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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

[Click here for a PDF of this post with nicer formatting]


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 ith) 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)


\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)

Equilibrium steady states

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)


  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


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