Emission and Absorption

1 Motivation - the Quasar 3C 273

A MERLIN radio image of the quasar 3C 273. This shows the point source at the nucleus and the jet.

High Energy Astrophysics: Emission and Absorption 1/114

HST optical image of 3C273 Note the very strong cen- tral point source and the less luminous jet. Objects such as 3C273 radiate as much energy from a region the size of the solar system as the entire galaxy.

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Set of 3 images of the jet of 3C273. Left: HST Middle: Chandra X-ray Right: Merlin radio Credits: Optical: NASA/STScI X-ray: NASA/CXC Radio: MERLIN

High Energy Astrophysics: Emission and Absorption 3/114

Integrated spectral energy distribu- tion of 3C 273 from 108 to 1019 Hz. From Elvis et al. 1994, ApJS, 95, 1 Note the rising spectrum in the ra- dio.

High Energy Astrophysics: Emission and Absorption 4/114

2 Radio-loud and radio quiet

Spectra of a typical radio loud quasar and a typical radio quiet object. Note that apart from the radio emission the major differences are: ¥ The slopes of the hard X-ray component ¥ The XUV excess in the radio quiet object

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3 Absorption in general

For every emission process, there is an absorption process, wherein an electromagnetic wave can affect the energy of particles thereby losing energy itself. We have dealt with the emission of synchrotron radiation in some detail. The corresponding process of synchrotron self-absorption whereby synchrotron emitting particles can absorb the radiation they emit is an important process in very compact sources. If it is present, it can be used to estimate the magnetic field in a source. There are two ways of calculating the absorption coefficient. 1. Calculate the dielectric tensor of the plasma and then use this ten- sor to calculate the effect on an electromagnetic wave.This method is valuable when more information is required such as when treat-

High Energy Astrophysics: Emission and Absorption 6/114

ing the radiative transfer of polarised radiation. 2. Use a remarkable set of generic relations discovered by Einstein in order to relate absorption to emission for any process. This method leads quickly to an expression for the absorption coeffi- cient and we shall use it here since it is of general interest.

4 Radiative transfer in a thermal gas

This section is important as a prelude to the treatment of the Einstein coefficients and is also important for our discussion of the emission from accretion disks.

High Energy Astrophysics: Emission and Absorption 7/114

4.1 The source function Consider the radiative transfer equation:

dIν ------= j Ð α I ds ν ν ν Divide through by αν

1 dIν jν ------= ------Ð Iν αν ds αν

dIν ------= S Ð I dτ ν ν

High Energy Astrophysics: Emission and Absorption 8/114 where

jν Sν = ------αν is the source function.

4.2 Thermodynamic equilibrium Now consider the situation where the matter and radiation are in thermodynamic equilibrium.

High Energy Astrophysics: Emission and Absorption 9/114

In this case we know that the equilibrium ra- diation field is described by the Planck func- tion, viz. hν Ð1 3 ------2hν kT Iν ==Bν()T ------e Ð 1 c2 Matter and radiation in an enclosure.

dIν Since ------= 0 inside the enclosure, the source function is given by ds

Sν = Bν.

High Energy Astrophysics: Emission and Absorption 10/114 The surface brightness (specific intensity) of a black body is given by Bν.

4.3 Kirchhoff’s law In a thermal plasma in which the matter is in thermal equilibrium, but not necessarily with the radiation, the coefficients of emission and absorption are functions of the temperature only so that the source function is given by

Sν = Bν()T

High Energy Astrophysics: Emission and Absorption 11/114 This holds irrespective of whether the matter and radiation are in thermal equilibrium or not. This then relates absorption to emission via hν 2 ------Ð1 c kT αν ==[]Bν()T εν ------e Ð 1 εν 2hν3 This relationship is known as Kirchhoff’s law. For a thermal plasma, in which the matter is in thermal equilibrium, Kirchhoff’s law is sufficient to characterise the source function and the absorption. There are two important cases where Kirchhoff’s law is insufficient: 1. The plasma is thermal but the levels of the various atoms are not in thermal equilibrium.

High Energy Astrophysics: Emission and Absorption 12/114 2.The plasma is nonthermal, i.e. no component of it is described in terms of a single temperature. In both of these cases, one is required to go one step beyond Kirch- hoff’s law to the Einstein relations.

5 Properties of blackbody radiation 5.1 Energy density Recall the expression for the energy density per unit frequency 4π 1 u = ------J J = ------∫ I dΩ ν c ν ν 4π ν 4π

High Energy Astrophysics: Emission and Absorption 13/114 The total energy density is: ∞ 4π ∞ ε ==∫ u dν ------∫ B ()νT d rad 0 ν c 0 ν 8πh ∞ ν3 = ------∫ ------dν 3 0 hν c exp------Ð 1 kt 8πk4T 4 ∞ ()hν ⁄ kt 3 hν = ------∫ ------d------3 3 0 hν kt h c exp------Ð 1 kt

High Energy Astrophysics: Emission and Absorption 14/114 The integral is π4 ⁄ 15 so that 8π5k4 ε ==------T 4 aT4 rad 15h3c3 8π5k4 a = ------15h3c3 This is known as Stefan’s Law.

5.2 Flux from the surface of a black body

The total (frequency integrated) flux is given by: ∞ c 4π ν ac F ==∫ πB ()νT d --- × ------∫ B ()νT d ==------T 4 σT 4 0 ν 4 c 0 ν 4

High Energy Astrophysics: Emission and Absorption 15/114 ac where σ = ------is the Stefan-Boltzmann constant. 4

Iν

θ n

2π π F ==∫ dφ∫ B ()T cos θsin θdθ πB ()T ν 0 0 ν ν

High Energy Astrophysics: Emission and Absorption 16/114 6 The Einstein relations 6.1 Definition of the Einstein coefficients These relations are arrived at in a similar manner to Kirchhoff’s law via the inclusion of a new process Ð stimulated emission. Einstein found it necessary to include stimulated emission in the analysis of radiation processes. Neglecting it led to inconsistencies.

2 Energy = E2 Spontaneous Stimulated Absorption emission emission

1 Energy = E1 Processes in a 2-level atom

High Energy Astrophysics: Emission and Absorption 17/114 The above diagram refers to a 2-level atom with energy levels E1 1 and E2. Emission corresponds to a transition of an electron from level 2 to level 1, with the emission of a photon with frequency given by ν h 21 = E2 Ð E1 where h is Planck’s constant. The transition is not exactly sharp as indicated. Absorption is the reverse process, whereby a photon with energy = E2 Ð E1 causes a transition from level 2 to level 1. In gen- eral the emitted and absorbed radiation is described by a profile func- φν() ν ν tion sharply peaked on = 21 . This function expresses the fact that in emission there is a range of photon energies resulting 1.Generalisation to a multilevel atom is trivial since we consider levels in pairs. High Energy Astrophysics: Emission and Absorption 18/114 from the transition. In absorption, photons with a frequency slightly ν φν() different from 21 also cause a transition. We assume that for emission and absorption are equal. This is an adequate assumption for the present purposes. The description of the various terms is as follows: Spontaneous emission This is the emission that occurs in the absence of a radiation field. This is what we calculate from the Quantum Mechanics of the atom in question or in the case of continuum radiation, this is what we cal- culate from the application of electromagnetic theory.

High Energy Astrophysics: Emission and Absorption 19/114 Let the probability that an atom makes a spontaneous transition from level 2 to 1 in time dt , emitting a photon within solid angle dΩ be Ω given by A21d dt . Another way of saying this is that the probabil- Ω ity per unit time is A21d .

Now let N2 be the number of atoms per unit volume in level 2. The contribution to the emissivity from spontaneous emission is then

jspont = N hν φν()A ν 2 21 21

High Energy Astrophysics: Emission and Absorption 20/114 Stimulated emission This is another component of the emission which occurs as a result of the radiation field. The presence of a photon field stimulates the production of additional photons. The stimulated photons have the same direction and polarisation as the original photons. Ω Let B21Iνd dt be the probability of stimulated emission in time dt into solid angle dΩ and let the number of atoms per unit volume in level 2 be N2 . The contribution to the emissivity from stimulated emission is then jstim = N hν φν()B I ν 2 21 21 ν Note that the stimulated photons are emitted in the same direction as the incident photons and with the same polarisation. High Energy Astrophysics: Emission and Absorption 21/114 Absorption Absorption occurs when a photon interacts with the atom and causes Ω a transition from level 1 to level 2. Let B12Iνd dt be the probabil- ity that an atom absorbs a photon from solid angle dΩ in time dt . Then the absorption is given by α ν φν() νIν = N1B12h 21 Iν.

The coefficients A21 , B21 and B12 are the Einstein coefficients.

High Energy Astrophysics: Emission and Absorption 22/114 Differences in notation and approach. The Einstein coefficients are sometimes described in terms of the mean intensity. This is valid when the emissivity is isotropic. They are also sometimes defined in terms of the photon energy density in which case there is a factor of 4π ⁄ c difference in the definition. In many treatments of the Einstein relations (including the original paper), it is assumed, either implicitly or explicitly, that the profile function, φν() is a delta function.

High Energy Astrophysics: Emission and Absorption 23/114 6.2 The radiative transfer equation in terms of the Ein- stein coefficients Taking into account spontaneous and stimulated emissivity, the total emissivity is j = N hν φν()A + N hν φν()B I ν 2 21 21 2 21 21 ν Hence the radiative transfer equation is:

dIν ------= j Ð α I ds ν ν ν = N hν φν()A 2 21 21 + N hν φν()B I Ð N B hν φν()I 2 21 21 ν 1 12 21 ν

High Energy Astrophysics: Emission and Absorption 24/114 Nett absorption The stimulated emission term has the same form as the absorption term and we therefore incorporate it into the absorption term as a negative absorption, giving:

dIν ------= N hν φν()A Ð []N B Ð N B hν φν()I ds 2 21 21 1 12 2 21 21 ν so that the absorption coefficient α* []ν φν() ν = N1B12 Ð N2B21 h 21

High Energy Astrophysics: Emission and Absorption 25/114 6.3 Derivation of the Einstein relations We proceed similarly to deriving the Kirchhoff relations. When ra- diation and matter are in thermal equilibrium, then hν Ð1 3 ------2hν kT Iν ==Bν()T ------e Ð 1 c2

dIν and ------= 0 in our blackbody cavity implies that ds N hν φν()A = []N B Ð N B hν φν()B 2 21 21 1 12 2 21 21 ν

High Energy Astrophysics: Emission and Absorption 26/114 Cancelling out common factors: hν Ð1 ν3 21 2h 21 ------N A = []N B Ð N B ------e kT Ð 1 2 21 1 12 2 21 c2 We know that when a system is in thermodynamic equilibrium, the population of the various energy levels is given by: E Ng∝ exp Ð------kT where g is the statistical weight.

High Energy Astrophysics: Emission and Absorption 27/114 Hence,

E2 exp Ð------() ν N2 g2 kt g2 E2 Ð E1 g2 h 21 ------==------exp Ð------=----- exp Ð------N g1 E1 g1 kT g1 kT 1 exp Ð------kT

High Energy Astrophysics: Emission and Absorption 28/114 The equation for radiative equilibrium can be written: hν Ð1 ν3 21 N1 2h 21 ------A = ------B Ð B ------e kT Ð 1 21 12 21 2 N2 c

hν hν Ð1 ------21- ν3 21 g1B12 2h 21 ------⇒ A = ------e kT Ð1 B × ------e kT Ð 1 21 21 2 g2B21 c These relationships are independent of the temperature if and only if ν3 g1B12 2h 21 ------= 1 A = B ------21 21 2 g2B21 c

High Energy Astrophysics: Emission and Absorption 29/114 That is, ν3 2h 21 A = ------B 21 c2 21

g2 B12 = ----- B21 g1 These are the Einstein relations. They have been derived for the spe- cial case where the matter and the radiation are all in thermal equi- librium. However, they represent general relationships between emission and absorption coefficients which are valid for all situa- tions. Important features ¥ The Einstein relations are independent of the temperature so that High Energy Astrophysics: Emission and Absorption 30/114 they are applicable to nonthermal as well as thermal distributions. ¥ The Einstein relations would be impossible without the presence of stimulated emission represented by the coefficient B21 . If B21 = 0 then both spontaneous emission and absorption coeffi- cients are zero.

High Energy Astrophysics: Emission and Absorption 31/114 Convenient representation Finally, for convenience, the relations are usually represented with- out the 21 subscript on the frequency: 2hν3 A = ------B 21 c2 21

g2 B12 = ----- B21 g1 Given any one coefficient, the others can be determined and the complete emission and absorption properties of the plasma can be specified.

High Energy Astrophysics: Emission and Absorption 32/114 6.4 Special cases The emissivity and absorption coefficient are given by j = N hν φν()A ν 2 21 21 α* []ν φν() ν = N1B12 Ð N2B21 h 21 Hence the source function, Ð1 jν N2A21 N1B12 A21 Sν ==α------[]------=------1Ð ------ν N1B12 Ð N2B21 N2B21 B21 Ð1 2hν3 N1g2 = ------Ð 1 2 c N2g1 where we have used the Einstein relations. High Energy Astrophysics: Emission and Absorption 33/114 The absorption coefficient

N1g2 α* = N B hν φν()------Ð 1 ν 2 21 21 N2g1 Local thermodynamic equilibrium => Kirchhoff’s Law In LTE we know that

N1g2 hν ------= exp------N2g1 kT so that for LTE, we recover Kirchhoff’s law, 2hν3 ⁄ c2 Sν = ------ν ⁄ - []eh kT Ð 1

High Energy Astrophysics: Emission and Absorption 34/114 Non Local Thermodynamic Equilibrium (Non LTE) Non Local Thermodynamic Equilibrium is a situation in which

N1g2 ∆E ------≠ exp ------N2g1 kt where ∆E is the difference in energy between any two levels. Masers and lasers If

N1g2 ------Ð0 1 < N2g1

High Energy Astrophysics: Emission and Absorption 35/114 i.e. g < 1 N1 ----- N2 g2 then, the absorption coefficient is negative. Radiation in this case is amplified, not absorbed. A population for which this is the case is called an inverted population because there are more particles in the upper level than in the higher (modulo the statistical weight.) Invert- ed populations like this in the laboratory give lasers. In astronomical contexts, they give masers. In LTE and in may other contexts, < ()⁄ N2 g1 g2 N1 so that the situations which give rise to masers are somewhat unusual. They arise from pumping of the upper level by

High Energy Astrophysics: Emission and Absorption 36/114 some source of radiation, usually in the infrared. The velocities of masers have proven of great importance in the detection of black holes.

High Energy Astrophysics: Emission and Absorption 37/114 7 The Einstein relations for continuum radiation

The Einstein relations can be generalised to polarised continuum ra- diation provided one makes the correct identification of the relevant number densities. Since we are dealing with continuum radiation there are no discrete energy levels and the emitted photons do not have discrete energies, i.e. there are no emission lines. We therefore consider the distribution of particles in momentum space and make an appropriate identification of the populations of the levels which we have previously called N1 and N2 .

High Energy Astrophysics: Emission and Absorption 38/114 7.1 The phase-space distribution function Remember the phase-space distribution function f ()xp, which is defined by The number of particles in = f ()xp, d3xd3 p an element of phase space Hence, The number density = f ()xp, d3 p in an element of momentum space We use this number density in discussing the Einstein relations for continuum emission.

High Energy Astrophysics: Emission and Absorption 39/114 Relation to the number per unit energy Taking polar coordinates in phase space, the element of volume is d3 pp==2dpdΩ p2dpsinθdθdφ For an isotropic distribution, in which f ()xp, = f ()x, p , the number density of electrons is ∞ ∞ n ==∫ dΩ∫ fp()p2dp 4π∫ p2 fp()dp 0 0 4π after integrating out the angular part. Hence, the number density of particles per unit momentum is Np()= 4π p2 f ()x, p

High Energy Astrophysics: Emission and Absorption 40/114 The relationship between f ()xp, and N()x, E is 4π p2 f ()x, p dp= N()x, E dE For relativistic particles, Ecp= and, dropping the explicit spatial dependence, 4π NE()= ------E2 fEc()⁄ c3

7.2 Einstein relations for continuum emission 7.2.1 Number densities We consider Einstein coefficients for polarised emission and absorp- tion of continuum photons as follows. Note that we consider each ra- diation mode separately.

High Energy Astrophysics: Emission and Absorption 41/114 E 2

()i ()i ()i A21 B21 B12

1 EhÐ ν Einstein coefficients for transitions between energy states in a plasma differing by the energy of the emit- ted photon, hν .

The wave vector of the emitted photon is k = kκ where k is the wave number and κ is the direction of the photon. The momentum is hk = ()hν ⁄ c κ. High Energy Astrophysics: Emission and Absorption 42/114 Consider a plasma in which the emission of a photon of momentum hk results in a change of the momentum of the emitting particle by the corresponding amount. The relevant number densities are ∆ () ∆ () N1 = N p Ð hk N2 = N p where ∆N()p = f ()xp, d3 p In focusing on ∆N()p we are considering pairs of particle momenta which are separated in momentum by the momentum of the photon. They are separated in energy by hν .

High Energy Astrophysics: Emission and Absorption 43/114 Momentum of p 3 py z d p1 emitted photon “Final” state

p hk 1 3 d p2

p2 “Excited” state

px

High Energy Astrophysics: Emission and Absorption 44/114 7.2.2 The Einstein coefficients for polarised continuum radiation

κ Illustrating the definition of the Einstein coefficients for the emis- sion and absorption of radiation in B a particular direction given by the κ dΩ unit vector .

High Energy Astrophysics: Emission and Absorption 45/114 The Einstein coefficients are defined by:

() Probability per unit time for spontaneous emission A i dνdΩ = 21 of a photon in mode i in the ranges dν and dΩ

()i Probability per unit time for the absorption B IνdνdΩ = 12 of a photon in mode i in the ranges dν and dΩ

()i Probability per unit time for stimulated emission B IνdνdΩ = 21 of a photon in mode i in the ranges dν and dΩ For polarised emission the Einstein relations are: 2 ()i ()i c ()i B ==B ------A 21 12 hν3 21

High Energy Astrophysics: Emission and Absorption 46/114 The factor of 2 difference is the result of the Planck function for ther- mal emission being halved for each mode of polarisation when one considers the detailed balance relations. For continuum states, the statistical weights of each level to be unity. The contribution to the absorption coefficient from states differing in momentum by the momentum of an emitted photon is: α()i []()i ()i ν []()i ν d ν ==N1B12 Ð N2B21 h N1 Ð N2 B21 h c2 () = []∆N()∆p Ð hk Ð N()p ------hνA i hν3 21 c2 () = []∆N()∆p Ð hk Ð N()p ------A i ν2 21

High Energy Astrophysics: Emission and Absorption 47/114 where the ()i refers to the two modes of polarisation of the emitted transverse waves. In this equation ∆ ()()3 ∆ () ()3 N p Ð hk = f p Ð hk d p1 N p = f p d p2

Relation between the respective volumes of momentum space 3 3 How do we relate d p1 and d p2 ? Remember that we are considering momentum states that are related by

p2 = p1 + hk

High Energy Astrophysics: Emission and Absorption 48/114 This can be thought of as a mapping between different regions of momentum space with the elementary volumes related by the Jaco- bean of the transformation which we can write out in full as

px, 2 = px, 1 + hkx

py, 2 = py, 1 + hky

pz, 2 = pz, 1 + hkx The Jacobean of this transformation is just 1. Hence 3 3 d p2 = d p1

High Energy Astrophysics: Emission and Absorption 49/114 Therefore, we can put for the populations in the different states ∆ ()()3 N1 ==N p Ð hk f p Ð hk d p ∆ () ()3 N2 ==N p f p d p where 3 3 3 d pd==p1 d p2 The contribution to the absorption coefficient from particles in this region of momentum space is therefore:

() c2 () dα i = []f ()p Ð hk Ð f ()p ------A i d3 p ν ν2 21

High Energy Astrophysics: Emission and Absorption 50/114 We assume that the momentum of the emitted photon is much less than that of the emitting particle. When hk= hν ⁄ c Ç p , we can expand the distribution function to first order: f ()p Ð hk = f ()p Ð hk ⋅ ∇ f ()p For an isotropic distribution of electrons f ()p = fp() ∂ df() p ⇒ ∇ f ()p ==fp()pˆ ------pˆ ∂ p dp df() p ⇒ hk ⋅ ∇ f ()p = ()hkp⋅ ˆ ------dp

High Energy Astrophysics: Emission and Absorption 51/114 Specific case of synchrotron emission We now utilise one of the features of synchrotron emission, namely that the photon is emitted in the direction of the particle to within an angle of γ Ð1 radians, i.e. hkp∝ . Hence, hν hkp⋅ ˆ ≈ hk = ------c The difference in the phase-space distribution functions at the two different momenta is: hνdf() p f ()p Ð hk Ð f ()p ≈ Ð------c dp

High Energy Astrophysics: Emission and Absorption 52/114 Since, 4π d d NE()=------E2 fp() and ------= c------c3 dp dE then hνc3 d NE() f ()p Ð hk Ð f ()p = Ð------4π dE E2

High Energy Astrophysics: Emission and Absorption 53/114 Contribution to the absorption coefficient The contribution to the absorption coefficient from this volume of momentum space is therefore: 3 2 ()i hνc d NE() c ()i 3 dαν = Ð------× ------A d p 4π dE E2 ν2 21 The total absorption coefficient is therefore: 3 2 ()i hνc d NE()c ()i 2 αν = Ð∫ ------A p dpdΩ 4π dE E2 ν2 21 p Ω where the integral is over momentum and solid angle p in momen- tum space. In order to calculate the absorption coefficient, all we ()i have to do now is evaluate A21 .

High Energy Astrophysics: Emission and Absorption 54/114 Emissivity and A21 Recall the definition of the Einstein coefficient:

() Probability per unit time for spontaneous emission A i dνdΩ = 21 of a photon in mode i in the ranges dν and dΩ ()i Therefore, the coefficient A21 is related to the single electron emis- ()i dPν sivity ------through dΩ ()i () dPν νA i dνdΩ = ------dνdΩ 21 dΩ

High Energy Astrophysics: Emission and Absorption 55/114 The quantity ()i dPν Power radiated per unit time ------= dΩ per unit frequency per unit solid angle In Synchrotron Radiation I, we calculated the single electron power emitted per unit circular frequency

() 3 q3Bsinα P i ()ω = ------[]Fx()±Gx() π2ε m 16 0c with the + sign for the perpendicular component and the - sign for the parallel component.

High Energy Astrophysics: Emission and Absorption 56/114 Now recall the above expression for the absorption coefficient 3 2 ()i hνc d NE()c ()i 2 αν = Ð∫ ------A p dpdΩ 4π dE E2 ν2 21 p

c3 d NE() () = Ð------∫ ------hνA i p2dpdΩ 4πν2 dE E2 21 p ()i 3 ()dPν c d NE 2 Ω = Ð------∫ ------Ω p dpd p 4πν2 dE E2 d

High Energy Astrophysics: Emission and Absorption 57/114 1 2 In this integral we take p2dp→ ----- E dE and c3 ()i () dPν α()i 1 d NE 2 Ω ν = Ð------∫ ------E dE∫Ω ------Ω d p 4πν2 E dE E2 p d Integrating over solid angle: ()i dPν () () ∫------dΩ ==P i 2πP i ()ω dΩ p ν

High Energy Astrophysics: Emission and Absorption 58/114 Since we use N()γ rather than NE() we also change the integral over E into one over γ , remembering that 1 NE()= ------N()γ 2 mec This gives

() 3 1 e2 α i = Ð------------()Ω sinθ ν 22 ε 0 32π ν 0mec γ ()γ × 2 γ 2 d N []γ()± () ∫γ γ ------Fx Gx d 1 d γ 2

High Energy Astrophysics: Emission and Absorption 59/114 We shall use the following below: α⊥ α||  γ ν + ν 3 1 e2 2 dN()γ ------= Ð------------()γΩ sinθ ∫ 2 ------Fx()dγ 22 ε 0 γ γ 2 2 32π ν 0mec 1 d γ α⊥ α||  γ ν Ð ν 3 1 e2 2 dN()γ ------= Ð------------()γΩ sinθ ∫ 2 ------Gx()dγ 22 ε 0 γ γ 2 2 32π ν 0mec 1 d γ

High Energy Astrophysics: Emission and Absorption 60/114 The first quantity is often referred to as the mean absorption coeffi- cient. The ratio γ ()γ 2 γ 2 dN ()γ α⊥ α|| ∫γ γ ------Gxd ν Ð ν 1 d γ 2 ------= ------α⊥ α|| γ ()γ ν + ν 2 γ 2 dN ()γ ∫γ γ ------Fxd 1 d γ 2

7.3 Absorption coefficients for a infinite power-law distri- bution For a power-law distribution: d N()γ () N()γ = Kγ Ða ⇒ γ 2 ------= Ð()a + 2 Kγ Ð a + 1 dγγ 2

High Energy Astrophysics: Emission and Absorption 61/114 and

() 3 1 e2 α i = ------()a + 2 ------------()Ω sinθ K ν 2 2 ε 0 32π ν 0mec γ × 2 γ Ð()a + 1 []γ()± () ∫γ Fx Gx d 1 As with computing the emission coefficient, we change the variable of integration to x , using: ω π ν γ 2 12/ Ð12/ 4 12/ Ð12/ ==---Ω------θ x ------Ω------θ x 3 0 sin 3 0 sin π ν γ 14 12/ Ð32/ d = Ð---------Ω θ x dx 2 3 0 sin

High Energy Astrophysics: Emission and Absorption 62/114 so that ()a + 2 ()a + 4 () 3 3 a/2 e2 ------Ð------α i = ------()a + 2 ------()Ω sinθ 2 Kν 2 ν 2π ε 0 64π 4 0mec ()a Ð 2 x ------× ∫ 1 x 2 []Fx()± Gx()dx x2

High Energy Astrophysics: Emission and Absorption 63/114 That is, ()a + 2 ()a + 4 ⊥ 3 3 a/2 e2 ------Ð------α = ------()a + 2 ------()Ω sinθ 2 Kν 2 ν 2π ε 0 64π 4 0mec ()a Ð 2 x ------× ∫ 1 x 2 []Fx()+ Gx()dx x2 ()a + 2 ()a + 4 3 3 a/2 e2 ------Ð------α|| = ------()a + 2 ------()Ω sinθ 2 Kν 2 ν 2π ε 0 64π 4 0mec ()a Ð 2 x ------× ∫ 1 x 2 []Fx()ÐG()x dx x2

High Energy Astrophysics: Emission and Absorption 64/114 Consequently, α⊥ α|| ()a + 2 ν + ν 3 3 a/2 e2 ------= ------()a + 2 ------()Ω sinθ 2 2π ε 0 2 64π 4 0mec ()a + 4 ()a Ð 2 Ð------x ------× Kν 2 ∫ 1 x 2 Fx()dx x2 and ()a Ð 2 x ------1 2 ⊥ || ∫ x Gx()dx αν Ð αν x ------= ------2 ⊥ || ()a Ð 2 αν + αν x ------∫ 1 x 2 Fx()dx x2 High Energy Astrophysics: Emission and Absorption 65/114 As before, when the frequency ν is well inside the limits determined by the upper and lower cutoff Lorentz factors, the limits of the inte- gral may be taken to be zero and infinity, respectively. Using the ex- pressions for the integrals of a power times Fx() and Gx() , we have, a + 2 ()a Ð 2 ------∞ ------2 2 a 11 a 1 ∫ x 2 Fx()dx = ------Γ --- + ------Γ --- + --- 0 a + 2 4 6 4 6 ()a Ð 2 ()a Ð 2 ∞ ------a 5 a 1 ∫ x 2 Gx()dx = 2 2 Γ --- + --- Γ --- + --- 0 4 6 4 6

High Energy Astrophysics: Emission and Absorption 66/114 Therefore, ()a Ð 2 ------2 Γa 5 Γa 1 α⊥ α|| 2 --- + --- --- + --- ν Ð ν 4 6 4 6 a + 2 ------==------α⊥ α|| a + 2 10 ν + ν ------a + ------2 2 a 11 a 1 3 ------Γ --- + ------Γ --- + --- a + 2 4 6 4 6 This gives for the ratio of absorption coefficients: α|| ν 2 ------= ------⊥ 3a + 8 αν

High Energy Astrophysics: Emission and Absorption 67/114 The mean absorption coefficient is: ⊥ || a + 2 ()a + 4 αν + αν 2 ------Ð------()e ()Ω θ 2 ν 2 ------= C3 a ------ε K 0 sin 2 0mec where ()()a + 4 a + 1 a + 10 ------Ð------Ða 11 a 1 C ()a = 3 2 2 2 π 2 Γ --- + ------Γ --- + --- 3 4 6 4 6 As with the expression for the emission coefficient, this expression is separated into a numerical coefficient which depends upon a , a factor involving physical constants, a part involving the non-relativ- istic gyrofrequency, a factor involving the parameter for the electron density and a factor involving a power of the frequency. This is often

High Energy Astrophysics: Emission and Absorption 68/114 the absorption coefficient which is quoted in text books. However, this coefficient alone does not convey the whole story since synchro- tron absorption unlike many other absorption processes is polarised.

High Energy Astrophysics: Emission and Absorption 69/114 () 7.4 Numerical value of C3 a At the left is a plot of the

Plot of the function C (a) appearing in function C ()a appearing 3 3 the expression for the synchrotron absorption 0.008 coefficient in the above expression for

0.006 the synchrotron absorption

(a) coefficient. 3 C 0.004

0.002

0.000 1.0 1.5 2.0 2.5 3.0 3.5 4.0

a

High Energy Astrophysics: Emission and Absorption 70/114 8 The transfer of polarised synchrotron radiation in an optically thick region

The transfer equations for the two modes of polarisation are: ⊥ dIν ⊥ ⊥ ⊥ ------= j Ðα I ds ν ν ν || dIν ------= j|| Ð α|| I || ds ν ν ν

High Energy Astrophysics: Emission and Absorption 71/114 Recall that the perpendicular and parallel components of intensity are related to the Stokes parameters by: ⊥ 1 I = ---()I + Q ν 2 ν ν 1 I || = ---()I Ð Q ν 2 ν ν The reverse transformation is: ⊥ || Iν = Iν + Iν ⊥ || Qν = Iν Ð Iν

High Energy Astrophysics: Emission and Absorption 72/114 We write the transfer equations in terms of source functions. ⊥ ⊥ ⊥ dIν ⊥ ⊥ ⊥ dIν jν ⊥ ⊥ ⊥ ------= j Ðα I ⇒ ------==------Ð I S Ð I ds ν ν ν ⊥ ⊥ ν ν ν dτν αν || || || dIν dIν jν ------= j|| Ð α|| I || ⇒ ------==------Ð I || S|| Ð I || ds ν ν ν || || ν ν ν dτν αν In a slab geometry with all parameters constant, the solutions are: ⊥ ⊥ ⊥ Iν = Sν ()1 Ð expÐτν || || || Iν = Sν()1 Ð expÐτν

High Energy Astrophysics: Emission and Absorption 73/114 ⊥ || For small optical depths τν , τν Ç 1 , these relations become the standard ones for optically thin emission. In the opposite limit of in- finite optical depth, ⊥ ⊥ ⊥ jν Iν ==Sν ------⊥ αν || jν I || ==S|| ------ν ν || αν

High Energy Astrophysics: Emission and Absorption 74/114 The ratio ⊥ || || || jν jν αν jν ------Ð ------Ð ------⊥ || α⊥ α|| α⊥ ⊥ Q Iν Ð Iν ν ν ν jν ---- ==------=------I ⊥ || ⊥ || || || Iν + Iν jν jν αν jν ------+ ------+ ----- ⊥ || ⊥ || αν αν αν jν We have: α|| || ν 2 jν 2 ------= ------= ------⊥ 3a + 8 ⊥ 3a + 5 αν jν

High Energy Astrophysics: Emission and Absorption 75/114 Using the previously determined ratios for these quantities

Qν 3 ------= Ð------Iν 6a + 13 An important result here is that the ratio is negative signifying that the parallel component of the intensity is the larger. This means that the major axis of the polarisation ellipse is parallel to the magnetic field. As one can see from the following plot the fractional polarisa- tion 3 Π = ------6a + 13 is lower for optically thick emission.

High Energy Astrophysics: Emission and Absorption 76/114 The fractional polarisation 0.20 of a self absorbed synchro- tron source. 0.15

Π 0.10

Polarisation of self absorbed 0.05 synchrotron emission

0.00 1.0 1.5 2.0 2.5 3.0 3.5 4.0

a

High Energy Astrophysics: Emission and Absorption 77/114 9 The spectral slope for optically thick emission

For optically thick emission, we have: ⊥ || ⊥ ⊥ jν jν I ==S ------I || ==S|| ------ν ν ⊥ ν ν || αν αν ⊥ || ⊥ jν jν ⇒ I ==S + S|| ------+ ------ν ν ν ⊥ || αν αν

High Energy Astrophysics: Emission and Absorption 78/114 Write this in terms of the total emissivity and the mean absorption coefficient in the following way: ⊥ || jν jν ⁄ jν jν ⁄ jν I = ------+ ------ν α ⊥ || ν αν ⁄ αν αν ⁄ αν ⊥ || αν + αν where α = ------is the mean absorption coefficient. ν 2 We have already determined the ratios || α|| jν 2 ν 2 s ==------r ==------⊥ 3a + 5 ⊥ 3a + 8 jν αν

High Energy Astrophysics: Emission and Absorption 79/114 Therefore,

jν 11⁄ ()+ s s ⁄ ()1 + s Iν = ------+ ------αν 21⁄ ()+ r 2r ⁄ ()1 + r

jν ()6a + 13 ()3a + 10 = ------αν 23()a + 8 ()3a + 7

High Energy Astrophysics: Emission and Absorption 80/114 Now use a + 1 ()a Ð 1 2 ------Ð------()e ()Ω θ 2 ν 2 jν = C1 a ------ε K 0 sin 0c a + 2 ()a + 4 2 ------Ð------α ()e ()Ω θ 2 ν 2 ν= C3 a ε------K 0 sin 0mec The intensity is then ()()C ()a 6a + 13 3a + 10 1 ()Ω θ Ð12/ ν52/ Iν = ------()()------()-me 0 sin 23a + 8 3a + 7 C3 a () ()Ω θ Ð12/ ν52/ = C5 a me 0 sin

High Energy Astrophysics: Emission and Absorption 81/114 where ()()C ()a () 6a + 13 3a + 10 1 C5 a = ------()()------()- . 23a + 8 3a + 7 C3 a 52/ The important result here is that Iν ∝ ν . The entire spectrum from optically thick to optically thin regimes is as indicated in the following plot for the case of a = 2.1

High Energy Astrophysics: Emission and Absorption 82/114 Plots of perpendicular and parallel components of inten- sity Write || || || jν αν jν jν αν jν S|| ==------× ------=------S ν || || j α || j ν αν αν ν ν αν ν ⊥ α ⊥ α ⊥ ⊥ jν ν jν jν ν jν S ==------× ------=------S ν ⊥ ⊥ j α ⊥ j ν αν αν ν ν αν ν

High Energy Astrophysics: Emission and Absorption 83/114 The mean source function Sν is given by: () jν C1 a S ==------m ()Ω sinθ Ð12/ ν52/ ν C ()a e 0 αν 3 From the above ⊥ ()3a + 5 ()3a + 10 ()3a + 10 S = ------S S|| = ------S ν 23()a + 8 ()3a + 7 ν ν 23()a + 7 ν ν We define a frequency 0 at which the mean optical depth is unity by a + 1 ()a + 4 2 ------Ð------α ()e ()Ω()θ 2 ν 2 νLC= 3 a ε------KL 0 sin 0 0mec

High Energy Astrophysics: Emission and Absorption 84/114 and then put ν 52/ Sν = A-----ν - 0 Then the perpendicular component is ⊥ ()3a + 5 ()3a + 10 ν 52/ Iν = ------()()A-----ν - 23a + 8 3a + 7 0 ()a + 4 ()ν Ð------× 23a + 8  2 1 Ð expÐ-----------ν - 3a + 10 0

High Energy Astrophysics: Emission and Absorption 85/114 and the parallel component is:

|| ()3a + 10 ν 52/ Iν = ------()A-----ν - 23a + 7 0 ()a + 4 ν Ð------× 4  2 1 Ð expÐ-----------ν - 3a + 10 0 These equations have used to produce the plots below.

High Energy Astrophysics: Emission and Absorption 86/114 1e+00 Iν, perp ν I Iν,par 1e-01

1e-02 1e-01 1e+00 1e+01 1e+02

ν/ ν 0

High Energy Astrophysics: Emission and Absorption 87/114 Polarisation

0.9

0.8

0.6

Π 0.4

0.2

-0.0

-0.2 1e-01 1e+00 1e+01

ν/ ν 0

High Energy Astrophysics: Emission and Absorption 88/114 The above curve shows the polarisation defined by

Qν Π = ------Iν Ð3 varying between the limits of ------(for the optically thick case) 6a + 13 a + 1 and ------for the optically thin regime. a + 73⁄ Synchrotron self-absorption is just one of the processes that can lead to a low frequency cutoff in the spectrum of a radio source. Others include free-free absorption (due to foreground ionised matter) and a process known as induced Compton scattering. A low energy cut- off in the low frequency spectrum can also lead to a low frequency cutoff.

High Energy Astrophysics: Emission and Absorption 89/114 10 When is synchrotron self absorption important?

Consider the optical depth based upon the mean absorption coeffi- cient:

τν = ∫ ανds ≈ ανL slab We also consider optically thin emission and consider the transition to the optically thick regime. Therefore, the surface brightness is

Iν = ∫ jνds ≈ jνL slab

High Energy Astrophysics: Emission and Absorption 90/114 We can therefore relate the optical depth to the surface brightness via:

τν ανL αν ----- ≈ ------= ------Iν jνL jν α ν 1 ⇒ τν ==------Iν ------× Iν jν Sν where Sν is the “mean” source function.

High Energy Astrophysics: Emission and Absorption 91/114 Using again the relations for emissivity and mean absorption coeffi- cient: a + 1 ()a Ð 1 2 ------Ð------()e ()Ω θ 2 ν 2 jν = C1 a ------ε K 0 sin 0c a + 2 ()a + 4 2 ------Ð------α ()e ()Ω θ 2 ν 2 ν= C3 a ε------K 0 sin 0mec we obtain α C ()a ν 3 ()Ω θ 12/ νÐ52/ ------= ------()- 0 sin jν C1 a

High Energy Astrophysics: Emission and Absorption 92/114 The mean optical depth is then given in terms of the optically thin surface brightness by: C ()a τ 3 Ð1()Ω θ 12/ × ()νÐ52/ ν = ------()-me 0 sin Iν C1 a

High Energy Astrophysics: Emission and Absorption 93/114 Note that this expression depends upon the magnetic field only and not on the parameter K . Now in the optically thin regime, we can write: ν Ðα Iν = Iν -----ν - 0 0 C ()a ν ()α ⁄ ⇒ τ 3 Ð1()Ω θ 12/ × νÐ52/ Ð + 52 ν = ------()-me 0 sin Iν 0 -----ν - C1 a 0 0 C ()a 12/ ν ()α ⁄ 3 e []θ 12/ νÐ52/ Ð + 52 = ------Bsin Iν ------C ()a 32/ 0 0 ν 1 me 0

High Energy Astrophysics: Emission and Absorption 94/114 Now let us put in some fiducial values: B ==1G 10Ð4 T 1 Jy Ð16 Ð2 Ð1 Ð1 Iν ==------4.3×10 W m Hz Sr 0 sq. arcsec ν 0 = 1 GHz The result is:

C ()a Iν ()α ⁄ Ð5 3 Bsinθ 12/ 0 ν Ð + 52 τ = 6.26×10 ------ν () Ð2  C1 a G Jy arcsec GHz

High Energy Astrophysics: Emission and Absorption 95/114 For typical kpc scale regions of radio galaxies, supernova remnants etc., we can see that there is no chance of plasma becoming optically thin, except at very low frequencies. Typical values would be Ð5 Ð2 B ∼ 10 G Iν ∼ 10 mJy arcsec 0 However, in the cores of radio galaxies and quasars, we can have Ð2 Ð2 Iν ∼ 1 Jy mas B ∼ 10 G 0 In which case the optical depth at 1 GHz would be

() ()α ⁄ C3 a Bsinθ 12/ I1GHz ν Ð + 52 τ = 6.3------ν () Ð2  C1 a 0.01G Jy mas Ghz

High Energy Astrophysics: Emission and Absorption 96/114 α ⁄ ≈ For a spectral index = 0.6 , a = 2.2 , C3 C1 29.8 and

θ 12/ I ν Ð()α + 52⁄ τ Bsin 1GHz  ν = 186 ------------0.01G Jy masÐ2 Ghz and the plasma can clearly be optically thick at frequencies much higher than a GHz. This is a characteristic feature of quasar spectra.

High Energy Astrophysics: Emission and Absorption 97/114 11 Brightness temperature 11.1 Definition A concept which commonly enters in to radio astronomy and which is tied in with the notion of self absorption is brightness temperature. Consider the blackbody spectrum: hν Ð1 3 ------2hν kT Iν = ------e Ð 1 c2 In the classical limit hν Ç kT , this becomes the Rayleigh-Jeans law:

2kT 2 Iν = ------ν c2

High Energy Astrophysics: Emission and Absorption 98/114 This is the emission at low frequencies (long wavelengths) from a blackbody (optically thick thermal emitter). This spectrum is the re- sult of the balance of emission and absorption in the emitting region. Astronomers use the Rayleigh-Jeans law to ascribe a brightness tem- perature to a source: 2 c Ð2 T = ------ν I b 2k ν For a nonthermal source, the brightness temperature is a strong func- tion of frequency: 2 ν ()α c νÐ2Ð + 2 Tb = ------Iν 0 -----ν - 2k 0 0

High Energy Astrophysics: Emission and Absorption 99/114 and increases rapidly at low frequency. Eventually, at low enough frequency, the brightness temperature exceeds the kinetic tempera- ture of the emitting electrons. The characteristic of a nonthermal source is that the particles have not come into an equilibrium distribution (i.e. a relativistic Max- wellian of a single temperature) because the collision times are too long to achieve this. However, we can ascribe a temperature to par- ticles of a given energy. Moreover, in a self-absorbed source, the photons which are absorbed are similar in energy to the photons which are emitted since the absorption process is the reverse of the emission process. Therefore, at each energy, we expect something similar to a blackbody equilibrium. A characteristic of such an equi- librium is that the brightness temperature of the radiation cannot ex-

High Energy Astrophysics: Emission and Absorption 100/114 ceed the equivalent temperature of the electrons. This gives a simple explanation of the frequency dependence of the surface brightness of a self-absorbed source.

11.2 Equivalent temperature For a gas in which the ratio of specific heats is χ and the number den- sity of particles is the energy density is given by NkT nKT u ==------3 =NkT χ Ð 1 43⁄ Ð 1

High Energy Astrophysics: Emission and Absorption 101/114 for a relativistic gas in which χ = 43⁄ . Hence, for electrons of Lorentz factor γ we have an equivalent temperature defined by: γ 2 N mec = 3NkT 2 1 mec ⇒ T = ---γ ------3 k We know from the theory of synchrotron emission that the circular frequency of emission is given by: Ω Ω 3 0 3 0 ω ∼ν------γ 2 ⇒ = ------γ 2 2 4π πν ⇒ γ ∼ 4 12/ ------Ω 3 0

High Energy Astrophysics: Emission and Absorption 102/114 Hence,

m c2 πν ∼ 1 e 4 12/ T ------------Ω 3 k 3 0

High Energy Astrophysics: Emission and Absorption 103/114 11.3 Condition for optically thick radiation - a thermody- namic argument As stated above, the brightness temperature of the radiation cannot exceed the kinetic temperature of the particles producing it. Hence, a synchrotron source becomes optically thick when ∼ TTb m c2 πν 2 ⇒ 1 e 4 12/ ∼ c νÐ2 ------------Ω ------Iν 3 k 3 0 2k 24π 12/ ⇒ νÐ52/ I ∼ ------m ΩÐ12/ ν 33 e 0

High Energy Astrophysics: Emission and Absorption 104/114 This defines a critical frequency at which the source becomes self- absorbed. For lower frequencies the processes of emission and ab- sorption must be balanced to guarantee the equilibration between brightness temperature and kinetic temperature, so that 2 I ∼ ---()2π 12/ m ΩÐ12/ ν52/ ν 3 e 0 This is close to the relation for optically thick sources that we de- rived above from the theory of synchrotron absorption. This deriva- tion also gives a physical explanation for the ν52/ dependence of a synchrotron self-absorbed source as opposed to the ν2 dependence for a thermal source. The extra factor of ν12/ arises from the fre- quency dependence of the kinetic temperature.

High Energy Astrophysics: Emission and Absorption 105/114 12 Preliminary estimate of the magnetic field and particle energy density in a quasar

We can use the above theory to estimate both the magnetic energy density and magnetic field in a self-absorbed source. This excursion into the estimation of observational parameters is only preliminary since relativistic effects mislead us in our estimation of the rest frame flux density. However, it is instructive to go through the exercise to see what deductions we can make. When a source is self-absorbed, we have two constraints on the number density and magnetic field, derived from the surface bright- ness and the optical depth.

High Energy Astrophysics: Emission and Absorption 106/114 The emissivity and absorption coefficient are given by: a + 1 ()a Ð 1 2 ------Ð------()e ()Ω θ 2 ν 2 jν = C1 a ------ε K 0 sin 0c

a + 2 ()a + 4 2 ------Ð------α ()e ()Ω θ 2 ν 2 ν= C3 a ε------K 0 sin 0mec

High Energy Astrophysics: Emission and Absorption 107/114 Let the path length through the source be L , so that at the peak of the spectrum a + 1 ()a Ð 1 2 ------Ð------≈ ()e ()Ω()θ 2 ν 2 Iν jνL = C1 a ------ε KL 0 sin 0c a + 2 ()a + 4 2 ------Ð------≈ α ()e ()Ω()θ 2 ν 2 1 νL = C3 a ------ε KL 0 sin 0mec Ω θ These are our two constraints on the parameters KL and 0 sin . We usually write the surface brightness as

Fν I = ------ν ψ2

High Energy Astrophysics: Emission and Absorption 108/114 where Fν is the flux density from a region of angular size ψ . We also use the relationship a = 2α + 1 in the above so that our equations read:

2 ()e ()Ω()θ 1 + ανÐα C1 a ------ε KL 0 sin = Iν 0c

2 ()e ()Ω()θ α + 32⁄ νÐ()α + 52⁄ C3 a ------ε KL 0 sin = 1 0mec

High Energy Astrophysics: Emission and Absorption 109/114 These equations can be solved (using Maple or brute force) to give Ω the solutions for 0 and KL : C 2 C 2 Ωθ 1 2 Ð2ν5 1 2 Ð2ψ4ν5 sin ==------me Iν ------me Fν C3 C3 ε 2 2α 0c C3 C3 KL = ------------I 3νÐ5 mÐ1------I νÐ2 2 2 3 ν e C ν e me C1 1 ε 2 2α 0c C3 C3 = ------------F 3ψ6νÐ5 mÐ1------F ψÐ2νÐ2 2 2 3 ν e C ν e me C1 1 Numerically,

High Energy Astrophysics: Emission and Absorption 110/114 2 C Fν Ð2 ψ 4 Ωθ × 31 ν5 sin= 4.6 10 ------------------9 C3 Jy mas 2 C Fν Ð2 ψ 4 ⇒ θ × Ð41 ν5 Bsin = 1.2 10 ------------------9 C3 Jy mas 2 3 21C3 Fν θ Ð6 KL = 9.62×10 ------νÐ5()2α 3 Jy mas 9 C1 Fν ψ Ð2 × 467 ------νÐ2 Jy mas 9

High Energy Astrophysics: Emission and Absorption 111/114 Units: KL ==mÐ3 × m mÐ2 B = Tesla (T) Let us estimate the parameters for a typical quasar with ψ ν α Fν = 1 Jy = 1 mas peak = 1 GHz = 0.6 × Ð3 For a = 2.2 , C1 = 4.44 10 and C3 = 0.132 , so that Ð7 30 Bsinθ = 1.4×10 T KL = 3.1×10 mÐ2 and for a typical path length through the source, 16 13 L ==1 pc 3.1×10 m, K = 9.8×10 mÐ3.

High Energy Astrophysics: Emission and Absorption 112/114 γ The energy density in particles for a lower cutoff Lorentz factor 1 is 2 Kmec () ε ==------γ Ð a Ð 2 40γ Ð0.2 J mÐ3 a Ð 2 1 1 γ Ð3 γ For 1 =10 this is approximately 25 J m and for 1 = 100 ε ≈ 16 J mÐ3. For comparison, the energy density in the magnetic field is: 2 B ∼ × Ð9 Ð3 ------µ 7.3 10 J m 2 0 approximately nine orders of magnitude lower and it would appear that the plasma is well removed from equipartition.

High Energy Astrophysics: Emission and Absorption 113/114 Note however the sensitive dependence of the magnetic field and particle energy on the flux density: Ð2 32+ α BF∝ ν KL∝ Fν so that the ratio of particle energy density to magnetic energy density 72+ α 8.2 is proportional to Fν ≈ Fν . It is this single factor which is most affected by relativistic beaming from the moving plasma.

High Energy Astrophysics: Emission and Absorption 114/114