I need to find the maximum of the Maxwell–Boltzmann distribution

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

I know that occurs when $\mathrm dP(v)/\mathrm dv = 0$, but can anyone walk through the steps and point out what you did? Thank you!


closed as off-topic by Mithoron, Tyberius, paracetamol, ron, Jan Aug 24 '17 at 15:34

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    $\begingroup$ Why, the steps are quite simple. (1) Take the derivative. (2) See where it turns to 0. Can you do that? $\endgroup$ – Ivan Neretin Aug 24 '17 at 11:19
  • $\begingroup$ To make things easy for yourself, substitute $v^2=\frac{2kT}{m}x$ and then take the derivative of P(x) with respect to x. $\endgroup$ – Chet Miller Aug 24 '17 at 12:15
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    $\begingroup$ In my opinion, this is fundamentally not chemistry, unless you plan to physically interpret the result. Are maths questions on topic, even if they're in a chemical context? $\endgroup$ – orthocresol Aug 24 '17 at 13:26

I don't think it's on topic, but it's no skin off my back to write a short answer. You need to use the chain rule $(fg)' = fg' + gf'$:

$$\begin{align} \frac{\mathrm dP}{\mathrm dv} &= \left(\frac{m}{2\pi kT}\right)^{3/2}4\pi \left[v^2\frac{\mathrm d}{\mathrm dv}(\mathrm e^{-mv^2/2kT}) + \mathrm e^{-mv^2/2kT}\frac{\mathrm d}{\mathrm dv}(v^2)\right] \\ &= \left(\frac{m}{2\pi kT}\right)^{3/2}4\pi \left[v^2\left(\frac{-2mv}{2kT}\mathrm e^{-mv^2/2kT}\right) + \mathrm e^{-mv^2/2kT}(2v)\right] \\ &= \left(\frac{m}{2\pi kT}\right)^{3/2}4\pi\mathrm e^{-mv^2/2kT}\left(\frac{-mv^3}{kT} + 2v\right) = 0 \end{align}$$

Since an exponential cannot be zero, the polynomial factor must be equal to zero:

$$\begin{align} \frac{-mv^3}{kT} + 2v &= 0 \\ v\left(2 - \frac{mv^2}{kT}\right) &= 0 \\ v &= 0, \pm \sqrt{\frac{2kT}{m}} \end{align}$$

You're not interested in the $v = 0$ stationary point, and the negative square root is unphysical (we're talking about a distribution of speeds, which must be non-negative). So the maximum occurs at $v = \sqrt{2kT/m}$.

From a chemist's point of view this is the physically important quantity, as it represents the most probable speed of a gas molecule. You could substitute it back into $P(v)$ and find out the value of $P$ for this value of $v$, but I don't know what circumstances this would be of interest in.


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