Neutron Stars

The collapsed core of a core-collapse supernova becomes a ball of neutrons with a radius of only about 12 km, holding more mass than the Sun. In 1967 Jocelyn Bell Burnell, a graduate student at Cambridge, found radio pulses repeating every 1.337 seconds with clock-like precision; they turned out to come from spinning neutron stars, now called pulsars. Two conservation laws explain why they spin so fast and carry such strong magnetic fields.

Why it works

Angular momentum. As in conservation of angular momentum, $I\omega$ stays the same when nothing twists the collapsing core. With $I \propto MR^2$, $R^2/P$ is conserved: shrinking the radius by a factor of 500 shortens the period by $500^2 = 250{,}000$. Magnetic flux. The core's hot plasma conducts electricity so well that the magnetic field lines are carried along with it, so the flux $\Phi \propto BR^2$ is conserved and $B$ grows by the same factor $R_i^2/R_f^2$.

Key idea

Collapse conserves angular momentum ($P \propto R^2$) and magnetic flux ($B \propto 1/R^2$). So neutron stars spin with periods from about a millisecond to seconds and have fields around $10^{8}$ T (magnetars up to about $10^{11}$ T), against about $10^{-4}$ T at the Sun's surface. Their density, about $4 \times 10^{17}$ kg/m³, matches that of atomic nuclei.

The same physics as a spinning skater who pulls in her arms: a smaller moment of inertia means a faster spin.

Worked examples

Worked example: Spinning up

A core of radius 6,000 km turns once every 30 minutes. It collapses to a 12 km neutron star. What is the new period?

$P_f = P_i(R_f/R_i)^2 = 1800\text{ s} \times (12/6000)^2$.
$(12/6000)^2 = 4.0 \times 10^{-6}$.

Answer: $P_f = 7.2 \times 10^{-3}$ s: about 140 turns per second.

Worked example: A teaspoon of neutron star

Find the mean density of a $1.4\,M_\odot$ neutron star of radius 12 km, and the mass of 5 cm³ of it.

$V = \tfrac43\pi(1.2 \times 10^{4})^3 = 7.24 \times 10^{12}$ m³.
$\bar\rho = 2.79 \times 10^{30}/7.24 \times 10^{12} = 3.8 \times 10^{17}$ kg/m³.

Answer: A teaspoon would hold about $2 \times 10^{12}$ kg, some 2 billion tonnes.

Worked example: How fast can it spin?

The equator of a spinning star must not move faster than orbital speed. What is the shortest period for $1.4\,M_\odot$ and $R = 12$ km?

$\omega^2R = GM/R^2$ gives $P_{\min} = 2\pi\sqrt{R^3/(GM)}$.
$R^3/(GM) = 1.73 \times 10^{12}/1.86 \times 10^{20} = 9.3 \times 10^{-9}$ s².

Answer: $P_{\min} \approx 0.6$ ms. The fastest known pulsar turns 716 times a second (1.4 ms); a white dwarf could never spin that fast without flying apart.

Common mistake

Thinking a neutron star spins slowly because the star it came from did. The Sun takes almost a month to turn, but shrink it to 12 km and conservation of angular momentum would spin it up by a factor of about $(7 \times 10^{5}/12)^2 \approx 3 \times 10^{9}$. Real cores lose some angular momentum along the way, yet young pulsars still spin many times a second. Check: $P \propto R^2$, so a large shrink means a huge spin-up.

Neutron Stars

Worked examples

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