Momentum average on phase space for free particle

Question

I'm studying from Greiner statistical mechanics, and he uses an approximation which I don't really understand.

On averaging over many phase-space points we have

$$\langle\vec{p}^2\rangle=3\langle p_x^2\rangle=3\langle p_y^2\rangle=3\langle p_z^2\rangle$$ since no direction in space is preferred, i.e., $$\sqrt{\langle\vec{p}^2\rangle}=\frac{\sqrt{3}}{3}\left(\sqrt{\langle p_x^2\rangle}+\sqrt{\langle p_y^2\rangle}+\sqrt{\langle p_z^2\rangle}\right).$$ Therefore, we make the approximation $$\epsilon=c\left(p_x^2+p_y^2+p_z^2\right)^{1/2}\approx\frac{c}{\sqrt{3}}\left(\vert p_x\vert+\vert p_y\vert+\vert p_z\vert\right)$$

Can someone please explain this approximation?

Answer 1

The claim on Greiner is

$$ \langle\vec{p}^2\rangle = 3\langle p_x^2\rangle = 3\langle p_y^2\rangle = 3\langle p_z^2\rangle $$

This claim follows from the fact that

$$ \text{E}[p^2] = \text{E}[p_x^2] + \text{E}[p_y^2] + \text{E}[p_z^2] $$

and that $(x,y,z)$ are indistinguishable:

$$ \text{E}[p^2] = 3 \text{E}[p_x^2] = 3 \text{E}[p_y^2] = 3 \text{E}[p_z^2] $$

The rest follows if you substitute all of $(x,y,z)$ with either one.