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Let us take an example: the ideal gas. I know for example that $\langle v \rangle^2$ is different than $\langle v^2 \rangle$, as $\langle v \rangle=0$ due to no preferred direction.
But know if I calculate the mean of the absolute value $\langle |v|\rangle$ I get:

$$\frac{m}{2}\langle |v| \rangle^2 = \frac{4}{\pi} k_B T \neq \frac{3}{2} k_BT =\frac{m}{2}\langle v^2\rangle $$ where $k_BT$ is the ''thermal energy'' and $m$ the mass of the particles. In simple terms: $$\frac{\langle v^2\rangle}{\langle |v|\rangle^2}=\frac{3\pi}{8}\approx1.18$$

Is there any physical or mathematical interpretation to this disagreement?

Edit: Maybe the ratio is not that of a big deal but I'm trying to clear out a difference in the literature related to the Lorenz number in the Drude model. Some authors use $\langle v^2 \rangle$ (Ashcroft,Mermin Solid State Physics) and others $\langle |v|\rangle^2$ (Kittel and others like Hyperphysics).

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  • $\begingroup$ The thing is that you are comparing $\langle v^2\rangle$ with $( \ \ \langle |v | \rangle\ \ ) ^2$. Those are different. The first one would be the same as $\langle |v|^2\rangle$, but not if the square is outside. $\endgroup$ – FGSUZ May 29 '18 at 13:01
  • $\begingroup$ @FGSUZ That I get. $\endgroup$ – Mauricio May 29 '18 at 13:22
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Forget that velocity is a vector $\vec{v}$ for the moment, and just let $v:=\vec{v}$ denote the speed (i.e. modulus of the velocity vector). This is just a non-negative real random variable, and you're asking why $\langle v^2\rangle\ne\langle v\rangle^2$. But famously, the difference between these two expressions is just the variance of $v$.

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  • $\begingroup$ Ok to clear this out, the variance is < v^2 > - < v>^2 or <v^2> - < vec(v) >•< vec(v) > ? Is there a difference? $\endgroup$ – Mauricio May 29 '18 at 14:04
  • $\begingroup$ @Mauricio I was discussing the variance of $|v|$, which is $\langle v^2\rangle -\langle v\rangle^2$. However, $\langle v^2\rangle-\langle \vec{v}\rangle\cdot\langle \vec{v}\rangle=\sum_i (\langle v_i^2\rangle-\langle v_i\rangle^2)$ is a sum of components' variances. $\endgroup$ – J.G. May 29 '18 at 15:00
  • $\begingroup$ Let us stick with 1D. In QM to calculate the variance of momentum P of a given state, we calculate it as <p^2> - <p>^2 not <p^2>- <|p|>^2 $\endgroup$ – Mauricio May 29 '18 at 17:01
  • $\begingroup$ @Mauricio The former is the variance of the momentum, the latter of the unsigned momentum. The radial-vectorial distinction exists even when the "vector" is 1D. $\endgroup$ – J.G. May 29 '18 at 17:04
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There is no reason for the identity to hold; in general the mean of the square root of a non-negative random variable is not the square root of the mean.

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  • $\begingroup$ But has this ratio any meaning? I mean the difference between the mean of the square of v and the mean of v squared is usually related to some standard deviation. In this case the question is about the difference between the mean of the square of v and the mean of |v| squared. $\endgroup$ – Mauricio May 28 '18 at 12:27
  • $\begingroup$ Any thing only makes sense if it arises in calculations naturally. $\endgroup$ – Vladimir Kalitvianski May 28 '18 at 13:01
  • $\begingroup$ @VladimirKalitvianski it does, see edit. $\endgroup$ – Mauricio May 28 '18 at 13:17
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One is defined as $$ \langle |v|\rangle=\frac{\int |v|f(v)\mathrm{d}v}{\int f(v)\mathrm{d}v} $$ while the other $$ \langle v^2\rangle=\frac{\int v^2f(v)\mathrm{d}v}{\int f(v)\mathrm{d}v}. $$ It is clear that $\langle |v|\rangle^2\neq\langle v^2\rangle$.

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