While doing some calculations on v_rms using the Maxwell-Boltzmann distribution, I noticed that v_rms and v_avg are pretty similar (https://casper.berkeley.edu/astrobaki/index.php/File:MaxwellSpeedDist.png).(adsbygoogle = window.adsbygoogle || []).push({});

In fact, really it's just the choice of using the 1-norm (|v|_avg) vs. 2-norm sqrt(v^2 avg). When deriving v_rms or <KE> from the Maxwell-Boltzmann distribution , we get <KE> = 0.5m<v^2> = 3kT/2; however, using v_avg from the Maxwell-Boltzmann distribution we get v_avg = sqrt(8kT/(pi*m)) and then 0.5m<v>^2 = 4kT/pi.

In computer science/statistics, people often choose the distance metric <|x|> vs. sqrt<(x^2)> based on their needs, and generally <|x|> is thought to be better and more robust to outliers (yet has far worse properties, e.g., no convergence). I'm familiar with the work-energy theorem and the derivation of KE = 0.5mv^2 from Newton's 2nd law, but am trying to figure out if there are alternative formulations using |v|. I'm also familiar with the equipartition theorem with energies of form Ax^2 being assigned 0.5kT energy and am not sure how this would be reformulated using |v|.

So two questions:

1. Can KE/classical laws be reformulated using |v| instead of |v^2|, or has anything been published using the 1-norm (absolute value) or any other norm besides the 2-norm (square)? I realize the units of energy won't make sense with just changing how KE is computed. But, for example, speed could be defined as (sum |v|) instead of (sum v^2), though I'm not sure if people do this.

2. Are there cases where v_avg is useful to use instead of v_rms?

Thanks in advance for the help/dicussion! This has been gnawing at me since in CS/stats we usually have the choice of using |v| or v^2 to compare distributions, but I'm not sure if we're "forced" to use v^2 in physics because of fundamental laws (and if so which ones?).

Image is from: https://casper.berkeley.edu/astrobaki/index.php/Maxwellian_velocity_distribution

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# A Can KE be reformulated using |v| instead of v^2?

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