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I am reading Maxwell-Boltzmann distribution for describing the velocities of ideal gas molecules. I went through the PSE question Derivation of the Maxwell-Boltzmann speed distribution and the Wikipedia article and several other resources for some clarification. The following is my understanding of the concept, derivation, and the discrepancy I am not able to resolve.

For discrete energy levels, the fraction of molecules $q(\epsilon)$, from statistical mechanics, of a particle having energy $\epsilon$ is proportional to $\exp{\dfrac{-\epsilon}{kT}}$. For continuous energy levels, the expression is similar, except that we have to multiply by a factor of $4\pi v^2$, which I will do in the end. The energy of an ideal gas particle is given by $\dfrac{1}{2}mv^2$ and therefore $q(v) = C\exp{\dfrac{-mv^2}{2kT}}$. I am trying to find $C$.


Since the particles may have speeds in one, $x$, $y$, or $z$, direction ranging from $0$ to $\infty$,

$$ \int_0^\infty C_x\exp{\dfrac{-m(v_x^2)}{2kT}}dv_x = 1\\ \implies C_x=\dfrac{1}{\int_0^\infty \exp{\dfrac{-m(v_x^2)}{2kT}}dv_x} $$

Discrepancy: Several books write that $\int_{-\infty}^\infty \exp{\dfrac{-mv_x^2}{2kT}}dv_x = 1$. I don't think that it is correct, so I am going to carry on with my derivation and give the result using this method in the end.

My Derivation

Since $\exp{\dfrac{-mv_x^2}{2kT}}$ is an even function of $v$, we can evaluate the integral as $\int_0^\infty \exp{\dfrac{-mv_x^2}{2kT}}dv = \dfrac{1}{2}\int_{-\infty}^\infty \exp{\dfrac{-mv_x^2}{2kT}}dv$. Therefore

$$ C_x=\dfrac{2}{\int_{-\infty}^\infty \exp{\dfrac{-m(v_x^2)}{2kT}}dv_x} $$

Using trick I learned from MSE question Integral of $e^{−x2}$ and the error function, which I verified using WolframAlpha, I obtained the following result.

$$ C_x = \dfrac{2\sqrt{\dfrac{m}{2kT}}}{\sqrt{\pi}} = \sqrt{\dfrac{2m}{\pi kT}}\\ q(v_x) = \sqrt{\dfrac{2m}{\pi kT}}\exp{\dfrac{-mv_x^2}{2kT}} $$

I understand $q(v_x)=q(v_y)=q(v_z)$ and $C_x=C_y=C_z$ and $q(v) = q(v_x)q(v_y)q(v_z)$:

$$ q(v) = \left(\dfrac{2m}{\pi kT}\right)^{\frac{3}{2}}\exp{\dfrac{-mv^2}{2kT}} $$

Speed is continuous, and we need to multiply by $4\pi v^2$ (Justification) to obtain the final result.

$$ q(v) = 4\pi\left(\dfrac{2m}{\pi kT}\right)^{\frac{3}{2}}v^2\exp{\dfrac{-mv^2}{2kT}} $$

Most textbooks and ChemLibreTexts follow the discrepancy and derive the following result.

$$ q'(v) = 4\pi\left(\dfrac{m}{2\pi kT}\right)^{\frac{3}{2}}v^2\exp{\dfrac{-mv^2}{2kT}} $$


Problems

This is problematic because my result would give different values of well-established $v_{\text{mean}}$, $v_{\text{average}}$, and $v_{\text{rms}}$, being greater by a factor of 8. Could someone help me out here?

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You are just wrong in asserting that those are speeds. Those are velocities, which is why everybody else integrates from $-\infty$ to $+\infty$. Your mistake of a factor of 8 is just coming from you making this mistake, since $0$ to $+\infty$ is a factor of 2 wrong, we have 3D space, and so $2^3=8$ factor mistake. You have to start with velocities integral, before converting to polar coördinates so as to work with speeds instead of velocities.

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  • $\begingroup$ Is this some kind of vectorial integration that I am unaware of? $\endgroup$
    – ananta
    Commented Feb 21 at 8:59
  • $\begingroup$ No, it is not. You simply misinterpreted those correct integrals. $\endgroup$ Commented Feb 21 at 8:59
  • $\begingroup$ Could you please elaborate? I am sorry I don't understand it. $\endgroup$
    – ananta
    Commented Feb 21 at 9:01
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    $\begingroup$ When it is written down as $v_x$, that is very clearly the $x$-component of velocity $\vec v$, and so you have to have both the positive and negative parts integrated. Only when you do the triple integral over all 3 components of the velocity will you get $4\pi v^2$ spherical shell, where $v=|\vec v|$ is the non-negative speed. If you only took the positive half of each direction, then you will not get $4\pi v^2$, and instead get $\frac18$ of that. $\endgroup$ Commented Feb 21 at 9:05
  • $\begingroup$ That explains it. Thank you. $\endgroup$
    – ananta
    Commented Feb 21 at 9:07

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