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diff --git a/content/know/concept/capillary-action/index.pdc b/content/know/concept/capillary-action/index.pdc
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+---
+title: "Capillary action"
+firstLetter: "C"
+publishDate: 2021-03-29
+categories:
+- Physics
+- Fluid mechanics
+
+date: 2021-03-07T20:42:28+01:00
+draft: false
+markup: pandoc
+---
+
+# Capillary action
+
+**Capillary action** refers to the movement of liquid
+through narrow spaces due to surface tension, often against gravity.
+It occurs when the [Laplace pressure](/know/concept/young-laplace-law/)
+from surface tension is much larger in magnitude than the
+[hydrostatic pressure](/know/concept/hydrostatic-pressure/) from gravity.
+
+Consider a spherical droplet of liquid with radius $R$.
+The hydrostatic pressure difference
+between the top and bottom of the drop
+is much smaller than the Laplace pressure:
+
+$$\begin{aligned}
+    2 R \rho g \ll 2 \frac{\alpha}{R}
+\end{aligned}$$
+
+Where $\rho$ is the density of the liquid,
+$g$ is the acceleration due to gravity,
+and $\alpha$ is the energy cost per unit surface area.
+Rearranging the inequality yields:
+
+$$\begin{aligned}
+    R^2 \ll \frac{\alpha}{\rho g}
+\end{aligned}$$
+
+From the right-hand side we define the **capillary length** $L_c$,
+so gravity is negligible if $R \ll L_c$:
+
+$$\begin{aligned}
+    \boxed{
+        L_c
+        = \sqrt{\frac{\alpha}{\rho g}}
+    }
+\end{aligned}$$
+
+In general, for a system with characteristic length $L$,
+the relative strength of gravity compared to surface tension
+is described by the **Bond number** $\mathrm{Bo}$
+or **Eötvös number** $\mathrm{Eo}$:
+
+$$\begin{aligned}
+    \boxed{
+        \mathrm{Bo}
+        = \mathrm{Eo}
+        = \frac{L^2}{L_c^2}
+        = \frac{m g}{\alpha L}
+    }
+\end{aligned}$$
+
+The right-most side gives an alternative way of understanding $\mathrm{Bo}$:
+$m$ is the mass of a cube with side $L$, such that the numerator is the weight force,
+and the denominator is the tension force of the surface.
+In any case, capillary action can be observed when $\mathrm{Bo \ll 1}$.
+
+The most famous example of capillary action is **capillary rise**,
+where a liquid "climbs" upwards in a narrow vertical tube with radius $R$,
+apparently defying gravity.
+Assuming the liquid-air interface is a spherical cap
+with constant [curvature](/know/concept/curvature/) radius $R_c$,
+then we know that the liquid is at rest
+when the hydrostatic pressure equals the Laplace pressure:
+
+$$\begin{aligned}
+    \rho g h
+    \approx \alpha \frac{2}{R_c}
+    = 2 \alpha \frac{\cos\theta}{R}
+\end{aligned}$$
+
+Where $\theta$ is the liquid-tube contact angle,
+and we are neglecting variations of the height $h$ due to the curvature
+(i.e. the [meniscus](/know/concept/meniscus/)).
+By isolating the above equation for $h$,
+we arrive at **Jurin's law**,
+which predicts the height climbed by a liquid in a tube with radius $R$:
+
+$$\begin{aligned}
+    \boxed{
+        h
+        = 2 \frac{L_c^2}{R} \cos\theta
+    }
+\end{aligned}$$
+
+Depending on $\theta$, $h$ can be negative,
+i.e. the liquid might descend below the ambient level.
+
+
+An alternative derivation of Jurin's law balances the forces instead of the pressures.
+On the right, we have the gravitational force
+(i.e. the energy-per-distance to lift the liquid),
+and on the left, the surface tension force
+(i.e. the energy-per-distance of the liquid-tube interface):
+
+$$\begin{aligned}
+     \pi R^2 \rho g h
+     \approx 2 \pi R (\alpha_{sg} - \alpha_{sl})
+\end{aligned}$$
+
+Where $\alpha_{sg}$ and $\alpha_{sl}$ are the energy costs
+of the solid-gas and solid-liquid interfaces.
+Thanks to the [Young-Dupré relation](/know/concept/young-dupre-relation/),
+we can rewrite this as follows:
+
+$$\begin{aligned}
+     R \rho g h
+     = 2 \alpha \cos\theta
+\end{aligned}$$
+
+Isolating this for $h$ simply yields Jurin's law again, as expected.
+
+
+
+## References
+1.  B. Lautrup,
+    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+    CRC Press.
diff --git a/content/know/concept/ghz-paradox/index.pdc b/content/know/concept/ghz-paradox/index.pdc
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+---
+title: "GHZ paradox"
+firstLetter: "G"
+publishDate: 2021-03-29
+categories:
+- Physics
+- Quantum mechanics
+- Quantum information
+
+date: 2021-03-29T15:15:41+02:00
+draft: false
+markup: pandoc
+---
+
+# GHZ Paradox
+
+The **Greenberger-Horne-Zeilinger** or **GHZ paradox**
+is an alternative proof of [Bell's theorem](/know/concept/bells-theorem/)
+that does not use inequalities,
+but the three-particle entangled **GHZ state** $\ket{\mathrm{GHZ}}$ instead,
+
+$$\begin{aligned}
+    \boxed{
+        \ket{\mathrm{GHZ}}
+        = \frac{1}{\sqrt{2}} \Big( \ket{000} + \ket{111} \Big)
+    }
+\end{aligned}$$
+
+Where $\ket{0}$ and $\ket{1}$ are qubit states,
+for example, the eigenvalues of the Pauli matrix $\hat{\sigma}_z$.
+
+If we now apply certain products of the Pauli matrices $\hat{\sigma}_x$ and $\hat{\sigma}_y$
+to the three particles, we find:
+
+
+$$\begin{aligned}
+    \hat{\sigma}_x \otimes \hat{\sigma}_x \otimes \hat{\sigma}_x \ket{\mathrm{GHZ}}
+    &= \frac{1}{\sqrt{2}} \Big( \hat{\sigma}_x \ket{0} \otimes \hat{\sigma}_x \ket{0} \otimes \hat{\sigma}_x \ket{0}
+    + \hat{\sigma}_x \ket{1} \otimes \hat{\sigma}_x \ket{1} \otimes \hat{\sigma}_x \ket{1} \Big)
+    \\
+    &= \frac{1}{\sqrt{2}} \Big( \ket{1} \otimes \ket{1} \otimes \ket{1} + \ket{0} \otimes \ket{0} \otimes \ket{0} \Big)
+    = \ket{\mathrm{GHZ}}
+    \\
+    \hat{\sigma}_x \otimes \hat{\sigma}_y \otimes \hat{\sigma}_y \ket{\mathrm{GHZ}}
+    &= \frac{1}{\sqrt{2}} \Big( \hat{\sigma}_x \ket{0} \otimes \hat{\sigma}_y \ket{0} \otimes \hat{\sigma}_y \ket{0}
+    + \hat{\sigma}_x \ket{1} \otimes \hat{\sigma}_y \ket{1} \otimes \hat{\sigma}_y \ket{1} \Big)
+    \\
+    &= \frac{1}{\sqrt{2}} \Big( \ket{1} \otimes i \ket{1} \otimes i \ket{1} + \ket{0} \otimes i \ket{0} \otimes i \ket{0} \Big)
+    = - \ket{\mathrm{GHZ}}
+\end{aligned}$$
+
+In other words, the GHZ state is a simultaneous eigenstate of these composite operators,
+with eigenvalues $+1$ and $-1$, respectively.
+Let us introduce two other product operators,
+such that we have a set of four observables,
+for which $\ket{\mathrm{GHZ}}$ gives these eigenvalues:
+
+$$\begin{aligned}
+    \hat{\sigma}_x \otimes \hat{\sigma}_x \otimes \hat{\sigma}_x
+    \quad &\implies \quad +1
+    \\
+    \hat{\sigma}_x \otimes \hat{\sigma}_y \otimes \hat{\sigma}_y
+    \quad &\implies \quad -1
+    \\
+    \hat{\sigma}_y \otimes \hat{\sigma}_x \otimes \hat{\sigma}_y
+    \quad &\implies \quad -1
+    \\
+    \hat{\sigma}_y \otimes \hat{\sigma}_y \otimes \hat{\sigma}_x
+    \quad &\implies \quad -1
+\end{aligned}$$
+
+According to any local hidden variable (LHV) theory,
+the measurement outcomes of the operators are predetermined,
+and the three particles $A$, $B$ and $C$ can be measured separately,
+or in other words, the eigenvalues can be factorized:
+
+$$\begin{aligned}
+    \hat{\sigma}_x \otimes \hat{\sigma}_x \otimes \hat{\sigma}_x
+    \quad &\implies \quad +1 = m_x^A m_x^B m_x^C
+    \\
+    \hat{\sigma}_x \otimes \hat{\sigma}_y \otimes \hat{\sigma}_y
+    \quad &\implies \quad -1 = m_x^A m_y^B m_y^C
+    \\
+    \hat{\sigma}_y \otimes \hat{\sigma}_x \otimes \hat{\sigma}_y
+    \quad &\implies \quad -1 = m_y^A m_x^B m_y^C
+    \\
+    \hat{\sigma}_y \otimes \hat{\sigma}_y \otimes \hat{\sigma}_x
+    \quad &\implies \quad -1 = m_y^A m_y^B m_x^C
+\end{aligned}$$
+
+Where $m_x^A = \pm 1$ etc.
+Let us now multiply both sides of these four equations together:
+
+$$\begin{aligned}
+    (+1) (-1) (-1) (-1)
+    &= (m_x^A m_x^B m_x^C) (m_x^A m_y^B m_y^C) (m_y^A m_x^B m_y^C) (m_y^A m_y^B m_x^C)
+    \\
+    -1
+    &= (m_x^A)^2 (m_x^B)^2 (m_x^C)^2 (m_y^A)^2 (m_y^B)^2 (m_y^C)^2
+\end{aligned}$$
+
+This is a contradiction: the left-hand side is $-1$,
+but all six factors on the right are $+1$.
+This means that we must have made an incorrect assumption along the way.
+
+Our only assumption was that we could factorize the eigenvalues,
+so that e.g. particle $A$ could be measured on its own
+without an "action-at-a-distance" effect on $B$ or $C$.
+However, because that leads us to a contradiction,
+we must conclude that action-at-a-distance exists,
+and that therefore all LHV-based theories are invalid.
+
+
+
+## References
+1.  N. Brunner,
+    *Quantum information theory: lecture notes*,
+    2019, unpublished.
+2.  J.B. Brask,
+    *Quantum information: lecture notes*,
+    2021, unpublished.
diff --git a/content/know/concept/quantum-gate/index.pdc b/content/know/concept/quantum-gate/index.pdc
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+---
+title: "Quantum gate"
+firstLetter: "Q"
+publishDate: 2021-03-29
+categories:
+- Quantum information
+
+date: 2021-03-29T21:37:57+02:00
+draft: false
+markup: pandoc
+---
+
+
+# Quantum gate
+
+In quantum computing, **quantum gates** are the equivalent
+of classical binary logic gates such as $\mathrm{NOT}$, $\mathrm{AND}$, etc.
+Because of the continuous nature of qubits,
+the number of possible quantum gates is uncountably infinite,
+so we only consider the most important examples here.
+
+
+## One-qubit gates
+
+As an example, consider the following must general single-qubit state $\ket{\psi}$:
+
+$$\begin{aligned}
+    \ket{\psi}
+    = \alpha \ket{0} + \beta \ket{1}
+    = \begin{bmatrix} \alpha \\ \beta \end{bmatrix}
+\end{aligned}$$
+
+Arguably the most famous and/or most fundamental quantum gates are the **Pauli matrices**:
+
+$$\begin{aligned}
+    \boxed{
+        X =
+        \begin{bmatrix}
+            0 & 1 \\
+            1 & 0
+        \end{bmatrix}
+    }
+    \qquad
+    \boxed{
+        Y =
+        \begin{bmatrix}
+            0 & -i \\
+            i & 0
+        \end{bmatrix}
+    }
+    \qquad
+    \boxed{
+        Z =
+        \begin{bmatrix}
+            1 & 0 \\
+            0 & -1
+        \end{bmatrix}
+    }
+\end{aligned}$$
+
+They have the following effect on $\ket{\psi}$.
+Note that $X$ is equivalent to the classical $\mathrm{NOT}$ gate
+(and is often given that name),
+and $Z$ is sometimes called the **phase-flip gate**:
+
+$$\begin{aligned}
+    X \ket{\psi}
+    = \begin{bmatrix} \beta \\ \alpha \end{bmatrix}
+    \qquad
+    Y \ket{\psi}
+    = \begin{bmatrix} -i \beta \\ i \alpha \end{bmatrix}
+    \qquad
+    Z \ket{\psi}
+    = \begin{bmatrix} \alpha \\ -\beta \end{bmatrix}
+\end{aligned}$$
+
+In fact, $Z$ is a specific case of the **phase shift gate** $R_\phi$,
+which modifies the qubit's phase without changing its amplitudes.
+For an angle $\phi$, it is given by:
+
+$$\begin{aligned}
+    \boxed{
+        R_\phi =
+        \begin{bmatrix}
+            1 & 0 \\
+            0 & e^{i \phi}
+        \end{bmatrix}
+    }
+\end{aligned}$$
+
+For $\phi = \pi$, we recover the Pauli-$Z$ gate.
+In general, the action of $R_\phi$ is as follows:
+
+$$\begin{aligned}
+    R_\phi \ket{\psi}
+    = \begin{bmatrix} \alpha \\ e^{i \phi} \beta \end{bmatrix}
+\end{aligned}$$
+
+Two common special cases of $R_\phi$
+are $\phi = \pi/2$ and $\phi = \pi/4$,
+respectively called $S$ and $T$:
+
+$$\begin{aligned}
+    \boxed{
+        S = R_{\pi/2} =
+        \begin{bmatrix}
+            1 & 0 \\
+            0 & i
+        \end{bmatrix}
+    }
+    \qquad \quad
+    \boxed{
+        T = R_{\pi/4} =
+        \frac{1}{\sqrt{2}}
+        \begin{bmatrix}
+            \sqrt{2} & 0 \\
+            0 & 1 + i
+        \end{bmatrix}
+    }
+\end{aligned}$$
+
+Finally, we have the **Hadamard gate** $H$,
+which is defined as follows:
+
+$$\begin{aligned}
+    \boxed{
+        H = \frac{1}{\sqrt{2}}
+        \begin{bmatrix}
+            1 & 1 \\
+            1 & -1
+        \end{bmatrix}
+    }
+\end{aligned}$$
+
+Its action consists of rotating the qubit
+by $\pi$ around the axis $(X + Z) / \sqrt{2}$ of the Bloch sphere:
+
+$$\begin{aligned}
+    H \ket{\psi}
+    = \frac{1}{\sqrt{2}} \begin{bmatrix} \alpha + \beta \\ \alpha - \beta \end{bmatrix}
+\end{aligned}$$
+
+Notably, it maps the eigenstates of $X$ and $Z$ to each other,
+and is its own inverse (i.e. unitary):
+
+$$\begin{aligned}
+    H \ket{0} = \ket{+}
+    \qquad
+    H \ket{1} = \ket{-}
+    \qquad
+    H \ket{+} = \ket{0}
+    \qquad
+    H \ket{-} = \ket{1}
+\end{aligned}$$
+
+The **Clifford gates** are a set including $X$, $Y$, $Z$, $H$ and $S$,
+or more generally any gates that rotate
+by multiples of $\pi/2$ around the Bloch sphere.
+This set is **not universal**, meaning that if we start from $\ket{0}$,
+we can only reach $\ket{0}$, $\ket{1}$, $\ket{+}$, $\ket{-}$, $\ket{+i}$ $\ket{-i}$ using these gates.
+
+If we add *any* non-Clifford gate, for example $T$,
+then we can reach any point on the Bloch sphere,
+which means that the set is **universal**.
+
+However, there is a problem: a qubit has an uncountable infinity of states,
+but a quantum circuit consists of a countably infinite sequence of gates, at most.
+Therefore, technically, we can never reach the whole Bloch sphere,
+but we *can* come up with circuits that approximate a target state to some degree $\varepsilon$.
+This is the definition of universality:
+any state can be approximated.
+
+
+## Two-qubit gates
+
+As an example, let us consider
+the following two pure one-qubit states $\ket{\psi_1}$ and $\ket{\psi_2}$:
+
+$$\begin{aligned}
+    \ket{\psi_1}
+    = \alpha_1 \ket{0} + \beta_1 \ket{1}
+    = \begin{bmatrix} \alpha_1 \\ \beta_1 \end{bmatrix}
+    \qquad \quad
+    \ket{\psi_2}
+    = \alpha_2 \ket{0} + \beta_2 \ket{1}
+    = \begin{bmatrix} \alpha_2 \\ \beta_2 \end{bmatrix}
+\end{aligned}$$
+
+The composite state of both qubits, assuming they are pure,
+is then their tensor product $\otimes$:
+
+$$\begin{aligned}
+    \ket{\psi_1 \psi_2}
+    = \ket{\psi_1} \otimes \ket{\psi_2}
+    &= \alpha_1 \alpha_2 \ket{00} + \alpha_1 \beta_2 \ket{01} + \beta_1 \alpha_2 \ket{10} + \beta_1 \beta_2 \ket{11}
+    \\
+    &= c_{00} \ket{00} + c_{01} \ket{01} + c_{10} \ket{10} + c_{11} \ket{11}
+\end{aligned}$$
+
+Note that a two-qubit system may be [entangled](/know/concept/quantum-entanglement/),
+in which case the coefficients $c_{00}$ etc. cannot be written as products,
+i.e. $\ket{\psi_2}$ cannot be expressed separately from $\ket{\psi_1}$, and vice versa.
+
+In other words, the general action of a two-qubit quantum gate
+can be expressed in the basis of $\ket{00}$, $\ket{01}$, $\ket{10}$ and $\ket{11}$,
+but not always in the basis of $\ket{0}_1$, $\ket{1}_1$, $\ket{0}_2$ and $\ket{1}_2$.
+
+With that said, the first two-qubit gate is $\mathrm{SWAP}$,
+which simply swaps $\ket{\psi_1}$ and $\ket{\psi_2}$:
+
+$$\begin{aligned}
+    \boxed{
+        \mathrm{SWAP} =
+        \begin{bmatrix}
+            1 & 0 & 0 & 0 \\
+            0 & 0 & 1 & 0 \\
+            0 & 1 & 0 & 0 \\
+            0 & 0 & 0 & 1
+        \end{bmatrix}
+    }
+\end{aligned}$$
+
+This matrix is given in the basis of $\ket{00}$, $\ket{01}$, $\ket{10}$ and $\ket{11}$.
+Note that $\mathrm{SWAP}$ cannot generate entanglement,
+so if its input is separable, its output is too.
+In any case, its effect is clear:
+
+$$\begin{aligned}
+    \mathrm{SWAP} \ket{\psi_1 \psi_2}
+    &= c_{00} \ket{00} + c_{10} \ket{01} + c_{01} \ket{10} + c_{11} \ket{11}
+\end{aligned}$$
+
+Next, there is the **controlled NOT gate** $\mathrm{CNOT}$,
+which "flips" (applies $X$ to) $\ket{\psi_2}$ if $\ket{\psi_1}$ is true:
+
+$$\begin{aligned}
+    \boxed{
+        \mathrm{CNOT} =
+        \begin{bmatrix}
+            1 & 0 & 0 & 0 \\
+            0 & 1 & 0 & 0 \\
+            0 & 0 & 0 & 1 \\
+            0 & 0 & 1 & 0
+        \end{bmatrix}
+    }
+\end{aligned}$$
+
+That is, it swaps the last two coefficients $c_{10}$ and $c_{11}$ in the composite state vector:
+
+$$\begin{aligned}
+    \mathrm{CNOT} \ket{\psi_1 \psi_2}
+    &= c_{00} \ket{00} + c_{01} \ket{01} + c_{11} \ket{10} + c_{10} \ket{11}
+\end{aligned}$$
+
+More generally, each one-qubit gate $U$ can be turned into a **controlled** $U$ **gate**:
+
+$$\begin{aligned}
+    \boxed{
+        \mathrm{CU} =
+        \begin{bmatrix}
+            1 & 0 & 0 & 0 \\
+            0 & 1 & 0 & 0 \\
+            0 & 0 & u_{00} & u_{01} \\
+            0 & 0 & u_{10} & u_{11}
+        \end{bmatrix}
+    }
+\end{aligned}$$
+
+Where the lower-right 2x2 block is simply $U$.
+The general action of this gate is given by:
+
+$$\begin{aligned}
+    \mathrm{CU} \ket{\psi_1 \psi_2}
+    &= c_{00} \ket{00} + c_{01} \ket{01} + (c_{10} u_{00} + c_{11} u_{01}) \ket{10} + (c_{10} u_{10} + c_{11} u_{11}) \ket{11}
+\end{aligned}$$
+
+A set of gates is **universal** if all possible mappings
+from $n$ to $n$ qubits can be approximated using only these gates.
+A minimal universal set is $\{\mathrm{CNOT}, T, S\}$,
+and there exist many others.
+
+
+## References
+1.  J.S. Neergaard-Nielsen,
+    *Quantum information: lectures notes*,
+    2021, unpublished.
+2.  S. Aaronson,
+    *Introduction to quantum information science: lecture notes*,
+    2018, unpublished.
diff --git a/content/know/concept/wetting/index.pdc b/content/know/concept/wetting/index.pdc
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+---
+title: "Wetting"
+firstLetter: "W"
+publishDate: 2021-03-29
+categories:
+- Physics
+- Fluid mechanics
+
+date: 2021-03-29T16:20:44+02:00
+draft: false
+markup: pandoc
+---
+
+# Wetting
+
+In fluid statics, **wetting** is the ability
+of a given liquid to touch a given surface.
+When a droplet of the liquid is placed on the surface,
+the **wettability** determines the contact angle $\theta$.
+
+If $\theta = 0$, we have **perfect** or **complete wetting**:
+the droplet spreads out over the entire surface.
+The other extreme is **dewetting** or **non-wetting**,
+where $\theta = \pi$, such that the droplet "floats" on the surface,
+which in the specific case of water is called **hydrophobia**.
+Furthermore, $\theta < \pi/2$ is **high wettability**,
+and $\pi/2 < \theta < \pi$ is **low wettability**.
+
+For a perfectly smooth homogeneous surface,
+$\theta$ is determined by
+the [Young-Dupré relation](/know/concept/young-dupre-relation/):
+
+$$\begin{aligned}
+    \alpha_{sg} - \alpha_{sl}
+    = \alpha_{gl} \cos\theta
+\end{aligned}$$
+
+In practice, however, surfaces can be rough and/or inhomogeneous.
+We start with the former.
+
+A rough surface has some structure, which may contain "gaps".
+There are two options:
+either the droplet fills those gaps (a **Wenzel state**),
+or it floats over them (a **Cassie-Baxter state**).
+
+For a Wenzel state, we define the **roughness ratio** $r$
+as the relative increase of the surface's area due to its rough structure,
+where $A_{real}$ and $A_{app}$ are the real and apparent areas:
+
+$$\begin{aligned}
+    r = \frac{A_{real}}{A_{app}}
+\end{aligned}$$
+
+The net energy cost $E$ of spreading the droplet over the surface is then given by:
+
+$$\begin{aligned}
+    E_{sl}
+    &= (\alpha_{sg} - \alpha_{sl}) A_{real}
+    = \alpha_{gl} A_{real} \cos\theta
+    \\
+    &= \alpha_{gl} A_{app} r \cos\theta
+    = \alpha_{gl} A_{app} \cos\theta^*
+\end{aligned}$$
+
+Where we have defined the **apparent contact angle** $\theta^*$
+as the correction to $\theta$ to account for the roughness.
+It is expressed as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \cos\theta^*
+        = r \cos\theta
+    }
+\end{aligned}$$
+
+For Cassie-Baxter states, where the gaps remain air-filled,
+we define $f$ as the "non-gap" fraction of the apparent surface, such that:
+
+$$\begin{aligned}
+    E
+    &= A_{app} \big( f (\alpha_{sg} - \alpha_{sl}) - (1 - f) \alpha_{gl} \big)
+    \\
+    &= A_{app} \alpha_{gl} \big( f \cos\theta + f - 1 \big)
+\end{aligned}$$
+
+Note the signs: for the solid-liquid interface,
+we "spend" $\alpha_{sg}$ and "get back" $\alpha_{sl}$,
+while for the gas-liquid interface, we spend nothing,
+but get $\alpha_{gl}$.
+The apparent angle $\theta^*$ is therefore:
+
+$$\begin{aligned}
+    \boxed{
+        \cos\theta^*
+        = f (\cos\theta + 1) - 1
+    }
+\end{aligned}$$
+
+We generalize this equation to inhomogeneous surfaces
+consisting of two materials with contact angles $\theta_1$ and $\theta_2$.
+The energy cost of the interface is then given by:
+
+$$\begin{aligned}
+    E
+    &= A \big( f_1 (\alpha_{s1g} - \alpha_{s1l}) + (1 - f_1) (\alpha_{s2g} - \alpha_{s2l}) \big)
+    \\
+    &= A \alpha_{gl} \big( f_1 \cos\theta_1 + (1 - f_1) \cos\theta_2 \big)
+\end{aligned}$$
+
+Such that $\theta^*$ for an inhomogeneous surface is given by this equation,
+called **Cassie's law**:
+
+$$\begin{aligned}
+    \boxed{
+        \cos\theta^*
+        = f_1 \cos\theta_1 + (1 - f_1) \cos\theta_2
+    }
+\end{aligned}$$
+
+Note that the materials need not be solids,
+for example, if one is air, we recover the previous case for rough surfaces.
+Cassie's law can also easily be generalized to three or more materials,
+and to include Wenzel-style roughness ratios $r_1$, $r_2$, etc.
+
diff --git a/content/know/concept/wkb-approximation/index.pdc b/content/know/concept/wkb-approximation/index.pdc
index 985bcec..b8ace29 100644
--- a/content/know/concept/wkb-approximation/index.pdc
+++ b/content/know/concept/wkb-approximation/index.pdc
@@ -29,17 +29,17 @@ $$\begin{aligned}
     m^2 (x')^2 = 2 m (E - V(x))
 \end{aligned}$$
 
-The left-hand side of the rearrangement is simply the momentum squared,
+The left-hand side of the rearranged version is simply the momentum squared,
 so we define the magnitude of the momentum $p(x)$ accordingly:
 
 $$\begin{aligned}
     p(x) = \sqrt{2 m (E - V(x))}
 \end{aligned}$$
 
-Note that this is under the assumption that $E > V$, which is always the
-case in classical mechanics, but not necessarily so in quantum
-mechanics, but we stick with it for now. We rewrite the Schrödinger
-equation:
+Note that this is under the assumption that $E > V$,
+which is always true in classical mechanics,
+but not necessarily in quantum mechanics.
+We rewrite the Schrödinger equation:
 
 $$\begin{aligned}
     0
@@ -172,19 +172,18 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-What if $E < V$? In classical mechanics, this is not allowed; a ball
+What if $E < V$? In classical mechanics, this is just not allowed; a ball
 cannot simply go through a potential bump without the necessary energy.
-However, in quantum mechanics, particles can **tunnel** through barriers.
+On the other hand, in quantum physics, particles can **tunnel** through barriers.
 
-Conveniently, all we need to change for the WKB approximation is to let
-the momentum take imaginary values:
+Luckily, the only thing we need to change for the WKB approximation
+is to let the momentum take imaginary values:
 
 $$\begin{aligned}
     p(x) = \sqrt{2 m (E - V(x))} = i \sqrt{2 m (V(x) - E)}
 \end{aligned}$$
 
-And then take the absolute value in the appropriate place in front of
-$\psi(x)$:
+And then take the absolute value in the appropriate place in front of $\psi(x)$:
 
 $$\begin{aligned}
     \boxed{
@@ -193,9 +192,9 @@ $$\begin{aligned}
 \end{aligned}$$
 
 In the classical region ($E > V$), the wave function oscillates, and
-in the quantum-mechanical region ($E < V$) it is exponential. Note that for
-$E \approx V$ the approximation breaks down, due to the appearance of
-$p(x)$ in the denominator.
+in the quantum-physical region ($E < V$) it is exponential.
+Note that for $E \approx V$ the approximation breaks down,
+because of the appearance of $p(x)$ in the denominator.
 
 
 ## References