Symmetry energy in nuclear physics

rahele
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I have a question about symmetry energy in semi-empirical mass formula,
According to semi-empirical mass formula as follows:
E=avA-asA2/3-acZ(Z-1)/A1/3-asym(N-Z)2/A
why in the symmetry energy only squared parameter symmetry are exist and there is not the first power of asymmetry parameter?
 
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(N-Z) would have nothing to do with symmetry, it would go from -infinity to +infinity (well, bounded by 0 neutrons and 0 protons of course). And the absolute proton and neutron numbers are in the total mass anyway (this is just the binding energy).

There could be |N-Z|, but experiments show this is not needed. And I don't see a physical reason for it.
 
I think absolute values are not so favored... :) they miss nice functional properties. So we wouldn't search for a fitting in | | but in ( )^2 if we knew a priori that something is happening, and see how that works
Also, I guess, it's because it fits the experiments as mfb said.
 
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Semi-empirical mass formula - Wikipedia has a derivation of the form of the symmetry-energy term. The derivation treats protons and neutrons as separate but overlapping Fermi liquids that both extend over the nucleus.

Ekinetic = EFermi/A2/3*(Z5/3 + N5/3)

One then sets Z = (A/2) + X and N = (A/2) - X and expands in X. The first term in X is a term in X2.

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The square function does not have that problem.
 
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