I am trying to understand the derivation of the asymmetry term in the semi-empirical mass formula. I have found a useful derivation on Wikipedia (https://en.wikipedia.org/wiki/Semi-empirical_mass_formula#:~:text=The%20imbalance%20between%20the%20number,basis%20for%20the%20asymmetry%20term), but can't get my head around the expansion below.

The actual form of the asymmetry term can again be derived by modelling the nucleus as a Fermi ball of protons and neutrons. Its total kinetic energy is $$E_k = C (N_p^{5/3} + N_n^{5/3})$$ for some constant ''C''. The leading expansion in the difference $$N_n - N_p$$ is then $$E_k = {C\over 2^{2/3}} \left((N_p+N_n)^{5/3} + {5\over 9}{(N_n-N_p)^2 \over (N_p+N_n)^{1/3}}\right) + O((N_n-N_p)^2).$$

It looks like a simple enough Taylor expansion, but I can't make the leap. Any clarification would be most appreciated!


1 Answer 1


Hint: $N_n = \frac{1}{2}\left((N_n + N_p) + (N_n - N_p)\right)$ and $N_p = \frac{1}{2}\left((N_n + N_p) - (N_n - N_p)\right)$, and by treating $N_n-N_p$ as small we can Taylor expand each of the two terms in the kinetic energy separately.

  • $\begingroup$ I knew I was missing a trick... thanks! $\endgroup$
    – Nik
    Dec 9, 2020 at 13:44

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