91影视

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. $2\mathrm{蟺惫}$, so we can give the probability as

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

$P\left(v\right)=C脳2{\mathrm{蟺惫别}}^{\raisebox{1ex}{$-{\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,

${鈭�}_0^{鈭�}P\left(v\right)-C脳2\mathrm{蟺}{鈭�}_0^{鈭�}{\mathrm{ve}}^{\raisebox{1ex}{$-{\mathrm{mv}}^2$}\!\left/ \!\raisebox{-1ex}{$2\mathrm{kT}$}\right.}-1$

But,

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

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

So,

$$\begin{gathered} C脳2蟺\left(\frac{kT}{m}\right){\left[e^{\raisebox{1ex}{$-{\mathrm{mv}}^2$}\!\left/ \!\raisebox{-1ex}{$2\mathrm{kT}$}\right.}\right]}_0^{鈭�}=1 \\ C脳2蟺\left(\frac{kT}{m}\right)\left[0-1\right]=1 \\ C脳2蟺\left(\frac{kT}{m}\right)=1 \\ C=\frac{m}{2\mathrm{蟺办罢}} \end{gathered}$$

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

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

As in 3D plane the probability fall exponentially $\left(e^{-v^2}\right)$as $v鈫�鈭�$.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,

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