💼 Case: 2 molecules with $\vec v_1$ and $\vec v_2$

$$ \vec v_r=\vec v_1-\vec v_2 \quad ; \quad |\vec v_r|^2 =(\vec v_1-\vec v_2)\cdot(\vec v_1-\vec v_2)=|\vec v_1|^2+|\vec v_2|^2 -2(\vec v_1\cdot \vec v_2) $$

• Since motion is uncorrelated $\left < \vec v_1 \cdot\vec v_2\right >=0$ (cf. Boltzmann equation)

$$ \left < |\vec v_r |^2 \right > =\left < | \vec v_1 |^2 \right > + \left < | \vec v_2 |^2 \right > $$

• If molecules have the same mass $\left < |\vec v_1|^2 \right > =\left < |\vec v_2 |^2 \right >$

$$ \left < | \vec v_r |^2 \right > = 2 \left < |\vec v_1|^2 \right > =2\left < |\vec v_2|^2 \right > $$

• We can approximate $\left < |\vec v_r |^2 \right > \approx \left < | \vec v_r |\right > ^2$ (this means we assume small variance)

$$ \left < v_r \right > =\sqrt{\left < v_r^2 \right >}=\sqrt{2\left < v^2\right >}=\sqrt{\frac{6k_BT}{m}} \quad ; \quad \left < v^2 \right >=\frac{3k_BT}{m} $$

Mean collision time and mean free path

💼 Case: consider 2 molecules of radius $r_m$ in rest frame of molecule 2. Molecule 1 has velocity $\vec v_r$ relative to molecule 2.

• Collision cross-section $\sigma=\pi(2r_m)^2$
• In a time $t\to \text dt$, molecule 1 goes through the volume = $\sigma v_r\text dt$
• Probability that the 2 molecule is in this volume is $\frac NV \sigma v_r\text dt$
• $P_s(t)$ is the probability of molecule 1 ‘surviving’ for a time $t$ without collisions

$$ \begin{aligned} -\text d P_s(t)&=P_s(t)\frac{N}{V}\sigma v_r \text dt \\ \frac{\text dP_s}{P_s}&=-\frac NV \sigma v_r \text dt \\ P_s(t)&= e^{-\frac NV \sigma v_r t}= e^{-\frac{t}{\tau}} \end{aligned} $$

where $\tau=\frac{1}{\frac nv \sigma v_r}=\frac 1{\frac NV \sigma \sqrt 2 \left < v \right >}$ this is the mean collision time

• Mean free path $\lambda = \tau \left < v\right > =\frac 1{\sqrt 2 \frac NV \sigma}$

Viscosity

💼 Case: consider a gas between 2 plates

• The gas exerts a stress on the top plate

$$ F_{zx}=-\eta \frac{\partial u_x}{\partial z} $$

where $\eta$ is the viscosity