# Natural Units and the Nuclear Radius

In a paper I'm looking at and also in a similar program, the authors note that the nuclear radius $R$ is given in "natural units" as $R = 3.11\times 10^{-3}\;A^{1/3}$. This has me slightly confused, as I seem to be missing a factor of 2. My understanding is that natural units usually entail $\hbar = c = 1$, so that length is measured in units of inverse energy. A typical estimate of the nuclear radius as a function of nucleon number $A$ is $R = 1.2\, \text{fm}$, so (I thought) $R_{natural} = R/\hbar c$ and thus $$R = \frac{1.2\,A^{1/3}\,\text{fm}}{197.327 \, \text{MeV-fm}} = 6.1 \times 10^{-3} \, A^{1/3}$$ in natural units, which is about twice what I expect to get. I'm not overly comfortable in these units yet, so I can't spot the error. Where is this factor of two coming from?

• You can't answer without knowing which natural units they are using--- they might not even set c=1, since nuclear stuff is nonrelativistic usually. Jun 16, 2012 at 6:45
• I thought the (nuclear physics) natural units for length were just fm. Can you post a reference to the paper? Jun 16, 2012 at 8:07
• @JohnRennie We tend to use femtometers because it is roughly the size of a single nucleon (which means about the length scale for the effective strong interaction), so it is "natural" in that sense, but there isn't a unique choice without specifying the energy of the probe. Also, read my comment on the OP's answer about the "size" of nucleons. Jun 16, 2012 at 17:28

I think the answer is found in another reference on beta decay, similar to where this question originated. It describes the nuclear radius R as being expressed in terms of the electron Compton wavelength. Then $$R_{natural} = \frac{R}{\hbar\,/\,m_ec} = R\frac{m_ec^2}{\hbar c} \approx \frac{(1.2\,\text{fm})\, (0.511 \,\text{MeV})}{197.327\,\text{MeV fm}}A^{1/3} \approx 3.3 \times 10^{-3} A^{1/3},$$ if I make the assumption the authors mean the reduced Compton wavelength and again set $\hbar = c = 1$.
This makes sense in the context of beta decay (which I did not mention I was looking at originally). Nuclear physics is normally non-relativistic, but beta decay results in relativistic electrons and neutrinos and we must handle them relativistically. Integrals are then often made dimensionless by expressing energies in the ratios $\epsilon = E\,/\,m_e$, so a dimensionless nuclear radius is logical here.
• BTW--Even the selection of $1.2\text{fm}$ for the radius of a single nucleon is a little tricky. I believe that value comes from a fit to make this power law work. The charge radius is around $1.5\text{ fm}$, and the nucleon--nucleon hard scattering range is about $1\text{ fm}$ suggesting a hard-core size of about $0.5\text{ fm}$. At these scales classical concepts really take a beating even where they are useful. Jun 16, 2012 at 17:25