The Quantum Limit

Part 2: Degeneracy Pressure

Deriving the Pressure (Scaling Argument)

We can derive the scaling of degeneracy pressure using the Heisenberg uncertainty principle from Reading 1. Consider $N$ electrons in a volume $V$.

Step 1: Find the Fermi momentum. The number density is $n_e = N/V$. Each electron is confined to a volume $\sim 1/n_e$, so $\Delta x \sim n_e^{-1/3}$, and by the uncertainty principle the momentum is $p_F \sim \hbar/\Delta x \sim \hbar\,n_e^{1/3}$.

Step 2: Find the kinetic energy. For non-relativistic electrons ($v \ll c$), the kinetic energy per electron is $E_F \sim p_F^2/(2m_e) \sim \hbar^2 n_e^{2/3}/(2m_e)$.

Step 3: Find the pressure. Pressure is energy per unit volume, $P \sim n_e \times E_F$:

Key Properties

This result reveals several remarkable properties:

1. No temperature dependence. $P_\text{deg}$ depends on $n_e$ (density), not $T$. The pressure exists at absolute zero. This is why white dwarfs don’t need an energy source — they are held up by quantum mechanics, not heat.
2. Scales as $\rho^{5/3}$. Since $n_e \propto \rho$ (for a fixed composition), $P_\text{deg} \propto \rho^{5/3}$. This is steeper than an ideal gas ($P_\text{gas} \propto \rho T$): degeneracy pressure grows faster with compression — a stiffer equation of state.
3. Depends on $1/m_e$. Lighter particles produce stronger degeneracy pressure at the same density. This is why electron degeneracy supports white dwarfs — electrons are $\sim 1{,}800$ times lighter than protons, so their degeneracy pressure is $\sim 1{,}800$ times stronger. Proton/neutron degeneracy matters only at much higher densities (neutron stars, Reading 5).

Comparing Thermal and Degeneracy Pressure

Which pressure dominates depends on density and temperature:

At high density and low temperature, $P_\text{deg}$ dominates — the gas is degenerate. At low density and high temperature, $P_\text{gas}$ dominates — the gas is classical. The boundary is exactly the condition $\lambda_\text{dB} \sim d$ from Reading 1.