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?

  • $\begingroup$ Hi Juan Pablo Arcila: Which page in Greiner? Ah found it: Example 6.2 p. 153. $\endgroup$
    – Qmechanic
    Commented May 14, 2019 at 8:07

1 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.

  • $\begingroup$ But how do you know the first claim? $\endgroup$ Commented May 13, 2019 at 16:40
  • $\begingroup$ I am not sure, looking into it right not $\endgroup$
    – acarturk
    Commented May 13, 2019 at 17:24
  • $\begingroup$ @JuanPabloArcila I redid the answer $\endgroup$
    – acarturk
    Commented May 13, 2019 at 17:39
  • $\begingroup$ Thanks a lot, it makes sense now :D $\endgroup$ Commented May 14, 2019 at 2:23

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