_i$$.
+You can convince yourself of this by defining $$\rho_i(\vb{r})$$
+as two opposite charges $$+q$$ and $$-q$$
+respectively located at $$+\vb{d}_i/2$$ and $$-\vb{d}_i/2$$;
+evaluating the integral then yields $$q \vb{d}_i = \vb{p}_i$$.
+Therefore:
+
+$$\begin{aligned}
+    V_i(\vb{r})
+    = \frac{1}{4 \pi \varepsilon_0} \frac{\vu{r} \cdot \vb{p}_i}{|\vb{r}|^2}
+\end{aligned}$$
+
+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:
+since $$\vb{E}_i(-\vb{r}) = -\vb{E}_i(\vb{r})$$,
+each dipole has a counterpart with the exact opposite field.
+We exclude the singular term $$\vb{E}_0(0)$$:
+
+$$\begin{aligned}
+    \vb{E}_\mathrm{loc}
+    \equiv \vb{E}(0)
+    = \vb{E}_\mathrm{int} + \sum_{i \neq 0} \vb{E}_i(-\vb{R}_i)
+    = \vb{E}_\mathrm{int}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    \vb{E}
+    = \vb{E}_\mathrm{loc}
+    + \frac{1}{\Omega} \int_\Omega \sum_i \vb{E}_i(\vb{r} - \vb{R}_i) \dd{\vb{r}}
+\end{aligned}$$
+
+Where $$\Omega$$ is an arbitrary averaging volume,
+large enough to contain many dipoles,
+and also small enough to assume that the polarization is uniform within.
+Thanks to linearity and symmetry,
+we only need to evaluate this volume integral for a single dipole:
+
+$$\begin{aligned}
+    \int_\Omega \vb{E}_i(\vb{r}) \dd{\vb{r}}
+    &= - \frac{1}{4 \pi \varepsilon_0} \int_\Omega \nabla \bigg( \frac{\vu{r} \cdot \vb{p}_i}{|\vb{r}|^2} \bigg) \dd{\vb{r}}
+    \\
+    &= - \frac{1}{4 \pi \varepsilon_0} \oint_{\partial \Omega} \frac{\vu{r} \cdot \vb{p}_i}{|\vb{r}|^2} \dd{\vu{n}}
+\end{aligned}$$
+
+Where we have used the divergence theorem.
+Let us define $$\Omega$$ as a sphere with radius $$R$$ centered at $$\vb{r} = 0$$.
+In [spherical coordinates](/know/concept/spherical-coordinates/),
+the surface integral then becomes:
+
+$$\begin{aligned}
+    \int_\Omega \vb{E}_i(\vb{r}) \dd{\vb{r}}
+    &= - \frac{1}{4 \pi \varepsilon_0} \int_0^\pi \int_0^{2 \pi}
+    \frac{\vu{r} \cdot \vb{p}_i}{|\vb{r}|^2} \: \vu{r} \: |\vb{r}|^2 \sin{\theta} \dd{\varphi} \dd{\theta}
+\end{aligned}$$
+
+The radial coordinate disappears, so in fact the radius $$R$$ is irrelevant.
+We choose our coordinate system such that $$\vb{p}_i$$ points
+in the positive $$z$$-direction, i.e. $$\vb{p}_i = (0, 0, |\vb{p}_i|)$$, leaving:
+
+$$\begin{aligned}
+    \int_\Omega \vb{E}_i(\vb{r}) \dd{\vb{r}}
+    &= - \frac{|\vb{p}_i|}{4 \pi \varepsilon_0} \int_0^\pi \int_0^{2 \pi}
+    \begin{pmatrix}
+        \sin{\theta} \cos{\varphi} \\
+        \sin{\theta} \sin{\varphi} \\
+        \cos{\theta}
+    \end{pmatrix}
+    \sin{\theta} \cos{\theta} \dd{\varphi} \dd{\theta}
+\end{aligned}$$
+
+Now, consider the following two straightforward indefinite integrals:
+
+$$\begin{aligned}
+    \int \cos{x} \sin^2{x} \dd{x}
+    &= \:\:\,\, \frac{1}{3} \sin^3{x}
+    \\
+    \int \cos^2{x} \sin{x} \dd{x}
+    &= -\frac{1}{3} \cos^3{x}
+\end{aligned}$$
+
+Applying these to our integral over $$\theta$$, the expression is reduced to just:
+
+$$\begin{aligned}
+    \int_\Omega \vb{E}_i(\vb{r}) \dd{\vb{r}}
+    &= \frac{|\vb{p}_i|}{12 \pi \varepsilon_0} \int_0^{2 \pi}
+    \begin{bmatrix}
+        \sin^3{\theta} \cos{\varphi} \\
+        \sin^3{\theta} \sin{\varphi} \\
+        \cos^3{\theta}
+    \end{bmatrix}_0^\pi
+    \dd{\varphi}
+    \\
+    &= - \frac{|\vb{p}_i|}{6 \pi \varepsilon_0} \int_0^{2 \pi}
+    \begin{pmatrix}
+        0 \\
+        0 \\
+        \:1\:
+    \end{pmatrix}
+    \dd{\varphi}
+    \\
+    &= - \frac{\vb{p}_i}{3 \varepsilon_0}
+\end{aligned}$$
+
+Inserting this result into the macroscopic field $$\vb{E}$$,
+and recognizing the second term as the linear polarization density $$\vb{P}$$,
+we arrive at the key result:
+
+$$\begin{aligned}
+    \vb{E}
+    = \vb{E}_\mathrm{loc}
+    - \frac{1}{3 \varepsilon_0} \sum_i \frac{\vb{p}_i}{\Omega}
+    \qquad \implies \qquad
+    \boxed{
+        \vb{E}_\mathrm{loc}
+        = \vb{E} + \frac{1}{3 \varepsilon_0} \vb{P}
+    }
+\end{aligned}$$
+
+Whereas the individual dipoles are polarized by $$\vb{E}_\mathrm{loc}$$,
+the macroscopic polarization density is defined from $$\vb{E}$$,
+so we now know that:
+
+$$\begin{aligned}
+    \vb{P}
+    = \varepsilon_0 \chi_e \vb{E}
+    = \varepsilon_0 N \alpha \vb{E}_\mathrm{loc}
+    = \varepsilon_0 N \alpha \bigg( \vb{E} + \frac{1}{3 \varepsilon_0} \vb{P} \bigg)
+\end{aligned}$$
+
+Isolating this equation for $$\vb{P}$$,
+and using that $$\vb{P} = \varepsilon_0 \chi_e \vb{E}$$,
+we get a more accurate expression for the electric susceptibility $$\chi_e$$:
+
+$$\begin{aligned}
+    \boxed{
+        \chi_e
+        = \frac{N \alpha}{1 - N \alpha / 3}
+    }
+\end{aligned}$$
+
+Notice that for $$N \alpha \ll 1$$ this reduces to our earlier naive estimate:
+if each dipole's neighbors are far away, they do not provide much shielding.
+
+The corresponding macroscopic [dielectric function](/know/concept/dielectric-function/)
+$$\varepsilon_r$$ is given by:
+
+$$\begin{aligned}
+    \boxed{
+        \varepsilon_r
+        = 1 + \chi_e
+        = \frac{3 + 2 N \alpha}{3 - N \alpha}
+    }
+\end{aligned}$$
+
+Experimentally, $$\varepsilon_r$$ can be measured directly, but $$\alpha$$ cannot.
+By isolating $$\varepsilon_r$$ for $$N \alpha$$,
+we finally arrive at the **Clausius-Mossotti relation** for calculating $$\alpha$$:
+
+$$\begin{aligned}
+    \boxed{
+        \frac{1}{3} N \alpha
+        = \frac{\varepsilon_r - 1}{\varepsilon_r + 2}
+    }
+\end{aligned}$$
+
+Since many theoretical models only calculate $$\alpha$$
+(e.g. the [Lorentz oscillator model](/know/concept/lorentz-oscillator-model/)),
+this result is useful for relating theory to experimental results.
+
+
+
+## References
+1.  M. Fox,
+    *Optical properties of solids*, 2nd edition,
+    Oxford.
+2.  D.E. Aspnes,
+    [Local-field effects and effective-medium theory: a microscopic perspective](https://doi.org/10.1119/1.12734),
+    1982, American Journal of Physics 50.
+3.  A. Zangwill,
+    *Modern Electrodynamics*,
+    Cambridge.
+4.  D.J. Griffiths,
+    *Introduction to Electrodynamics*, 5th edition,
+    Cambridge.
diff --git a/source/know/concept/lorentz-oscillator-model/index.md b/source/know/concept/lorentz-oscillator-model/index.md
index 61bbf6b..580ba99 100644
--- a/source/know/concept/lorentz-oscillator-model/index.md
+++ b/source/know/concept/lorentz-oscillator-model/index.md
@@ -61,20 +61,22 @@ The polarization density $$\vb{P}(t)$$ is therefore as shown below,
 where $$N$$ is the number of atoms per unit of volume.
 Note that the dipole moment vector $$\vb{p}$$ is defined
 as pointing from negative to positive,
-whereas the electric field $$\vb{E}$$ goes from positive to negative:
+whereas the electric field $$\vb{E}$$ points from positive to negative:
 
 $$\begin{aligned}
     \vb{P}(t)
-    = N \vb{p}(t)
+    \approx N \vb{p}(t)
     = N q \vb{x}(t)
     = \frac{N q^2}{m (\omega_0^2 - \omega^2 - i \gamma \omega)} \vb{E}(t)
 \end{aligned}$$
 
+Also note that $$\vb{P}$$ is not equal to $$N \vb{p}$$;
+this will be clarified later.
 From the definition of the electric displacement field
 $$\vb{D} = \varepsilon_0 \vb{E} + \vb{P} = \varepsilon_0 \varepsilon_r \vb{E}$$,
-we find that the material's
+we see that the material's
 [dielectric function](/know/concept/dielectric-function/)
-$$\varepsilon_r(\omega)$$ is given by:
+$$\varepsilon_r(\omega)$$ must be given by:
 
 $$\begin{aligned}
     \boxed{
@@ -84,7 +86,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 You may recognize the Drude model's plasma frequency $$\omega_p$$ here,
-but the concept of plasma oscillation does not apply
+but the concept of plasma oscillation does not apply,
 because there are no conduction electrons.
 
 When the light's driving frequency $$\omega$$ is far from the resonance $$\omega_0$$,
@@ -108,23 +110,28 @@ $$\begin{aligned}
 
 In reality, atoms have multiple spectral lines,
 so we should treat them as if they have multiple oscillators
-with different resonances $$\omega_\nu$$.
+with different resonances $$\omega_n$$.
 In that case, the relative permittivity $$\varepsilon_r$$ becomes:
 
-
 $$\begin{aligned}
     \boxed{
         \varepsilon_r(\omega)
-        = 1 + \frac{N q^2}{\varepsilon_0 m} \sum_{\nu} \frac{1}{(\omega_\nu^2 - \omega^2 - i \gamma_\nu \omega)}
+        = 1 + \frac{N q^2}{\varepsilon_0 m} \sum_{n} \frac{1}{(\omega_n^2 - \omega^2 - i \gamma_n \omega)}
     }
 \end{aligned}$$
 
 This gives $$\varepsilon_r$$ the shape of a staircase,
-descending from low to high $$\omega$$ in clear steps at each $$\omega_\nu$$.
+descending from low to high $$\omega$$ in clear steps at each $$\omega_n$$.
 Around each such resonance there is a distinctive "squiggle" in $$\Real\{\varepsilon_r\}$$
 corresponding to a peak in the material's reflectivity,
 and there is an absorption peak in $$\Imag\{\varepsilon_r\}$$.
-The damping from $$\gamma_\nu$$ broadens those peaks and reduces their amplitude.
+The damping from $$\gamma_n$$ broadens those peaks and reduces their amplitude.
+
+Finally, recall that $$\vb{P}$$ was not exactly equal to $$N \vb{p}$$.
+This is because each atomic dipole generates its own electric field,
+affecting the response of its neighbors.
+There exists a formula to correct for this effect:
+the [Clausius-Mossotti relation](/know/concept/clausius-mossotti-relation/).
 
 
 
diff --git a/source/know/concept/step-index-fiber/index.md b/source/know/concept/step-index-fiber/index.md
index 2d049a1..e80480b 100644
--- a/source/know/concept/step-index-fiber/index.md
+++ b/source/know/concept/step-index-fiber/index.md
@@ -147,7 +147,7 @@ and check if we can then arrive at a non-trivial $$\Phi$$ for each case.
         &&= B
     \end{alignedat}$$
 
-    $$B$$ can be nonzero, so this a valid solution.
+    $$B$$ can be nonzero, so this is a valid solution.
     We conclude that $$\ell^2 = 0$$ is the ground state.
 
 * For $$\ell^2 > 0$$, all solutions have the form