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Why do we use root mean square velocity ($v_{rms}$) and not the mean of the absolute values (|$\bar{v}$|) of the velocities in kinetic theory?

Rickson
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    Because we're really finding the average energy i.e. the average of $\tfrac{1}{2}mv^2$. – John Rennie Apr 20 '18 at 15:38
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    Hmmm ... keep in mind that the version of Kinetic Theory we do in the intro class leaves out some very hard work: it originally took Boltzmann about 60 pages to show that the (vector) velocity distribution is stationary for equilibrium gasses. (Of course Boltzmann had a reputation for simply spewing out math and not bothering to go back and find a short and elegant version before publishing, but still.) So there are a lot of details that have precise arguments that are simple stated and accepted in the classroom version. – dmckee --- ex-moderator kitten Apr 20 '18 at 16:27

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As I discuss at https://physics.stackexchange.com/a/517508/59023, the speed distribution is different than the velocity distribution. For instance, a 3D velocity distribution would look like: $$ f\left( \mathbf{v} \right) = \frac{ 1 }{ \pi^{3/2} \ V_{T x} \ V_{T y} \ V_{T z} } \ e^{\left[ - \left( \frac{ v_{x} - v_{o, x} }{ V_{T x} } \right)^{2} - \left( \frac{ v_{y} - v_{o, y} }{ V_{T y} } \right)^{2} - \left( \frac{ v_{z} - v_{o, z} }{ V_{T z} } \right)^{2} \right]} \tag{0} $$ whereas the speed distribution looks like: $$ \tilde{f}\left( v \right) = \frac{ 4 \ \pi \ v^{2} }{ \pi^{3/2} \ V_{T x} \ V_{T y} \ V_{T z} } \ e^{\left[ - \left( \frac{ v_{x} - v_{o, x} }{ V_{T x} } \right)^{2} - \left( \frac{ v_{y} - v_{o, y} }{ V_{T y} } \right)^{2} - \left( \frac{ v_{z} - v_{o, z} }{ V_{T z} } \right)^{2} \right]} \tag{1} $$ where $v^{2} = \sum_{j} \ v_{j}^{2}$, $v_{o, j}$ is the displacement of the distribution peak along the jth direction, and $V_{T j}^{2} = \tfrac{ 2 \ k_{B} \ T_{j} }{ m }$ is the most probable speed of a one-dimensional Gaussian.

Why do we use root mean square velocity ($v_{rms}$) and not the mean of the absolute values ($\lvert \bar{v} \rvert$) of the velocities in kinetic theory?

Aside from the mess this causes mathematically1, even the speed distribution function is not a function of the random variable $\langle \lvert \mathbf{v} \rvert \rangle$. Note that your question has the norm and average operators switched from what the text implies.

Footnotes

  1. As I started to illustrate at https://physics.stackexchange.com/a/517508/59023, the change of variables from the component velocities to a scalar speed for the integration becomes extremely messy unless the distribution is completely isotropic, i.e., all $v_{o, j} = 0$ and $V_{T x} = V_{T y} = V_{T z}$. Then one can integrate Equation 1 as if it were a 1D equation and using symmetry change the limits of integrate from ($- \infty$, $+ \infty$) to (0, $+ \infty$).
honeste_vivere
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