Rotation-Powered PulsarsAs we discussed last time, the extreme properties of neutron stars (gravity, density,magnetic field) make them excellent objects to study to see how physics operates in unusualenvironments. However, until about forty years ago it was not clear that they would everbe observed. It is useful to backtrack a bit and go over some of the early history of thoughtabout neutron stars.The neutron was discovered in 1932 by Chadwick (evidence had been seen earlier byIrene and Frederic Joliot-Curie, but they thought it was a high-energy photon). As soon asnews of this reached Moscow, Lev Landau and friends had a dinner meeting in which Landausuggested that, just as white dwarfs are supported by electron degeneracy pressure, “neutronstars” might exist that are supported by neutron degeneracy pressure. Just two years later,Baade and Zwicky suggested that supernovae represent the transition from normal stars toneutron stars (and, incidentally, that supernovae are the source of cosmic rays). However,how could these be seen? Such objects would be extremely dim, even though hot, becausethey have such small areas. There was some work in the early 1960s on whether they mightbe observable in the X-rays (since X-ray sounding rockets were just then being launched),but it didn’t look really feasible.Into this milieu in 1967 came Jocelyn Bell, a graduate student at Cambridge Universityworking with Anthony Hewish. Her thesis project had to do with scintillation observationsof quasars, which had just been discovered in 1963. She and the other members of Hewish’steam constructed a radio array, then she sat down to take the data. In the fall of 1967 shesaw a strange item on one of the recording charts: a bit of “scruff” that was distinct fromthe usual types of noise that she had seen before. At the suggestion of Hewish, she put ahigh-speed recorder on the instrument, and incredibly the scruff was resolved as a series ofvery regular pulses, about 4/3 seconds apart. After eliminating artificial sources, it was clearthat this source (and a couple others like it) were natural, coming from somewhere in theheavens. But what could cause this?The Nature of PulsarsWe know now that they are rotation-powered pulsars, but it is very instructive to gothrough the chain of logic leading to that conclusion. In 1968, there were three importantfacts established about pulsars:1. They were fast, with periods between 0.033 s and 3 s. The pulses themselves could beeven shorter, <0.01 s in some cases.2. They were extremely stable, with P/˙P equal to thousands to millions of years.3. They always increased in period slightly, never decreased.Let’s consider what this implies. Ask class: what does the first fact imply about thenature of the object? The fast periods and short pulses imply something small. Barringbeaming, the size of the object would have to be < ct, if t is the time of the pulse. Thatgives 3,000 km for 0.01 s. Barring relativistic beaming with γ ∼ 100 or more, this demandsa compact object (white dwarf, neutron star, black hole) instead of a normal star. The nextquestion is Ask class: what are ways in which one can get a periodic signal from a star orstars? Generically one might think of rotation, vibration, or a binary orbit (indeed, all threeare seen in various circumstances). Let’s focus first on white dwarfs. Can they, by rotating,vibrating, or pulsating, give a 0.033 s period? No: for all three processes, the minimumperiod is of the order 2π(Gρ)−1/2, which is a minimum of ∼seconds for ρ ∼ 108g cm−3maximum densities of white dwarfs. Therefore, white dwarfs can’t do it. Ask class: whatcould one say about black holes? Can they produce periodic pulsations? No, but the answerisn’t as obvious as it used to be. Getting a periodic signal out implies some fixed structure,and there just isn’t any such place on a black hole. Of course, quasiperiodic oscillations areseen, just nothing as coherent as from a pulsar. That means we’re down to neutron stars. Isit rotation, vibration, or a binary orbit? Ask class: what could one say against vibration?Here the ultrahigh density means that the vibration period, of order 2π(Gρ)−1/2, is a fewmilliseconds, much too fast for the majority of pulsars. Ask class: what can be said againstbinary orbits? Two things. First, inspiral due to gravitational waves would cause the periodto decrease, not increase. Second, a binary orbit with a period of seconds would cause thetwo to spiral into each other in at most a few hours, whereas many pulsars have now beenobserved for thirty years.This line of argument, due originally to Tommy Gold, indicates that pulsars must berotating neutron stars. Therefore, remarkably, neutron stars are very observable.Pulsar Physics: Rotating DipoleInterestingly, before the discovery of pulsars Franco Pacini had suggested that a rotatingpulsar might emit radio waves. This was a promising start, but our understanding of whypulsars emit radio waves is very incomplete even now, more than thirty years later.Let’s start with the overall energetics. Suppose that we have a neutron star of radiusR with a centered dipolar magnetic field of strength Bpat the magnetic pole, and that theaxis of the dipole is offset by an angle α from the rotational axis. The magnitude of themagnetic moment is then|m| = BpR3/2 . (1)The nonalignment of the magnetic moment with the rotation axis means that the magneticFig. 1.— The lighthouse model of pulsars. A beam of radio waves, tiedto the surface of the star, sweeps past us as the star rotates. Fromhttp://www.astro.cornell.edu/academics/courses/astro201/images/pulsar.gifmoment changes with time. Just as an accelerated charge radiates, so does an acceleratedmagnetic moment, and the rate is˙E = −23c3| ¨m|2= −B2pR6Ω4sin2α6c3. (2)This radiation is (at least initially) emitted at the rotational frequency Ω. This energy comesfrom the rotational energy of the star, E =12IΩ2, where I ∼ 1045g cm2is the moment ofinertia. Therefore we also have˙E = IΩ˙Ω. The pulsar slows down as a result of the torqueexerted by the radiation. If it starts with spin frequency Ωi, then if we consider the currentslowdown time T = −Ω/˙Ω = P/˙P and say that the current spin frequency is Ω0, then thespin frequency at time t isΩ = Ωiµ1 +2Ω2iΩ20tT¶−1/2, (3)so the present age of the pulsar is (Ω = Ω0)t =T2¡1 − Ω20/Ω2i¢. (4)If the current spin frequency is