Expand knowledge base

7 files changed, 511 insertions, 21 deletions

diff --git a/content/know/concept/bernoullis-theorem/index.pdc b/content/know/concept/bernoullis-theorem/index.pdc
index 5ff5225..fbabff2 100644
--- a/content/know/concept/bernoullis-theorem/index.pdc
+++ b/content/know/concept/bernoullis-theorem/index.pdc
@@ -27,13 +27,8 @@ $$\begin{aligned}
     = \va{g} - \frac{\nabla p}{\rho}
 \end{aligned}$$
 
-Assuming that $\va{v}$ and $\va{g}$ are constant in $t$,
-it becomes clear that a higher $\va{v}$ requires a lower $p$:
-
-$$\begin{aligned}
-    \frac{1}{2} \nabla \va{v}^2
-    = \va{g} - \frac{\nabla p}{\rho}
-\end{aligned}$$
+Assuming that $\va{v}$ is constant in $t$,
+it becomes clear that a higher $\va{v}$ requires a lower $p$.
 
 
 ## Simple form
diff --git a/content/know/concept/density-of-states/index.pdc b/content/know/concept/density-of-states/index.pdc
new file mode 100644
index 0000000..195ac2a
--- /dev/null
+++ b/content/know/concept/density-of-states/index.pdc
@@ -0,0 +1,156 @@
+---
+title: "Density of states"
+firstLetter: "D"
+publishDate: 2021-05-08
+categories:
+- Physics
+- Statistics
+
+date: 2021-05-08T18:35:46+02:00
+draft: false
+markup: pandoc
+---
+
+# Density of states
+
+The **density of states** $g(E)$ of a physical system is defined such that
+$g(E) \dd{E}$ is the number of states which could be occupied
+with an energy in the interval $[E, E + \dd{E}]$.
+In fact, $E$ need not be an energy;
+it should just be something that effectively identifies the state.
+
+In its simplest form, the density of states is as follows,
+where $\Gamma(E)$ is the number of states with energy
+less than or equal to the argument $E$:
+
+$$\begin{aligned}
+    g(E)
+    = \dv{\Gamma}{E}
+\end{aligned}$$
+
+If the states can be treated as waves,
+which is often the case,
+then we can calculate the density of states $g(k)$ in
+$k$-space, i.e. as a function of the wavenumber $k = |\vb{k}|$.
+Once we have $g(k)$, we use the dispersion relation $E(k)$ to find $g(E)$,
+by demanding that:
+
+$$\begin{aligned}
+    g(k) \dd{k} = g(E) \dd{E}
+    \quad \implies \quad
+    g(E)
+    = g(k) \dv{k}{E}
+\end{aligned}$$
+
+Inverting the dispersion relation $E(k)$ to get $k(E)$ might be difficult,
+in which case the left-hand equation can be satisfied numerically.
+
+
+Define $\Omega_n(k)$ as the number of states with
+a $k$-value less than or equal to the argument,
+or in other words, the volume of a hypersphere with radius $k$.
+Then the $n$-dimensional density of states $g_n(k)$
+has the following general form:
+
+$$\begin{aligned}
+    \boxed{
+        g_n(k)
+        = \frac{D}{2^n k_{\mathrm{min}}^n} \: \dv{\Omega_n}{k}
+    }
+\end{aligned}$$
+
+Where $D$ is each state's degeneracy (e.g. due to spin),
+and $k_{\mathrm{min}}$ is the smallest allowed $k$-value,
+according to the characteristic length $L$ of the system.
+We divide by $2^n$ to limit ourselves to the sector where all axes are positive,
+because we are only considering the magnitude of $k$.
+
+In one dimension $n = 1$, the number of states within a distance $k$ from the
+origin is the distance from $k$ to $-k$
+(we let it run negative, since its meaning does not matter here), given by:
+
+$$\begin{aligned}
+    \Omega_1(k)
+    = 2 k
+\end{aligned}$$
+
+To get $k_{\mathrm{min}}$, we choose to look at a rod of length $L$,
+across which the function is a standing wave, meaning that
+the allowed values of $k$ must be as follows, where $m \in \mathbb{N}$:
+
+$$\begin{aligned}
+    \lambda = \frac{2 L}{m}
+    \quad \implies \quad
+    k = \frac{2 \pi}{\lambda} = \frac{m \pi}{L}
+\end{aligned}$$
+
+Take the smallest option $m = 1$,
+such that $k_{\mathrm{min}} = \pi / L$,
+the 1D density of states $g_1(k)$ is:
+
+$$\begin{aligned}
+    \boxed{
+        g_1(k)
+        = \frac{D L}{2 \pi} \: 2
+        = \frac{D L}{\pi}
+    }
+\end{aligned}$$
+
+In 2D, the number of states within a range $k$ of the
+origin is the area of a circle with radius $k$:
+
+$$\begin{aligned}
+    \Omega_2(k)
+    = \pi k^2
+\end{aligned}$$
+
+Analogously to the 1D case,
+we take the system to be a square of side $L$,
+so $k_{\mathrm{min}} = \pi / L$ again.
+The density of states then becomes:
+
+$$\begin{aligned}
+    \boxed{
+        g_2(k)
+        = \frac{D L^2}{4 \pi^2} \:2 \pi k
+        = \frac{D L^2 k}{2 \pi}
+    }
+\end{aligned}$$
+
+In 3D, the number of states is the volume of a sphere with radius $k$:
+
+$$\begin{aligned}
+    \Omega_3(k)
+    = \frac{4 \pi}{3} k^3
+\end{aligned}$$
+
+For a cube with side $L$, we once again find $k_{\mathrm{min}} = \pi / L$.
+We thus get:
+
+$$\begin{aligned}
+    \boxed{
+        g_3(k)
+        = \frac{D L^3}{8 \pi^3} \:4 \pi k^2
+        = \frac{D L^3 k^2}{2 \pi^2}
+    }
+\end{aligned}$$
+
+All these expressions contain the characteristic length/area/volume $L^n$,
+and therefore give the number of states in that region only.
+Keep in mind that $L$ is free to choose;
+it need not be the physical size of the system.
+In fact, we typically want the density of states
+per unit length/area/volume,
+so we can just set $L = 1$ in our preferred unit of distance.
+
+If the system is infinitely large, or if it has periodic boundaries,
+then $k$ becomes a continuous variable and $k_\mathrm{min} \to 0$.
+But again, $L$ is arbitrary,
+so a finite value can be chosen.
+
+
+
+## References
+1.  H. Gould, J. Tobochnik,
+    *Statistical and thermal physics*, 2nd edition,
+    Princeton.
diff --git a/content/know/concept/hydrostatic-pressure/index.pdc b/content/know/concept/hydrostatic-pressure/index.pdc
index 90e57ce..001a198 100644
--- a/content/know/concept/hydrostatic-pressure/index.pdc
+++ b/content/know/concept/hydrostatic-pressure/index.pdc
@@ -141,8 +141,10 @@ $$\begin{aligned}
 With this, the equilibrium condition is turned into the following equation:
 
 $$\begin{aligned}
-    \nabla \Phi + \frac{\nabla p}{\rho}
-    = 0
+    \boxed{
+        \nabla \Phi + \frac{\nabla p}{\rho}
+        = 0
+    }
 \end{aligned}$$
 
 In practice, the density $\rho$ of the fluid
@@ -156,7 +158,7 @@ the indefinite integral of the density:
 
 $$\begin{aligned}
     w(p)
-    = \int \frac{1}{\rho(p)} \dd{p}
+    \equiv \int \frac{1}{\rho(p)} \dd{p}
 \end{aligned}$$
 
 Using this, we can rewrite the equilibrium condition as a single gradient like so:
@@ -172,9 +174,7 @@ From this, let us now define the
 **effective gravitational potential** $\Phi^*$ as follows:
 
 $$\begin{aligned}
-    \boxed{
-        \Phi^* = \Phi + w(p)
-    }
+    \Phi^* \equiv \Phi + w(p)
 \end{aligned}$$
 
 This results in the cleanest form yet of the equilibrium condition, namely:
diff --git a/content/know/concept/maxwell-boltzmann-distribution/index.pdc b/content/know/concept/maxwell-boltzmann-distribution/index.pdc
new file mode 100644
index 0000000..082d2db
--- /dev/null
+++ b/content/know/concept/maxwell-boltzmann-distribution/index.pdc
@@ -0,0 +1,219 @@
+---
+title: "Maxwell-Boltzmann distribution"
+firstLetter: "M"
+publishDate: 2021-05-08
+categories:
+- Physics
+- Statistics
+
+date: 2021-05-08T18:35:37+02:00
+draft: false
+markup: pandoc
+---
+
+# Maxwell-Boltzmann distribution
+
+The **Maxwell-Boltzmann distributions** are a set of closely related
+probability distributions with applications in classical statistical physics.
+
+
+## Velocity vector distribution
+
+In the canonical ensemble
+(where a fixed-size system can exchange energy with its environment),
+the probability of a microstate with energy $E$ is given by the Boltzmann distribution:
+
+$$\begin{aligned}
+    f(E)
+    \:\propto\: \exp\!\big(\!-\! \beta E\big)
+\end{aligned}$$
+
+Where $\beta = 1 / k_B T$.
+We split $E = K + U$,
+where $K$ and $U$ are the total kinetic and potential energy contributions.
+If there are $N$ particles in the system,
+with positions $\tilde{r} = (\vec{r}_1, ..., \vec{r}_N)$
+and momenta $\tilde{p} = (\vec{p}_1, ..., \vec{p}_N)$,
+then $K$ only depends on $\tilde{p}$,
+and $U$ only depends on $\tilde{r}$,
+so the probability of a specific microstate
+$(\tilde{r}, \tilde{p})$ is as follows:
+
+$$\begin{aligned}
+    f(\tilde{r}, \tilde{p})
+    \:\propto\: \exp\!\Big(\!-\! \beta \big( K(\tilde{p}) + U(\tilde{r}) \big) \Big)
+\end{aligned}$$
+
+Since this is classical physics,
+we can split the exponential.
+In quantum mechanics,
+the canonical commutation relation would prevent that.
+Anyway, splitting yields:
+
+$$\begin{aligned}
+    f(\tilde{r}, \tilde{p})
+    \:\propto\: \exp\!\big(\!-\! \beta K(\tilde{p}) \big) \exp\!\big(\!-\! \beta U(\tilde{r}) \big)
+\end{aligned}$$
+
+Classically, the probability
+distributions of the momenta and positions are independent:
+
+$$\begin{aligned}
+    f_K(\tilde{p})
+    \:\propto\: \exp\!\big(\!-\! \beta K(\tilde{p}) \big)
+    \qquad
+    f_U(\tilde{r})
+    \:\propto\: \exp\!\big(\!-\! \beta U(\tilde{r}) \big)
+\end{aligned}$$
+
+We cannot evaluate $f_U(\tilde{r})$ further without knowing $U(\tilde{r})$ for a system.
+We thus turn to $f_K(\tilde{p})$, and see that the total kinetic
+energy $K(\tilde{p})$ is simply the sum of the particles' individual
+kinetic energies $K_n(\vec{p}_n)$, which are well-known:
+
+$$\begin{aligned}
+    K(\tilde{p})
+    = \sum_{n = 1}^N K_n(\vec{p}_n)
+    \qquad \mathrm{where} \qquad
+    K_n(\vec{p}_n)
+    = \frac{|\vec{p}_n|^2}{2 m}
+\end{aligned}$$
+
+Consequently, the probability distribution $f(p_x, p_y, p_z)$ for the
+momentum vector of a single particle is as follows,
+after normalization:
+
+$$\begin{aligned}
+    f(p_x, p_y, p_z)
+    = \Big( \frac{1}{2 \pi m k_B T} \Big)^{3/2} \exp\!\Big( \!-\!\frac{(p_x^2 + p_y^2 + p_z^2)}{2 m k_B T} \Big)
+\end{aligned}$$
+
+We now rewrite this using the velocities $v_x = p_x / m$,
+and update the normalization, giving:
+
+$$\begin{aligned}
+    \boxed{
+        f(v_x, v_y, v_z)
+        = \Big( \frac{m}{2 \pi k_B T} \Big)^{3/2} \exp\!\Big( \!-\!\frac{m (v_x^2 + v_y^2 + v_z^2)}{2 k_B T} \Big)
+    }
+\end{aligned}$$
+
+This is the **Maxwell-Boltzmann velocity vector distribution**.
+Clearly, this is a product of three exponentials,
+so the velocity in each direction is independent of the others:
+
+$$\begin{aligned}
+    f(v_x)
+    = \sqrt{\frac{m}{2 \pi k_B T}} \exp\!\Big( \!-\!\frac{m v_x^2}{2 k_B T} \Big)
+\end{aligned}$$
+
+The distribution is thus an isotropic gaussian with standard deviations given by:
+
+$$\begin{aligned}
+    \sigma_x = \sigma_y = \sigma_z
+    = \sqrt{\frac{k_B T}{m}}
+\end{aligned}$$
+
+
+## Speed distribution
+
+We know the distribution of the velocities along each axis,
+but what about the speed $v = |\vec{v}|$?
+Because we do not care about the direction of $\vec{v}$, only its magnitude,
+the [density of states](/know/concept/density-of-states/) $g(v)$ is not constant:
+it is the rate-of-change of the volume of a sphere of radius $v$:
+
+$$\begin{aligned}
+    g(v)
+    = \dv{v} \Big( \frac{4 \pi}{3} v^3 \Big)
+    = 4 \pi v^2
+\end{aligned}$$
+
+Multiplying the velocity vector distribution by $g(v)$
+and substituting $v^2 = v_x^2 + v_y^2 + v_z^2$
+then gives us the **Maxwell-Boltzmann speed distribution**:
+
+$$\begin{aligned}
+    \boxed{
+        f(v)
+        = 4 \pi \Big( \frac{m}{2 \pi k_B T} \Big)^{3/2} v^2 \exp\!\Big( \!-\!\frac{m v^2}{2 k_B T} \Big)
+    }
+\end{aligned}$$
+
+Some notable points on this distribution are
+the most probable speed $v_{\mathrm{mode}}$,
+the mean average speed $v_{\mathrm{mean}}$,
+and the root-mean-square speed $v_{\mathrm{rms}}$:
+
+$$\begin{aligned}
+    f'(v_\mathrm{mode})
+    = 0
+    \qquad
+    v_\mathrm{mean}
+    = \int_0^\infty v \: f(v) \dd{v}
+    \qquad
+    v_\mathrm{rms}
+    = \bigg( \int_0^\infty v^2 \: f(v) \dd{v} \bigg)^{1/2}
+\end{aligned}$$
+
+Which can be calculated to have the following exact expressions:
+
+$$\begin{aligned}
+    \boxed{
+        v_{\mathrm{mode}}
+        = \sqrt{\frac{2 k_B T}{m}}
+    }
+    \qquad
+    \boxed{
+        v_{\mathrm{mean}}
+        = \sqrt{\frac{8 k_B T}{\pi m}}
+    }
+    \qquad
+    \boxed{
+        v_{\mathrm{rms}}
+        = \sqrt{\frac{3 k_B T}{m}}
+    }
+\end{aligned}$$
+
+
+## Kinetic energy distribution
+
+Using the speed distribution,
+we can work out the kinetic energy distribution.
+Because $K$ is not proportional to $v$,
+we must do this by demanding that:
+
+$$\begin{aligned}
+    f(K) \dd{K}
+    = f(v) \dd{v}
+    \quad \implies \quad
+    f(K)
+    = f(v) \dv{v}{K}
+\end{aligned}$$
+
+We know that $K = m v^2 / 2$,
+meaning $\dd{K} = m v \dd{v}$
+so the energy distribution $f(K)$ is:
+
+$$\begin{aligned}
+    f(K)
+    = \frac{f(v)}{m v}
+    = \sqrt{\frac{2 m}{\pi}} \: \bigg( \frac{1}{k_B T} \bigg)^{3/2} v \exp\!\Big( \!-\!\frac{m v^2}{2 k_B T} \Big)
+\end{aligned}$$
+
+Substituting $v = \sqrt{2 K/m}$ leads to
+the **Maxwell-Boltzmann kinetic energy distribution**:
+
+$$\begin{aligned}
+    \boxed{
+        f(K)
+        = 2 \sqrt{\frac{K}{\pi}} \: \bigg( \frac{1}{k_B T} \bigg)^{3/2} \exp\!\Big( \!-\!\frac{K}{k_B T} \Big)
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  H. Gould, J. Tobochnik,
+    *Statistical and thermal physics*, 2nd edition,
+    Princeton.
diff --git a/content/know/concept/newtons-bucket/index.pdc b/content/know/concept/newtons-bucket/index.pdc
new file mode 100644
index 0000000..3f074f5
--- /dev/null
+++ b/content/know/concept/newtons-bucket/index.pdc
@@ -0,0 +1,97 @@
+---
+title: "Newton's bucket"
+firstLetter: "N"
+publishDate: 2021-05-13
+categories:
+- Physics
+- Fluid mechanics
+- Fluid statics
+
+date: 2021-05-13T17:06:45+02:00
+draft: false
+markup: pandoc
+---
+
+# Newton's bucket
+
+**Newton's bucket** is a cylindrical bucket
+that rotates at angular velocity $\omega$.
+Due to [viscosity](/know/concept/viscosity/),
+any liquid in the bucket is affected by the rotation,
+eventually achieving the exact same $\omega$.
+
+However, once in equilibrium, the liquid's surface is not flat,
+but curved upwards from the center.
+This is due to the centrifugal force $\va{F}_\mathrm{f} = m \va{f}$ on a molecule with mass $m$:
+
+$$\begin{aligned}
+    \va{f}
+    = \omega^2 \va{r}
+\end{aligned}$$
+
+Where $\va{r}$ is the molecule's position relative to the axis of rotation.
+This (fictitious) force can be written as the gradient
+of a potential $\Phi_\mathrm{f}$, such that $\va{f} = - \nabla \Phi_\mathrm{f}$:
+
+$$\begin{aligned}
+    \Phi_\mathrm{f}
+    = - \frac{\omega^2}{2} r^2
+    = - \frac{\omega^2}{2} (x^2 + y^2)
+\end{aligned}$$
+
+In addition, each molecule feels a gravitational force $\va{F}_\mathrm{g} = m \va{g}$,
+where $\va{g} = - \nabla \Phi_\mathrm{g}$:
+
+$$\begin{aligned}
+    \Phi_\mathrm{g}
+    = \mathrm{g} z
+\end{aligned}$$
+
+Overall, the molecule therefore feels an "effective" force
+with a potential $\Phi$ given by:
+
+$$\begin{aligned}
+    \Phi
+    = \Phi_\mathrm{g} + \Phi_\mathrm{f}
+    = \mathrm{g} z - \frac{\omega^2}{2} (x^2 + y^2)
+\end{aligned}$$
+
+At equilibrium, the [hydrostatic pressure](/know/concept/hydrostatic-pressure/) $p$
+in the liquid is the one that satisfies:
+
+$$\begin{aligned}
+    \frac{\nabla p}{\rho}
+    = - \nabla \Phi
+\end{aligned}$$
+
+Removing the gradients gives integration constants $p_0$ and $\Phi_0$,
+so the equilibrium equation is:
+
+$$\begin{aligned}
+    p - p_0
+    = - \rho (\Phi - \Phi_0)
+\end{aligned}$$
+
+We isolate this for $p$ and rewrite $\Phi_0 = \mathrm{g} z_0$,
+where $z_0$ is the liquid height at the center:
+
+$$\begin{aligned}
+    p
+    = p_0 - \rho \mathrm{g} (z - z_0) + \frac{\omega^2}{2} \rho (x^2 + y^2)
+\end{aligned}$$
+
+At the surface, we demand that $p = p_0$, where $p_0$ is the air pressure.
+The $z$-coordinate at which this is satisfied is as follows,
+telling us that the surface is parabolic:
+
+$$\begin{aligned}
+    z
+    = z_0 + \frac{\omega^2}{2 \mathrm{g}} (x^2 + y^2)
+\end{aligned}$$
+
+
+
+## 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/stokes-law/index.pdc b/content/know/concept/stokes-law/index.pdc
index acc98af..dee40d6 100644
--- a/content/know/concept/stokes-law/index.pdc
+++ b/content/know/concept/stokes-law/index.pdc
@@ -356,8 +356,10 @@ This is an equation for the **terminal velocity** $U_t$,
 which we find to be as follows:
 
 $$\begin{aligned}
-    U_t
-    = \frac{2 a^2 (\rho_s - \rho_f) g_0}{9 \eta}
+    \boxed{
+        U_t
+        = \frac{2 a^2 (\rho_s - \rho_f) g_0}{9 \eta}
+    }
 \end{aligned}$$
 
 The falling sphere will accelerate until $U_t$,
diff --git a/content/know/concept/superdense-coding/index.pdc b/content/know/concept/superdense-coding/index.pdc
index e50be2b..1a9337c 100644
--- a/content/know/concept/superdense-coding/index.pdc
+++ b/content/know/concept/superdense-coding/index.pdc
@@ -32,12 +32,33 @@ 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 |
-|:--:|:--|:---------|
-| $00$ | $\hat{I}$ | $\ket*{\Phi^{+}} = \frac{1}{\sqrt{2}} \Big(\ket{0}_A \ket{0}_B + \ket{1}_A \ket{1}_B \Big)$ |
-| $01$ | $\hat{\sigma}_z$ | $\ket*{\Phi^{-}} = \frac{1}{\sqrt{2}} \Big(\ket{0}_A \ket{0}_B - \ket{1}_A \ket{1}_B \Big)$ |
-| $10$ | $\hat{\sigma}_x$ | $\ket*{\Psi^{+}} = \frac{1}{\sqrt{2}} \Big(\ket{0}_A \ket{1}_B + \ket{1}_A \ket{0}_B \Big)$ |
-| $11$ | $\hat{\sigma}_x \hat{\sigma}_z$ | $\ket*{\Psi^{-}} = \frac{1}{\sqrt{2}} \Big(\ket{0}_A \ket{1}_B - \ket{1}_A \ket{0}_B \Big)$ |
+<table style="width:70%;margin:auto;text-align:center;">
+    <tr>
+        <th>$(a_1, a_2)$</th>
+        <th>Operator</th>
+        <th>Result</th>
+    </tr>
+    <tr>
+        <td>$00$</td>
+        <td>$\hat{I}$</td>
+        <td>$\ket*{\Phi^{+}} = \frac{1}{\sqrt{2}} \Big(\ket{0}_A \ket{0}_B + \ket{1}_A \ket{1}_B \Big)$</td>
+    </tr>
+    <tr>
+        <td>$01$</td>
+        <td>$\hat{\sigma}_z$</td>
+        <td>$\ket*{\Phi^{-}} = \frac{1}{\sqrt{2}} \Big(\ket{0}_A \ket{0}_B - \ket{1}_A \ket{1}_B \Big)$</td>
+    </tr>
+    <tr>
+        <td>$10$</td>
+        <td>$\hat{\sigma}_x$</td>
+        <td>$\ket*{\Psi^{+}} = \frac{1}{\sqrt{2}} \Big(\ket{0}_A \ket{1}_B + \ket{1}_A \ket{0}_B \Big)$</td>
+    </tr>
+    <tr>
+        <td>$11$</td>
+        <td>$\hat{\sigma}_x \hat{\sigma}_z$</td>
+        <td>$\ket*{\Psi^{-}} = \frac{1}{\sqrt{2}} \Big(\ket{0}_A \ket{1}_B - \ket{1}_A \ket{0}_B \Big)$</td>
+    </tr>
+</table>
 
 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,