What holds up a white dwarf? Not heat: it has no fusion and slowly cools forever. The answer is quantum mechanics, the same exclusion principle that builds the periodic table, applied to the electrons squeezed together in its dense matter.
Why it works
The exclusion principle allows only two electrons (spin up and spin down) in each quantum state. Squeeze $n_e$ electrons per cubic meter into a box and the states fill up from the lowest momentum: the uncertainty principle says an electron confined to a spacing $\Delta x \sim n_e^{-1/3}$ must have momentum $p \sim \hbar n_e^{1/3}$. Counting states exactly, every state up to the Fermi momentum $p_F = \hbar(3\pi^2n_e)^{1/3}$ is filled, even at zero temperature. Pressure is momentum flux, roughly $n_e \times p \times v$ with $v = p/m_e$, so $$P \sim \frac{n_e\,p_F^2}{m_e} \propto \frac{\hbar^2n_e^{5/3}}{m_e}, \qquad \text{exactly } P = \frac{(3\pi^2)^{2/3}\hbar^2}{5m_e}\,n_e^{5/3}.$$ Electrons dominate rather than protons because the pressure goes as $1/m$: at the same momentum, the light electrons move about 1,800 times faster.
Key idea
Electron degeneracy pressure comes from the exclusion principle, not from heat. It grows as $n_e^{5/3}$ and does not depend on temperature. Matter is degenerate when the Fermi energy $p_F^2/(2m_e)$ is much larger than $kT$.
Worked examples
Worked example: Electrons in white-dwarf matter
Carbon–oxygen matter at $1.0 \times 10^{9}$ kg/m³ has $\mu_e = 2$ nucleons per electron. Find $n_e$ and $p_F$.
- $n_e = \rho/(2m_p) = 1.0 \times 10^{9}/3.34 \times 10^{-27} = 3.0 \times 10^{35}$ m⁻³.
- $p_F = \hbar(3\pi^2n_e)^{1/3} = (1.055 \times 10^{-34})(2.07 \times 10^{12})$.
Answer: $p_F \approx 2.2 \times 10^{-22}$ kg·m/s, about $0.8\,m_ec$: the fastest electrons move at a large fraction of the speed of light.
Worked example: Degenerate versus thermal pressure
Compare the degeneracy pressure at that density with the ideal-gas pressure $n_ekT$ at $T = 10^{7}$ K.
- $P_{\text{deg}} = \dfrac{(3\pi^2)^{2/3}\hbar^2}{5m_e}n_e^{5/3} = 3.1 \times 10^{21}$ Pa.
- $n_ekT = (3.0 \times 10^{35})(1.38 \times 10^{-23})(10^{7}) = 4.1 \times 10^{19}$ Pa.
Answer: The degeneracy pressure is about 75 times larger. Equivalently, $E_F \approx 160$ keV against $kT \approx 0.9$ keV.
Worked example: Squeezing harder
If white-dwarf matter is compressed to 8 times its density, how does the (nonrelativistic) degeneracy pressure change?
- $P \propto n_e^{5/3}$.
- $8^{5/3} = (8^{1/3})^5 = 2^5$.
Answer: 32 times larger, with no change in temperature.
Common mistake
Thinking a white dwarf is held up by hot gas pressure. A white dwarf has no energy source; it radiates away its heat and cools for billions of years, yet it does not shrink, because its support does not depend on temperature. The pressure comes from electrons forced into high-momentum states by the exclusion principle. Check with numbers: $E_F$ is hundreds of times $kT$ inside, so cooling barely changes the pressure.