# Show, for a sphere, $\langle r^2\rangle = \frac{3}{5}r_o ^2$

The following equation is used as an inference but not explained in my solid state physics book (Economou, "The Physics of Solids"). I figure it's a math/geometry problem so I ask here. Can anyone show this?

$$\langle r^2\rangle = \frac{3}{5} r_o^2$$

where $r_o$ is the radius of a sphere and $\langle r^2\rangle$ is the average square distance from the center (e.g. for some particle in this volume ($\frac{4}{3}\pi r_o^3$), assuming equal probability distribution).

I would have thought, naively, that $\langle r\rangle = \frac{1}{2}r_o$ thus $\langle r^2\rangle = \frac{1}{4}r_o^2$

If interested this is the context:

• Take a look at math.berkeley.edu/~preskill/math53/worksheet_vecfield_soln.pdf. Substitute $\rho^2$ for $\rho$ in number 1 to get your answer. Intuitively, the average position is not $\frac{1}{2}r_0$ because the further you get from the origin the more positions the particle can occupy. Jan 20, 2016 at 3:49
• Please don't write $<r^2>$ if you mean $\langle r^2\rangle$. $\qquad$ Jan 20, 2016 at 4:03

The amount of the sphere at a distance of $r$ is $$4\pi r^2\,\mathrm{d}r$$ so the average of $r^2$ would be \begin{align} \frac{\int_0^{r_0}4\pi r^2r^2\,\mathrm{d}r}{\int_0^{r_0}4\pi r^2\,\mathrm{d}r} &=\frac{\frac{4\pi}5r_0^5}{\frac{4\pi}3r_0^3}\\[3pt] &=\frac35r_0^2 \end{align}
• You mean $\frac{4}{3} \pi r^2\,\mathrm{d}r$ right? Though the $\frac{1}{3}$ will cancel in the calculation. Jan 20, 2016 at 3:57
• @khaverim: no, I mean $4\pi r^2\,\mathrm{d}r$ since the area of the sphere is $4\pi r^2$ and $\mathrm{d}r$ is the thickness of a shell.