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?

• Hi Juan Pablo Arcila: Which page in Greiner? Ah found it: Example 6.2 p. 153. – Qmechanic May 14 at 8:07

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.

• But how do you know the first claim? – Juan Pablo Arcila May 13 at 16:40
• I am not sure, looking into it right not – acarturk May 13 at 17:24
• @JuanPabloArcila I redid the answer – acarturk May 13 at 17:39
• Thanks a lot, it makes sense now :D – Juan Pablo Arcila May 14 at 2:23