7.1 Introduction

   Cosmic-ray physicists demonstrated that our Galaxy is filled with very energetic cosmic rays, mostly protons. Observations in the radio and x-ray range have made astronomers realize that much of the universe is filled with energetic electrons. Synchrotron radiation from relativistic electrons is a very common cosmic phenomenon (problems 7.2, 7.3). During the last decades, as observations were expanded from the x-rays toward ever higher energy gamma rays, the highest deduced energies of electrons have grown dramatically. Inverse-Compton radiation, which used to be mostly a topic of theoretical physics, has become very real (problem 7.4). Just how the electrons acquire their high energies is not clear, but magnetic energy is surely involved, and one part of the process is observed in the Crab Nebula (problem 7.5).

   The radiation process are derived here in a heuristic manner, using only rather simple concepts from radiation theory and relativity. It is much more important, in astrophysics, to keep in mind the relevant physics rather than worry about the factors of two arising from exact theory.

An overview of radiation processes. The radiation from the hot gases in the solar corona, in accretion disks, and between the galaxies located in clusters of galaxies mimics the shape of the Planck spectrum, though not its intensity. This radiation is considered "thermal" radiation because the particles have a thermal Boltzmann velocity distribution. Collisions among these particles both create the Boltzmann velocity distribution and cause the emission of the "thermal" radiation.

   In many astrophysical situations such as solar flares, supernova remnants, or quasars, electrons are accelerated by electromagnetic fields to energies orders of magnitude higher than their original thermal energies. Typically, these electrons acquire a power-law distribution in energies. In solar flares, the electrons reach mildly relativistic energies. When these electrons migrate from the flare in the solar corona to the underlying denser gases, they collide with local protons and electrons and emit a power-law spectrum of x-rays. Such radiation is referred to as "non-thermal" x-rays. Both thermal and nonthermal x-rays are caused by electron collisions, and in both cases the emission process is often called Bremsstrahlung.

   More frequently in astrophysics, the electrons reach highly relativistic energies. These electrons easily radiate on their own, without any collisions with other particles. They radiate via the cyclotron, synchrotron, and inverse-Compton radiation mechanisms. The cyclotron and synchrotron radiation mechanisms depend on the energy density in magnetic fields. These fields are present everywhere, but they tend to be stronger in those places where electrons are accelerated to high energies, thus helping the high efficiency with which these electrons emit cyclotron or synchrotron radiation. The inverse-Compton mechanism depends on the energy density of electromagnetic radiation, which is very high in powerful objects like quasars.

7.2 The Crab Nebula
      Synchrotron radiation - relativistic beaming

Physics needed: Lorentz transformations in special relativity; relativistic gyrofrequency.

Physics Introduction. A non-relativistic electron in a magnetic field of strength B has an angular frequency qB/m rad/s (q=1.6x10-19C) or a frequency fo = 2.8x104B MHz where B is in Teslas. The electron emits electromagnetic radiation at frequency fo. When an electron is highly relativistic with energy E = gammamc2, gamma>>1, the gyrofrequency is fo/gamma. [mc2 = 0.8x10-13j = 0.5MeV, where eV = electron Volt.] The most intense radiation we observe from this electron is at a frequency different from the gyrofrequency. First, a Lorentz transformation alters the apparent angle at which a photon propagates. If photons are emitted roughly isotropically in the frame of an electron, they are observed by us as a beam of photons concentrated in the forward direction, with opening angle theta proportional 1/gamma radians. In addition, there is the equivalent of a Doppler shift. Let the beam be emitted while the electron travels the distance 2Rtheta between points a and b in time deltat(el)=2Rtheta/v. This would be the duration of the beam we would observe if the speed of light were infinite. In reality, while the electrons travel from a to b, the first part of the beam, emitted when the electrons were at a, travels the distance cdeltat(el) > 2Rtheta. Thus the front of the beam is ahead of the back of the beam, emitted when the electrons are at b, by (c-v) deltat(el) and the duration of the beam we observe when it arrives here is (1-v/c) deltat(el). Therefore, the duration of the beam is shortened by a factor 1-v/c = (1- v2/c2)/(1+v/c) = 1/2gamma2, where we have set 1+v/c = 2 in the denominator. Accordingly we see a beam of a duration shortened by a factor of 2gamma2relative to deltat(el). Using theta proportional 1/gamma, we find that the pulse lasts only a fraction proportional1/gamma 3 of the gyroperiod 2piR/v = gamma /fo, that is, it lasts only about 1/(fogamma2). Such a pulse, when Fourier analyzed, has its main signal at a frequency proportional fogamma2.

   Computed in more detail, the main emission is observed at about f = 104gamma2B MHz. Any one electron also radiates at somewhat higher and lower frequencies, say from 0.3 to 1.5 f. If we observe an object containing many electrons with a large range of energies, we actually detect at our selected observing frequency contributions from electrons with gamma whose f is within 1/1.5 to 1/0.3 times the observing frequency. If we gradually change the observing frequency, we detect radiation from electrons with a gradually changing range of energies. The received radiation changes smoothly with observing frequency. We observe a continuous spectrum. This type of radiation is named synchrotron radiation.

The problem: We observe the Crab Nebula (B = 4x10-8Tesla) in the radio range at 300 MHz. What is the energy of the electrons whose radiation we observe? Compare to the energy 3/2 kT for T=108oK. We also observe the Crab Nebula in the x-ray range, say at wavelengths of 1 nm. What is the energy of the x-radiating electrons? Compare the energy to that of a flying mosquito.

The astrophysical setting: The Crab Nebula is the gaseous remnant of a supernova seen by humans in the year 1054 A.D. = 433 A.H. = 4815 A.M. (Picture in problem 3.4.) The Crab is named after a pattern of filaments, most obvious in red light due to atomic hydrogen emission at 656 nm. In addition, there is continuum radiation from all parts of the roughly circular nebula. The continuum is synchrotron radiation. This radiation was first detected and identified in the 1950's in the radio range, but now it is observed at frequencies up to the x-ray range. The pulsar which is left from the exploded star energizes the nebula (problems 3.4, 7.5) and indirectly accelerates electrons to the high energies evaluated here.

The solution: f = 4x10-4gamma2 MHz. Radio range: gamma2 = 0.8x106, energy 0.5x109eV = 0.8x10-10j >> 3/2 kT = 2x10-15j. X-ray range: f = c/lambda = 3x1017Hz = 3x1011MHz, gamma2 = 8x1014, energy 1.4x1013eV = 2.3x10-6j. This is for just one electron. For the mosquito, estimate v = 1 km/hr, mass 10-2gm , kinetic energy = 1/2 10-5kg (0.3m/s) 2= 0.5x10-6j, comparable.

Interpretation: These electrons have energies far above any plausible thermal energies. These electrons radiate so powerfully that they lose their energy in a time shorter than the age of the Crab Nebula (see next problem). Therefore, such electrons must be accelerated in recent times. This was the first good evidence for what is now generally accepted : electrons are accelerated much more easily than skeptical theoreticians had expected.

Didactics: The derivation of f is only heuristic. Most advanced students in astrophysicists must suffer through a detailed, quite mathematical derivation and they (almost) all decide that the heuristic derivation is sufficient for their needs.


  7.3 The Crab Nebula
      Synchrotron radiation - total power

Physics needed: Larmor formula for rate of radiation; relativistic time dilation and gyrofrequency. The student must first do problem 7.2.

Physics Introduction: Classically, the energy loss by radiation from an electron spiraling in a magnetic field of strength B is dE/dt = 2/3 kq2adota/c3 where non-relativistically a = dv/dt with mdv/dt = -qvxB, leading to a gyrofrequency qB/m rad/s. (Here k = muc2/4pi = 9x109, q = 1.6x10-19Coulomb.) For very energetic electrons, there are two relativistic effects. First, the gyrofrequency becomes qB/gammam, introducing a factor 1/gamma for each factor a. Second, since the time appears squared in a, time dilation introduces a factor gamma2 for each factor a. The above formula becomes dE/dt = 8pi/3 re2cgamma2 (B2/mu)(vperp/c) 2, where re = kq2/mc2 = 2.8x10-15 m is the classical electron radius. For an isotropic velocity distribution, averaging (vperp/c) 2over angles yields an additional factor 2/3, so that dE/dt = 4/3 sigmaT c gamma2UB, where sigmaT =8pi/3 re2 is the Thomson crossection, UB = B2/2mu is the energy density in the magnetic field, and v = c for highly relativistic electrons. This last form of dE/dt gives the total power radiated per electron in an isotropic distribution of electrons with energy gammamc2.

The problem: The electrons lose kinetic energy by radiation. Derive the time scale tau = E/(dE/dt) using dE/dt averaged over isotropic electrons. Compute tau for both the electrons yielding the radio emissions and the electrons yielding the x-ray emissions in the Crab Nebula (see problem 7.2). Can the relevant electrons have existed for the entire age of the Crab Nebula proportional 950 years?

The solution: tau = 3mc/(4sigmaT UBgamma) = 0.48x1016/gamma seconds. For the radio emitting electrons, tau = 1.7x105years >> 103years; yes; for the x-ray emitting electrons, tau = 5 years << 103years; no.

Interpretation: The radio emitting electrons have lost very little of their energy. They may have been dispersed throughout the nebula at the time of the supernova explosion. But the x-ray emitting electrons lose their energy very rapidly. Any such energetic electrons accelerated at the time of the explosion would have lost their energy long ago. They must have been accelerated during the last few years. Evidently the pulsar is still a very active object and the energy flowing from it (problems 3.4, 7.5) continuously accelerates the electrons.


  1. Students may ask why we bother to express dE/dt in terms of reand the Thomson cross section, since we deal with the spiraling of an electron in a magnetic field, while the Thomson cross section measures an electron's response to the electric field of passing electromagnetic radiation. Physically, the magnetic field is just a low-frequency electromagnetic field. In fact, we shall use this knowledge in the next problem. Moreover, the use of a cross section also leads to the use of an energy density. It is always good in astrophysics to express quantities in terms of energies, since energies directly provide physical implications. Mathematically, the formula is in a form where one can easily check that the dimensions are correct, energy per second, without getting into the dimensions of electric charge.

  2. The formula for dE/dt was derived assuming constant electron energy over the time of a few gyrations around the magnetic field, that is, during tau >> R/v. This is very well satisfied. The result for dE/dt given here is rigorous, although the derivation is abbreviated. Radiation tends to make the electrons anisotropic, but they are scattered and kept nearly isotropic by Alfvén waves.

  3. Most students will first evaluate dE/dt, then E, then divide. Let them do this. Afterwards suggest the simpler way of writing down the combined formula for tau before evaluating.

7.4 Gamma rays from quasars
      Inverse Compton radiation

Physics Introduction: Highly relativistic electrons may collide with low-energy photons to create high-energy photons, the inverse-Compton (IC) effect. The resulting photon energy can be derived quite simply by two transformations of reference frames: a photon of frequency v is "seen" by an electron with energy gammamc2as having a frequency proportional gammav; in the frame of the electron, the electron scatters the photon at constant energy (true if the photon energy in this frame is less than mc2 so that the electron suffers no recoil); finally, back in our frame of reference, we see the photon as having a frequency proportional gamma2v. This process can easily turn a radio photon into a gamma ray. The rate at which a fast electron turns low-energy photons into high-energy photons is best given by analogy to the rate of synchrotron loss, dE/dt = 4/3 sigmaT c gamma2Urad, where Urad is now the energy density in the low-energy photons. The low-energy photons may belong to the cosmic background radiation (often called 3oK radiation but actually T = 2.7oK). This radiation has an energy density of 0.4x10-13j m-3. Alternatively, the low-energy photons may be the synchrotron photons emitted by the energetic electrons. The latter process is called the synchrotron self-Compton process.

The problem: The quasar 3C279 emits photons up to the high gamma-ray range, 4 GeV (GeV = 109 eV). Suppose that the 4 GeV photons have been IC-boosted, starting as microwave photons of 1011Hz. What is the gamma of the electrons? Suppose that the 4 GeV photons have been IC-boosted twice by the same electrons, starting from microwave photons of 1011Hz. What is the required gamma of the electrons?

The setting: A number of quasars and "active galactic nuclei" emit several components of radiation. One component may be due to synchrotron radiation. Another, often the most energetic component, may be inverse-Compton radiation. Some quasars produce astoundingly intense high-energy radiation. The quasar 3C279 has been observed with the satellite called Compton Gamma-Ray Observatory up to photon energies of 4 GeV. Its spectrum between 1011Hz and 4 GeV is close to a single power law f-athat is flatter than a = 1. The object emits most of its energy in the gamma-ray range.

The solution: hv = 6.6x10-23j; 4x109eV/4x10-4eV = 1013. For a single IC boost in energy, gamma2 = 1013, gamma = 3x106. For a double IC boost, gamma4 =1013, gamma =2x103.

Interpretation: The spectrum of 3C279 rises rapidly through the radio range and becomes much flatter at 1011Hz (microwave range). Thus, in this problem, the microwave photons are taken to be the most abundant photons available for IC-boosting. But a single IC boost needs electrons at extremely high energies, for which there is no independent evidence. No theory yields sufficiently rapid acceleration to these very high energies. Therefore, double-IC-boosting has been investigated. It functions with electrons of more normal energies. So far, however, the details of the radio and gamma-ray spectrum of this particular quasar do not fit the theory of double IC boosting. (The theoretical details include the electron recoil at the second scattering.) Perhaps the compact jet emanating from this object provides a local intense source of higher-frequency photons that can be IC-boosted.

   The Crab Nebula emits radiation up to 5x1013eV! Perhaps this is IC radiation, but again the theory does not fit well to observations.


  1. Ask your students to find the total power loss by integrating the spectrum f-ato ever higher f, using a < 1. They will, of course, find that the integral diverges. In some range of very large frequencies, the spectrum must decrease, a > 1. We do not know at what frequencies this occurs.

  2. Quasars and related objects (with a host of names) are exciting because they still surprise us. Thirty years ago, their total power output seemed unbelievable, but astrophysicists gradually became used to the power released near a black hole, once some black holes (with either star or galaxy masses) were clearly identified. Today the excitement focuses on the extreme energy to which at least a fraction of all the fast electrons are accelerated and the powerful gas flows that make this acceleration possible. The work is challenging both observationally - to make and combine observations spanning fifteen orders of magnitude in frequency - and theoretically - to understand the electron-photon interactions that can focus so much energy into a few electrons, to understand the violent gas flows, especially jets, which probably provide the energy source for the acceleration and, ultimately, to identify how all this relates to the central power source surrounding the probable black hole.

Relativistic motion (jets): When radio telescopes first resolved the angular structure of quasars on the sky, some of the radio sources were resolved into parts that appeared to separate on the sky faster than the speed of light. Indeed, a source moving toward us, nearly along our line of sight, at a speed v close to c can appear to move across the sky at a speed up to gammav with gamma = 1/(1-v2/c2)1/2. When this explanation was first suggested, such fast objects seemed incredible. Now, however, relativistic motions with gamma in the range 2 to 10 are also used to explain why electron energies in very powerful quasars and in apparently rapidly varying (thus apparently very compact) quasars are evidently not quickly drained by inverse - Compton radiation. (The deduced energy density of radiation is lower and tau proportional to 1/Urad is longer if one invokes relativistic motion.) Relativistic jets now seem to be common to many quasars and centers of active galaxies. In our Galaxy, we find relativistic jets on a stellar scale. The large variety of powerful objects may perhaps all be reduced to one kind of object, with power emanating from a central "engine", and with jets that are merely seen from different directions.


  7.5 Magnetic field of pulsars
      low-frequency magnetic "dipole" radiation

Didactic purpose: An example of approximations that retain the essential physics.

Physics needed: Energy density and propagation speed of an electromagnetic wave.

The setting: The Crab nebula needs a prodigious source of energy, about 1.2x105Lo. Indeed, the central rotating neutron star is slowing down and provides the energy needed for the nebula (problem 3.4). But how is the energy carried from the pulsar to the nebula? The rotating pulsar causes the surrounding magnetic field to vary in time. Thus the pulsar emits electromagnetic radiation at the pulsar rotation period. The pulsar radiates as much energy as the kinetic energy lost by the pulsar's slow-down if the magnetic field at the pulsar's surface has the appropriate value. This problem leads the student through the appropriate (approximate) steps to evaluate the necessary pulsar magnetic field.

The problem: A neutron star (of mass M, radius R, rotation period P) carries a dipole-like magnetic field with strength Bp at the surface, tilted with respect to the rotation axis. Radio radiation is emitted in a cone centered on the magnetic axis. As the cone sweeps past Earth, we observe a pulse of radiation. The low-frequency radiation at period P removes rotational energy from the pulsar and causes a slow-down in the rotation rate, dP/dt > 0. Evaluate the low-frequency radiation loss by the pulsar, as follows: Near the pulsar, write B(r)=Bp(r/R) n. What is n for a dipole-like field? Assume that this relation is valid out to the "light cylinder", that value of r where an object circling the neutron star with period P moves at the speed of light, c. Evaluate B at the light cylinder, roughly at the equator, in terms of Bp, R, P, and c.

   Now assume that, exactly at the light cylinder, B makes a transition from a static dipole inside the light cylinder to a propagating electromagnetic field outside. The electromagnetic wave consists mostly of magnetic field and very little electric field. The wave moves from the light cylinder into space at speed c. What is the energy density of the electromagnetic field at the light cylinder, expressed in terms of Bp, R, P, and c? What is the energy flux (i.e. energy crossing unit area per second)?

   Suppose this radiation is emitted from the light cylinder all around the equator and for a distance R above and below the equator. What is the total energy, dE/dt, emitted over this surface, expressed in terms of Bp, R, c, and P?

   Now equate your formula for dE/dt to the kinetic energy lost by the neutron star as its rotation slows down (derived in problem 3.4), dE/dt = -8/5 pi2MR2 (dP/dt) / P3. Derive an equation for Bp involving M, R, c, P and dP/dt. Evaluate Bp using the same numerical values appropriate for the Crab Nebula pulsar : P = 0.0333 sec , dP/dt = 4.21 x 10-13 sec/sec, R = 10 km and, as used in problem 3.4, M = 1.4 Mo .

The solution: n=3. The radius of the light cylinder is Pc/2pi. B at the light cylinder is Bp(2piR/Pc) 3. The energy density there is [Bp(2piR/Pc) 3]2/2mu, the flux is energy density x c, the area is 4pi(Pc/2pi)2, and so dE/dt = -(2pic/mu)Bp2 (2pi/Pc) 4 R6.

   On equating the two expressions for dE/dt, Bp2 = (mu/20pi3)(Mc3P dP/dt /R4). The numbers for the Crab pulsar yield Bp = 4.6x108Tesla.

Interpretation: The space surrounding the neutron star is clearly not a vacuum but is permeated by a plasma, possibly electrons and protons. After a few theoretical improvements for the radiation loss and plasma flow, the most widely accepted value for the Crab pulsar is Bp = 4x108Tesla, remarkably close to our estimate. Magnetic fields estimated in this way for various radio pulsars range from 3x104to 4x109Tesla. The lower values are for the very fast (millisecond) pulsars that have acquired gases from a companion star.

   One obvious simplification here is the assumed dipole magnetic field inside the light cylinder. A dipole field involves no electrical current, it cannot do work, and thus it cannot emit radiation. It is more accurate, but much more complicated, to describe the field as nearly dipole near the neutron star and becoming gradually more distorted, current-carrying, time-dependent and electromagnetic at larger distances from the pulsar. If one attempts this, one must promptly ask: "What is "near", relative to what distance?". The electromagnetic nature of the desired waves then implicates the distance of the light cylinder, and that distance is then chosen for the assumed sudden transition.

   Even the surface of the neutron star is by no means free of electrical current. Because of the rapid rotation, the electric field there is so strong that it pulls either ions or electrons from the surface. The details of the resulting magnetosphere, including the electrical currents, are not yet solved self-consistently. (MHD is not adequate.) Certainly, particle energies are high enough for emission of gamma rays, which then can create electron-positron pairs. Probably an electron-positron wind, aided by the low-frequency magnetic waves, permeates the entire Crab Nebula (except the filaments for which the Crab is named). Perhaps shocks in this wind accelerate a fraction of the electrons and positrons to sufficiently high energies so that they emit the observed synchrotron radiation. There is not yet any observational evidence whether the synchrotron radiation is emitted by only electrons or by both electrons and positrons. The low-frequency waves, being mostly magnetic, somehow also maintain the magnetic field of the Crab Nebula.

Didactics: A great art in frontier physics, and especially in astrophysics, is to make those simplifications that retain the important physics. The example here concerns the assumed sudden transition from the dipole to the radiation field. There have been many other examples earlier. These simplifications allow some understanding of new phenomena and allows asking questions for new observations. Of course, afterwards it is important to verify the simplifications in detail. Sometimes, the simplifications turn out to be wrong. But not in this example of the Crab pulsar's energy loss.