Semi Empirical Mass Formula and Neutron Stars

[city2113]

Homework Statement

Using the liquid drop model but adding a term for gravity to apply it to a neutron star, with only a large number of neutrons, rewrite the binding energy equation for this case

Relevant Equations

Semi Empirical Mass Formula: B(A,Z) = avA - asA^(2/3)-aa(A-2Z)^2/A+ap(delta/A^(1/2))
Gravity Term to add: (3/5)((G/ro)A^2*mn^2)/A^(-1/3)

Sorry, the equations are messy. I already know the answer to the actual homework problem, but I don't really know why certain terms are neglected.
I know that the equation will only include the volume term, symmetry term and the extra added gravity term. I just want to understand why

I know that the coulomb term is ignored because if you have only neutrons, there is no charge to consider.
I also know that the surface term and the pairing term become negligible at very large N values, but I don't know why.

Could someone explain why the surface and pairing terms get neglected in this case?

 

[Steve4Physics]

Hi @city2113 and welcome to PF.

I don’t understand enough to know why the pairing term becomes negligible. But see if this helps with the surface term...

The volume term depends on the total number of nucleons and the number of (very short range strong nuclear force) interactions of each nucleon with its neighbours.

The surface term is simply a correction to the volume term to allow for the fact that surface nucleons have fewer neighbours than the internal nucleons.

Suppose there are $A$ nucleons and $f$ is the fraction of the nucleons which are at the surface.

What do you think happens to $f$ as $A$ gets big?

 

[city2113]

Ohh ok, so as the number of nucleons (A) gets bigger, the fraction of nucleons on the surface would be smaller compared to the total number of nucleons. Once A becomes really bug the effect of the ones on the surface having fewer neighbors becomes so small that it's negligible because there are far more in the middle.

Thank you, that helps a lot!