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.

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


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?


1 Answer 1


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.