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If, according to the Maxwell-Boltzmann distribution

$${f(v)}\propto\exp\left(-\frac{\varepsilon}{kT}\right),\tag{1}$$

which is the equation from which the whole final equation is derived, then why is the final equation

$$f(v) = 4\pi v^2\left(\frac{m}{2{\pi}kT}\right)^{3/2}\exp\left(-\frac{mv^2}{2kT}\right)?\tag{2}$$

Clearly, here

$${f(v)}\propto v^2\exp\left(-\frac{mv^2}{2kT}\right).\tag{3}$$

Where have I made a mistake in understanding this?

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    $\begingroup$ The link you shared goes to a not-so-informative website. What's $\epsilon$? Let's say that $\epsilon = mv^2$. The proportionality statement still holds. $\endgroup$ Apr 20, 2022 at 12:52
  • $\begingroup$ The 'additional' terms are obtained when transforming the probability distribution from 3-D distribution into magnitude of the velocity vector (averaged over direction) using spherical coordinates. Referenced from ocw.mit.edu/courses/5-62-physical-chemistry-ii-spring-2008/… $\endgroup$
    – user4723
    Apr 20, 2022 at 13:28
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    $\begingroup$ Closely related, nearly duplicate // You may confuse Boltzmann and Maxwell-Boltzmann distributions, with the latter based on the former, adding the v^2 factor from dV=4pi.r^2 dr, being integrated for the radius spherical coordinate. $\endgroup$
    – Poutnik
    Apr 20, 2022 at 13:28
  • $\begingroup$ The funny thing is both posts share the same OP. $\endgroup$
    – Poutnik
    Apr 20, 2022 at 13:43
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    $\begingroup$ @Poutnik I thought it was familiar... $\endgroup$ Apr 20, 2022 at 13:58

1 Answer 1

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The $v$ in your first expression (that $f(v) \propto \exp(-\varepsilon/kT)$) most likely refers to the true velocity which is a vector

$$\vec{v} = (v_x, v_y, v_z),$$

whereas the $v$ in the Maxwell–Boltzmann distribution refers to the magnitude of the velocity

$$v = |\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}.$$

The latter is more useful to a chemist because we don't really care which direction the particle is moving in, only its speed. However, note that there are many possible combinations of $(v_x, v_y, v_z)$ which yield the same magnitude $v$; so the formula needs to be adjusted for this. The factor ends up being $4\pi v^2$, which is the surface area of a sphere with radius $v$ (you can think of the surface of this sphere as representing all possible velocities with magnitude $v$).

My suggestion would be to look for a derivation in a good physical chemistry textbook, which should explain this more thoroughly.

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  • $\begingroup$ Actually, I did read the first expression that I gave, in Physical Chemistry by Peter Atkins, but I really couldn't share the link because I have a hard copy, and I couldn't find any similar derivation on the internet. So, I had to link that website. But the book did in fact give $f(v)\propto e^{\frac{-\epsilon}{kT}}$ $\endgroup$ Apr 20, 2022 at 14:08
  • $\begingroup$ Also, I found that the derivation in the same mentioned book is wrong in many places such as equating $f(v)$ with $f(v_x).f(v_y).f(v_z)$. $\endgroup$ Apr 20, 2022 at 14:10
  • $\begingroup$ I think I understand now. We were taking $\vec{v}$ and therefore only one combination of $v_x$, $v_y$ and $v_z$ is present(due to only one direction). Thus we need not add the $4πv^2$ part to accommodate for the total volume in the velocity space. Is this what you meant @orthocresol? $\endgroup$ Apr 20, 2022 at 14:32
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    $\begingroup$ I think that's pretty much what I said, yeah. $\endgroup$ Apr 20, 2022 at 14:36

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