0
$\begingroup$

Let's take into consideration the Maxwell-Boltzmann distribution of molecular speeds in one dimension, say

$$P(v_x)dv_x=\left(\frac{m}{2\pi K_BT}\right)^{1/2}e^{-mv^2_x/{2K_BT}}dv_x$$

As we can see, the factor $m$ appears twice.

It is commonly said that the role of such factor is to narrow (for higher values - i.e. $\frac{1}{2}mv^2_x$ is a constant quantity, so increasing mass means decreasing speed) or broaden (for lower ones) the distribution, as we can see in the following graphic, which depicts the distribution for various gases:

enter image description here

This is consistent with the graph of $m^{1/2}e^{-m}$, for values of order of an hydrogen atom mass:

enter image description here

However, if we graph $m^{1/2}e^{-m}$ also for bigger values, we get:

enter image description here

My question is: what could be the physical meaning behind such behaviour? Why the distribution starts again to broaden after a certain value?

$\endgroup$
  • 2
    $\begingroup$ (1) Your first equation is the Boltzmann distribution in 1D (i.e. the distribution in the variable $v_x$, the "$x$" component of the velocity), but your first plot shows the Boltzmann distribution in 3D (i.e. the distribution in the variable $v=\sqrt{v_x^2+v_y^2+v_z^2}$. Are you aware of this? $\endgroup$ – Arturo don Juan Nov 11 '16 at 21:05
  • 2
    $\begingroup$ (2) When you say "This is consistent with the graph $m^{1/2}e^{-m}$", you do realize the function you are plotting is nonsense, right? You're plotting the $v_x$ probability distribution as a function of $m$, which is pointless. You should be looking at meaningful functions, like the average energy or position of the maximum of the Boltzmann distribution, as a function of the mass parameter. $\endgroup$ – Arturo don Juan Nov 11 '16 at 21:05
  • $\begingroup$ (1) I know, that's true. I've just googled for an image instead of create my own. Do you mean that the monodimensional distribution has a roughly different shape? $\endgroup$ – Lo Scrondo Nov 11 '16 at 21:27
  • $\begingroup$ (2) It's obvious that the function $m^{1/2}/e^{-m}$ is not connected per se with speed. However, what I wanted was to grasp its effect on the whole distribution. $\endgroup$ – Lo Scrondo Nov 11 '16 at 21:33
  • 2
    $\begingroup$ In response to (1), they are absolutely different. Just plot it! The distribution in $v_x$ is a pure Gaussian, whereas the distribution in $v$ is a chi-distribution with 3 degrees of freedom. In response to (2), the first graph in your post shows exactly the relevant dependence of the distribution on $m$ - it broadens it out. You could go further and plot the peak of $f(v)$ as a function of $m$, where you fix all the other parameters. $\endgroup$ – Arturo don Juan Nov 11 '16 at 21:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.