
#1
Dec513, 10:21 AM

P: 3

I have a question about symmetry energy in semiempirical mass formula,
According to semiempirical mass formula as follows: E=a_{v}Aa_{s}A^{2/3}a_{c}Z(Z1)/A^{1/3}a_{sym}(NZ)^{2}/A why in the symmetry energy only squared parameter symmetry are exist and there is not the first power of asymmetry parameter? 



#2
Dec513, 01:56 PM

Mentor
P: 10,853

(NZ) 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 NZ, but experiments show this is not needed. And I don't see a physical reason for it. 



#3
Dec513, 02:48 PM

P: 377

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. 



#4
Dec513, 09:37 PM

P: 1,273

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.




#5
Dec1313, 07:28 AM

P: 514

Semiempirical mass formula  Wikipedia has a derivation of the form of the symmetryenergy term. The derivation treats protons and neutrons as separate but overlapping Fermi liquids that both extend over the nucleus.
E_{kinetic} = E_{Fermi}/A^{2/3}*(Z^{5/3} + N^{5/3}) 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 X^{2}. The absolutevalue 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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