Imagine a world in which space is two-dimensional, but the laws of physics are otherwise the same. Derive the speed distribution formula for an ideal gas of nonrelativistic particles in this fictitious world, and sketch this distribution. Carefully explain the similarities and differences between the two-dimensional and three-dimensional cases. What is the most likely velocity vector? What is the most likely speed?

We need to find out the probability and most likely velocity for a world in which space is two-dimensional, but the laws of physics are otherwise the same.

The probability of a molecule in 2D plane is proportional to Boltzmann factor $e^{\raisebox{1ex}{$-m{v}^2$}\!\left/ \!\raisebox{-1ex}{$2kT$}\right.}$, in 2D plane the velocity life is inside the circle of radius $v$, which means that the no. of the velocity vectors is proportional to the circumference of this circle, i.e. $2\mathrm{\pi v}$, so we can give the probability as

role="math" localid="1650192711584" $P\left(v\right)\propto 2{\mathrm{\pi ve}}^{\raisebox{1ex}{$-{\mathrm{mv}}^2$}\!\left/ \!\raisebox{-1ex}{$2\mathrm{kT}$}\right.}$

$P\left(v\right)=C\times 2{\mathrm{\pi ve}}^{\raisebox{1ex}{$-{\mathrm{mv}}^2$}\!\left/ \!\raisebox{-1ex}{$2\mathrm{kT}$}\right.}$ Let it be equation 1

Where C is constant

To find this constant we use the fact that the integration of the probability all over space is unity,

So,

${\int }_0^{\infty }P\left(v\right)-C\times 2\mathrm{\pi }{\int }_0^{\infty }{\mathrm{ve}}^{\raisebox{1ex}{$-{\mathrm{mv}}^2$}\!\left/ \!\raisebox{-1ex}{$2\mathrm{kT}$}\right.}-1$

But,

$\int x{e}^{-a{x}^2}=-\frac{e^{-a{x}^2}}{2a}$

Here, we have $a=\frac{m}{2kT}$

So,

$$\begin{gathered} C\times 2\pi \left(\frac{kT}{m}\right){\left[e^{\raisebox{1ex}{$-{\mathrm{mv}}^2$}\!\left/ \!\raisebox{-1ex}{$2\mathrm{kT}$}\right.}\right]}_0^{\infty }=1 \\ C\times 2\pi \left(\frac{kT}{m}\right)\left[0-1\right]=1 \\ C\times 2\pi \left(\frac{kT}{m}\right)=1 \\ C=\frac{m}{2\mathrm{\pi kT}} \end{gathered}$$

Now, substitute above value of C in (1) to get,

$P\left(v\right)=\left(\frac{m}{2\mathrm{\pi kT}}\right)2{\mathrm{\pi ve}}^{\raisebox{1ex}{$-{\mathrm{mv}}^2$}\!\left/ \!\raisebox{-1ex}{$2\mathrm{kT}$}\right.}$

As in 3D plane the probability fall exponentially $\left(e^{-v^2}\right)$as $v\to \infty$.However, at low velocities the linear term dominate, so instead of the parabolic we have linear, plot the probability we set:

$v\text{'}=v\sqrt{\frac{m}{kT}}$

In terms of $v\text{'}$the probability can be given as,

$P\left(v\right)=\sqrt{\frac{m}{kT}}v\text{'}{e}^{-\raisebox{1ex}{$v\text{'}2$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

So, we plot a graph between $P\left(v\right)$and $v\text{'}$, were the slope of the linear part of the graph is,

Slope= $\sqrt{\frac{m}{kT}}$

To find the most likely speed, we take the derivative of the probability then set it equals to zero,

$$\begin{gathered} \frac{dP}{dv}=0 \\ \frac{d}{dv}\left(v{e}^{-a{v}^2}\right)=0 \end{gathered}$$

Here, $a=\frac{m}{2kT}$

So,

$$\begin{gathered} -2a{v}^2{e}^{-a{v}^2}+e^{-a{v}^2}=0 \\ 2a{v}^2=1 \\ v=\sqrt{\frac{1}{2a}} \\ v=\sqrt{\frac{kT}{m}} \end{gathered}$$