Symmetry energy

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In nuclear physics, the symmetry energy reflects the variation of the binding energy of the nucleons in the nuclear matter depending on its neutron to proton ratio as a function of baryon density. Symmetry energy is an important parameter in the equation of state describing the nuclear structure of heavy nuclei and neutron stars.^[1][2][3][4]

Let ${\displaystyle n_{p}}$ and ${\displaystyle n_{n}}$ be the number density of protons and neutrons in nuclear matter, and ${\displaystyle n=n_{p}+n_{n}}$. Let ${\displaystyle E_{0}(n)}$ be the binding energy per nucleon in symmetric matter, with equally many protons as neutrons, as a function of density. The binding energy per nucleon ${\displaystyle E}$ of non-symmetric matter is then a function that also depends on the isospin asymmetry,

${\displaystyle \delta ={\frac {n_{p}-n_{n}}{n}}}$

so to lowest order the energy per baryon is

${\displaystyle E(n,\delta )=E_{0}(n)+S(n)\delta ^{2}+O(\delta ^{4}),}$

where ${\displaystyle S}$ is the symmetry energy.^[2] There are no odd powers of ${\displaystyle \delta }$ in the expansion because the nuclear force acts the same between two protons as between two neutrons.^[5] At saturation density ${\displaystyle n_{0}}$, the symmetry energy is 32.0±1.1 MeV.^[4]

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