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authorPrefetch2022-12-20 20:11:25 +0100
committerPrefetch2022-12-20 20:11:25 +0100
commit1d700ab734aa9b6711eb31796beb25cb7659d8e0 (patch)
treeefdd26b83be1d350d7c6c01baef11a54fa2c5b36 /source/know/concept/maxwell-boltzmann-distribution
parenta39bb3b8aab1aeb4fceaedc54c756703819776c3 (diff)
More improvements to knowledge base
Diffstat (limited to 'source/know/concept/maxwell-boltzmann-distribution')
-rw-r--r--source/know/concept/maxwell-boltzmann-distribution/index.md85
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**:
$$\begin{aligned}
\boxed{
f(K)
- = 2 \sqrt{\frac{K}{\pi}} \: \bigg( \frac{1}{k_B T} \bigg)^{3/2} \exp\!\Big( \!-\!\frac{K}{k_B T} \Big)
+ = 2 \sqrt{\frac{K}{\pi}} \Big( \frac{1}{k_B T} \Big)^{3/2} \exp\!\bigg( \!-\!\frac{K}{k_B T} \bigg)
}
\end{aligned}$$