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## PHY452H1S Basic Statistical Mechanics. Lecture 4: Maxwell distribution for gases. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on January 17, 2013

# Disclaimer

Peeter’s lecture notes from class. May not be entirely coherent.

# Review: Lead up to Maxwell distribution for gases

For the random walk, after a number of collisions $N_{\mathrm{c}}$, we found that a particle (labeled the $i$th) will have a velocity

\begin{aligned}\mathbf{v}_i(N_{\mathrm{c}}) = \mathbf{v}_i(0) + \sum_{l = 1}^{N_{\mathrm{c}}} \Delta \mathbf{v}_i(l)\end{aligned} \hspace{\stretch{1}}(1.1.1)

We argued that the probability distribution for finding a velocity $\mathbf{v}$ was as in

\begin{aligned}\mathcal{P}_{N_{\mathrm{c}}}(\mathbf{v}_i) \propto \exp\left(-\frac{(\mathbf{v}_i - \mathbf{v}_i(0))^2}{ 2 N_{\mathrm{c}}}\right)\end{aligned} \hspace{\stretch{1}}(1.1.2)

Fig1: Velocity distribution found without considering kinetic energy conservation

# What went wrong?

However, we know that this must be wrong, since we require

\begin{aligned}T = \frac{1}{{2}} \sum_{i = 1}^{n} \mathbf{v}_i^2 = \mbox{conserved}\end{aligned} \hspace{\stretch{1}}(1.2.3)

Where our argument went wrong is that when the particle has a greater than average velocity, the effect of a collision will be to slow it down. We have to account for

1. Fluctuations $\rightarrow$ “random walk”
2. Dissipation $\rightarrow$ “slowing down”

There were two ingredients to diffusion (the random walk), these were

1. Conservation of particles

\begin{aligned}\frac{\partial {c}}{\partial {t}} + \frac{\partial {j}}{\partial {x}} = 0\end{aligned} \hspace{\stretch{1}}(1.2.4)

We can also think about a conservation of a particles in a velocity space

\begin{aligned}\frac{\partial {c}}{\partial {t}}(v, t) + \frac{\partial {j_v}}{\partial {v}} = 0\end{aligned} \hspace{\stretch{1}}(1.2.5)

where $j_v$ is a probability current in this velocity space.

2. Fick’s law in velocity space takes the form

\begin{aligned}j_v = -D \frac{\partial {c}}{\partial {v}}(v, t)\end{aligned} \hspace{\stretch{1}}(1.2.6)

The diffusion results in an “attempt” to flatten the distribution of the concentration as in (Fig 2).

Fig2: A friction like term is require to oppose the diffusion pressure

We’d like to add to the diffusion current an extra frictional like term

\begin{aligned}j_v = -D \frac{\partial {c}}{\partial {v}}(v, t) - \eta v c(v)\end{aligned} \hspace{\stretch{1}}(1.2.7)

We want something directed opposite to the velocity and the concentration

\begin{aligned}\text{Diffusion current} \equiv -D \frac{\partial {c}}{\partial {v}}(v, t)\end{aligned} \hspace{\stretch{1}}(1.0.8a)

\begin{aligned}\text{Dissipation current} \equiv - \eta v c(v, t)\end{aligned} \hspace{\stretch{1}}(1.0.8b)

This gives

\begin{aligned}\frac{\partial {c}}{\partial {t}}(v, t) &= -\frac{\partial {j_v}}{\partial {v}} \\ &= -\frac{\partial {}}{\partial {v}} \left(-D \frac{\partial {c}}{\partial {v}}(v, t) - \eta v c(v)\right) \\ &= D \frac{\partial^2 {{c}}}{\partial {{v}}^2}(v, t) + \eta \frac{\partial {}}{\partial {v}}\left(v c(v, t)\right)\end{aligned} \hspace{\stretch{1}}(1.0.8b)

Can we find a steady state solution to this equation when $t \rightarrow \infty$? For such a steady state we have

\begin{aligned}0 = \frac{d^2 c}{dv^2} + \eta \frac{d}{dv} \left(v c\right)\end{aligned} \hspace{\stretch{1}}(1.0.10)

Integrating once we have

\begin{aligned}\frac{d c}{dv} = -\eta v c + \text{constant}\end{aligned} \hspace{\stretch{1}}(1.0.11)

supposing that $dc/dv = 0$ at $v = 0$, integrating once more we have

\begin{aligned}\boxed{c(v) \propto \exp\left(- \frac{\eta v^2}{2 D}\right).}\end{aligned} \hspace{\stretch{1}}(1.0.12)

This is the Maxwell-Boltzmann distribution, illustrated in (Fig3).

Fig3: Maxwell-Boltzmann distribution

The concentration $c(v)$ has a probability distribution.

Calculating $\left\langle{{v^2}}\right\rangle$ from this distribution we can identify the $D/\eta$ factor.

\begin{aligned}\left\langle{{v^2}}\right\rangle &= \frac{\int v^2 e^{- \eta v^2/2D} dv}{\int e^{- \eta v^2/2D} dv} \\ &= \frac{\frac{D}{\eta} \int v \frac{d}{dv} \left( -e^{- \eta v^2/2D} \right) dv}{\int e^{- \eta v^2/2D} dv} \\ &= -\frac{D}{\eta} \frac{\int -e^{- \eta v^2/2D} dv}{\int e^{- \eta v^2/2D} dv} \\ &= \frac{D}{\eta}.\end{aligned} \hspace{\stretch{1}}(1.13)

This also happens to be the energy in terms of temperature (we can view this as a definition of the temperature for now), writing

\begin{aligned}\frac{1}{{2}} m \left\langle{{\mathbf{v}^2}}\right\rangle = \frac{1}{{2}} m \left( \frac{D}{\eta} \right) = \frac{1}{{2}} k_{\mathrm{B}} T\end{aligned} \hspace{\stretch{1}}(1.0.14)

Here

\begin{aligned}k_{\mathrm{B}} = \mbox{Boltzmann constant}\end{aligned} \hspace{\stretch{1}}(1.0.15a)

\begin{aligned}T = \mbox{absolute temperature}\end{aligned} \hspace{\stretch{1}}(1.0.15b)

Fluctuations $\leftrightarrow$ Dissipation

\begin{aligned}\boxed{\frac{D}{\eta} = \frac{ k_{\mathrm{B}} T}{m}}\end{aligned} \hspace{\stretch{1}}(1.0.16)

This is a specific example of the more general Fluctuation-Dissipation theorem.

Generalizing to 3D

Fick’s law and the continuity equation in 3D are respectively

\begin{aligned}\mathbf{j} = -D \boldsymbol{\nabla}_\mathbf{v} c(\mathbf{v}, t) - \eta \mathbf{v} c(\mathbf{v}, t)\end{aligned} \hspace{\stretch{1}}(1.0.17a)

\begin{aligned}\frac{\partial {}}{\partial {t}} c(\mathbf{v}, t) + \boldsymbol{\nabla}_\mathbf{v} \cdot \mathbf{j}(\mathbf{v}, t) = 0\end{aligned} \hspace{\stretch{1}}(1.0.17b)

As above we have for the steady state

\begin{aligned}0 &= \frac{\partial {}}{\partial {t}} c(\mathbf{v}, t) \\ &= \boldsymbol{\nabla}_\mathbf{v} \cdot \left( -D \boldsymbol{\nabla}_\mathbf{v} c(\mathbf{v}, t) - \eta \mathbf{v} c(\mathbf{v}, t)\right) \\ &= -D \boldsymbol{\nabla}^2_\mathbf{v} c - \eta \boldsymbol{\nabla}_\mathbf{v} \cdot (\mathbf{v} c)\end{aligned} \hspace{\stretch{1}}(1.0.17b)

Integrating once over all space

\begin{aligned}D \boldsymbol{\nabla}_\mathbf{v} c = -\eta \mathbf{v} c + \text{vector constant, assumed zero}\end{aligned} \hspace{\stretch{1}}(1.0.19)

This is three sets of equations, one for each component $v_\alpha$ of $\mathbf{v}$

\begin{aligned}\frac{\partial {c}}{\partial {v_\alpha}} = -\frac{\eta}{D} v_\alpha c \end{aligned} \hspace{\stretch{1}}(1.0.20)

So that our steady state equation is

\begin{aligned}c(\mathbf{v}, t \rightarrow \infty) \propto \exp\left(-\frac{v_x^2 +v_y^2 +v_z^2 }{ 2 (D/\eta) }\right).\end{aligned} \hspace{\stretch{1}}(1.0.21)

Computing the average 3D squared velocity for this distribution, we have

\begin{aligned}\frac{1}{{2}} m \left\langle{{\mathbf{v} \cdot \mathbf{v}}}\right\rangle &= \frac{ \int dv_x dv_y dv_z \left( v_x^2 + v_y^2 + v_z^2 \right) \exp \left( -\frac{ v_x^2 + v_y^2 + v_z^2 }{ 2 (D/\eta) } \right)}{ \int dv_x dv_y dv_z \exp \left( -\frac{ v_x^2 + v_y^2 + v_z^2 }{ 2 (D/\eta) } \right)} \\ &= \frac{ \int dv_x v_x^2 \exp \left( -\frac{ v_x^2 }{ 2 (D/\eta) } \right)}{ \int dv_x \exp \left( -\frac{ v_x^2 }{ 2 (D/\eta) } \right)}+\frac{ \int dv_y v_y^2 \exp \left( -\frac{ v_y^2 }{ 2 (D/\eta) } \right)}{ \int dv_y \exp \left( -\frac{ v_y^2 }{ 2 (D/\eta) } \right)}+\frac{ \int dv_z v_z^2 \exp \left( -\frac{ v_z^2 }{ 2 (D/\eta) } \right)}{ \int dv_z \exp \left( -\frac{ v_z^2 }{ 2 (D/\eta) } \right)} \\ &= 3 \left( \frac{1}{2} k_{\mathrm{B}} T\right) \\ &= \frac{3}{2} k_{\mathrm{B}} T\end{aligned} \hspace{\stretch{1}}(1.0.21)

For each component $v_\alpha$ the normalization kills off all the contributions for the other components, leaving us with the usual $3 k_{\mathrm{B}} T/2$ ideal gas law kinetic energy.

# Phase space

Now let’s switch directions a bit and look at how to examine a more general system described by the phase space of generalized coordinates

\begin{aligned}\{ x_{i_\alpha}(t),p_{i_\alpha}(t) \}\end{aligned} \hspace{\stretch{1}}(1.0.23)

Here

1. $i = \mbox{molecule or particle number}$
2. $\alpha \in \{x, y, z\}$
3. $\mbox{Dimension} = N_{\text{particles}} \times 2 \times d$, where $d$ is the physical space dimension.

The motion in phase space will be governed by the knowledge of how each of these coordinates change for all the particles. Assuming a Hamiltonian $H$ and recalling that $H = \dot{x} p - \mathcal{L}$, gives us

\begin{aligned}\frac{\partial {H}}{\partial {p}} = \dot{x}\end{aligned} \hspace{\stretch{1}}(1.0.24a)

\begin{aligned}\frac{\partial {H}}{\partial {x}} = -\frac{\partial {\mathcal{L}}}{\partial {x}} = - \frac{d{{p}}}{dt},\end{aligned} \hspace{\stretch{1}}(1.0.24b)

we have for the following set of equations describing the entire system

\begin{aligned}\frac{d{{}}}{dt} x_{i_\alpha}(t) = \frac{\partial {H}}{\partial {p_{i_\alpha}}}\end{aligned} \hspace{\stretch{1}}(1.0.25a)

\begin{aligned}\frac{d{{}}}{dt} p_{i_\alpha}(t)= -\frac{\partial {H}}{\partial {x_{i_\alpha}}}\end{aligned} \hspace{\stretch{1}}(1.0.25b)

Example, 1D SHO

\begin{aligned}H = \frac{1}{{2}} \frac{p^2}{m} + \frac{1}{{2}} k x^2\end{aligned} \hspace{\stretch{1}}(1.0.26)

This has phase space trajectories as in (Fig4).

Fig4: Classical SHO phase space trajectories