Expand knowledge base

5 files changed, 563 insertions, 10 deletions

diff --git a/content/know/concept/archimedes-principle/index.pdc b/content/know/concept/archimedes-principle/index.pdc
index 6335a77..0837cc9 100644
--- a/content/know/concept/archimedes-principle/index.pdc
+++ b/content/know/concept/archimedes-principle/index.pdc
@@ -44,16 +44,19 @@ on the surface $S$ of $V$:
 $$\begin{aligned}
     \va{F}_p
     = - \oint_S p \dd{\va{S}}
+    = - \int_V \nabla p \dd{V}
 \end{aligned}$$
 
-We rewrite this using Gauss' theorem,
-and replace $\nabla p$ by demanding
-[hydrostatic equilibrium](/know/concept/hydrostatic-pressure/):
+The last step follows from Gauss' theorem.
+We replace $\nabla p$ by assuming
+[hydrostatic equilibrium](/know/concept/hydrostatic-pressure/),
+leading to the definition of the **buoyant force**:
 
 $$\begin{aligned}
-    \va{F}_p
-    = - \int_V \nabla p \dd{V}
-    = - \int_V \va{g} \rho_\mathrm{f} \dd{V}
+    \boxed{
+        \va{F}_p
+        = - \int_V \va{g} \rho_\mathrm{f} \dd{V}
+    }
 \end{aligned}$$
 
 For the body to be at rest, we require $\va{F}_g + \va{F}_p = 0$.
@@ -66,10 +69,9 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-It is commonly assumed that $\va{g}$ has a constant direction
-and magnitude $\mathrm{g}$ everywhere.
-If we also assume that $\rho_\mathrm{b}$ and $\rho_\mathrm{f}$ are constant,
-and only integrate over the "submerged" part, we find:
+It is commonly assumed that $\va{g}$ is constant everywhere, with magnitude $\mathrm{g}$.
+If we also assume that $\rho_\mathrm{f}$ is constant on the "submerged" side,
+and zero on the "non-submerged" side, we find:
 
 $$\begin{aligned}
     0
diff --git a/content/know/concept/hagen-poiseuille-equation/index.pdc b/content/know/concept/hagen-poiseuille-equation/index.pdc
new file mode 100644
index 0000000..b33e085
--- /dev/null
+++ b/content/know/concept/hagen-poiseuille-equation/index.pdc
@@ -0,0 +1,203 @@
+---
+title: "Hagen-Poiseuille equation"
+firstLetter: "H"
+publishDate: 2021-04-13
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+
+date: 2021-04-13T10:42:46+02:00
+draft: false
+markup: pandoc
+---
+
+# Hagen-Poiseuille equation
+
+The **Hagen-Poiseuille equation**, or simply the **Poiseuille equation**,
+describes the flow of a fluid with nonzero [viscosity](/know/concept/viscosity/)
+through a cylindrical pipe.
+Due to its viscosity, the fluid clings to the sides,
+limiting the amount that can pass through, for a pipe with radius $R$.
+
+Consider the [Navier-Stokes equations](/know/concept/navier-stokes-equations/)
+of an incompressible fluid with spatially uniform density $\rho$.
+Assuming that the flow is steady $\pdv*{\va{v}}{t} = 0$,
+and that gravity is negligible $\va{g} = 0$, we get:
+
+$$\begin{aligned}
+    (\va{v} \cdot \nabla) \va{v}
+    = - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v}
+    \qquad \quad
+    \nabla \cdot \va{v} = 0
+\end{aligned}$$
+
+Into this, we insert the ansatz $\va{v} = \vu{e}_z \: v_z(r)$,
+where $\vu{e}_z$ is the $z$-axis' unit vector.
+In other words, we assume that the flow velocity depends only on $r$;
+not on $\phi$ or $z$.
+Plugging this into the Navier-Stokes equations,
+$\nabla \cdot \va{v}$ is trivially zero,
+and in the other equation we multiply out $\rho$, yielding this,
+where $\eta = \rho \nu$ is the dynamic viscosity:
+
+$$\begin{aligned}
+    \nabla p
+    = \vu{e}_z \: \eta \nabla^2 v_z
+\end{aligned}$$
+
+Because only $\vu{e}_z$ appears on the right-hand side,
+only the $z$-component of $\nabla p$ can be nonzero.
+However, $v_z(r)$ is a function of $r$, not $z$!
+The left thus only depends on $z$, and the right only on $r$,
+meaning that both sides must equal a constant,
+which we call $-G$:
+
+$$\begin{aligned}
+    \dv{p}{z}
+    = -G
+    \qquad \quad
+    \eta \frac{1}{r} \dv{r} \Big( r \dv{v_z}{r} \Big)
+    = - G
+\end{aligned}$$
+
+The former equation, for $p(z)$, is easy to solve.
+We get an integration constant $p(0)$:
+
+$$\begin{aligned}
+    p(z)
+    = p(0) - G z
+\end{aligned}$$
+
+This gives meaning to the **pressure gradient** $G$:
+for a pipe of length $L$,
+it describes the pressure difference $\Delta p = p(0) - p(L)$
+that is driving the fluid,
+i.e. $G = \Delta p / L$
+
+As for the latter equation, for $v_z(r)$,
+we start by integrating it once, introducing a constant $A$:
+
+$$\begin{aligned}
+    \dv{r} \Big( r \dv{v_z}{r} \Big)
+    = - \frac{G}{\eta} r
+    \quad \implies \quad
+    \dv{v_z}{r}
+    = - \frac{G}{2 \eta} r + \frac{A}{r}
+\end{aligned}$$
+
+Integrating this one more time,
+thereby introducing another constant $B$,
+we arrive at:
+
+$$\begin{aligned}
+    v_z
+    = - \frac{G}{4 \eta} r^2 + A \ln{r} + B
+\end{aligned}$$
+
+The velocity must be finite at $r = 0$, so we set $A = 0$.
+Furthermore, the Navier-Stokes equation's *no-slip* condition
+demands that $v_z = 0$ at the boundary $r = R$,
+so $B = G R^2 / (4 \eta)$.
+This brings us to the **Poiseuille solution** for $v_z(r)$:
+
+$$\begin{aligned}
+    \boxed{
+        v_z(r)
+        = \frac{G}{4 \eta} (R^2 - r^2)
+    }
+\end{aligned}$$
+
+How much fluid can pass through the pipe per unit time?
+This is denoted by the **volumetric flow rate** $Q$,
+which is the integral of $v_z$ over the circular cross-section:
+
+$$\begin{aligned}
+    Q
+    = 2 \pi \int_0^R v_z(r) \: r \dd{r}
+    = \frac{\pi G}{2 \eta} \int_0^R R^2 r - r^3 \dd{r}
+    = \frac{\pi G}{2 \eta} \bigg[ \frac{R^2 r^2}{2} - \frac{r^4}{4} \bigg]_0^R
+\end{aligned}$$
+
+We thus arrive at the main Hagen-Poiseuille equation,
+which predicts $Q$ for a given setup:
+
+$$\begin{aligned}
+    \boxed{
+        Q
+        = \frac{\pi G R^4}{8 \eta}
+    }
+\end{aligned}$$
+
+Consequently, the average flow velocity $\expval{v_z}$
+is simply $Q$ divided by the cross-sectional area:
+
+$$\begin{aligned}
+    \expval{v_z}
+    = \frac{Q}{\pi R^2}
+    = \frac{G R^2}{8 \eta}
+\end{aligned}$$
+
+The fluid's viscous stickiness means it exerts a drag force $D$
+on the pipe as it flows. For a pipe of length $L$ and radius $R$,
+we calculate $D$ by multiplying the internal area $2 \pi R L$
+by the [shear stress](/know/concept/cauchy-stress-tensor/)
+$-\sigma_{zr}$ on the wall
+(i.e. the wall applies $\sigma_{zr}$, the fluid responds with $- \sigma_{zr}$):
+
+$$\begin{aligned}
+    D
+    = - 2 \pi R L \: \sigma_{zr} \big|_{r = R}
+    = - 2 \pi R L \eta \dv{v_z}{r}\Big|_{r = R}
+    = 2 \pi R L \eta \frac{G R}{2 \eta}
+    = \pi R^2 L G
+\end{aligned}$$
+
+We would like to get rid of $G$ for being impractical,
+so we substitute $R^2 G = 8 \eta \expval{v_z}$, yielding:
+
+$$\begin{aligned}
+    \boxed{
+        D
+        = 8 \pi \eta L \expval{v_z}
+    }
+\end{aligned}$$
+
+Due to this drag, the pressure difference $\Delta p = p(0) - p(L)$
+does work on the fluid, at a rate $P$,
+since power equals force (i.e. pressure times area) times velocity:
+
+$$\begin{aligned}
+    P
+    = 2 \pi \int_0^R \Delta p \: v_z(r) \: r \dd{r}
+\end{aligned}$$
+
+Because $\Delta p$ is independent of $r$,
+we get the same integral we used to calculate $Q$.
+Then, thanks to the fact that $\Delta p = G L$
+and $Q = \pi R^2 \expval{v_z}$, it follows that:
+
+$$\begin{aligned}
+    P
+    = \Delta p \: Q
+    = G L \pi R^2  \expval{v_z}
+    = D \expval{v_z}
+\end{aligned}$$
+
+In conclusion, the power $P$,
+needed to drive a fluid through the pipe at a rate $Q$,
+is given by:
+
+$$\begin{aligned}
+    \boxed{
+        P
+        = 8 \pi \eta L \expval{v_z}^2
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  B. Lautrup,
+    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+    CRC Press.
diff --git a/content/know/concept/navier-stokes-equations/index.pdc b/content/know/concept/navier-stokes-equations/index.pdc
new file mode 100644
index 0000000..7256b7e
--- /dev/null
+++ b/content/know/concept/navier-stokes-equations/index.pdc
@@ -0,0 +1,134 @@
+---
+title: "Navier-Stokes equations"
+firstLetter: "N"
+publishDate: 2021-04-12
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+
+date: 2021-04-12T13:14:09+02:00
+draft: false
+markup: pandoc
+---
+
+# Navier-Stokes equations
+
+While the [Euler equations](/know/concept/euler-equations/) govern *ideal* "dry" fluids,
+the **Navier-Stokes equations** govern *nonideal* "wet" fluids,
+i.e. fluids with nonzero [viscosity](/know/concept/viscosity/).
+
+
+## Incompressible fluid
+
+First of all, we can reuse the incompressibility condition for ideal fluids, without modifications:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cdot \va{v} = 0
+    }
+\end{aligned}$$
+
+Furthermore, from the derivation of the Euler equations,
+we know that Newton's second law can be written as follows,
+for an infinitesimal particle of the fluid:
+
+$$\begin{aligned}
+    \rho \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
+    = \va{f^*}
+\end{aligned}$$
+
+$\mathrm{D}/\mathrm{D}t$ is the [material derivative](/know/concept/material-derivative/),
+$\rho$ is the density, and $\va{f^*}$ is the effective force density,
+expressed in terms of an external body force $\va{f}$ (e.g. gravity)
+and the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{\sigma}$:
+
+$$\begin{aligned}
+    \va{f^*}
+    = \va{f} + \nabla \cdot \hat{\sigma}^\top
+\end{aligned}$$
+
+From the definition of viscosity,
+the stress tensor's elements are like so for a Newtonian fluid:
+
+$$\begin{aligned}
+    \sigma_{ij}
+    = - p \delta_{ij} + \eta (\nabla_i v_j + \nabla_j v_i)
+\end{aligned}$$
+
+Where $\eta$ is the dynamic viscosity.
+Inserting this, we calculate $\nabla \cdot \hat{\sigma}^\top$ in index notation:
+
+$$\begin{aligned}
+    \big( \nabla \cdot \hat{\sigma}^\top \big)_i
+    = \sum_{j} \nabla_j \sigma_{ij}
+    &= \sum_{j} \Big( \!-\! \delta_{ij} \nabla_j p + \eta (\nabla_i \nabla_j v_j + \nabla_j^2 v_i) \Big)
+    \\
+    &= - \nabla_i p + \eta \nabla_i \sum_{j} \nabla_j v_j + \eta \sum_{j} \nabla_j^2 v_i
+\end{aligned}$$
+
+Thanks to incompressibility $\nabla \cdot \va{v} = 0$,
+the middle term vanishes, leaving us with:
+
+$$\begin{aligned}
+    \va{f^*}
+    = \va{f} - \nabla p + \eta \nabla^2 \va{v}
+\end{aligned}$$
+
+We assume that the only body force is gravity $\va{f} = \rho \va{g}$.
+Newton's second law then becomes:
+
+$$\begin{aligned}
+    \rho \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
+    = \rho \va{g} - \nabla p + \eta \nabla^2 \va{v}
+\end{aligned}$$
+
+Dividing by $\rho$, and replacing $\eta$
+with the kinematic viscosity $\nu = \eta/\rho$,
+yields the main equation:
+
+$$\begin{aligned}
+    \boxed{
+        \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
+        = \va{g} - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v}
+    }
+\end{aligned}$$
+
+Finally, we can optionally allow incompressible fluids
+with an inhomogeneous "lumpy" density $\rho$,
+by demanding conservation of mass,
+just like for the Euler equations:
+
+$$\begin{aligned}
+    \boxed{
+        \frac{\mathrm{D} \rho}{\mathrm{D} t}
+        = 0
+    }
+\end{aligned}$$
+
+Putting it all together, the Navier-Stokes equations for an incompressible fluid are given by:
+
+$$\begin{aligned}
+    \boxed{
+        \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
+        = \va{g} - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v}
+        \qquad
+        \nabla \cdot \va{v} = 0
+        \qquad
+        \frac{\mathrm{D} \rho}{\mathrm{D} t}
+        = 0
+    }
+\end{aligned}$$
+
+Due to the definition of viscosity $\nu$ as the molecular "stickiness",
+we have boundary conditions for the velocity field $\va{v}$:
+at any interface, $\va{v}$ must be continuous.
+Likewise, Newton's third law demands that the normal component
+of stress $\hat{\sigma} \cdot \vu{n}$ is continuous there.
+
+
+
+## References
+1.  B. Lautrup,
+    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+    CRC Press.
diff --git a/content/know/concept/renyi-entropy/index.pdc b/content/know/concept/renyi-entropy/index.pdc
new file mode 100644
index 0000000..073d773
--- /dev/null
+++ b/content/know/concept/renyi-entropy/index.pdc
@@ -0,0 +1,114 @@
+---
+title: "Rényi entropy"
+firstLetter: "R"
+publishDate: 2021-04-11
+categories:
+- Cryptography
+
+date: 2021-04-11T15:13:07+02:00
+draft: false
+markup: pandoc
+---
+
+# Rényi entropy
+
+In information theory, the **Rényi entropy** is a measure
+(or family of measures) of the "suprise" or "information"
+contained in a random variable $X$.
+It is defined as follows:
+
+$$\begin{aligned}
+    \boxed{
+        H_\alpha(X)
+        = \frac{1}{1 - \alpha} \log\!\bigg( \sum_{i = 1}^N p_i^\alpha \bigg)
+    }
+\end{aligned}$$
+
+Where $\alpha \ge 0$ is a free parameter.
+The logarithm is usually base-2, but variations exist.
+
+The case $\alpha = 0$ is known as the **Hartley entropy** or **max-entropy**,
+and quantifies the "surprise" of an event from $X$,
+if $X$ is uniformly distributed:
+
+$$\begin{aligned}
+    \boxed{
+        H_0(X)
+        = \log N
+    }
+\end{aligned}$$
+
+Where $N$ is the cardinality of $X$; the number of different possible events.
+The most famous case, however, is $\alpha = 1$.
+Since $H_\alpha$ is problematic for $\alpha \to 1$, we must take the limit:
+
+$$\begin{aligned}
+    H_1(X)
+    = \lim_{\alpha \to 1} H_\alpha(X)
+    = \lim_{\alpha \to 1} \frac{\log\!\left( \sum_i p_i^\alpha \right)}{1 - \alpha}
+\end{aligned}$$
+
+We then apply L'Hôpital's rule to evaluate this limit,
+and use the fact that all $p_i$ sum to $1$:
+
+$$\begin{aligned}
+    H_1(X)
+    = \lim_{\alpha \to 1} \frac{\dv{\alpha} \log\!\left( \sum_i p_i^\alpha \right)}{\dv{\alpha} (1 - \alpha)}
+    = \lim_{\alpha \to 1} \frac{\sum_i p_i^\alpha \log p_i}{- \sum_i p_i^\alpha}
+    = - \sum_{i = 1}^N p_i \log p_i
+\end{aligned}$$
+
+This quantity is the **Shannon entropy**,
+which is the most general measure of "surprise":
+
+$$\begin{aligned}
+    \boxed{
+        H_1(X)
+        = \lim_{\alpha \to 1} H_\alpha(X)
+        = - \sum_{i = 1}^N p_i \log p_i
+    }
+\end{aligned}$$
+
+Next, for $\alpha = 2$, we get the **collision entropy**, which describes
+the surprise of two independent and identically distributed variables
+$X$ and $Y$ yielding the same event:
+
+$$\begin{aligned}
+    \boxed{
+        H_2(X)
+        = - \log\!\bigg( \sum_{i = 1}^N p_i^2 \bigg)
+        = - \log P(X = Y)
+    }
+\end{aligned}$$
+
+Finally, in the limit $\alpha \to \infty$,
+the largest probability dominates the sum,
+leading to the definition of the **min-entropy** $H_\infty$,
+describing the surprise of the most likely event:
+
+$$\begin{aligned}
+    \boxed{
+        H_\infty(X)
+        = \lim_{\alpha \to \infty} H_\alpha(x)
+        = - \log\!\big( \max_{i} p_i \big)
+    }
+\end{aligned}$$
+
+It is straightforward to convince yourself that these entropies
+are ordered in the following way:
+
+$$\begin{aligned}
+    H_0 \ge H_1 \ge H_2 \ge H_\infty
+\end{aligned}$$
+
+In other words, from left to right,
+they go from permissive to conservative, roughly speaking.
+
+
+## References
+1.  P.A. Bromiley, N.A. Thacker, E. Bouhova-Thacker,
+    [Shannon entropy, Rényi entropy, and information](https://www.researchgate.net/publication/253537416_Shannon_Entropy_Renyi_Entropy_and_Information),
+    2010, University of Manchester.
+2.  J.B. Brask,
+    *Quantum information: lecture notes*,
+    2021, unpublished.
diff --git a/content/know/concept/viscosity/index.pdc b/content/know/concept/viscosity/index.pdc
new file mode 100644
index 0000000..d7abd7a
--- /dev/null
+++ b/content/know/concept/viscosity/index.pdc
@@ -0,0 +1,100 @@
+---
+title: "Viscosity"
+firstLetter: "V"
+publishDate: 2021-04-12
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+
+date: 2021-04-12T13:14:16+02:00
+draft: false
+markup: pandoc
+---
+
+# Viscosity
+
+The **viscosity** of a fluid describes how
+"sticky" its constituent molecules are;
+when one part of the fluid moves, it "drags"
+neighbouring parts by an amount proportional to the viscosity.
+
+Imagine a liquid in a canal,
+flowing in the $x$-direction at a velocity $v(z)$
+as a function of depth $z$.
+Due to the liquid's viscosity,
+its molecules are "stuck" to the bottom of the canal $z = 0$,
+such that it is stationary there $v(0) = 0$.
+However, at the surface $z = z_s$, there is a flow at $v(z_s) = v_s$.
+
+This difference in $v$ means that there is a velocity gradient across $z$.
+Each infinitesimal layer of the liquid
+is dragging on the layers above and below it,
+meaning there is a nonzero shear stress $\sigma_{xz}$
+(see [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/)).
+Formally, the **dynamic viscosity** $\eta$ is defined as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \sigma_{xz}
+        = \eta \dv{v}{z}
+    }
+\end{aligned}$$
+
+This is **Newton's law of viscosity**,
+and fluids obeying it are known as **Newtonian**.
+In a Newtonian fluid *at rest*, there are no such shear stresses,
+and the Cauchy stress tensor $\hat{\sigma}$ is diagonal:
+
+$$\begin{aligned}
+    \sigma_{ij} = - p \delta_{ij}
+\end{aligned}$$
+
+Where $p$ is the pressure, and $\delta_{ij}$ is the Kronecker delta.
+If the fluid flows according to a velocity field $\va{v}$,
+then a more general definition of $\eta$ is as follows,
+in index notation with $\nabla_i \!=\! \pdv*{x_i}$:
+
+$$\begin{aligned}
+    \boxed{
+        \sigma_{ij}
+        = - p \delta_{ij} + \eta (\nabla_i v_j + \nabla_j v_i)
+    }
+\end{aligned}$$
+
+The double term $\nabla_i v_j + \nabla_j v_i$ comes from the fact that
+the stress tensor of a Newtonian fluid is always symmetric;
+this definition of $\sigma_{ij}$ enforces that.
+
+Another quantity is the **kinematic viscosity** $\nu$,
+which is simply $\eta$ divided by the density $\rho$:
+
+$$\begin{aligned}
+    \boxed{
+        \nu
+        \equiv \frac{\eta}{\rho}
+    }
+\end{aligned}$$
+
+With this, Newton's law of viscosity is written
+using the momentum density $P = \rho v$:
+
+$$\begin{aligned}
+    \sigma_{xz}
+    = \nu \dv{P}{z}
+\end{aligned}$$
+
+Because momentum is "more fundamental" than velocity,
+is $\nu$ often more useful than $\eta$.
+However, this comes at the cost of our intuition:
+for example, as you would expect, $\eta_\mathrm{water} > \eta_\mathrm{air}$,
+but you may be surprised that $\nu_\mathrm{water} < \nu_\mathrm{air}$.
+Since air is less dense, it is easier to set in motion,
+hence we expect it to be less viscous than water,
+but in fact air's molecules are stickier than water's.
+
+
+## References
+1.  B. Lautrup,
+    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+    CRC Press.