In almost every reference site I can find, the Sommerfeld parameter $\eta$ is explicitly stated as

$$\eta=\frac{Z_1Z_2e^2}{\hbar \nu}=\frac{Z_1Z_2e^2}{\hbar }\sqrt{\frac{\mu}{2E}}$$

Where $Z_1$ and $Z_2$ are the atomic numbers of the two nuclei involved, $\mu$ is the reduced mass of the two nuclei, and $E$ is the approximate center of mass energy of the the nuclei, and $v$ is the relative velocity of the nuclei.

The problem I have with this is that it is not dimensionless as it should be. The units of what I have just wrote down are


However, when I check numerically with other references, it seems like it is defined to be

$$\eta=\frac{Z_1Z_2e^2}{4\pi\epsilon_{0}\hbar c}\sqrt{\frac{\mu}{2E}}=\alpha Z_1Z_2\sqrt{\frac{\mu}{2E}}$$

Where $\alpha$ is the fine-structure constant, and $\mu$ is the reduced mass in terms of energy/$c^2$. This makes sense, but is not what is explicitly stated as the Sommerfeld parameter.

What exactly is the definition of the Sommerfeld parameter? Must I go through the solution of the Schrodinger equation for a two-nuclei system (woods-saxon + yukawa potential) to derive this for myself?



2 Answers 2


The first formula is written in the Gaussian unit system, while the second one is in the SI system. In the Gaussian system, the unit of electric charge is $statC =g^{1/2}cm^{3/2}s^{-1}$. So, the Sommefeld parameter in the Gaussian unit system is dimensionless as it would be.

  • 1
    $\begingroup$ As a general rule, many of these unit issues can be fixed by a factor of $4\pi\epsilon_0$, which is unity in CGS units but has dimension $\rm (coulomb)^2 (joule)^{-1}(meter)^{-1}$ in SI. $\endgroup$
    – rob
    Commented Apr 13, 2015 at 12:12

Simply multiply by $d/d$, where $d$ is a arbitrary distance, then the unit of $d/(hv)$ will be (Joule)^(-1), and $\frac{Z_1Z_2e^2}{d}$ will be in $J$. Then the product is dimensionless.

  • 1
    $\begingroup$ Hello and welcome to StackExchange! Please consider using LaTeX for better readability. You can introduce simple LaTeX formula between dollar signs. $\endgroup$
    – Martin
    Commented Sep 28, 2015 at 11:41
  • $\begingroup$ To extend Martin's comment, see this link on using MathJax. $\endgroup$
    – Kyle Kanos
    Commented Sep 28, 2015 at 11:49

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