1
$\begingroup$

I am trying to integrate over the Maxwell-Boltzmann velocity distribution to find the $v_{\text{rms}}$.

The Maxwell-Boltzmann velocity distribution is given by:

$$P(v) =\sqrt{\frac{m}{2\pi kT}} e^{-mV^2 / 2kT}$$

To find the average square velocity I multiply each possible value for $v^2$ by the fraction of molecules with velocity $v$ and sum by integration like so:

$$v_{rms} = \int^{\infty}_{-\infty} V^2 P(v) = \sqrt{\frac{m}{2 \pi kT}} \int^{\infty}_{-\infty} V^2 e^{-mV^2 / 2kT}$$

I have attempted to integrate by parts with

$u = V^2$, $dv = e^{-mV^2 / 2kT}$, $du= 2V$, and $v = \sqrt{\frac{2kT \pi}{m}}$ ($dv$ is a Gaussian integral)

Then using the integration by parts formula:

$$ \begin{align} v_{rms} &= \sqrt{\frac{m}{2 \pi kT}} \left[ uv - \int v du\right] \\ &= \sqrt{\frac{m}{2 \pi kT}} \left[ V^2 \sqrt{\frac{2 \pi kT}{m}} - \int{\infty}_{-\infty} \sqrt{\frac{2 \pi kT}{m}} 2V\right] \\ &= \sqrt{\frac{m}{2 \pi kT}} \sqrt{\frac{2 \pi kT}{m}} [V^2 - \int^{\infty}_{-\infty} 2 V] \\ &= [V^2 - \frac{2}{3} V^3]^{\infty}_{-\infty} \end{align} $$

which is undefined. The expected answer was $v_{rms} = \frac{kT}{m}$ (which is valid for an ideal monatomic gas). Did I make a mistake in my integration?

$\endgroup$
7
  • $\begingroup$ You must have; it's no longer dimensionally consistent. $\endgroup$
    – J.G.
    Nov 1, 2021 at 17:54
  • $\begingroup$ when you integrate by parts, you defined dv, but the v you wrote does not come from that dv. You are also missing differentials on many integrals. $\endgroup$
    – user65081
    Nov 1, 2021 at 18:03
  • $\begingroup$ @Wolphramjonny I used the formula for a Gaussian integral: $\int e^{-ax^2} dx = \sqrt{\frac{\pi}{a}}$. Setting $a = \frac{m}{2kT}$ and $x = v$ gives $\sqrt{\frac{2kT \pi}{m}}$ $\endgroup$
    – physBa
    Nov 1, 2021 at 18:14
  • $\begingroup$ You should refresh your knowledge on how to integrate $\endgroup$
    – user65081
    Nov 1, 2021 at 18:17
  • $\begingroup$ @Wolphramjonny Could you please be a bit more specific? Is my formula for the Gaussian integral incorrect? The way I have applied it? $\endgroup$
    – physBa
    Nov 1, 2021 at 18:23

2 Answers 2

2
$\begingroup$

Mistakes include integrating from $-\infty$, conflating the PDF of $\vec{V}$ in $\Bbb R^3$ with that of $V:=|\vec{V}|$ in $\Bbb R$, and replacing $v$ in IBP with $\int_0^\infty vdV$, which doesn't even have the same dimension. Let $\alpha:=m/(2kT)$. Differentiating $\int_0^\infty e^{-\alpha V^2}dV=\frac{\sqrt{\pi}}{2}\alpha^{-1/2}$ twice with respect to $-\alpha$ gives $\int_0^\infty V^2e^{-\alpha V^2}dV=\frac{\sqrt{\pi}}{4}\alpha^{-3/2},\,\int_0^\infty V^4e^{-\alpha V^2}dV=\frac{3\sqrt{\pi}}{8}\alpha^{-5/2}$, so$$v_\text{rms}^2=\frac{\int_0^\infty V^4e^{-\alpha V^2}dV}{\int_0^\infty V^2e^{-\alpha V^2}dV}=\frac{3}{2\alpha}=\frac{3kT}{m}.$$Unfortunately, each integral is hard to evaluate by IBP because the power of $V$ is even.

$\endgroup$
0
$\begingroup$

Note that:

$$d^3v = v^2dvd\Omega $$

is the differential in the integral, so you are evaluating $\langle 1 \rangle$, not $\langle v \rangle$.

Also: $0\le v\lt\infty$.

$\endgroup$

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.