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Symmetry energy in nuclear physics

by rahele
Tags: energy, nuclear, physics, symmetry
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rahele
#1
Dec5-13, 10:21 AM
P: 3
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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mfb
#2
Dec5-13, 01:56 PM
Mentor
P: 11,925
(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.
ChrisVer
#3
Dec5-13, 02:48 PM
P: 919
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.

dauto
#4
Dec5-13, 09:37 PM
Thanks
P: 1,948
Symmetry energy in nuclear physics

Calculate the average potential energy of a brick in a brick wall of height N. Calculate the same for a wall of height Z. Keep the sum of the height A = Z + N fixed but allow their difference (N - Z) to be a free parameter. Find out the dependency of the total energy on that free parameter.
lpetrich
#5
Dec13-13, 07:28 AM
P: 530
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.

The absolute-value function has a problem: it has a singularity at 0. Its first derivative is a step function and its second derivative a Dirac delta function.

The square function does not have that problem.


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