# Discrepancy in the Derivation of Maxwell-Boltzmann Distribution

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?

## 1 Answer

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.

• Is this some kind of vectorial integration that I am unaware of? Commented Feb 21 at 8:59
• No, it is not. You simply misinterpreted those correct integrals. Commented Feb 21 at 8:59
• Could you please elaborate? I am sorry I don't understand it. Commented Feb 21 at 9:01
• 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. Commented Feb 21 at 9:05
• That explains it. Thank you. Commented Feb 21 at 9:07