Asymmetry Term from the Fermi-Gas Model

Derivation

Asymmetry Term from the Fermi-Gas Model

D-267 Home PU-304 Threads energy · matter · chance Depends on fermi-gas-degeneracy-pressure, density-of-states-3d-box

Statement

Treating the nucleus as two independent, fully degenerate Fermi gases of neutrons and protons sharing a common volume held at fixed nucleon density, we derive that the total ground-state kinetic energy carries, in addition to a bulk term proportional to \(A\), an isospin-restoring piece \(E_{\text{sym}}=\dfrac{1}{3}E_F^{0}\,\dfrac{(N-Z)^2}{A}\), where \(E_F^{0}\) is the Fermi energy of symmetric nuclear matter. This is the \((N-Z)^2/A\) term of the semi-empirical mass formula, and the Fermi-gas model fixes its kinetic coefficient at \(a_{\text{sym}}^{\text{kin}}=E_F^{0}/3\approx13\ \text{MeV}\).

Why it matters

The symmetry term is the reason stable nuclei hug the \(N\approx Z\) valley for light \(A\) and drift to neutron excess for heavy \(A\): it is a purely quantum-statistical energy cost for making the two Fermi seas unequal in depth. The Fermi-gas derivation shows this cost exists even with no interaction between nucleons — it is Pauli pressure, not force.

Quantitatively the model delivers only about half of the empirical coefficient (\(\sim23\) MeV), and that shortfall is itself the physics: it isolates how much of the symmetry energy is kinetic (degeneracy) and how much must come from the isospin dependence of the nuclear interaction. The same term governs the density dependence of the symmetry energy that controls neutron-star radii.

Assumptions

Nucleons are a non-interacting, non-relativistic Fermi gas in a rigid box.If dropped, the mean-field potential and effective mass \(m^\*\neq m\) rescale \(E_F^{0}\), and correlations shift the coefficient; the \((N-Z)^2/A\) form survives but its value is renormalised.

Both species are fully degenerate (temperature \(T=0\), all states below \(p_F\) filled).If dropped, a Sommerfeld \(T^2\) correction and partial occupation smear the sharp Fermi surface, adding thermal terms absent here — valid only for \(k_BT\ll E_F^{0}\sim38\) MeV.

Nuclear matter saturates: the nucleon number density \(\rho_0=A/V\) is a constant, independent of \(A\) and of the asymmetry.If dropped, \(V\) no longer scales as \(A\), the leading term stops being linear in \(A\), and the coefficient acquires a density dependence \(a_{\text{sym}}(\rho)\).

Neutrons and protons occupy the same spatial volume with equal effective mass, differing only in Fermi momentum.If dropped (e.g. proton skin from Coulomb, or \(m^\*_n\neq m^\*_p\) in isospin-asymmetric matter), separate volumes and masses break the clean \((N^{5/3}+Z^{5/3})\) structure and mix in surface and Coulomb contributions.

Derivation

\[ N_q=\frac{V\,p_{F,q}^{3}}{3\pi^{2}\hbar^{3}}\quad\Longrightarrow\quad p_{F,q}=\hbar\left(3\pi^{2}\frac{N_q}{V}\right)^{1/3} \]

Count spin-\(\tfrac12\) states (degeneracy \(g=2\)) filling momentum space up to \(p_{F,q}\) using the 3D box density of states; invert for each species \(q\in\{n,p\}\). A

\[ E_q=\int_0^{p_{F,q}}\frac{p^{2}}{2m}\,\frac{V}{\pi^{2}\hbar^{3}}\,p^{2}\,dp=\frac{V\,p_{F,q}^{5}}{10\,m\,\pi^{2}\hbar^{3}} \]

Sum the single-particle kinetic energy \(p^2/2m\) weighted by the momentum-space density of states \(dN=(V/\pi^2\hbar^3)p^2\,dp\); the integral of \(p^4\) gives \(p_{F,q}^5/5\). B

\[ E_q=\frac{\hbar^{2}}{10\,m\,\pi^{2}}\,(3\pi^{2})^{5/3}\,\frac{N_q^{5/3}}{V^{2/3}}=\frac{3}{5}N_q\,E_{F,q},\qquad E_{F,q}=\frac{p_{F,q}^{2}}{2m} \]

Substitute \(p_{F,q}\) from Step 1 into Step 2; the mean kinetic energy per particle is \(\tfrac35 E_{F,q}\), a standard degenerate-gas check. B

\[ E=E_n+E_p=\frac{\hbar^{2}(3\pi^{2})^{5/3}}{10\,m\,\pi^{2}}\;V^{-2/3}\left(N^{5/3}+Z^{5/3}\right) \]

Add the two independent Fermi gases; they share \(V\) and (by assumption) the same nucleon mass \(m\), so only the population powers \(N^{5/3},Z^{5/3}\) differ. A

\[ V=\frac{A}{\rho_0}\quad\Longrightarrow\quad V^{-2/3}=\rho_0^{2/3}\,A^{-2/3} \]

Impose saturation: hold the density \(\rho_0=A/V\) fixed so the volume grows linearly with mass number. This is the physical input that turns the bulk term into one \(\propto A\). B

\[ N=\frac{A+\delta}{2},\quad Z=\frac{A-\delta}{2},\quad \delta\equiv N-Z,\qquad N^{5/3}+Z^{5/3}=\left(\frac{A}{2}\right)^{5/3}\!\left[\left(1+\tfrac{\delta}{A}\right)^{5/3}+\left(1-\tfrac{\delta}{A}\right)^{5/3}\right] \]

Change variables to the total \(A\) and the asymmetry \(\delta=N-Z\); factor out \((A/2)^{5/3}\) to expose a symmetric function of \(\delta/A\). A

\[ \left(1+\varepsilon\right)^{5/3}+\left(1-\varepsilon\right)^{5/3}=2+\frac{10}{9}\,\varepsilon^{2}+O(\varepsilon^{4}),\qquad \varepsilon=\frac{\delta}{A} \]

Taylor-expand to second order: odd powers cancel by the \(\varepsilon\!\to\!-\varepsilon\) symmetry, and the quadratic coefficient is \(\tfrac12\cdot\tfrac53\cdot\tfrac23\cdot2=\tfrac{10}{9}\). Valid because \(|N-Z|\ll A\); the \(O(\varepsilon^4)\) terms are the leading correction. C

\[ E=\underbrace{\frac{3}{5}E_F^{0}\,A}_{\text{bulk}}\;+\;\underbrace{K\,\frac{(N-Z)^2}{A}}_{\text{symmetry}},\qquad K=\frac{\hbar^{2}(3\pi^{2})^{5/3}}{10\,m\,\pi^{2}}\,\rho_0^{2/3}\cdot\frac{10}{9}\,2^{-5/3} \]

Insert Steps 5–7 into Step 4. The "2" gives the leading term \(2^{-2/3}\!\cdot\!(\ldots)A=\tfrac35E_F^{0}A\) (mean KE per nucleon); the \(\tfrac{10}{9}\varepsilon^2\) piece gives the \((N-Z)^2/A\) term since \(\varepsilon^2 A=(N-Z)^2/A\). B

\[ E_F^{0}=\frac{\hbar^{2}}{2m}\left(\frac{3\pi^{2}\rho_0}{2}\right)^{2/3}\ \Rightarrow\ \frac{\hbar^{2}(3\pi^{2})^{5/3}}{10\,m\,\pi^{2}}\rho_0^{2/3}=\frac{3}{5}\,2^{2/3}E_F^{0}\ \Rightarrow\ K=\frac{3}{5}\,2^{2/3}\!\cdot\!\frac{10}{9}2^{-5/3}E_F^{0}=\frac{1}{3}E_F^{0} \]

Identify the symmetric-matter Fermi energy (each species at density \(\rho_0/2\)) and collect powers of 2: \(2^{2/3-5/3}=2^{-1}\), and \(\tfrac35\cdot\tfrac{10}{9}\cdot\tfrac12=\tfrac13\). The coefficient reduces to exactly \(E_F^{0}/3\). C

Result

\[ E_{\text{sym}}=\frac{1}{3}\,E_F^{0}\,\frac{(N-Z)^2}{A},\qquad a_{\text{sym}}^{\text{kin}}=\frac{E_F^{0}}{3}\approx 13\ \text{MeV} \]

Reading. Making the neutron and proton Fermi seas unequal — deepening one, shallowing the other at fixed total number — always raises the kinetic energy, and the penalty is quadratic in the imbalance \(N-Z\) and inversely proportional to \(A\) (a larger system dilutes the cost per unit imbalance). The scale is set by the Fermi energy: a third of it. No force enters; this is the Pauli exclusion tax on isospin asymmetry.

Units check. \(E_F^{0}\) is an energy (MeV); \((N-Z)^2/A\) is a pure number. Their product is an energy. With \(E_F^{0}=\hbar^2 k_F^2/2m\): \([\text{J}\,\text{s}]^2[\text{m}^{-1}]^2/[\text{kg}]=\text{kg}\,\text{m}^2\text{s}^{-2}=\text{J}\). Confirmed.

Limiting cases

Symmetric matter \(N=Z\): \(E_{\text{sym}}=0\); only the bulk term \(\tfrac35E_F^{0}A\) survives, as it must by neutron–proton symmetry.
Pure neutron matter \(Z=0,\ N=A\): \((N-Z)^2/A=A\), so \(E_{\text{sym}}=\tfrac13E_F^{0}A\); combined with the bulk term the energy per neutron is \(2^{2/3}\) times the symmetric value — the exact Fermi-gas result for a single species.
Small asymmetry \(|N-Z|\ll A\): the quadratic approximation is excellent; the neglected \(O(\varepsilon^4)\) term is smaller by \(\sim(N-Z)^2/A^2\).
Heavy-nucleus limit \(A\to\infty\) at fixed \(N/Z\): \(E_{\text{sym}}\propto A\), so it scales like the bulk term and remains a fixed fraction of the total energy.

Breaks when

Large asymmetry (\(|N-Z|\sim A\), e.g. the neutron drip line): the second-order Taylor expansion of Step 7 fails; the full \(N^{5/3}+Z^{5/3}\) must be kept and \(E_{\text{sym}}\) is no longer a clean \((N-Z)^2/A\).
Density leaves saturation (surface region, dilute matter, or compressed neutron-star cores): \(\rho_0\) is not constant, \(E_F^{0}\to E_F(\rho)\), and the coefficient becomes density-dependent — the model gives no surface symmetry energy.
Relativistic Fermi momenta (\(p_F\sim mc\), ultradense matter): the non-relativistic \(p^2/2m\) dispersion is wrong; \(E_F\) and the \(5/3\) power change, altering the coefficient.
Interaction-dominated regime: the kinetic value \(E_F^{0}/3\approx13\) MeV is only \(\sim55\%\) of the empirical \(\sim23\) MeV; the isospin-dependent potential energy, absent here, cannot be neglected.

Failure modes

Using one Fermi energy for both species. Students set \(p_{F,n}=p_{F,p}\); the whole effect lives in \(p_{F,n}\neq p_{F,p}\). You must keep \(N^{5/3}\) and \(Z^{5/3}\) distinct before expanding.
Forgetting the \(5/3\) power. Writing \(E\propto N^2+Z^2\) (as if energy \(\propto\) number squared) gives the wrong quadratic coefficient and misses the density factor.
Fixing \(V\) instead of \(\rho_0\). Holding volume constant while varying \(A\) destroys the linear-in-\(A\) bulk term and the \(1/A\) in the symmetry term.
Dropping the spin degeneracy \(g=2\). Halving the density of states misplaces \(k_F\) by \(2^{1/3}\) and \(E_F\) by \(2^{2/3}\), poisoning every subsequent number.
Expanding to first order. Stopping at \(O(\varepsilon)\) gives zero (odd terms cancel); the effect is intrinsically second order — a common "the term vanishes" mistake.
Claiming the model reproduces \(a_{\text{sym}}\approx23\) MeV. It gives \(\approx13\) MeV; treating the discrepancy as an error rather than as the interaction contribution misreads the physics.

Discussion

The symmetry energy is the archetype of a Pauli-pressure effect. At fixed total nucleon number, converting a proton into a neutron forces that nucleon into a higher, previously empty neutron level while the vacated proton level lies below the proton Fermi surface — a net upward transfer of a particle from one Fermi sea to another. Because both Fermi energies scale as \(N_q^{2/3}\), the cost of unbalancing is quadratic near equality, which is exactly why the term appears as \((N-Z)^2\) and not \(|N-Z|\).

The factor \(1/A\) has a clean interpretation: for a system of fixed density, a fixed absolute imbalance \(N-Z\) is "spread thinner" in a larger volume, so the Fermi-surface splitting it induces shrinks. This is the same reasoning that makes the symmetry term a volume effect (\(\propto A\) when \((N-Z)\propto A\)) rather than a surface effect in the leading Fermi-gas treatment — and the absence of a surface piece is precisely one of the model's known deficiencies.

Comparing \(E_F^{0}/3\approx13\) MeV to the empirical \(a_{\text{sym}}\approx23\) MeV is one of the most instructive numbers in nuclear physics: roughly half the symmetry energy is kinetic (degeneracy pressure) and half is potential, arising from the isospin dependence of the nucleon–nucleon interaction, which is more attractive in the \(T=0\) (np) channel than in \(T=1\) (nn, pp). Realistic energy-density functionals (Skyrme, relativistic mean field) split \(a_{\text{sym}}(\rho)=a_{\text{kin}}(\rho)+a_{\text{pot}}(\rho)\) exactly along this line, and the density slope \(L=3\rho_0\,\partial a_{\text{sym}}/\partial\rho\) — which the pure Fermi gas fixes at \(L_{\text{kin}}=2a_{\text{kin}}\) — is the single most important nuclear input to neutron-star radius predictions.

Common misconceptions. The symmetry energy is not Coulomb repulsion (that is a separate \(Z^2/A^{1/3}\) term and would favour more neutrons without limit); nor is it a mass difference between neutron and proton. It is a kinetic/statistical energy that would exist for two flavours of identical-mass, chargeless fermions. And it does not vanish at first order in \(N-Z\): the linear terms cancel by symmetry, leaving the quadratic as the leading effect.

Worked examples

Example 1 — The kinetic symmetry coefficient from \(k_F\).

\[ E_F^{0}=\frac{(\hbar c\,k_F)^2}{2mc^2},\qquad k_F=1.36\ \text{fm}^{-1},\ \hbar c=197.3\ \text{MeV fm},\ mc^2=939\ \text{MeV} \]

Write the Fermi energy in natural units to keep everything in MeV and fm. A

\[ \hbar c\,k_F=197.3\times1.36=268.3\ \text{MeV},\qquad E_F^{0}=\frac{(268.3)^2}{2\times939}=\frac{71\,985}{1878}=38.3\ \text{MeV} \]

Substitute numbers; the fm units cancel, leaving MeV. B

\[ a_{\text{sym}}^{\text{kin}}=\frac{E_F^{0}}{3}=\frac{38.3}{3}=12.8\ \text{MeV} \]

Apply the boxed result. A

\[ E_F^{0}\approx38\ \text{MeV},\qquad a_{\text{sym}}^{\text{kin}}\approx12.8\ \text{MeV} \]

Reading. The Fermi-gas kinetic coefficient is about \(13\) MeV — a little over half the empirical \(\sim23\) MeV, quantifying the interaction's share.

Units check. \((\text{MeV fm}\cdot\text{fm}^{-1})^2/\text{MeV}=\text{MeV}\). Consistent.

Example 2 — Symmetry energy of \(^{208}\text{Pb}\).

\[ N=126,\quad Z=82,\quad A=208,\quad N-Z=44 \]

Read off the neutron and proton numbers of lead-208. A

\[ \frac{(N-Z)^2}{A}=\frac{44^2}{208}=\frac{1936}{208}=9.31 \]

Form the dimensionless asymmetry factor. A

\[ E_{\text{sym}}^{\text{kin}}=a_{\text{sym}}^{\text{kin}}\,\frac{(N-Z)^2}{A}=12.8\times9.31=119\ \text{MeV} \]

Multiply the coefficient from Example 1 by the asymmetry factor. B

\[ E_{\text{sym}}^{\text{kin}}(^{208}\text{Pb})\approx119\ \text{MeV}\quad(\text{empirical }\sim214\ \text{MeV using }a_{\text{sym}}\approx23) \]

Reading. The kinetic term alone accounts for roughly \(120\) MeV of the \(\sim210\) MeV empirical symmetry energy of lead; the remaining \(\sim90\) MeV is the isospin-dependent potential energy.

Units check. MeV \(\times\) (dimensionless) \(=\) MeV. Consistent.

Problems

Show from the density of states that the mean kinetic energy per particle in a degenerate Fermi gas is \(\tfrac35E_F\).
Solution
\(\langle E\rangle=\dfrac{\int_0^{p_F}(p^2/2m)p^2\,dp}{\int_0^{p_F}p^2\,dp}=\dfrac{p_F^5/5}{2m\cdot p_F^3/3}\cdot\dfrac{1}{1}=\dfrac{p_F^5/10m}{p_F^3/3}=\dfrac{3}{5}\dfrac{p_F^2}{2m}=\dfrac35E_F.\) The common \((V/\pi^2\hbar^3)\) factors cancel in the ratio.
Compute the Fermi energy of pure neutron matter at the same total density \(\rho_0\) as symmetric matter, and express it as a multiple of \(E_F^{0}\).
Solution
In symmetric matter each species has density \(\rho_0/2\); in pure neutron matter the neutrons have density \(\rho_0\). Since \(E_F\propto\rho_q^{2/3}\), \(E_{F,n}^{\text{PNM}}=E_F^{0}\,(\rho_0/(\rho_0/2))^{2/3}=2^{2/3}E_F^{0}=1.587\times38.3=60.8\ \text{MeV}.\)
Verify the leading-order coefficient \(\tfrac{10}{9}\) in Step 7 by expanding \(f(\varepsilon)=(1+\varepsilon)^{5/3}+(1-\varepsilon)^{5/3}\) about \(\varepsilon=0\).
Solution
\(f(0)=2\). \(f'=\tfrac53[(1+\varepsilon)^{2/3}-(1-\varepsilon)^{2/3}]\Rightarrow f'(0)=0\). \(f''=\tfrac53\cdot\tfrac23[(1+\varepsilon)^{-1/3}+(1-\varepsilon)^{-1/3}]\Rightarrow f''(0)=\tfrac{10}{9}\cdot2=\tfrac{20}{9}\). So \(f\approx2+\tfrac12 f''(0)\varepsilon^2=2+\tfrac{10}{9}\varepsilon^2\).
For \(^{56}\text{Fe}\) (\(N=30,\ Z=26\)) estimate the kinetic symmetry energy, and compare with the empirical value using \(a_{\text{sym}}=23\) MeV.
Solution
\(N-Z=4\), \(A=56\); \((N-Z)^2/A=16/56=0.286\). Kinetic: \(E_{\text{sym}}^{\text{kin}}=12.8\times0.286=3.7\ \text{MeV}\). Empirical: \(23\times0.286=6.6\ \text{MeV}\). The kinetic term is again \(\sim56\%\) of the total, consistent with lead.
The most stable charge \(Z\) of an isobar minimises the total energy at fixed \(A\). Ignoring Coulomb, minimise \(E=a_{\text{sym}}(A-2Z)^2/A\) and interpret. Then state qualitatively how Coulomb repulsion shifts the minimum.
Solution
Write \(N-Z=A-2Z\). \(\dfrac{\partial E}{\partial Z}=a_{\text{sym}}\dfrac{2(A-2Z)(-2)}{A}=0\Rightarrow A-2Z=0\Rightarrow Z=A/2\), i.e. symmetry alone favours \(N=Z\). Adding the Coulomb term \(a_C Z^2/A^{1/3}\), \(\partial/\partial Z=0\) gives \(Z=\dfrac{A/2}{1+(a_C/4a_{\text{sym}})A^{2/3}}\), so the stable charge falls below \(A/2\) — heavier nuclei become progressively neutron-rich, exactly the observed drift of the valley of stability.