Here is a schematic picture, due to Newton et al. (2011) that shows how the pasta phases develop with increasing density. As I show in some more detail below, the departure from an ordered lattice of spherical nuclei is expected to occur when the nuclei fill more than abut 1/10 of the volume.

Here is a schematic picture, due to Newton et al. (2011) that shows how the pasta phases develop with increasing density. As I show in some more detail below, the departure from an ordered lattice of spherical nuclei is expected to occur when the nuclei fill more than abut 1/10 of the volume.

The inner crust of a neutron star has densities $>4 \times 10^{14}$ kg m$^{-3}$ and consists of a lattice of extremely neutron-rich nuclei, surrounded by degenerate gases of neutrons and ultrarelativistic electrons that behave like fluids. The equilibrium composition and structure is found by forming the sum of the energy densities of these components and then minimising it with respect to the number densities of the nuclei, neutrons and protons, and the numbers of neutrons and protons in the nuclei, subject to baryon and charge conservation.

The energy density sum must include the rest mass energy of the nuclei, which is a function of the number of protons and neutrons, their binding energy plus modifications that account for the presence of the neutron fluid surrounding the nuclei that reduces the surface energy term and a negative term due the binding energy of the nuclear lattice.

Above densities of $\simeq 3\times10^{16}$ kg m$^{-3}$, nuclei probably start to lose their separate identities and spherical nature, forming elongated rods or planes of nuclear material -- nuclear pasta. Initially the nuclei are thought to group into "spaghetti" and then into "lasagna", where the sauce is provided by the gaseous free neutrons. As densities exceed $10^{17}$ kg m$^{-3}$, it is thought these roles reverse; the gaseous neutrons form into spaghetti like structures sourrounded by nuclear material and then finally, shortly before dissolution, the crust consists of nuclear material enclosing spherical bubbles of neutron gas.

Here is a schematic picture, due to Newton et al. (2011) that shows how the pasta phases develop with increasing density. As I show in some more detail below, the departure from an ordered lattice of spherical nuclei is expected to occur on the nuclei fill more than abut 1/10 of the volume.

However, in terms of the energtics of the gas as a whole, the nuclei are not isolated, but sit in a neutral Wigner-Seitz sphere, accompanied by $Z$ electrons distributed pseudo-uniformly, where $q_e(r) = -Ze(r/r_0)^3$, is the charge distribution of the electrons, and $r_0$ is the radius of the neutral Wigner-Seitz sphere. From this, it can be shown that the total Coulomb energy is $$ E_C = \frac{Z^2 E^2}{4\pi \epsilon_0}\left( \frac{3}{5r_N} + \frac{3}{5r_0} - \frac{3}{2r_0}\right)\, . $$ If we write the fraction of the volume occupied by the nuclei as $f = (r_N/r_0)^3$, we can rearrange this to give $$ E_C = E_{C}^{(0)}\left( 1 - \frac{3}{2} f^{1/3} \right)\, . \tag{1} $$

Now since $E_{C}^{(0)} \propto Z^2/r_N$ and $r_N \propto A^{1/3}$, where $A$ is the number of nucleons in each nucleus, then for a certain value of $Z/A$, the Coulomb energy per nucleon $E_C/A \propto A^{2/3}$. The surface energy term is proportional to the area of the nucleus, so $E_S/A \propto A^{-1/3}$. If we minimise the sum of these two terms with respect to $A$ as follows $$ \frac{\partial (E_{\rm tot}/A)}{\partial A} = \frac{\partial }{\partial A} \left( \alpha A^{2/3} + \beta A^{-1/3} \right) = 0\, \, $$ where $\alpha$ and $\beta$ are just arbitrary constants of proportionality. Solving this gives $2\alpha A^{2/3} = \beta A^{-1/3}$ and hence $2E_C = E_S$ at equilibrium.

The inner crust of a neutron star has densities $>4 \times 10^{14}$ kg m$^{-3}$ and consists of a lattice of extremely neutron-rich nuclei, surrounded by degenerate gases of neutrons and ultra-relativistic electrons that behave like fluids. The equilibrium composition and structure is found by forming the sum of the energy densities of these components and then minimising it with respect to the number densities of the nuclei, neutrons and protons, and the numbers of neutrons and protons in the nuclei, subject to baryon and charge conservation.

The energy density sum must include the rest mass energy of the nuclei, which is a function of the number of protons and neutrons, their binding energy, plus modifications that account for the presence of the neutron fluid surrounding the nuclei that reduces the surface energy term, and a negative term due the binding energy of the nuclear lattice.

Above densities of $\simeq 3\times10^{16}$ kg m$^{-3}$, nuclei probably start to lose their separate identities and spherical nature, forming elongated rods or planes of nuclear material -- nuclear pasta. Initially the nuclei are thought to group into "spaghetti" and then into "lasagna", where the sauce is provided by the gaseous free neutrons. As densities exceed $10^{17}$ kg m$^{-3}$, it is thought these roles reverse; the gaseous neutrons form into spaghetti like structures surrounded by nuclear material and then finally, shortly before dissolution, the crust consists of nuclear material enclosing spherical bubbles of neutron gas.

Here is a schematic picture, due to Newton et al. (2011) that shows how the pasta phases develop with increasing density. As I show in some more detail below, the departure from an ordered lattice of spherical nuclei is expected to occur when the nuclei fill more than abut 1/10 of the volume.

However, in terms of the energetics of the gas as a whole, the nuclei are not isolated, but sit in a neutral Wigner-Seitz sphere, accompanied by $Z$ electrons distributed pseudo-uniformly, where $q_e(r) = -Ze(r/r_0)^3$, is the charge distribution of the electrons, and $r_0$ is the radius of the neutral Wigner-Seitz sphere. From this, it can be shown that the total Coulomb energy is $$ E_C = \frac{Z^2 E^2}{4\pi \epsilon_0}\left( \frac{3}{5r_N} + \frac{3}{5r_0} - \frac{3}{2r_0}\right)\, . $$ If we write the fraction of the volume occupied by the nuclei as $f = (r_N/r_0)^3$, we can rearrange this to give $$ E_C = E_{C}^{(0)}\left( 1 - \frac{3}{2} f^{1/3} \right)\, . \tag{1} $$

Now, since $E_{C}^{(0)} \propto Z^2/r_N$ and $r_N \propto A^{1/3}$, where $A$ is the number of nucleons in each nucleus, then for a certain value of $Z/A$, the Coulomb energy per nucleon $E_C/A \propto A^{2/3}$. The surface energy term is proportional to the area of the nucleus, so $E_S/A \propto A^{-1/3}$. If we minimise the sum of these two terms with respect to $A$ as follows $$ \frac{\partial (E_{\rm tot}/A)}{\partial A} = \frac{\partial }{\partial A} \left( \alpha A^{2/3} + \beta A^{-1/3} \right) = 0\, \, $$ where $\alpha$ and $\beta$ are just arbitrary constants of proportionality, then solving this gives $2\alpha A^{2/3} = \beta A^{-1/3}$ and hence $2E_C = E_S$ at equilibrium.