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Diffstat (limited to 'source/know/concept/boltzmann-equation')
| -rw-r--r-- | source/know/concept/boltzmann-equation/index.md | 85 |
1 files changed, 44 insertions, 41 deletions
diff --git a/source/know/concept/boltzmann-equation/index.md b/source/know/concept/boltzmann-equation/index.md index 20399df..9ed2fd2 100644 --- a/source/know/concept/boltzmann-equation/index.md +++ b/source/know/concept/boltzmann-equation/index.md @@ -10,17 +10,17 @@ layout: "concept" --- Consider a collection of particles, -each with its own position $\vb{r}$ and velocity $\vb{v}$. -We can thus define a probability density function $f(\vb{r}, \vb{v}, t)$ -describing the expected number of particles at $(\vb{r}, \vb{v})$ at time $t$. -Let the total number of particles $N$ be conserved, then clearly: +each with its own position $$\vb{r}$$ and velocity $$\vb{v}$$. +We can thus define a probability density function $$f(\vb{r}, \vb{v}, t)$$ +describing the expected number of particles at $$(\vb{r}, \vb{v})$$ at time $$t$$. +Let the total number of particles $$N$$ be conserved, then clearly: $$\begin{aligned} N = \iint_{-\infty}^\infty f(\vb{r}, \vb{v}, t) \dd{\vb{r}} \dd{\vb{v}} \end{aligned}$$ At equilibrium, all processes affecting the particles -no longer have a net effect, so $f$ is fixed: +no longer have a net effect, so $$f$$ is fixed: $$\begin{aligned} \dv{f}{t} @@ -35,7 +35,7 @@ $$\begin{aligned} = \bigg(\! \pdv{f}{t} \!\bigg)_\mathrm{\!col} \end{aligned}$$ -Where the right-hand side simply means "all changes in $f$ due to collisions". +Where the right-hand side simply means "all changes in $$f$$ due to collisions". Applying the chain rule to the left-hand side then yields: $$\begin{aligned} @@ -49,8 +49,8 @@ $$\begin{aligned} &= \pdv{f}{t} + \vb{v} \cdot \nabla f + \vb{a} \cdot \pdv{f}{\vb{v}} \end{aligned}$$ -Where we have introduced the shorthand $\ipdv{f}{\vb{v}}$. -Inserting Newton's second law $\vb{F} = m \vb{a}$ +Where we have introduced the shorthand $$\ipdv{f}{\vb{v}}$$. +Inserting Newton's second law $$\vb{F} = m \vb{a}$$ leads us to the **Boltzmann equation** or **Boltzmann transport equation** (BTE): @@ -64,41 +64,41 @@ $$\begin{aligned} But what about the collision term? Expressions for it exist, which are almost exact in many cases, but unfortunately also quite difficult to work with. -In addition, $f$ is a 7-dimensional function, +In addition, $$f$$ is a 7-dimensional function, so the BTE is already hard to solve without collisions. We only present the simplest case, known as the **Bhatnagar-Gross-Krook approximation**: -if the equilibrium state $f_0(\vb{r}, \vb{v})$ is known, -then each collision brings the system closer to $f_0$: +if the equilibrium state $$f_0(\vb{r}, \vb{v})$$ is known, +then each collision brings the system closer to $$f_0$$: $$\begin{aligned} \pdv{f}{t} + \vb{v} \cdot \nabla f + \frac{\vb{F}}{m} \cdot \pdv{f}{\vb{v}} = \frac{f_0 - f}{\tau} \end{aligned}$$ -Where $\tau$ is the average collision period. +Where $$\tau$$ is the average collision period. The right-hand side is called the **Krook term**. ## Moment equations -From the definition of $f$, -we see that integrating over all $\vb{v}$ yields the particle density $n$: +From the definition of $$f$$, +we see that integrating over all $$\vb{v}$$ yields the particle density $$n$$: $$\begin{aligned} n(\vb{r}, t) = \int_{-\infty}^\infty f(\vb{r}, \vb{v}, t) \dd{\vb{v}} \end{aligned}$$ -Consequently, a purely velocity-dependent quantity $Q(\vb{v})$ can be averaged like so: +Consequently, a purely velocity-dependent quantity $$Q(\vb{v})$$ can be averaged like so: $$\begin{aligned} \Expval{Q} = \frac{1}{n} \int_{-\infty}^\infty Q(\vb{r}, \vb{v}, t) \: f(\vb{r}, \vb{v}, t) \dd{\vb{v}} \end{aligned}$$ -With that in mind, we multiply the collisionless BTE equation by $Q(\vb{v})$ and integrate, -assuming that $\vb{F}$ does not depend on $\vb{v}$: +With that in mind, we multiply the collisionless BTE equation by $$Q(\vb{v})$$ and integrate, +assuming that $$\vb{F}$$ does not depend on $$\vb{v}$$: $$\begin{aligned} 0 @@ -110,10 +110,10 @@ $$\begin{aligned} + \frac{\vb{F}}{m} \cdot \int \bigg( \pdv{}{\vb{v}} (Q f) - f \pdv{Q}{\vb{v}} \bigg) \dd{\vb{v}} \end{aligned}$$ -The first integral is simply $n \Expval{Q}$. -In the second integral, note that $\vb{v}$ is a coordinate -and hence not dependent on $\vb{r}$, so $\nabla \cdot \vb{v} = 0$. -Since $f$ is a probability density, $f \to 0$ for $\vb{v} \to \pm\infty$, +The first integral is simply $$n \Expval{Q}$$. +In the second integral, note that $$\vb{v}$$ is a coordinate +and hence not dependent on $$\vb{r}$$, so $$\nabla \cdot \vb{v} = 0$$. +Since $$f$$ is a probability density, $$f \to 0$$ for $$\vb{v} \to \pm\infty$$, so the first term in the third integral vanishes after it is integrated: $$\begin{aligned} @@ -134,9 +134,9 @@ $$\begin{aligned} } \end{aligned}$$ -If we set $Q = m$, then the mass density $\rho = n \Expval{Q}$, +If we set $$Q = m$$, then the mass density $$\rho = n \Expval{Q}$$, and we find that the **zeroth moment** of the BTE describes conservation of mass, -where $\vb{V} \equiv \Expval{\vb{v}} = \int \vb{v} f \dd{\vb{v}}$ is the fluid velocity: +where $$\vb{V} \equiv \Expval{\vb{v}} = \int \vb{v} f \dd{\vb{v}}$$ is the fluid velocity: $$\begin{aligned} \boxed{ @@ -150,8 +150,8 @@ $$\begin{aligned} <label for="proof-moment0">Proof</label> <div class="hidden" markdown="1"> <label for="proof-moment0">Proof.</label> -We insert $Q = m$ into our prototype, -and since $m$ is constant, the rest is trivial: +We insert $$Q = m$$ into our prototype, +and since $$m$$ is constant, the rest is trivial: $$\begin{aligned} 0 @@ -159,12 +159,13 @@ $$\begin{aligned} \\ &= \pdv{\rho}{t} + \nabla \cdot \big(\rho \Expval{\vb{v}}\big) - 0 \end{aligned}$$ + </div> </div> -If we instead choose the momentum $Q = m \vb{v}$, +If we instead choose the momentum $$Q = m \vb{v}$$, we find that the **first moment** of the BTE describes conservation of momentum, -where $\hat{P}$ is the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/): +where $$\hat{P}$$ is the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/): $$\begin{aligned} \boxed{ @@ -178,7 +179,7 @@ $$\begin{aligned} <label for="proof-moment1">Proof</label> <div class="hidden" markdown="1"> <label for="proof-moment1">Proof.</label> -We insert $Q = m \vb{v}$ into our prototype and recognize $\rho$ wherever possible: +We insert $$Q = m \vb{v}$$ into our prototype and recognize $$\rho$$ wherever possible: $$\begin{aligned} 0 @@ -189,11 +190,11 @@ $$\begin{aligned} - \vb{F} \cdot \bigg( n \Expval{\pdv{\vb{v}}{\vb{v}}} \bigg) \end{aligned}$$ -With $\vb{v} \vb{v}$ being a dyadic product. +With $$\vb{v} \vb{v}$$ being a dyadic product. To give it a physical interpretation, -we split $\vb{v} = \vb{V} \!+\! \vb{w}$, -where $\vb{V}$ is the average velocity vector, -and $\vb{w}$ is the local deviation from $\vb{V}$: +we split $$\vb{v} = \vb{V} \!+\! \vb{w}$$, +where $$\vb{V}$$ is the average velocity vector, +and $$\vb{w}$$ is the local deviation from $$\vb{V}$$: $$\begin{aligned} \Expval{\vb{v} \vb{v}} @@ -202,7 +203,7 @@ $$\begin{aligned} = \vb{V} \vb{V} + 2 \vb{V} \Expval{\vb{w}} + \Expval{\vb{w} \vb{w}} \end{aligned}$$ -Since $\vb{w}$ represents a deviation from the mean, $\Expval{\vb{w}} = 0$. +Since $$\vb{w}$$ represents a deviation from the mean, $$\Expval{\vb{w}} = 0$$. We define the pressure tensor: $$\begin{aligned} @@ -212,19 +213,20 @@ $$\begin{aligned} \end{aligned}$$ This leads to the expected result, -where $\nabla \cdot (\rho \vb{V}\vb{V})$ represents the fluid momentum, -and $\nabla \cdot \hat{P}$ the viscous/pressure momentum: +where $$\nabla \cdot (\rho \vb{V}\vb{V})$$ represents the fluid momentum, +and $$\nabla \cdot \hat{P}$$ the viscous/pressure momentum: $$\begin{aligned} 0 &= \pdv{}{t}\big(\rho \vb{V}\big) + \nabla \cdot \big(\rho \vb{V} \vb{V} + \hat{P}\big) - n \vb{F} \end{aligned}$$ + </div> </div> -Finally, if we choose the kinetic energy $Q = m |\vb{v}|^2 / 2$, +Finally, if we choose the kinetic energy $$Q = m |\vb{v}|^2 / 2$$, we find that the **second moment** gives conservation of energy, -where $U$ is the thermal energy density and $\vb{J}$ is the heat flux: +where $$U$$ is the thermal energy density and $$\vb{J}$$ is the heat flux: $$\begin{aligned} \boxed{ @@ -240,7 +242,7 @@ $$\begin{aligned} <label for="proof-moment2">Proof</label> <div class="hidden" markdown="1"> <label for="proof-moment2">Proof.</label> -We insert $Q = m |\vb{v}|^2 / 2$ into our prototype and recognize $\rho$ wherever possible: +We insert $$Q = m |\vb{v}|^2 / 2$$ into our prototype and recognize $$\rho$$ wherever possible: $$\begin{aligned} 0 @@ -253,7 +255,7 @@ $$\begin{aligned} - \frac{\vb{F}}{2} \cdot \bigg( n \Expval{\pdv{|\vb{v}|^2}{\vb{v}}} \bigg) \end{aligned}$$ -We handle these terms one by one. Substituting $\vb{v} = \vb{V} + \vb{w}$ in the first gives: +We handle these terms one by one. Substituting $$\vb{v} = \vb{V} + \vb{w}$$ in the first gives: $$\begin{aligned} \Expval{|\vb{v}|^2} @@ -265,7 +267,7 @@ $$\begin{aligned} \end{aligned}$$ And likewise for the second term, -where we recognize the stress tensor $\Expval{\vb{w} \vb{w}}$: +where we recognize the stress tensor $$\Expval{\vb{w} \vb{w}}$$: $$\begin{aligned} \Expval{|\vb{v}|^2 \vb{v}} @@ -294,7 +296,7 @@ $$\begin{aligned} \end{aligned}$$ To clarify the physical interpretation, -we define $U$, $\vb{J}$ and $\hat{P}$ as follows: +we define $$U$$, $$\vb{J}$$ and $$\hat{P}$$ as follows: $$\begin{aligned} U @@ -347,6 +349,7 @@ $$\begin{aligned} \end{bmatrix} = \sum_{i=1}^{3} \sum_{j=1}^{3} \pdv{P_{ij}}{x_j} V_i \end{aligned}$$ + </div> </div> |