Minor rewrites and corrections

1 files changed, 153 insertions, 49 deletions

diff --git a/content/know/concept/fabry-perot-cavity/index.pdc b/content/know/concept/fabry-perot-cavity/index.pdc
index d40852f..50b7c62 100644
--- a/content/know/concept/fabry-perot-cavity/index.pdc
+++ b/content/know/concept/fabry-perot-cavity/index.pdc
@@ -14,45 +14,154 @@ markup: pandoc
 # Fabry-Pérot cavity
 
 In its simplest form, a **Fabry-Pérot cavity**
-is a region of light-transmitting medium
-surrounded by two mirrors,
+is a region of light-transmitting medium surrounded by two mirrors,
 which may transmit some of the incoming light.
 Such a setup can be used as e.g. an interferometer or a laser cavity.
 
+Below, we calculate its quasinormal modes in 1D.
+We divide the $x$-axis into three domains: left $L$, center $C$, and right $R$.
+The cavity $C$ has length $\ell$ and is centered on $x = 0$.
+Let $n_L$, $n_C$ and $n_R$ be the respective domains' refractive indices:
 
-## Modes of macroscopic cavity
+<a href="cavity.png">
+<img src="cavity.png" style="width:70%;display:block;margin:auto;">
+</a>
 
-Consider a Fabry-Pérot cavity large enough
-that we can neglect the mirrors' thicknesses,
-which have reflection coefficients $r_L$ and $r_R$.
-Let $\tilde{n}_C$ be the complex refractive index inside,
-and $\tilde{n}_L$ and $\tilde{n}_R$ be the indices outside.
-The cavity has length $L$, centered on $x = 0$.
 
-To find the quasinormal modes,
-we make the following ansatz, with mode number $m$:
+## Microscopic cavity
+
+In its simplest "microscopic" form, the reflection at the boundaries
+is simply caused by the index differences there.
+Consider this ansatz for the [electric field](/know/concept/electric-field/) $E_m(x)$,
+where $m$ is the mode:
+
+$$\begin{aligned}
+    E_m(x)
+    = \begin{cases}
+        A_1 e^{- i k_m n_L x} & \mathrm{for}\; x < -\ell/2 \\
+        A_2 e^{- i k_m n_C x} + A_3 e^{i k_m n_C x} & \mathrm{for}\; \!-\!\ell/2 < x < \ell/2 \\
+        A_4 e^{i k_m n_R x} & \mathrm{for}\; x > \ell/2
+    \end{cases}
+\end{aligned}$$
+
+The goal is to find the modes' wavenumbers $k_m$.
+First, we demand that $E_m$ and its derivative $\dv*{E_m}{x}$
+are continuous at the boundaries $x = \pm \ell/2$:
+
+$$\begin{aligned}
+    A_1 e^{i k_m n_L \ell/2}
+    &= A_2 e^{i k_m n_C \ell/2} + A_3 e^{- i k_m n_C \ell/2}
+    \\
+    A_4 e^{i k_m n_R \ell/2}
+    &= A_2 e^{- i k_m n_C \ell/2} + A_3 e^{i k_m n_C \ell/2}
+\end{aligned}$$
+$$\begin{aligned}
+    - i k_m n_L A_1 e^{i k_m n_L \ell/2}
+    &= - i k_m n_C A_2 e^{i k_m n_C \ell/2} + i k_m n_C A_3 e^{- i k_m n_C \ell/2}
+    \\
+    i k_m n_R A_4 e^{i k_m n_R \ell/2}
+    &= - i k_m n_C A_2 e^{- i k_m n_C \ell/2} + i k_m n_C A_3 e^{i k_m n_C \ell/2}
+\end{aligned}$$
+
+Rearranging the four equations above yields the following linear system:
+
+$$\begin{aligned}
+    0
+    &= A_1 - A_2 e^{i k_m (n_C - n_L) \ell/2} - A_3 e^{- i k_m (n_C + n_L) \ell/2}
+    \\
+    0
+    &= A_2 e^{- i k_m (n_C + n_R) \ell/2} + A_3 e^{i k_m (n_C - n_R) \ell/2} - A_4
+    \\
+    0
+    &= n_L A_1 + n_C \big( A_3 e^{- i k_m (n_C + n_L) \ell/2} - A_2 e^{i k_m (n_C - n_L) \ell/2} \big)
+    \\
+    0
+    &= n_C \big( A_3 e^{i k_m (n_C - n_R) \ell/2} - A_2 e^{- i k_m (n_C + n_R) \ell/2} \big) - n_R A_4
+\end{aligned}$$
+
+Which can be rewritten in matrix form as follows, with the system matrix on the left:
+
+$$\begin{aligned}
+    \begin{bmatrix}
+        1 & -e^{i k_m (n_C - n_L) \ell/2} & -e^{- i k_m (n_C + n_L) \ell/2} & 0 \\
+        0 & e^{- i k_m (n_C + n_R) \ell/2} & e^{i k_m (n_C - n_R) \ell/2} & -1 \\
+        n_L & -n_C e^{i k_m (n_C - n_L) \ell/2} & n_C e^{- i k_m (n_C + n_L) \ell/2} & 0 \\
+        0 & -n_C e^{- i k_m (n_C + n_R) \ell/2} & n_C e^{i k_m (n_C - n_R) \ell/2} & -n_R
+    \end{bmatrix}
+    \cdot
+    \begin{bmatrix}
+        A_1 \\ A_2 \\ A_3 \\ A_4
+    \end{bmatrix}
+    =
+    \begin{bmatrix}
+        0 \\ 0 \\ 0 \\ 0
+    \end{bmatrix}
+\end{aligned}$$
+
+We want non-trivial solutions, where we
+cannot simply satisfy the system by setting $A_1$, $A_2$, $A_3$ and
+$A_4$; this constraint will give us an equation for $k_m$. Therefore, we
+demand that the system matrix is singular, i.e. its determinant is zero:
+
+$$\begin{aligned}
+    0 =
+    &- n_C (n_L + n_R) \big( e^{i k_m (2 n_C - n_L - n_R) \ell/2} + e^{- i k_m (2 n_C + n_L + n_R) \ell/2} \big)
+    \\
+    &+ (n_C^2 + n_L n_R) \big( e^{i k_m (2 n_C - n_L - n_R) \ell/2} - e^{- i k_m (2 n_C + n_L + n_R) \ell/2} \big)
+\end{aligned}$$
+
+We multiply by $e^{i k_m (n_L + n_R) \ell / 2}$ and
+decompose the exponentials into sines and cosines:
+
+$$\begin{aligned}
+    0
+    = i 2 (n_C^2 + n_L n_R) \sin\!(k_m n_C \ell)
+    - 2 n_C (n_L + n_R) \cos\!(k_m n_C \ell)
+\end{aligned}$$
+
+Finally, some further rearranging gives a convenient transcendental equation:
+
+$$\begin{aligned}
+    \boxed{
+        0
+        = \tan\!(k_m n_C \ell) + i \frac{n_C (n_L + n_R)}{n_C^2 + n_L n_R}
+    }
+\end{aligned}$$
+
+Thanks to linearity, we can choose one of the amplitudes
+$A_1$, $A_2$, $A_3$ or $A_4$ freely,
+and then the others are determined by $k_m$ and the field's continuity.
+
+
+## Macroscopic cavity
+
+Next, consider a "macroscopic" Fabry-Pérot cavity
+with complex mirror structures at boundaries, e.g. Bragg reflectors.
+If the cavity is large enough, we can neglect the mirrors' thicknesses,
+and just use their reflection coefficients $r_L$ and $r_R$.
+We use the same ansatz:
 
 $$\begin{aligned}
     E_m(x)
     =
     \begin{cases}
-        A_m \exp\!(-i \tilde{n}_L \tilde{k}_m x) & \mathrm{if}\; x < -L/2 \\
-        B_m \exp\!(i \tilde{n}_C \tilde{k}_m x) + C_m \exp\!(-i \tilde{n}_C \tilde{k}_m x) & \mathrm{if}\; -\!L/2 < x < L/2 \\
-        D_m \exp\!(i \tilde{n}_R \tilde{k}_m x) & \mathrm{if}\; L/2 < x
+        A_1 e^{-i k_m n_L x} & \mathrm{for}\; x < -\ell/2 \\
+        A_2 e^{-i k_m n_C x} + A_3 e^{i k_m n_C x} & \mathrm{for}\; \!-\!\ell/2 < x < \ell/2 \\
+        A_4 e^{i k_m n_R x} & \mathrm{for}\; \ell/2 < x
     \end{cases}
 \end{aligned}$$
 
-On the left, $B_m$ is the reflection of $C_m$,
-and on the right, $C_m$ is the reflection of $B_m$,
-where the reflected amplitude is determined
-by the coefficients $r_L$ and $r_L$, respectively:
+On the left, $A_3$ is the reflection of $A_2$,
+and on the right, $A_2$ is the reflection of $A_3$,
+where the reflected amplitudes are determined
+by the coefficients $r_L$ and $r_R$, respectively:
 
 $$\begin{aligned}
-    B_m \exp\!(-i \tilde{n}_C \tilde{k}_m L/2)
-    &= r_L C_m \exp\!(i \tilde{n}_C \tilde{k}_m L/2)
+    A_3 e^{- i k_m n_C \ell/2}
+    &= r_L A_2 e^{i k_m n_C \ell/2}
     \\
-    C_m \exp\!(-i \tilde{n}_C \tilde{k}_m L/2)
-    &= r_R B_m \exp\!(i \tilde{n}_C \tilde{k}_m L/2)
+    A_2 e^{-i k_m n_C \ell/2}
+    &= r_R A_3 e^{i k_m n_C \ell/2}
 \end{aligned}$$
 
 These equations might seem to contradict each other.
@@ -60,12 +169,12 @@ We recast them into matrix form:
 
 $$\begin{aligned}
     \begin{bmatrix}
-        1 & - r_L \exp\!(i \tilde{n}_C \tilde{k}_m L) \\
-        - r_R \exp\!(i \tilde{n}_C \tilde{k}_m L) & 1
+        1 & - r_R e^{i k_m n_C \ell} \\
+        - r_L e^{i k_m n_C \ell} & 1
     \end{bmatrix}
     \cdot
     \begin{bmatrix}
-        B_m \\ C_m
+        A_2 \\ A_3
     \end{bmatrix}
     =
     \begin{bmatrix}
@@ -73,58 +182,53 @@ $$\begin{aligned}
     \end{bmatrix}
 \end{aligned}$$
 
-Now, we do not want to be able to find values for $B_m$ and $C_m$
-that satisfy this for a given $\tilde{k}_m$.
-Instead, we only want specific values of $\tilde{k}_m$ to be allowed,
-corresponding to the cavity's modes.
-We thus demand that the determinant to zero:
+Again, we demand that the determinant is zero, in order to get non-trivial solutions:
 
 $$\begin{aligned}
     0
-    &= 1 - r_L r_R \exp\!(i 2 \tilde{n}_C \tilde{k}_m L)
+    &= 1 - r_L r_R e^{i 2 k_m n_C \ell}
 \end{aligned}$$
 
-Isolating this for $\tilde{k}_m$ yields the following modes,
+Isolating this for $k_m$ yields the following modes,
 where $m$ is an arbitrary integer:
 
 $$\begin{aligned}
     \boxed{
-        \tilde{k}_m
-        = - \frac{\ln\!(r_L r_R) + i 2 \pi m}{i 2 \tilde{n}_C L}
+        k_m
+        = - \frac{\ln\!(r_L r_R) + i 2 \pi m}{i 2 n_C \ell}
     }
 \end{aligned}$$
 
-These $\tilde{k}_m$ satisfy the matrix equation above.
-Thanks to linearity, we can choose one of $B_m$ or $C_m$,
-and then the other is determined by the corresponding equation.
+These $k_m$ satisfy the matrix equation above.
+Thanks to linearity, we can choose one of $A_2$ or $A_3$,
+and then the other is determined by the corresponding reflection equation.
 
 Finally, we look at the light transmitted through the mirrors,
 according to $1 \!-\! r_L$ and $1 \!-\! r_R$:
 
 $$\begin{aligned}
-    A_m \exp\!(i \tilde{n}_L \tilde{k}_m L/2)
-    &= (1 - r_L) C_m \exp\!(i \tilde{n}_C \tilde{k}_m L/2)
+    A_1 e^{i k_m n_L \ell/2}
+    &= (1 - r_L) A_2 e^{i k_m n_C \ell/2}
     \\
-    D_m \exp\!(i \tilde{n}_R \tilde{k}_m L/2)
-    &= (1 - r_R) B_m \exp\!(i \tilde{n}_C \tilde{k}_m L/2)
+    A_4 e^{i k_m n_R \ell/2}
+    &= (1 - r_R) A_3 e^{i k_m n_C \ell/2}
 \end{aligned}$$
 
-We simply isolate for $A_m$ and $D_m$ respectively,
+We simply isolate for $A_1$ and $A_4$ respectively,
 yielding the following amplitudes:
 
 $$\begin{aligned}
-    A_m
-    &= (1 - r_L) C_m \exp\!\big( i (\tilde{n}_C \!-\! \tilde{n}_L) \tilde{k}_m L/2 \big)
+    A_1
+    &= (1 - r_L) A_2 e^{i k_m (n_C - n_L) \ell/2}
     \\
-    D_m
-    &= (1 - r_R) B_m \exp\!\big( i (\tilde{n}_C \!-\! \tilde{n}_R) \tilde{k}_m L/2 \big)
+    A_4
+    &= (1 - r_R) A_3 e^{i k_m (n_C - n_R) \ell/2}
 \end{aligned}$$
 
 Note that we have not demanded continuity of the electric field.
 This is because the mirrors are infinitely thin "magic" planes;
-if we had instead used a full physical mirror structure,
-then the we would have demanded continuity,
-as you might have expected.
+had we instead used the full mirror structure,
+then we would have demanded continuity, as you maybe expected.