Expand knowledge base

10 files changed, 583 insertions, 5 deletions

diff --git a/content/know/concept/bb84-protocol/index.pdc b/content/know/concept/bb84-protocol/index.pdc
index 9f7c7db..32d3666 100644
--- a/content/know/concept/bb84-protocol/index.pdc
+++ b/content/know/concept/bb84-protocol/index.pdc
@@ -231,4 +231,9 @@ which is equivalent to her guessing at random.
 
 
 ## References
-1.  N. Brunner, *Quantum information theory: lecture notes*, 2019.
+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+    2019, unpublished.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+    2021, unpublished.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/bloch-sphere/index.pdc b/content/know/concept/bloch-sphere/index.pdc
index 843de1f..f0c48a9 100644
--- a/content/know/concept/bloch-sphere/index.pdc
+++ b/content/know/concept/bloch-sphere/index.pdc
@@ -127,3 +127,12 @@ $$\begin{aligned}
     = \frac{1}{2} \Tr\big( r_z \hat{I} \big)
     = r_z
 \end{aligned}$$
+
+
+## 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+    2019, unpublished.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+    2021, unpublished.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/curvature/index.pdc b/content/know/concept/curvature/index.pdc
new file mode 100644
index 0000000..50c61b2
--- /dev/null
+++ b/content/know/concept/curvature/index.pdc
@@ -0,0 +1,395 @@
+---
+title: "Curvature"
+firstLetter: "C"
+publishDate: 2021-03-07
+categories:
+- Mathematics
+
+date: 2021-03-04T09:46:31+01:00
+draft: false
+markup: pandoc
+---
+
+# Curvature
+
+Given a curve or surface, its **curvature** $\kappa$
+describes how sharply it is bending at a given point.
+It is defined as the inverse of the **radius of curvature** $R$,
+which is the radius of the tangent circle
+that **osculates** (i.e. best approximates)
+the curve/surface at that point:
+
+$$\begin{aligned}
+    \kappa = \frac{1}{R}
+\end{aligned}$$
+
+Typically, $\kappa$ is positive for convex curves/surfaces,
+and negative for concave ones, although this distinction is somewhat arbitrary.
+Below, we calculate the curvature in several general cases.
+
+
+## 2D height functions
+
+We start with a specialized case: height functions,
+where one coordinate is a function of the other one (2D) or two (3D).
+In this case, we can use the
+[calculus of variations](/know/concept/calculus-of-variations/)
+to find the curvature.
+
+This approach relies on the fact that a circle
+has the highest area-perimeter ratio of any 2D shape,
+and a sphere has the highest volume-surface ratio of any 3D body.
+By the definition of curvature, these shapes have constant $\kappa$.
+
+We will thus minimize the perimeter/surface while keeping the area/volume fixed,
+which will give us a shape with constant curvature,
+and from that we can extrapolate an expression for $\kappa$.
+
+In 2D, for a single-variable height function $h(x)$,
+the length of a small segment of the curve is:
+
+$$\begin{aligned}
+    \sqrt{\dd{x}^2 + \dd{h}^2}
+    = \dd{x} \sqrt{\Big( \dv{x}{x} \Big)^2 + \Big( \dv{h}{x} \Big)^2}
+    = \dd{x} \sqrt{1 + h_x^2}
+\end{aligned}$$
+
+Which leads us to define the following Lagrangian $\mathcal{L}$
+describing the "energy cost" of the curve:
+
+$$\begin{aligned}
+    \mathcal{L}
+    = \sqrt{1 + h_x^2}
+\end{aligned}$$
+
+Furthermore,
+we demand that the area under the curve (i.e. the "volume") is constant:
+
+$$\begin{aligned}
+    V
+    = \int_{x_0}^{x_1} h(x) \dd{x}
+\end{aligned}$$
+
+By putting these things together,
+we arrive at the following energy functional $E[h]$,
+where $\kappa$ is an ominously-named [Lagrange multiplier](/know/concept/lagrange-multiplier/):
+
+$$\begin{aligned}
+    E[h]
+    = \int (\mathcal{L} + \kappa h) \dd{x}
+\end{aligned}$$
+
+Minimizing this functional leads to the following
+Lagrange equation of the first kind:
+
+$$\begin{aligned}
+    0
+    = \pdv{\mathcal{L}}{h} - \dv{x} \Big( \pdv{\mathcal{L}}{h_x} \Big) + \kappa
+\end{aligned}$$
+
+We evaluate the terms of this equation
+to arrive at an expression for the curvature $\kappa$:
+
+$$\begin{aligned}
+    \boxed{
+        \kappa
+        = \frac{h_{xx}}{\big(1 + h_x^2\big)^{3/2}}
+    }
+\end{aligned}$$
+
+In this optimization problem, $\kappa$ is a constant,
+but in fact the statement above is valid for variable curvatures too,
+in which case $\kappa$ is a function of $x$.
+
+
+## 2D in general
+
+We can parametrically describe an arbitrary plane curve
+as a function of the arc length $s$:
+
+$$\begin{aligned}
+    \big( x(s), y(s) \big)
+    \qquad \mathrm{where} \qquad
+    \dd{s}^2 = \dd{x}^2 + \dd{y}^2
+\end{aligned}$$
+
+If we choose the horizontal $x$-axis as a reference,
+we can furthermore define the **elevation angle** $\theta(s)$
+as the angle between the reference and the curve's tangent vector $\vu{t}$:
+
+$$\begin{aligned}
+    \vu{t}
+    = \big( x_s(s), y_s(s) \big)
+    = \big( \cos\theta(s), \sin\theta(s) \big)
+\end{aligned}$$
+
+Where $x_s(s) = \dv*{x}{s}$.
+The curvature $\kappa$ is defined as
+the $s$-derivative of this elevation angle:
+
+$$\begin{aligned}
+    \kappa
+    = \dv{\theta}{s}
+    = \theta_s(s)
+\end{aligned}$$
+
+We have two ways of writing $\vu{t}$:
+using the derivatives $x_s$ and $y_s$,
+or the elevation angle $\theta$.
+Now, let us take the $s$-derivative of both expressions,
+and equate them:
+
+$$\begin{aligned}
+    \big( x_{ss}, y_{ss} \big)
+    = \dv{\vu{t}}{s}
+    = \theta_s \: \big( \!-\!\sin\theta, \cos\theta \big)
+    = \kappa \big( \!-\!y_s, x_s \big)
+\end{aligned}$$
+
+$$\begin{aligned}
+    x_{ss} = - \kappa y_s
+    \qquad
+    y_{ss} = \kappa x_s
+\end{aligned}$$
+
+We multiply these equation by $y_s$ and $x_s$, respectively,
+and subtract the first from the last:
+
+$$\begin{aligned}
+    y_{ss} x_s - x_{ss} y_s = \kappa x_s^2 + \kappa y_s^2
+\end{aligned}$$
+
+Isolating this for $\kappa$ and using the fact that $x_s^2 + y_s^2 = 1$
+thanks to $s$ being the arc length:
+
+$$\begin{aligned}
+    \kappa
+    = \frac{y_{ss} x_s - x_{ss} y_s}{x_s^2 + y_s^2}
+    = y_{ss} x_s - x_{ss} y_s
+\end{aligned}$$
+
+While this result is correct,
+we would like to generalize it to cases where the curve
+is parametrized by some other $t$, not necessarily the arc length.
+Let prime denote the $t$-derivative:
+
+$$\begin{aligned}
+    x_s
+    = x' t_s
+    \qquad
+    x_{ss}
+    = x'' t_s^2 + x' t_{ss}
+    \\
+    y_s
+    = y' t_s
+    \qquad \:
+    y_{ss}
+    = y'' t_s^2 + x' t_{ss}
+\end{aligned}$$
+
+By inserting these expression into the earlier formula for $\kappa$, we find:
+
+$$\begin{aligned}
+    \kappa
+    = y_{ss} x_s - x_{ss} y_s
+    &= x' t_s (y'' t_s^2 + y' t_{ss}) - y' t_s (x'' t_s^2 + x' t_{ss})
+    \\
+    &= t_s t_{ss} (x' y' - y' x') + t_s^3 (x' y'' - y' x'')
+    \\
+    &= t_s^3 (x' y'' - y' x'')
+\end{aligned}$$
+
+Since $x_s^2 + y_s^2 = 1$, we know that $(x')^2 + (y')^2 = 1 / t_s^2$,
+which leads us to the following general expression for
+the curvature $\kappa$ of a plane curve:
+
+$$\begin{aligned}
+    \boxed{
+        \kappa
+        = \frac{y'' x' - x'' y'}{\big((x')^2 + (y')^2\big)^{3/2}}
+    }
+\end{aligned}$$
+
+If the curve happens to be a height function, i.e. $y(x)$,
+then $x' = 1$ and $x'' = 0$, and we arrive at our previous result again.
+
+
+## 3D height functions
+
+The generalization to a 3D height function $h(x, y)$ is straightforward:
+the cost of an infinitesimal portion of the surface is as follows,
+using the same reasoning as before:
+
+$$\begin{aligned}
+    \mathcal{L}
+    = \sqrt{1 + h_x^2 + h_y^2}
+\end{aligned}$$
+
+Keeping the volume $V$ constant,
+we get the following energy functional $E$ to minimize:
+
+$$\begin{aligned}
+    E[h]
+    = \iint (\mathcal{L} + \lambda h) \dd{x} \dd{y}
+\end{aligned}$$
+
+Which gives us an Euler-Lagrange equation
+involving the Lagrange multiplier $\lambda$:
+
+$$\begin{aligned}
+    0
+    = \pdv{\mathcal{L}}{h} - \dv{x} \Big( \pdv{\mathcal{L}}{h_x} \Big) - \dv{y} \Big( \pdv{\mathcal{L}}{h_y} \Big) + \lambda
+\end{aligned}$$
+
+Inserting $\mathcal{L}$ into this and evaluating all the derivatives
+yields a result for the (variable) curvature:
+
+$$\begin{aligned}
+    \boxed{
+        \lambda
+        = \kappa_1 + \kappa_2
+        = \frac{(1 + h_y^2) h_{xx} - 2 h_x h_y h_{xy} + (1 + h_x^2) h_{yy}}{\big(1 + h_x^2 + h_y^2\big)^{3/2}}
+    }
+\end{aligned}$$
+
+What are $\kappa_1$ and $\kappa_2$?
+Well, the problem in 3D is that the curvature of an osculating circle
+depends on the orientation of that circle.
+The **principal curvatures** $\kappa_1$ and $\kappa_2$
+are the largest and smallest curvatures at a given point,
+but finding their values and the corresponding **principal directions** is not so easy.
+Fortunately, in practice, we are often only interested in their sum:
+
+$$\begin{aligned}
+    \lambda
+    = \kappa_1 + \kappa_2
+    = \frac{1}{R_1} + \frac{1}{R_2}
+\end{aligned}$$
+
+These **principal radii** $R_1$ and $R_2$ are important
+for e.g. the [Young-Laplace law](/know/concept/young-laplace-law/).
+
+
+## 3D in general
+
+To find a general expression for the mean curvature of an arbitrary surface,
+we "cut off" a small part of the surface that we can regard as a height function.
+We call the "cutting" reference plane $(x, y)$,
+and the surface it describes $h(x, y)$.
+We then define the unit tangent vectors $\vu{t}_x$ and $\vu{t}_y$
+to be parallel to the $x$-axis and $y$-axis, respectively:
+
+$$\begin{aligned}
+    \vu{t}_x
+    = \frac{1}{\sqrt{1 + (h_x)^2}}
+    \begin{bmatrix}
+        1 \\ 0 \\ h_x
+    \end{bmatrix}
+    \qquad
+    \vu{t}_y
+    = \frac{1}{\sqrt{1 + (h_y)^2}}
+    \begin{bmatrix}
+        0 \\ 1 \\ h_y
+    \end{bmatrix}
+\end{aligned}$$
+
+Since they were chosen to lie along the axes,
+these vectors are not necessarily orthogonal,
+so we need to normalize the resulting normal vector $\vu{n}$:
+
+$$\begin{aligned}
+    \vu{n}
+    = \vu{t}_x \cross \vu{t}_y
+    = \frac{1}{\sqrt{1 + (h_x)^2 + (h_y)^2}}
+    \begin{bmatrix}
+        - h_x \\ - h_y \\ 1
+    \end{bmatrix}
+\end{aligned}$$
+
+Let us take a look at the divergence of $\vu{n}$,
+or to be precise, its *projection* onto the reference plane
+(although this distinction is not really important for our purposes):
+
+$$\begin{aligned}
+    \nabla \cdot \vu{n}
+    = - \dv{x} \bigg( \frac{h_x}{\sqrt{1 + (h_x)^2 + (h_y)^2}} \bigg) - \dv{y} \bigg( \frac{h_y}{\sqrt{1 + (h_x)^2 + (h_y)^2}} \bigg)
+\end{aligned}$$
+
+Compare this with the expression for $\lambda$ we found earlier,
+with the help of variational calculus:
+
+$$\begin{aligned}
+    \lambda
+    &= \dv{x} \Big( \pdv{\mathcal{L}}{h_x} \Big) + \dv{y} \Big( \pdv{\mathcal{L}}{h_y} \Big)
+    \\
+    &= \dv{x} \bigg( \frac{h_x}{\sqrt{1 + (h_x)^2 + (h_y)^2}} \bigg) + \dv{y} \bigg( \frac{h_y}{\sqrt{1 + (h_x)^2 + (h_y)^2}} \bigg)
+\end{aligned}$$
+
+The similarity is clearly visible.
+This leads us to the following general expression:
+
+$$\begin{aligned}
+    \boxed{
+        \kappa_1 + \kappa_2
+        = - \nabla \cdot \vu{n}
+    }
+\end{aligned}$$
+
+A useful property is that
+the principal directions of curvature are always orthogonal.
+To show this, consider the most general second-order approximating surface,
+in polar coordinates:
+
+$$\begin{aligned}
+    h(x, y)
+    &= \frac{1}{2} a x^2 + \frac{1}{2} b y^2 + c x y
+    \\
+    &= \frac{1}{2} a r^2 \cos^2\varphi + \frac{1}{2} b r^2 \sin^2\varphi + c r^2 \cos\varphi \sin\varphi
+\end{aligned}$$
+
+Sufficiently close to the extremum, where $h_x$ and $h_y$ are negligible,
+the curvature along a certain direction $\varphi$ is given by
+our earlier formula for a 2D height function:
+
+$$\begin{aligned}
+    \kappa(\varphi)
+    \approx \pdv[2]{h}{r}
+    = a \cos^2\varphi + b \sin^2\varphi + c \sin(2 \varphi)
+\end{aligned}$$
+
+To find the extremes of $\kappa$,
+we differentiate with respect to $\varphi$ and demand that it is zero:
+
+$$\begin{aligned}
+    0
+    &= - 2 a \cos\varphi \sin\varphi + 2 b \sin\varphi \cos\varphi + 2 c \cos(2 \varphi)
+    \\
+    &= - a \sin(2 \varphi) + b \sin(2 \varphi) + 2 c \cos(2 \varphi)
+\end{aligned}$$
+
+After rearranging this a bit, we arrive at the following transcendental equation:
+
+$$\begin{aligned}
+    \frac{2 c}{a - b}
+    = \frac{\sin(2 \varphi)}{\cos(2 \varphi)}
+    = \tan(2 \varphi)
+\end{aligned}$$
+
+Since the $\tan$ function is $\pi$-periodic,
+this has two solutions, $\varphi_0$ and $\varphi_0 + \pi/2$,
+which are clearly orthogonal,
+hence the principal directions are at an angle of $\pi/2$.
+
+Finally, it is also worth mentioning that
+the principal directions always lie in planes
+containing the normal of the surface.
+
+
+
+## References
+1.  T. Bohr,
+    *Curvature of plane curves and surfaces*,
+    2020, unpublished.
+2.  B. Lautrup,
+    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+    CRC Press.
diff --git a/content/know/concept/meniscus/index.pdc b/content/know/concept/meniscus/index.pdc
index d11c2d8..c3d1c6a 100644
--- a/content/know/concept/meniscus/index.pdc
+++ b/content/know/concept/meniscus/index.pdc
@@ -92,7 +92,8 @@ there is no curvature in the orthogonal principal direction.
 
 Just below the liquid surface in the meniscus,
 we expect the hydrostatic pressure
-and the Young-Laplace law agree about the pressure $p$,
+and the [Young-Laplace law](/know/concept/young-laplace-law/)
+to agree about the pressure $p$,
 where $p_0$ is the external air pressure:
 
 $$\begin{aligned}
diff --git a/content/know/concept/no-cloning-theorem/index.pdc b/content/know/concept/no-cloning-theorem/index.pdc
index e253464..38bf50c 100644
--- a/content/know/concept/no-cloning-theorem/index.pdc
+++ b/content/know/concept/no-cloning-theorem/index.pdc
@@ -68,4 +68,9 @@ which implies that such a general cloning machine cannot ever exist.
 
 
 ## References
-1.  N. Brunner, *Quantum information theory: lecture notes*, 2019.
+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-entanglement/index.pdc b/content/know/concept/quantum-entanglement/index.pdc
index f8d0b76..4fd1275 100644
--- a/content/know/concept/quantum-entanglement/index.pdc
+++ b/content/know/concept/quantum-entanglement/index.pdc
@@ -150,3 +150,8 @@ $$\begin{aligned}
     = \Tr_A\big(\hat{\rho}_A \hat{O}_A\big)
 \end{aligned}$$
 
+
+## References
+1.  J.B. Brask,
+    *Quantum information: lecture notes*,
+    2021, unpublished.
diff --git a/content/know/concept/quantum-teleportation/index.pdc b/content/know/concept/quantum-teleportation/index.pdc
index bf33a11..027635f 100644
--- a/content/know/concept/quantum-teleportation/index.pdc
+++ b/content/know/concept/quantum-teleportation/index.pdc
@@ -146,4 +146,6 @@ Bell state $\ket*{\Phi^{+}}_{AB}$, which contains nothing of $\ket{q}$.
 
 
 ## References
-1.  N. Brunner, *Quantum information theory: lecture notes*, 2019.
+1.  J.B. Brask,
+    *Quantum information: lecture notes*,
+    2021, unpublished.
diff --git a/content/know/concept/rayleigh-plateau-instability/index.pdc b/content/know/concept/rayleigh-plateau-instability/index.pdc
index f7e10d4..ae0a21d 100644
--- a/content/know/concept/rayleigh-plateau-instability/index.pdc
+++ b/content/know/concept/rayleigh-plateau-instability/index.pdc
@@ -21,7 +21,8 @@ eventually breaks into droplets as it falls.
 
 Consider an infinitely long cylinder of liquid
 with radius $R_0$ and surface tension $\alpha$.
-In this case, the Young-Laplace equation states that its internal pressure
+In this case, the [Young-Laplace equation](/know/concept/young-laplace-law/)
+states that its internal pressure
 is a constant $p_i$ expressed as follows,
 where $p_o$ is the exterior air pressure:
 
diff --git a/content/know/concept/superdense-coding/index.pdc b/content/know/concept/superdense-coding/index.pdc
new file mode 100644
index 0000000..e50be2b
--- /dev/null
+++ b/content/know/concept/superdense-coding/index.pdc
@@ -0,0 +1,56 @@
+---
+title: "Superdense coding"
+firstLetter: "S"
+publishDate: 2021-03-07
+categories:
+- Quantum information
+
+date: 2021-03-07T20:30:41+01:00
+draft: false
+markup: pandoc
+---
+
+# Superdense coding
+
+In quantum information, **(super)dense coding**
+is a protocol to enhance classical communication.
+It uses a quantum communication channel and
+[entanglement](/know/concept/quantum-entanglement/)
+to send two bits of classical data with just one qubit.
+It is conceptually similar to [quantum teleportation](/know/concept/quantum-teleportation/).
+
+Suppose that Alice wants to send two bits of classical data to Bob,
+but she can only communicate with him over a quantum channel.
+She could send a qubit, which has a larger state space than a classical bit,
+but it can only be measured once, thereby yielding only one bit of data.
+
+However, they are already sharing an entangled pair of qubits
+in the [Bell state](/know/concept/bell-state/) $\ket*{\Phi^{+}}_{AB}$,
+where $A$ and $B$ are qubits belonging to Alice and Bob, respectively.
+
+Based on the values of the two classical bits $(a_1, a_2)$,
+Alice performs the following operations on her side $A$
+of the Bell state:
+
+| $(a_1, a_2)$ | Operator $\qquad$ | Result |
+|:--:|:--|:---------|
+| $00$ | $\hat{I}$ | $\ket*{\Phi^{+}} = \frac{1}{\sqrt{2}} \Big(\ket{0}_A \ket{0}_B + \ket{1}_A \ket{1}_B \Big)$ |
+| $01$ | $\hat{\sigma}_z$ | $\ket*{\Phi^{-}} = \frac{1}{\sqrt{2}} \Big(\ket{0}_A \ket{0}_B - \ket{1}_A \ket{1}_B \Big)$ |
+| $10$ | $\hat{\sigma}_x$ | $\ket*{\Psi^{+}} = \frac{1}{\sqrt{2}} \Big(\ket{0}_A \ket{1}_B + \ket{1}_A \ket{0}_B \Big)$ |
+| $11$ | $\hat{\sigma}_x \hat{\sigma}_z$ | $\ket*{\Psi^{-}} = \frac{1}{\sqrt{2}} \Big(\ket{0}_A \ket{1}_B - \ket{1}_A \ket{0}_B \Big)$ |
+
+Her actions affect the state on Bob's side $B$ due to entanglement.
+Alice then sends her qubit $A$ to Bob over the quantum channel,
+so he has both sides of the entangled pair.
+
+Finally, Bob performs a measurement of his pair in the Bell basis,
+which will yield a Bell state that he can then look up in the table above
+to recover the values of the bits $(a_1, a_2)$.
+In the end, Alice only sent a single qubit,
+and the rest of the information transfer was via entanglement.
+
+
+## References
+1.  J.B. Brask,
+    *Quantum information: lecture notes*,
+    2021, unpublished.
diff --git a/content/know/concept/young-laplace-law/index.pdc b/content/know/concept/young-laplace-law/index.pdc
new file mode 100644
index 0000000..505125e
--- /dev/null
+++ b/content/know/concept/young-laplace-law/index.pdc
@@ -0,0 +1,99 @@
+---
+title: "Young-Laplace law"
+firstLetter: "Y"
+publishDate: 2021-03-11
+categories:
+- Physics
+- Fluid mechanics
+
+date: 2021-03-07T14:54:41+01:00
+draft: false
+markup: pandoc
+---
+
+# Young-Laplace law
+
+In liquids, the **Young-Laplace law** governs surface tension:
+it describes the tension forces on a surface
+as a pressure difference between the two sides of the liquid.
+
+Consider a small rectangle on the surface with sides $\dd{\ell_1}$ and $\dd{\ell_2}$,
+orientated such that the sides are parallel to the (orthogonal)
+principal directions of the surface' [curvature](/know/concept/curvature/).
+
+Surface tension then pulls at the sides with a force
+of magnitude $\alpha \dd{\ell_2}$ and $\alpha \dd{\ell_2}$,
+where $\alpha$ is the energy cost per unit of area,
+which is the same as the force per unit of distance.
+However, due to the surface' curvature,
+those forces are not quite in the same plane as the rectangle.
+
+Along both principal directions,
+if we treat this portion of the surface as a small arc of a circle
+with a radius equal to the principal radius of curvature $R_1$ or $R_2$,
+then the tension forces are at angles $\theta_1$ and $\theta_2$
+calculated from the arc length:
+
+$$\begin{aligned}
+    \theta_1 R_1
+    = \frac{1}{2} \dd{\ell_2}
+    \qquad \qquad
+    \theta_2 R_2
+    = \frac{1}{2} \dd{\ell_1}
+\end{aligned}$$
+
+Pay attention to the indices $1$ and $2$:
+to get the angle of the force pulling at $\dd{\ell_1}$,
+we need to treat $\dd{\ell_2} / 2$ as an arc,
+and vice versa.
+
+Since the forces are not quite in the plane,
+they have a small component acting *perpendicular* to the surface,
+with the following magnitudes $\dd{F_1}$ and $\dd{F_2}$
+along the principal axes:
+
+$$\begin{aligned}
+    \dd{F_1}
+    &= 2 \alpha \dd{\ell_1} \sin\theta_1
+    \approx 2 \alpha \dd{\ell_1} \theta_1
+    = \alpha \dd{\ell_1} \frac{\dd{\ell_2}}{R_1}
+    = \frac{\alpha}{R_1} \dd{A}
+    \\
+    \dd{F_2}
+    &= 2 \alpha \dd{\ell_2} \sin\theta_2
+    \approx 2 \alpha \dd{\ell_2} \theta_2
+    = \alpha \dd{\ell_2} \frac{\dd{\ell_1}}{R_2}
+    = \frac{\alpha}{R_2} \dd{A}
+\end{aligned}$$
+
+The initial factor of $2$ is there since
+the same force is pulling at opposide sides of the rectangle.
+We end up with $\alpha / R_{1,2}$ multiplied by
+the surface area $\dd{A} = \dd{\ell_1} \dd{\ell_2}$ of the rectangle.
+
+Adding together $\dd{F_1}$ and $\dd{F_2}$ and
+dividing out $\dd{A}$ gives us the force-per-area (i.e. the pressure)
+added by surface tension,
+which is given by the **Young-Laplace law**:
+
+$$\begin{aligned}
+    \boxed{
+        \Delta p
+        = \alpha \Big( \frac{1}{R_1} + \frac{1}{R_2} \Big)
+    }
+\end{aligned}$$
+
+The total excess pressure $\Delta p$ is called the **Laplace pressure**,
+and fully determines the effects of surface tension:
+a certain interface shape leads to a certain $\Delta p$,
+and the liquid will flow (i.e. the surface will move)
+to try to reach an equilibrium.
+
+
+## References
+1.  B. Lautrup,
+    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+    CRC Press.
+2.  T. Bohr,
+    *Surface tension and Laplace pressure*, 2021,
+    unpublished.