Summary of Chapter 5 1 The Big Bang produced
Exercise 7.4 (a) What is the maximum radius for a 1 M & object rotating with
a period of 1 s?
(b) What is the minimum rotation period for a 1 M&neutron star with a radius of
10 km? ■
The suggestion that pulsars could be rotating neutron stars was made in 1968 by Thomas Gold. He suggested that the radio pulses are produced in the envelope of ions and electrons that surrounds the neutron star and rotates with it. In the outer regions of this envelope, the speed of individual electrons will be approaching the speed of light. As they are breaking away at such speeds from the magnetic field of the star, they can emit radio waves by the mechanism of synchrotron radiation. Although many working details of such a model are still a matter of discussion, one important consequence of this model was considered by Gold: if a rotating neutron star must continually support all the losses of energy due to the emission of fast particles and electromagnetic waves, its rate of rotation must be slowing down as it loses mass. Thus, if the period of radio pulses is determined by the period of rotation, it too must be changing – the pulsars should be slowing down.
7.3 Pulsars
Strong support for this theory was provided in November 1968 by the discovery of a pulsar in the central region of the Crab nebula. The Crab nebula is a supernova remnant (see Figure 7.3) which is the result of a supernova explosion that was observed by Chinese astronomers in 1054. The rotation period of the Crab pulsar is about 33 milliseconds, but more importantly, it was found that this pulsar was indeed slowing down. In 24 hours the interval between 2 successive pulses was longer by some 36 nanoseconds (the period would, at this rate, double in about a million days, or roughly every 2500 years).
Figure 7.3 An optical
image of the Crab nebula, the gaseous remnant of a supernova in 1054 (i.e.
≈ 1000 years ago). At
the heart of the gaseous remnant lies a pulsar with a 33 ms period, being the neutron-star remnant of the collapsed core of the massive parent star (progenitor).
● If the period of the Crab pulsar is 33 ms and it increases in period by 36 ns in 24 hours, what are its angular frequency ω, and its rate of change of angular frequency dω/dt (often written as ˙ω)?
❍ Angular frequency and period are related by ω = 2π/P , so
ω = 2π/(33 × 10−3 s) = 190 s−1.
By the chain rule, the rate of change of angular frequency is dω/dt = dω/dP ×dP /dt. Now, dω/dP = −2π/P2, so ˙ω = (−2π/P2) × ˙P .
An increase in period by 36 ns in 24 hours corresponds to a rate of change of period of ˙P = (36 × 10−9s)/(24 × 3600 s) = 4.2 × 10−13.
So the rate of change of angular frequency is dω/dt ≡ ˙ω = (−2π/(33 × 10−3s)2) × 4.2 × 10−13= −2.4 × 10−9 s−2.
Notice that if the pulsar is slowing down its period derivative ˙P is positive and dimensionless (or could be given units of s s−1), but the derivative of its angular
frequency ˙ω is negative and has units of s−2. The period derivative ˙P and the
angular deceleration ˙ω are sometime’s referred to as the pulsar’s spin-down rate. The supernova that produced the Crab nebula happened around 1000 years ago, and conditions within the nebula are no longer favourable for nuclear reactions to
occur. So the question arises, how does it shine? Suspicion naturally falls on the pulsar, which is clearly not conserving energy, and this leads to two further questions: (i) what is the mechanism by which the pulsar loses energy and (ii) is this energy loss rate enough to power the nebula? We address these questions in the following subsections.
7.3.3 The energy loss from a rotating pulsar
To test whether the slowing down of the pulsar in the Crab nebula is powering the observed nebula we begin by considering the moment of inertia of a neutron star. For a uniform sphere, the moment of inertia is I = 2MR2/5. The rotational
energy of such as sphere is then Erot = 12Iω2.
Worked Example 7.4
The angular frequency of the Crab pulsar is ω = 190 s−1and the rate of
change of angular frequency ˙ω = −2.4 × 10−9s−2.
(a) Calculate the moment of inertia of the Crab pulsar assuming it to have a mass of 1.35 M&and a radius of 10 km.
(b) Calculate the rotational energy of the Crab pulsar.
(c) Differentiate the equation for rotational energy to find an expression for the rate of change of rotational energy with time as a function of the rate of change of angular frequency with time (assuming the moment of inertia is constant).
(d) Calculate the current rate of loss of energy from the Crab pulsar.
Solution
(a) The moment of inertia is I = 2MR2/5 =
2 × 1.35 × 1.99 × 1030kg × (104m)2/5 ∼ 1038kg m2.
(b) The rotational energy is Erot= 12Iω2= 21 × 1038kg m2× (190 s−1)2
= 1.8 × 1042J.
(c) Differentiating the expression for rotational energy with respect to time gives ˙Erot = Iω ˙ω (where we have used the ‘dot’ notation to signify a
derivative with respect to time). (d) So in this case
˙Erot= (1038kg m2) × (190 s−1) × (−2.4 × 10−9s−2) = −4.6 × 1031J s−1.
The observed luminosity of the Crab nebula is about 5 × 1031W, so this ties in
extremely well with the observed rate of energy loss from the Crab pulsar. The loss of rotational energy from the Crab pulsar powers the Crab nebula.
7.3.4 The magnetic field strength of a pulsar
The rotational energy of pulsars is believed to be lost as a result of magnetic dipole radiation. If the magnetic axis of the neutron star is offset from its spin The magnetic axis of the star is
parallel to the direction of the star’s magnetic dipole moment.
7.3 Pulsars
axis by an angle θ (much as the Earth’s magnetic axis is offset from its spin axis), then it will emit electromagnetic radiation at a rate
˙Erad= 3c234πµ0m2ω4sin2θ (7.6)
where m is the magnitude of the neutron star’s magnetic dipole moment and µ0is
the permeability of free space.
● Assuming that all the rotational energy lost by the Crab pulsar is turned into magnetic dipole radiation, calculate the magnetic dipole moment m of the Crab pulsar.
❍ In this case ˙Erad = − ˙Erot. So rearranging Equation 7.6, we have m sin θ = * − ˙Eωrot4 3c23 4πµ 0 "1/2 .
So in the case of the Crab pulsar
m sin θ = ( 4.6 × 1031W (190 s−1)4 × 3 × (2.998 × 108m s−1)3 2 × 4π 4π × 10−7T m A−1 41/2 ∼ 4 × 1027J1/2m T−1/2A1/2 ∼ 4 × 1027(T A m2)1/2m T−1/2A1/2 ∼ 4 × 1027A m2.
Since sin θ must be ≤ 1, we can say m ≥ 4 × 1027A m2.
The magnitude B of the surface magnetic field of a neutron star is related to the magnitude of its magnetic dipole moment m and radius R by
B =4πRµ0m3. (7.7)
● Calculate the surface magnetic field strength of the Crab pulsar, assuming it to have a radius of 10 km.
❍ We have
B ≥ 4π × 10−7T m A4π × (10−1× 4 × 104m)3 27A m2 B ≥ 4 × 108T.
Hence the surface magnetic field strength of the Crab pulsar is in excess of 400 megateslas.
Exercise 7.5 Combine the equation for the rotational energy of a uniform