More improvements to knowledge base

1 files changed, 42 insertions, 43 deletions

diff --git a/source/know/concept/maxwell-boltzmann-distribution/index.md b/source/know/concept/maxwell-boltzmann-distribution/index.md
index 318e659..946525c 100644
--- a/source/know/concept/maxwell-boltzmann-distribution/index.md
+++ b/source/know/concept/maxwell-boltzmann-distribution/index.md
@@ -13,6 +13,7 @@ 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](/know/concept/canonical-ensemble/)
@@ -24,55 +25,51 @@ $$\begin{aligned}
     \:\propto\: \exp\!\big(\!-\! \beta E\big)
 \end{aligned}$$
 
-Where $$\beta = 1 / k_B T$$.
-We split $$E = K + U$$,
-with $$K$$ and $$U$$ 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 $$\tilde{r}$$,
+Where $$\beta \equiv 1 / k_B T$$. We split $$E = K + U$$,
+where $$K$$ and $$U$$ are the total contributions
+from the kinetic and potential energies of the system.
+For $$N$$ particles
+with positions $$\va{r} \equiv (\vb{r}_1, ..., \vb{r}_N)$$
+and momenta $$\va{p} = (\vb{p}_1, ..., \vb{p}_N)$$,
+then $$K$$ only depends on $$\va{p}$$ and $$U$$ only on $$\va{r}$$,
 so the probability of a specific microstate
-$$(\tilde{r}, \tilde{p})$$ is as follows:
+$$(\va{r}, \va{p})$$ is as follows:
 
 $$\begin{aligned}
-    f(\tilde{r}, \tilde{p})
-    \:\propto\: \exp\!\Big(\!-\! \beta \big( K(\tilde{p}) + U(\tilde{r}) \big) \Big)
+    f(\va{r}, \va{p})
+    \:\propto\: \exp\!\Big(\!-\! \beta \big( K(\va{p}) + U(\va{r}) \big) \Big)
 \end{aligned}$$
 
-Since this is classical physics,
-we can split the exponential.
-In quantum mechanics,
-the canonical commutation relation would prevent that.
-Anyway, splitting yields:
+Since this is classical physics, we can split the exponential
+(in quantum mechanics, the canonical commutation relation would prevent that):
 
 $$\begin{aligned}
-    f(\tilde{r}, \tilde{p})
-    \:\propto\: \exp\!\big(\!-\! \beta K(\tilde{p}) \big) \exp\!\big(\!-\! \beta U(\tilde{r}) \big)
+    f(\va{r}, \va{p})
+    \:\propto\: \exp\!\big(\!-\! \beta K(\va{p}) \big) \exp\!\big(\!-\! \beta U(\va{r}) \big)
 \end{aligned}$$
 
 Classically, the probability
 distributions of the momenta and positions are independent:
 
 $$\begin{aligned}
-    f_K(\tilde{p})
-    \:\propto\: \exp\!\big(\!-\! \beta K(\tilde{p}) \big)
+    f_K(\va{p})
+    \:\propto\: \exp\!\big(\!-\! \beta K(\va{p}) \big)
     \qquad \qquad
-    f_U(\tilde{r})
-    \:\propto\: \exp\!\big(\!-\! \beta U(\tilde{r}) \big)
+    f_U(\va{r})
+    \:\propto\: \exp\!\big(\!-\! \beta U(\va{r}) \big)
 \end{aligned}$$
 
-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:
+We cannot evaluate $$f_U(\va{r})$$ further without knowing $$U(\va{r})$$ for a system.
+We thus turn to $$f_K(\va{p})$$, and see that the total kinetic
+energy $$K(\va{p})$$ is simply the sum of the particles' individual
+kinetic energies $$K_n(\vb{p}_n)$$, which are well-known:
 
 $$\begin{aligned}
-    K(\tilde{p})
-    = \sum_{n = 1}^N K_n(\vec{p}_n)
+    K(\va{p})
+    = \sum_{n = 1}^N K_n(\vb{p}_n)
     \qquad \mathrm{where} \qquad
-    K_n(\vec{p}_n)
-    = \frac{|\vec{p}_n|^2}{2 m}
+    K_n(\vb{p}_n)
+    = \frac{|\vb{p}_n|^2}{2 m}
 \end{aligned}$$
 
 Consequently, the probability distribution $$f(p_x, p_y, p_z)$$ for the
@@ -100,10 +97,10 @@ so the velocity in each direction is independent of the others:
 
 $$\begin{aligned}
     f(v_x)
-    = \sqrt{\frac{m}{2 \pi k_B T}} \exp\!\Big( \!-\!\frac{m v_x^2}{2 k_B T} \Big)
+    = \sqrt{\frac{m}{2 \pi k_B T}} \exp\!\bigg( \!-\!\frac{m v_x^2}{2 k_B T} \bigg)
 \end{aligned}$$
 
-The distribution is thus an isotropic gaussian with standard deviations given by:
+The distribution is thus an isotropic Gaussian with standard deviations given by:
 
 $$\begin{aligned}
     \sigma_x = \sigma_y = \sigma_z
@@ -111,17 +108,18 @@ $$\begin{aligned}
 \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,
+That was the distribution of the velocities along each axis,
+but what about the speed $$v = |\vb{v}|$$?
+Because we do not care about the direction of $$\vb{v}$$, 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 $$v$$:
 
 $$\begin{aligned}
     g(v)
-    = \dv{}{v}\Big( \frac{4 \pi}{3} v^3 \Big)
+    = \dv{}{v} \bigg( \frac{4 \pi}{3} v^3 \bigg)
     = 4 \pi v^2
 \end{aligned}$$
 
@@ -132,7 +130,7 @@ then gives us the **Maxwell-Boltzmann speed distribution**:
 $$\begin{aligned}
     \boxed{
         f(v)
-        = 4 \pi \Big( \frac{m}{2 \pi k_B T} \Big)^{3/2} v^2 \exp\!\Big( \!-\!\frac{m v^2}{2 k_B T} \Big)
+        = 4 \pi \Big( \frac{m}{2 \pi k_B T} \Big)^{3/2} v^2 \exp\!\bigg( \!-\!\frac{m v^2}{2 k_B T} \bigg)
     }
 \end{aligned}$$
 
@@ -144,10 +142,10 @@ and the root-mean-square speed $$v_{\mathrm{rms}}$$:
 $$\begin{aligned}
     f'(v_\mathrm{mode})
     = 0
-    \qquad
+    \qquad \quad
     v_\mathrm{mean}
     = \int_0^\infty v \: f(v) \dd{v}
-    \qquad
+    \qquad \quad
     v_\mathrm{rms}
     = \bigg( \int_0^\infty v^2 \: f(v) \dd{v} \bigg)^{1/2}
 \end{aligned}$$
@@ -159,12 +157,12 @@ $$\begin{aligned}
         v_{\mathrm{mode}}
         = \sqrt{\frac{2 k_B T}{m}}
     }
-    \qquad
+    \qquad \quad
     \boxed{
         v_{\mathrm{mean}}
         = \sqrt{\frac{8 k_B T}{\pi m}}
     }
-    \qquad
+    \qquad \quad
     \boxed{
         v_{\mathrm{rms}}
         = \sqrt{\frac{3 k_B T}{m}}
@@ -172,6 +170,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 
+
 ## Kinetic energy distribution
 
 Using the speed distribution,
@@ -194,7 +193,7 @@ so the energy distribution $$f(K)$$ is:
 $$\begin{aligned}
     f(K)
     = \frac{f(v)}{m v}
-    = \sqrt{\frac{2 m}{\pi}} \: \bigg( \frac{1}{k_B T} \bigg)^{3/2} v \exp\!\Big( \!-\!\frac{m v^2}{2 k_B T} \Big)
+    = \sqrt{\frac{2 m}{\pi}} \Big( \frac{1}{k_B T} \Big)^{3/2} v \exp\!\bigg( \!-\!\frac{m v^2}{2 k_B T} \bigg)
 \end{aligned}$$
 
 Substituting $$v = \sqrt{2 K/m}$$ leads to
@@ -203,7 +202,7 @@ the **Maxwell-Boltzmann kinetic energy distribution**:
 $$\begin{aligned}
     \boxed{
         f(K)
-        = 2 \sqrt{\frac{K}{\pi}} \: \bigg( \frac{1}{k_B T} \bigg)^{3/2} \exp\!\Big( \!-\!\frac{K}{k_B T} \Big)
+        = 2 \sqrt{\frac{K}{\pi}} \Big( \frac{1}{k_B T} \Big)^{3/2} \exp\!\bigg( \!-\!\frac{K}{k_B T} \bigg)
     }
 \end{aligned}$$