There are several ways to define the radius of a nucleus, one of which is the mean square radius: $$ \langle r^2 \rangle = \frac{\int r^2 \rho (r) d\tau}{\int \rho (r) d\tau}$$

What is the convenience of this opposed to: $$ \langle r \rangle = \frac{\int r\rho(r)d\tau}{\int \rho(r)d\tau}$$

  • $\begingroup$ The root mean square radius is $\sqrt{\langle r^2\rangle}$, which is not necessarily equal to $\langle r\rangle$. $\endgroup$ Apr 2, 2019 at 8:52
  • $\begingroup$ Yes, that is exactly my question, I know that they are not equal; but why is the mean square radius preferred over the mean radius. $\endgroup$ Apr 2, 2019 at 8:55
  • $\begingroup$ You should edit your title. The first is the root mean square radius, also sometimes called the mean square radius for brevity. The second is the mean radius. You should also put a square root radical over <r^2>. $\endgroup$
    – garyp
    Apr 2, 2019 at 11:01
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    $\begingroup$ It has to do with the form factor, which is the Fourier Xform of the charge distribution. The mean-square radius is simply related to the derivative of the form factor. $\endgroup$ Apr 2, 2019 at 12:06
  • $\begingroup$ In other words, what you get in some physical calculations is the right "definition"; the others are just a play of mind. $\endgroup$ Jul 28, 2021 at 16:51

2 Answers 2


In principle you can define the nuclear radius in many ways, but we have a preference for definitions that can be related to experiment in a straightforward way. First of all, notice that text books typically discuss electric charge radii, that means $\rho(r)$ is the density of electric charge, not the baryon density. This is a short coming, because $\rho(r)$ is not sensitive to neutrons, but it has the advantage that the electric charge distribution is easily probed by electron scattering. In the the low energy limit electron scattering simply measures the charge form factor $$ F(q^2) = \int dr^3 e^{i\vec{q}\cdot\vec{r}}\rho(\vec{r}) $$ For small $q$ I can expand this out. Because of rotational invariance there is no linear term, and the leading correction to the total charge $F(0)=Q$ is the slope of the form factor $$ F'(0) = -\frac{1}{6}\int dr^3 r^2\rho(\vec{r}) $$ which is related to the mean square radius.


We consider the nuclear charge distribution to be uniform i.e., $$\rho(r)=Constant$$

Also, Elmentary volume in spherical-polar co-ordiantes $$d\tau= r^2 \sin\theta drd\theta d\phi$$

Now, $$\left< r^2 \right> =\frac{\iiint r^2 \rho(r)d\tau}{\iiint \rho(r)d\tau}$$

$$\left< r^2 \right> =\frac{\iiint r^2\rho(r) \times r^2 \sin\theta drd\theta d\phi}{\iiint r^2 \rho(r) \sin\theta drd\theta d\phi}$$

$$r:0\to R$$ $$\theta:0\to \pi$$ $$\phi : 0 \to 2\pi$$

$$\left< r^2 \right> =\frac{\int_0^R r^4 dr}{\int_0^R r^2 dr}$$

$$\left< r^2 \right> =\frac{R^5/5}{R^3/3}=\frac{3R^2}{5}$$

Similarly, $$\left< r \right> =\frac{\iiint r\rho(r) \times r^2 \sin\theta drd\theta d\phi}{\iiint r^2\rho(r) \sin\theta drd\theta d\phi}$$ $$\left< r \right> =\frac{\int_0^R r^3dr}{\int_0^R r^2 dr}=\frac{R^4/4}{R^3/3}=\frac{3R}{4}$$

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    $\begingroup$ Welcome! Note that you can use $\left< r^2 \right>$ or $\langle r^2 \rangle$ to get angle brackets that have the proper spacing. $\endgroup$
    – rob
    Jul 28, 2021 at 17:39

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