Pulsars

Builds on: Sections 10.1, 4.1 and S4
Reading time: 45 min
Exercises: 3
Olympiad problems: 1

In 1967 Jocelyn Bell, a research student at Cambridge, noticed a scrap of interference on the paper charts of a new radio telescope. It came back night after night from the same place in the sky, and when it was recorded in detail it turned out to be a train of short pulses, one every 1.3373 s, as regular as a good clock. The source is now called PSR B1919+21. The next year a source flashing about 30 times a second was found inside the Crab Nebula, the cloud left by a supernova that Chinese astronomers saw in Taurus in 1054 (Section O1.12). These sources are called pulsars. In this section we see what they are, and what their ticking tells us.

10.3.1What can flash so fast

Before you readMake a guess first

A source in the Crab Nebula flashes 30 times a second. What could it be?

A regular flash suggests something that turns, sending out a beam that sweeps past us once in each turn. A spinning body can only turn so fast before it flies apart, because the gravity at its equator must supply the pull towards the centre that keeps the surface going round. With $Ω = 2 π / P_{rot}$ the rate of spin, this needs $Ω^{2} R$ to be less than $G M / R^{2}$ , and with $M = \frac{4}{3} π R^{3} \bar{ρ}$ the radius drops out, leaving a limit set by the density alone.

ResultThe shortest period of a spinning star(10.6)

P_{rot} \geq \sqrt{\frac{3 π}{G \bar{ρ}}}

$P_{rot}$	the period of rotation
$\bar{ρ}$	the mean density of the star

This is the period at which a sphere would start to shed matter from its equator. A real star flattens as it spins, so its equator bulges out to where gravity is weaker, and the true limit is somewhat longer. The order of size is still right.

For the Sun the limit is 2.8 hours. Sirius B, with the density of $2.71 \times 10^{9}$ kg/m³ found in Section 10.1, could spin no faster than once every 7.2 s, so PSR B1919+21, with its period of 1.3373 s, is too fast for an ordinary white dwarf. Only a white dwarf very close to 1.4 solar masses, shrunk to about 1900 km, is dense enough to spin once every 1.2 s. The Crab pulsar, which spins every 33.8 ms, rules out every white dwarf. A neutron star, with its density of $3.85 \times 10^{17}$ kg/m³, can spin once every 0.61 ms. The fastest pulsar known spins 716 times a second, once every 1.40 ms, so every pulsar fits.

Quick checkHow fast the Sun could spin

The Sun's mean density is 1408 kg/m³. What is the shortest period with which it could spin without shedding matter, in hours?

hours

10.3.2The lighthouse model

A young neutron star such as the Crab pulsar has a magnetic field of about $10^{8}$ T, and its magnetic axis is usually tilted from its spin axis. Charged particles near the magnetic poles are flung along the field lines and give out radio waves in two narrow beams, one from each pole. As the star spins, each beam sweeps round the sky like the beam of a lighthouse, and we see a pulse each time a beam crosses Earth. The star itself shines steadily.

Let the magnetic axis be tilted by $χ$ from the spin axis, and each beam be a cone of half-width $θ_{b}$ . As the star turns, a pole stays at the angle $χ$ from the spin axis, so its beam sweeps a band of the sky between the angles $χ - θ_{b}$ and $χ + θ_{b}$ from the spin axis, and the other beam sweeps the matching band round the opposite pole. We see pulses only if our line of sight, at the angle $ζ$ from the spin axis, lies in one of these bands. If the beam is so wide that our line of sight never leaves it during a turn, the light we receive barely changes and there are no pulses.

ResultThe fraction of directions that see a pulsar(10.7)

F_{b} = \cos (χ - θ_{b}) - \cos (χ + θ_{b})

$F_{b}$	the beaming fraction, the fraction of all directions from which pulses are seen
$χ$	the tilt of the magnetic axis from the spin axis
$θ_{b}$	the half-width of each beam

This holds when the two bands do not overlap, that is when

χ - θ_{b}

is more than 0 and

χ + θ_{b}

is less than 90°. Each band covers

2 π [\cos (χ - θ_{b}) - \cos (χ + θ_{b})]

of the

4 π

steradians round the star, and there are two of them.

Figure 10.3The lighthouse model of a pulsar

Tilt of the magnetic axis

χ

Line of sight

ζ

Beam half-width

θ_{b}

Watch

Pulses at Earth
The beams sweep over

The figure shows a neutron star seen from the side, spinning about the dotted axis S, with its two beams along the magnetic axis M, tilted by

χ

, and the direction E to Earth at the angle

ζ

from the spin axis. The strip on the right runs from 0° to 180° from the spin axis and shades the bands that the two beams sweep, with Earth's line of sight marked E. The graph below shows the pulses that reach Earth over two turns. Spinning the star turns the beams, and the line to Earth turns heavy whenever a beam is crossing it. The figure can be set to , , , or .

A pulsar whose magnetic axis is at right angles to its spin axis, seen from the side, shows two pulses in each turn, one from each pole. Unless the beams are wide, the two bands cover only a small part of the sky, so we see a pulsar only if we happen to lie in one of them. For every pulsar that we see there are several that we cannot.

10.3.3Slowing down

The Crab pulsar spins once every 33.8 ms, and its period grows by $4.2 \times 10^{- 13}$ s every second, which adds a millisecond in about 75 years. The energy of its spin, $\frac{1}{2} I Ω^{2}$ , is falling. For a neutron star the moment of inertia $I$ is usually taken as $10^{38}$ kg m², which gives a store of $1.73 \times 10^{42}$ J. The spin is taken away by the star's spinning magnetic field, which drives a wind of particles and low-frequency waves out into the nebula.

ResultSpin-down luminosity and characteristic age(10.8)

L = - \frac{d}{d t} (\frac{1}{2} I Ω^{2}) = \frac{4 π^{2} I \dot{P}}{P^{3}}, τ = \frac{P}{2 \dot{P}}

$L$	the power lost by the spin
$I$	the moment of inertia of the neutron star, about $10^{38}$ kg m²
$P, \dot{P}$	the period and the rate at which it grows
$τ$	the characteristic age

The characteristic age is the true age if the pulsar is slowed by a spinning magnetic dipole, for which

\dot{P}

goes as

1 / P

, and if it was born spinning much faster than it does now. Otherwise it is only an estimate, too large if the pulsar was born spinning slowly.

Show the derivation

With $Ω = 2 π / P$ , $\dot{Ω} = - 2 π \dot{P} / P^{2}$ , so $L = - I Ω \dot{Ω} = 4 π^{2} I \dot{P} / P^{3}$ . A spinning magnetic dipole loses energy as $Ω^{4}$ , so $\dot{Ω}$ goes as $- Ω^{3}$ , which means that $P \dot{P}$ stays the same as the pulsar ages. Then $d (P^{2}) / d t = 2 P \dot{P}$ is constant, and $P^{2} = P_{0}^{2} + 2 P \dot{P} t$ , where $P_{0}$ is the period at birth. If $P_{0}$ is much less than $P$ , the age is $t = P^{2} / (2 P \dot{P}) = P / (2 \dot{P})$ .

Example 10.5The age of the Crab pulsar

Find the characteristic age of the Crab pulsar from its period of 33.8 ms and its rate of slowing of $4.2 \times 10^{- 13}$ , and compare it with the supernova of 1054.

Solution

By (10.8),

τ = \frac{0.0338}{2 \times 4.2 \times 10^{- 13}} = 4.02 \times 10^{10} s,

which is 1280 years.

AnswerThe characteristic age is 1280 years, against a true age of 972 years in 2026. The two agree in size, and the difference shows that the Crab pulsar was born spinning only about twice as fast as now, at a period of about 16 ms in this model, so

P_{0}

is not much less than

P

Example 10.6What powers the Crab Nebulafinish it

Find the power lost by the spin of the Crab pulsar, with $P = 33.8$ ms, $\dot{P} = 4.2 \times 10^{- 13}$ and $I = 10^{38}$ kg m².

The start

By (10.8), $L = 4 π^{2} I \dot{P} / P^{3}$ , with $P^{3} = {0.0338}^{3}$ s³.

Your stepNow work out

L

× 10³¹ W

10.3.4The dispersion measure

The pulses of a pulsar travel to us through the thin ionised gas between the stars. Radio waves of low frequency are slowed by the free electrons in this gas more than those of high frequency, so each pulse reaches us first at high frequencies and later at low ones. The delay grows with the number of electrons along the path, which is called the dispersion measure.

ResultThe delay of a pulse by dispersion(10.9)

Δ t = \frac{e^{2}}{8 π^{2} ε_{0} m_{e} c} \frac{DM}{f^{2}}, DM = \int_{0}^{d} n_{e} d l

$Δ t$	the delay of a pulse at the radio frequency $f$ compared with one at infinite frequency
$DM$	the dispersion measure, the number of free electrons in a column of unit area from the pulsar to us
$n_{e}$	the number of free electrons in each cubic metre along the path
$e, m_{e}, ε_{0}$	the charge and mass of the electron and the permittivity of free space

The constant is

1.34 \times 10^{- 7}

in SI units. Astronomers give the dispersion measure in pc cm⁻³, and then the delay is 4.15 ms times DM divided by the square of the frequency in GHz. The result holds for frequencies far above the plasma frequency of the gas, which is true for radio waves between the stars.

Show the derivation

In a gas of free electrons, waves below the plasma frequency $f_{p}$ , given by $f_{p}^{2} = n_{e} e^{2} / (4 π^{2} ε_{0} m_{e})$ , cannot travel at all, and a pulse at a higher frequency $f$ travels at $c \sqrt{1 - f_{p}^{2} / f^{2}}$ . When $f$ is much more than $f_{p}$ this is about $c (1 - f_{p}^{2} / (2 f^{2}))$ , so each metre of the path takes longer than it would in a vacuum by $f_{p}^{2} / (2 c f^{2})$ . Adding this up along the path gives the result.

The Crab pulsar has a dispersion measure of 56.8 pc cm⁻³. Its pulses at 400 MHz arrive 1.47 s later than they would at infinite frequency, and 1.35 s later than its pulses at 1400 MHz, so 40 pulses pass between the two. Dispersion also measures distance. Between the stars there are about 0.03 free electrons in each cubic centimetre, so the Crab's dispersion measure puts it about $56.8 / 0.03 = 1890$ pc away, close to the 2.0 kpc found in other ways.

*10.3.5The magnetic field of a pulsarOptional extension

The slowing of a pulsar also gives the strength of its magnetic field. A magnetic dipole whose field at the magnetic equator is $B$ , of radius $R$ and spinning at $Ω$ , with its axis at right angles to the spin axis, radiates the power $8 π B^{2} R^{6} Ω^{4} / (3 μ_{0} c^{3})$ . Setting this equal to the spin-down luminosity of (10.8) gives the field.

ResultThe magnetic field of a pulsar(10.10)

B = \sqrt{\frac{3 μ_{0} c^{3} I P \dot{P}}{32 π^{3} R^{6}}} \approx 3.2 \times 10^{15} T \times \sqrt{P \dot{P}}

$B$	the strength of the magnetic field at the surface, at the magnetic equator, half the field at the poles
$μ_{0}$	the permeability of free space
$I, R$	the moment of inertia and the radius of the neutron star
$P, \dot{P}$	the period in seconds and its rate of growth

The number

3.2 \times 10^{15}

T is for

I = 10^{38}

kg m² and

R = 10

km, the values usually taken in this formula. With the radius of 12 km used in Section 10.1 it would be

1.85 \times 10^{15}

For the Crab pulsar, $\sqrt{0.0338 \times 4.2 \times 10^{- 13}}$ makes the field $3.8 \times 10^{8}$ T, and for PSR B1919+21, which slows by $1.35 \times 10^{- 15}$ s each second, about $1.4 \times 10^{8}$ T. These match the fields that a collapsing core reaches by keeping its magnetic flux, as we estimated in Section 10.1.

Common errorsSection 10.3

A pulsar does not switch on and off. It shines steadily in two beams, and the pulses come from the beams sweeping across Earth.
The shortest period of a spinning star depends on its mean density, not on its mass or radius separately.
The characteristic age $P / (2 \dot{P})$ is the true age only if the pulsar was born spinning much faster than now and is slowed as a magnetic dipole.
The dispersion delay goes as $1 / f^{2}$ , so the delay between two frequencies is proportional to $1 / f_{1}^{2} - 1 / f_{2}^{2}$ , not the delay at the lower frequency alone.

ExercisesSection 10.33 questions

10.3.1
A pulsar has a dispersion measure of 30 pc cm⁻³. By how much does its pulse at 1.0 GHz lag behind its pulse at 1.5 GHz, in milliseconds?
ms
10.3.2
Why can PSR B1919+21, with its period of 1.3373 s, not be a white dwarf like Sirius B, spinning?
10.3.3
A pulsar has its magnetic axis at right angles to its spin axis and beams of half-width 10°. What percentage of all directions see its pulses?
%

Olympiad problemsPast IOAA problems, restated1 problem

2012Will we see the pulsar?10 points

Adapted from IOAA 2012 (Rio de Janeiro), theory problem S8. We have restated it in our own words and kept the original data. The official solution reaches −3.24, which corresponds to taking the Sun's absolute bolometric magnitude as 4.80, and with the course's 4.74 it is −3.30. The solution was last checked on 23 October 2026.

A pulsar 1000 pc from Earth has a luminosity of 10 000 times the Sun's, all of it sent out in two opposite beams. Each beam is a uniform cone with a full opening angle of 4°. The beams are tilted by 30° from the spin axis, and the spin axis points in a random direction. What is the probability that we see its pulses? If we see it, what is its apparent bolometric magnitude?

I solved it without the solution

AnswerThe probability is 3.5%, and the pulsar would have an apparent bolometric magnitude of −3.30, brighter than any star in the sky.