🎈 Ideal gas

Two-dimensional Boltzmann gas

Maxwell-Boltzmann 2D vis-à-vis 3D

Maxwell-Boltzmann velocity distributions for $\overrightarrow{v}$

The velocity distribution for $f(\overrightarrow{v}) , d^n\overrightarrow{v}$ is given by

3D case

$$f(\overrightarrow{v}) d^3\overrightarrow{v} = \left(\frac{m}{2 \pi k_B T}\right)^{3/2} \exp\Big(-\frac{m v^2}{2 k_B T}\Big) d^3\overrightarrow{v}$$

2D case

$$f(\overrightarrow{v}) d^2\overrightarrow{v} = \left(\frac{m}{2 \pi k_B T}\right) \exp\Big(-\frac{m v^2}{2 k_B T}\Big) d^2\overrightarrow{v}$$

• We have $v^2 = v_x^2 + v_y^2 + v_z^2$ in 3 dimensions and $v^2 = v_x^2 + v_y^2$ in 2 dimensions.
• Normalisation because the integral volume $d^n\overrightarrow{v}$ depends on the dimension.

Maxwell-Boltzmann velocity distributions for $\|\overrightarrow{v}\|$

In the graphs we reduce to a radial distribution, so we take the “surface of the cirkel/sphere”:

$$d^n\overrightarrow{v} = v^{n-1} dv , d\Omega_n$$

where $d\Omega_n$ denotes the angular part. This amounts to an additional factor $v$ for 2D, $v^2$ for 3D.

3D case

$$f(v) dv = 4\pi \left(\frac{m}{2 \pi k_B T}\right)^{3/2} v^2 \exp\Big(-\frac{m v^2}{2 k_B T}\Big) dv$$

2D case

$$f(v) dv = \frac{m}{k_B T} , v , \exp\Big(-\frac{m v^2}{2 k_B T}\Big) dv$$

• Factor $v^{n-1}$ originates from the velocities with the same speeds (surface of a cirkel/sphere).
• This is what is used in the code, e.g. value = (v / T) * exp(-v*v / (2 * T)) in 2D.