Äîêóìåíò âçÿò èç êýøà ïîèñêîâîé ìàøèíû. Àäðåñ îðèãèíàëüíîãî äîêóìåíòà : http://www.mso.anu.edu.au/~geoff/HEA/6_Compton_Scattering_II.pdf Äàòà èçìåíåíèÿ: Tue May 27 07:04:57 2003 Äàòà èíäåêñèðîâàíèÿ: Tue Oct 2 05:39:51 2012 Êîäèðîâêà: Ïîèñêîâûå ñëîâà: m 101

Compton Scattering II
1 Introduction In the previous chapter we considered the total power produced by a single electron from inverse Compton scattering. This is useful but limited information. Here we calculate the resulting spectra. 1.1 Importance This emission mechanism is frequently encountered in a number of areas, for example: · The X-ray emission from the lobes of radio galaxies can
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be used, along with the synchrotron emission to estimate the magnetic field and the energy density of particles and magnetic field. The additional information available from X-ray observations unlocks the degeneracy between particle estimates and fields involved in the expression for, say, the surface brightness of synchrotron emission. Such estimates can then be compared with the minimum energy estimates obtained from synchrotron theory. · Blazars emit X-ray synchrotron and GeV or TeV -rays. The simultaneous measurement of these spectra enable us to estimate the properties of blazar jets, even though they are unresolved. It is well to keep in mind two possible scenarios:
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1.The "soft" photons which are scattered originate from an external source, e.g. starlight or the microwave background. This is known as External Inverse Compton (EIC) emission. 2.The soft photons originate from within the source itself and are scattered by the same population of relativistic electrons. This is known as Synchrotron Self Compton (SSC) emission.

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1.2 Example The western hot spots in Cygnus A. The radio image is shown as a greyscale; the X-ray image is shown as contours overlaid on the radio image. From Wilson et al. (2000) astro-ph 0009308
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2 Inverse compton spectrum from a single relativistic electron 2.1 Relative importance of SSC and EIC emission One can readily ascertain the relative importance of SSC and EIC emission from the expression for the Compton power: 4 P Compton = -3 4 = -3 T c 2 2 U ph
(1)

T c 2 U ph for relativistic particles

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Since the power is proportional to the radiation energy density, SSC is going to be more important than EIC when Radiation energy density Radiation energy density > (2) of the synchrotron emission of background emission

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2.2 Calculation of the emitted spectrum
Electron rest frame Lab frame 1 = h
1

1

S

1

v

S



1

= h Electron scattering in laboratory frame and rest frame of electron

Note that all angles are measured clockwise from the positive x -axis as defined by the electron velocity.
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Outline of what happens in practice

Incident photons v Scattered photons 2 v Lab frame Electron rest frame
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Photon-electron collisions in the rest and lab frames One of the simplifications of inverse compton emission from relativistic electrons is due to the aberration of the photons in the rest frame of the electron. Let µ = cos for the incoming photon in the rest frame and µ = cos for the incoming photon in the lab frame. Then the relativistic aberration formula gives: µ­ µ = --------------1 ­ µ
(3)

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For 1 and most values of µ : µ­1 µ ----------- ­ 1 1­µ The only exception is when µ ­ 1 ­ µ µ 1 and then µ 1
(6) (5) (4)

Thus the values of µ cluster around µ = ­ 1 for high Lorentz factors. This corresponds to virtually head on collisions between electrons and photons in the rest frame. The only exception is for collisions which are close to overtaking.
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1.0 µ (electron rest frame)

0.5

0.0 = 2 = 10 -1.0 -1.0 -0.5 0.0 µ (lab frame) 0.5 1.0

The plot at the left shows the result of the aberration formula for µ = cos , for 2 different values of . It is well to remember that the lab frame distribution is isotropic so that µ ­ 1 is going to be the most common result.

-0.5

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Electron energy On the other hand, the ratio of the photon energy in the rest frame to that in the lab frame is given by Rest frame photon energy ------------------------------------------------------------- = --- = ( 1 ­ µ ) Lab frame photon energy
(7)

and varies linearly with µ . There is no specific value of -- which is favoured.

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Minimum and maximum values of rest frame photon energy The minimum value is for 1 1 ­ 1 + -------- = ----µ = 1 = ( 1 ­ ) 2 2 2 The maximum value is at µ = ­ 1 = ( 1 + ) 2
(9) (8)

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That is, ----- < < 2 2
(10)

2.3 Evaluation of rate of scattering We consider the situation 1 first in the rest frame. The S complexity of the scattering is considerably simpli1 fied by the fact that the photons are effectively in
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cident on the electron at one angle = . This means that we only have to consider one scattering angle for the scattered photons. In the rest frame: No of photons scattered dN ------------------- = per unit time per unit solid angle dt d 1 in the electron rest frame d T = ------------ cf ( p ) d 3 p d 1
p

(11)

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The differential cross section is given by Thomson scattering:
2 r0 d T ------------ = ---- ( 1 + µ 1 2 ) 2 d 1 (12)

We can write the cross-section this way because we are approximating the photons as all coming in along the x -axis.

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Relation to distribution in lab frame As we did for the total power calculation, we transform the integral over the rest frame momentum space into one over the lab frame momentum space. Note that d 3 p = p 2 sin d pd d = ­ p 2 dpd ( cos ) d = ­ p 2 dpd µ d
­1 1 0 sin d ­ 1 dµ ­1 dµ (13)

so that integrations over become integrations over µ . Also
(14)

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Hence, d 3 p p 2 dpd µ d in integrations over momentum space. In Compton Scattering I, we found that: d 3 p = ( 1 ­ µ ) d 3 p Hence, f ( p ) d 3 p = f ( p ) ( 1 ­ µ ) d 3 p
(17) (16) (15)

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We also have = ( 1 ­ µ ) 4 f ( p ) p 2 dp = n ( ) d And
3 p = f ( p ) d 3 p = f ( p ) p 2 dpd µ d ----f( p)d (18)

Since the distribution in the lab frame is isotropic. Therefore,
(19)





1 = ----4

--- n ( ) d d µ d

(20)

(21)
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The integral over is easy: d T 1 dN ------------------- = ----- ------------ c --- n ( ) d dµ d 4 d 1 dt d 1 p d T 1 = -- ------------ c --- n ( ) d dµ 2 d 1
( µ, )

(22)

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Change of variables from ( , µ ) to ( , ) We now make a change of variables from ( , µ ) to ( , ) utilising the transformation = = ( 1 ­ µ )
(23)

That is we are replacing µ by the photon energy ( ) seen by the electron in the rest frame. Our reason for doing this is that, later on, we relate to the energy of the scattered photon and thus derive the differential no of photons scattered per unit energy of the final scattered energy ­ that is, the energy we receive at the telescope.
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Remember: = Energy of incident photon in lab frame = Energy of incident photon in rest frame The Jacobean ( , ) ----------------- = ( , µ ) 1 0 ( 1 ­ µ ) ­ =
(25) (24)

1 1 d d µ = -------- d d ---- d d

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Hence, 1 dN ------------------- = -- 2 dt d 1 ( µ, d T ------------ c --- n ( ) d dµ d 1

)

d T ----------- n ( ) d d = ------------ c d 1 2 2 ( , ) We now write this in the form: d T dN --------------------------------- = c -----------d 1 dt d 1 d d ----------- n ( ) 2 2

(26)

(27)

(28)

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Scattering in the Thomson limit Now use the Thomson limit approximation for the energy of the scattered photon: 1 =
(29)

As we saw in Compton Scattering I, the condition for this is that « m e c 2
(30)

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Using 1 = gives: d T dN ------------------------------------ = c -----------d 1 dt d 1 d d 1 1 ----------- n ( ) 2 2 1 ----------- n ( ) 2 2
(31)

2 cr 0 2) = ------- ( 1 + µ 1 2

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Relationship to the scattering rate in the lab frame The next step is to relate this differential expression to quantities in the lab frame. The two main variables to transform are t and 1 . We use the Lorentz transformation for time t ­ x- t= ------ c
(32)

We also use the Lorentz transformation relating photon energy in the lab frame to photon energy in the rest frame given that the scattered photon has a direction given by µ 1 = cos 1 .
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This is: 1 = 1 ( 1 + µ 1 ) These transformations imply that: dt = dt dt =
­1 (33)

dt

d 1 = ( 1 + µ 1 ) d 1 d 1 d 1 = --------------------------- ( 1 + µ 1 )

(34)

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Also, since dN is a number it is a Lorentz invariant and dN dt d 1 dN --------------------------------- = ------------------------------------ â ------ â ---------dt d 1 d d 1 dt d 1 dtd 1 d d 1
2 cr 0 1 + µ 1 2 = ------- --------------------------- 2 ( 1 + µ 1 ) 2

1 n ( ) ----------- --------- 4 2

(35)

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Rate of scattering in all directions We now calculate the rate of scattering of photons into all directions. We integrate over solid angle in the rest frame, with d 1 d µ 1 d
2 r0 dN ------------------- = c ----td d 1 2

1 n ( ) 2 µ 1, max ( 1 + µ 1 2 ) ------------------------ dµ 1(36) d ----------- ---------- µ 1, min ( 1 + µ 1 ) 2 4 2 0

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Limits on scattered energy in lab frame In order to complete the above integral, we have to determine the limits on µ 1 . We first calculate the limits on 1 . We have the relation between rest frame incident soft photon energy and lab frame scattered energy 1 : 1 = ( 1 + µ 1 ) = ( 1 + µ 1 ) and the previously derived limits on : ----- < < 2 2
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(37)

(38)

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For a given the upper limit on 1 is obtained for µ 1 = 1 . This gives 1 2 Therefore, from the upper limit on 1 < 4 2 maximum energy, 4 2 of the scattered photons.
(39)

That is, the kinematics of Compton scattering establishes a

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Later we make use of the variable: 1 Final photon energy x = ------------------------------------------------------------ = ----------Maximum photon energy 4 2 Limits on µ 1 for a fixed 1 and Since 1 = ( 1 + µ 1 ) < 2 2 ( 1 + µ 1 )
(41) (40)

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then, for a given , 1 1 1 + µ 1 > ----------2 2 This provides the lower limit on the integral over µ 1 . We also have from the lower limit on : 1 = ( 1 + µ 1 ) > ----- ( 1 + µ 1 ) = -- ( 1 + µ 1 ) 2 2
(43) (42)

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Therefore, 1 ( 1 + µ 1 ) < 2 ---
(44)

We also have the obvious constraint on µ 1 , namely µ 1 < 1 so that ( 1 + µ1 ) < 2
(45)

Since we are normally interested in 1 / » 1 then the latter limit is usually the most relevant.

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Summary of limits 1 ----------- < ( 1 + µ 1 ) < 2 2 2 Evaluation of integral The indefinite integral ( 1 + µ1 2 ) ------------------------ dµ 1 ( 1 + µ1 ) 2 ( 1 + µ1 ) 1 + µ1 ­1 1 + µ1 = 2 --------------------- ­ 2 ln ---------------- ­ ---------------- 22 2
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(46)

(47)


Evaluation between the lower and upper limits gives: 1 1 1 ­1 1 ­ 2 ----------- + 2 ln ----------- + ----------- 4 2 4 2 4 2
(48)

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Putting it all together:
2 r 0 1 n ( ) 2 µ 1, max ( 1 + µ 1 2 ) dN ------------------------ dµ 1 d -------------------- = c ----- ----------- ---------- µ dtd d 1 2 4 2 0 1, min ( 1 + µ ) 2 1 2 2 cr 0 n ( ) = -------------- --------- 2 (49)

1 1 1 2 1 â 1 + ----------- + 2 ----------- ln ----------- ­ 2 ----------- 4 2 4 2 4 2 4 2

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We put 1 x = ----------- 4 2 to obtain 3 ( c T ) n ( ) dN -------------------- = -- -------------- ---------- F C ( x ) 4 2 dtd d 1 F C ( x ) = 2 x ln x + x + 1 ­ 2 x 2
2 = 3 -- T r0 8 (50)

(51)

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1.0

0.8 The function F C (x) used in the theory of Inverse Compton emission

A plot of the funcFC( x) tion is shown at the left. This shows that there is a broad distribution of electron energies resulting from Compton scattering.

0.6 FC (x) 0.4 0.2 0.0 0.0

0.2

0.4

0.6

0.8

1.0

2 x = ( 1 /4 )

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Rate of energy of scattered photons The above expression is more informative if we compute the rate of energy scattered per unit final photon energy per unit incident energy. This is dN 3 ( c T ) n ( ) 1 -------------------- = 1 -- -------------- ---------- F C ( x ) dtd 1 d 4 2 = 3 ( c T ) n ( ) xF C ( x )

(52)

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0.20

The function xF C ( x ) is plotted at the left. This function gives a better idea of the rate of photon energy scattered to high energies as a result of Compton scattering.

0.15

x FC (x

0.10

0.05

0.00 0.0

0.2

0.4

0.6

0.8

1.0

x=( /4 2 ) 1

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As you can see this function indicates a peak rate of scattered energy at 1 0.6 â 4 2 2.4 2 Check on calculation As a check on this expression, one can evaluate the total energy radiated against the previous expression by integrating with respect to 1 .
(53)

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Since x = 1 / ( 4 2 ) , then dN 2 ( c ) n ( ) 1 xF ( x ) dx 1 -------------------- d 1 = 12 0 C T dtd 1 d The value of the integral is 1 / 9 and Energy emitted per unit 4 = -- ( c T ) n ( ) 2 3 soft photon energy 4 P = -- ( c T ) 2 n ( ) d 3
High Energy Astrophysics: Compton Scattering II

(54)

(55)

The total power emitted comes from integrating over and is
(56)

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in agreement with the ultrarelativistic limit derived earlier. 3 Inverse Compton spectrum from a distribution of electrons 3.1 Direction of emission The goal of the following is to determine the emissivity of a distribution of electrons that are emitting photons via the inverse Compton process. This calculation is simplified by the fact that, in the lab frame, the emitted photons are scattered in the direction of motion of the electron.

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To see this consider a photon that is scattered at an angle cos­ 1 µ 1 in the rest frame. In the lab frame µ1 + µ 1 = cos 1 = -------------------1 + µ 1 When 1 , for most angles 1 , µ1 1
(58) (57)

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The only exception is when µ 1 ­ 1 , then µ1 ­1
(59)

Since Thomson scattering in the rest frame is quasi-isotropic, most photons will be scattered in the forward direction given by µ 1 = 1, = 0 .

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This plot effectively illustrates this point. The consequence of this is that most of the photons are emitted in the direction of motion of the electron. This is similar to synchrotron emission.

1.0 = 10 0.5 µ(Lab frame) = 2

0.0 Relationship between µ= cos in lab and electron rest frames for relativistic scattering -0.5

-1.0 -1.0

-0.5

0.0 µ(Electron rest frame)

0.5

1.0

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B

d

As before let N ( ) be the number of photons per unit Lorentz factor. Then the number of electrons with Lorentz factors in the range d pointing in the direction of the observer is: N ( ) ----------- d d 4
(60)

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Hence the rate of energy per unit time per unit scattered photon energy radiated into this solid angle, produced by soft photons in the range d is dN N ( ) 1 -------------------- ----------- d d d 1 d dtd d 1 4
(61)

(Note that we have multiplied by 1 to determine the energy associated with the number of photons.)

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Hence, the emissivity of inverse Compton scattered radiation is 1 dN j = ----- 1 -------------------- N ( ) d d 4 dtd d 1 1


3 ( c T ) N ( ) 1 n() = ----------------------- ---------- F C ( x ) ----------- d d 16 2

(62)

The emissivity therefore involves a double integration over the particle distribution and the input photon distribution.

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Integration over Lorentz factor We can do this part first since we generally have a good idea of what the particle distribution is. We assume, as before, that the particle distribution is given by N ( ) = K e ­a min < < max
(63)

Remember that the energy of the incident and scattered photons satisfy: 1 1 -------- < ---- < 4 2 2 2
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(64)

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so that if we are considering values of 1 / that approach ei2 2 ther 1 / ( 2 min ) or 4 max then there will be "end effects" associated with the integral. We assume that this is not the case and estimate the inner integral

N ( ) (65) F C ( x ) ------------ d 2 We calculate the integral by changing the integration variable to x = 1 / ( 4 2 ) .
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(66)

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Then 1 1 / 2 = ----- x ­1 / 2 4 1 d = ­ -2 4 1 1 / 2 ----- x ­ 3 / 2 dx 4
(67)

(a + 2) a + 2 1 ­ ---------------- ----------2 ­ 2 N ( ) = K ----- x2

The initial limits on x are defined by min and max implying that 1 1 ------------------ < x < ----------------2 2 4 max 4 min
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(68)

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For the upper limit however, we also have the requirement x < 1 , so that x< 1 If ---- > 4 ing very 1 / 4 -------------2 min , 1
(69)

2 min , as is usually the case, (that is we are consider-

high energies 1 ) then we have x < 1 .

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Making the above substitutions and using the derived limits:
­( a + 1 ) ) K e 1 ------------------- 1 ( a ­ 1 -------------- max N ( ) 2 f ( x ) ----------- d = ------ ----- x 2 F C ( x ) dx 0 2 2 4 min (70) ­( a + 1 ) 1 ------------------2 + 4 a + 11 ) 2(a ----- 2 = K e -----------------------------------------------------( a + 1 ) ( a + 3 ) 2 ( a + 5 ) 4

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Then, 3 ( c T ) N ( ) 1 n() j = ----------------------- ---------- F C ( x ) ----------- d d 16 1 2
(a ­ 1) ­ ---------------2 + 4 a + 11 ) 3 (a ­ 2) (a 2 (71) ------------------------------------------------------ ( c T ) K e = -- 2 1 ( a + 1 )( a + 3 )2( a + 5 ) (a ­ 1) max --------------- 2 n()d â min

W m ­3 J ­1

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We write this more compactly as:
(a ­ 1) (a ­ 1) ­ ---------------- max ---------------2 j = f ( a ) ( c T ) K e 2 n()d 1 1 min

(72)

3 (a ­ 2) ( a 2 + 4 a + 11 ) -----------------------------------------------------f ( a ) = -- 2 ( a + 1 )( a + 3 )2( a + 5 )

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3.2 Emissivity in terms of photon number High energy observers count photons and photon-counting statistics come into the estimation of errors. Hence, the results of X-ray or -ray observations are often represented in terms of a photon flux: number of photons per second per unit area. Accordingly, the emissivity can also be represented in terms of number. Let Rate of emission of photons per unit time dN -------------------------- = (73) dtdVd d per unit volume per unit energy per Steradian

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Then dN -------------------------- = ­ 1 j dtdVd d 1
(74) (a ­ 1) (a + 1) ­ ---------------- max ---------------2 = f ( a ) ( c T ) K e 2 n()d 1 min

An important feature is a difference of 1 between the photon index and the spectral index.

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4 Features This expression for the emissivity has some important features: · The expression for the emissivity is proportional to the Thomson cross-section. When one calculates the surface brightness one multiplies by the length, L , through the source to obtain the Thomson optical depth T L . This always figures in applications involving electron scattering. · The emissivity is a power-law with exactly the same spectral index as the synchrotron emission. This is a tell-tale signature of inverse Compton emission. · The dependence on the soft photon spectrum enters
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max ( a ­ 1 ) / 2 n ( ) d . The energy through the factor min

dependence of the inverse Compton emission is independent of the type of input photon spectrum. 5 Applications There are 2 traditional main applications for this formula. · When the incident photon spectrum is a black body, e.g. the microwave background or approximating the emission from starlight by a black body · When the synchrotron emission from a source produces a quasi-isotropic spectrum in the emitting volume
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5.1 Evaluation of the soft photon integral We require
(a ­ 1) / 2 n()d 0 (75)

where n ( ) is the number density of photons per unit energy. The energy density of photons per unit energy is u ( ) = n ( ) n ( ) = ­1 u ( )
(76)

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We had previously defined the energy density per unit frequency by 1 u = -- I d c
4 (77)

For the isotropic distribution we have used here, 4 u = ----- I c Moreover d u d = u ( ) d u ( ) = u ----- = h ­ 1 u d
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(78)

(79)

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Therefore 4 ­ 1 u ( ) = -------- ­ 1 u n() = 2c h
(80)

5.2 Blackbody distribution When the soft photon field is an isotropic black body distribution such as the microwave background, then the specific intensity is just given by the black body value:
­1 h 2 h 3 -----I = B = ------------ e kT ­ 1 c2 (81)

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The energy density per unit energy is then:
­1 h 8 2 -----n ( ) = -------- e kT ­ 1 hc 3 (82)

Hence,
(a ­ 1) / 2 n ( ) d 0

8 ( a + 5 ) / 2 x ( a + 3 ) / 2 [ e x ­ 1 ] ­1 dx = ----------- ( kT ) 0 3c3 h

(83)

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The integral
(a + 3) / 2 x a+5 ­ 1 d x = ----------- a + 5 ----------[e ­ 1] 0 x 22 (84)

Hence, 8 T a + 5 ----------- a + 5 ( kT ) ----------j = f ( a ) ------------- 1 h3c2 2 2
(a + 5) ---------------2

K e

(a ­ 1) ­ ---------------2 1

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Group the numerical and a -dependent terms into, f bb ( a ) = a + 5 a + 5 (85) 3.2 a + 1 ( a 2 + 4 a + 11 ) ------------------------------------------------------- â ----------- ----------22 ( a + 1 )( a + 3 )2( a + 5 ) The emissivity from a black body soft photon field and a relativistic electron distribution is therefore:
(a ­ 1) a+5 T ­ -------------------------j = ----------- f bb ( a ) ( kT ) 2 K e 1 2 1 h3c2 (86)

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25

20

The function f bb ( a ) is plotted at the left. This can be used in estimates of the inverse Compton emission produced by the microwave background, for example.
(87)

15 f bb (a) 10 5 0 0.0

0.5

1.0

1.5

2.0 a

2.5

3.0

3.5

4.0

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5.3 Synchrotron Self-Compton emission In this case, the soft photons are synchrotron photons emitted from the same volume by the same electrons. Therefore, the power-law index of the soft photon input is ( a ­ 1 ) / 2 and we can take u ( ) ­ ( a ­ 1 ) / 2 . We suppose that the power-law holds between soft photon frequencies of l and u (l for lower and u for upper). Thus, ­ 1 u ( ) = [ ­ 1 u ( ) ] ---- ­ ( a + 1 ) / 2 n() = 0 0 - 0
(88)

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where 0 is some fiducial soft photon energy. The integral,
u u ( ( a ­ 1 ) / 2 n ( ) d = 0a ­ 1 ) / 2 u ( 0 ) l l

---- ­ 1 d --- 0 0

(89)

( = 0a ­ 1 ) / 2 u ( 0 ) ln ( u / l )

Notes: · Because of the assumed power-law, the parameter
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rithm. The upper energy can be determined from observation. The lower frequency may not be as well determined, although synchrotron self-absorption (to be treated later) sets a lower limit on l . Fortunately, the evaluation of the emissivity is not sensitive to either the upper or lower cutoff since they appear in a logarithmic term.

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Final form in terms of energy In the case of SSC emission, then, the emissivity is:
(a ­ 1) (a ­ 1) ­ ---------------- max ---------------2 j = f ( a ) ( c T ) K e 2 n()d 1 1 min

= f ( a ) ( c T ) K e ( 1 / 0 ) ­ ( a ­ 1 ) / 2 u ( 0 ) ln

u ---- l

(90)

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When we compare this with the expression for synchrotron emission from, say the isotropic version resulting from a tangled field
syn = C ( a ) j 2

e2 ( ------- K e 0a + 1 ) / 2 ­ ( a ­ 1 ) / 2 0 c

(91)

then we can see that the ratio of SSC to synchrotron emissivity can give us an estimate of the magnetic field since the factor of K e occurs in both expressions and therefore cancels out.

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Emissivity in terms of frequency Note that, for the inverse Compton emissivity, j = hj 1 1
(92)

and Planck's constant combines with the factor u ( 0 ) as follows: hu ( 0 ) = u 0
(93)

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Hence, a useful expression for SSC emissivity, when comparing different frequencies is SSC = f ( a ) ( c ) K ( / ) ­ ( a ­ 1 ) / 2 u ln j Te10 0 1 u ---- (94) l

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5.4 Using synchrotron and self-Compton fluxes to estimate the magnetic field The flux density from a stationary source at a luminosity distance, D L , is: 1 F = ------- j dV 2 DL V
(95)

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We assume a spherical source with constant electron density and uniform synchrotron emissivity. The flux density from the source at frequency 1 is then 1 F = ------- f ( a ) ( c T ) K e ( 1 / 0 ) ­ ( a ­ 1 ) / 2 ln ( u / l ) 2 1 DL â u dV V0

(96)

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We could simply assume at this point that the photon energy density is constant throughout the volume. However, one of the exercises shows that: 1 1 ­1 1 + -u ( r ) = u v ( 0 ) -- + -- ( ­ ) ln ---------- 1 ­ 24 r = -R where R is the radius of the volume.

(97)

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Hence,

V



R 1 1 ­1 1+ ----------- r 2 dr u dV = 4 u ( 0 ) -- + -- ( ­ ) ln - 1 ­ 02 4 (98)

1+ 3 1 1 + 1 ( ­ 1 ­ ) ln ----------- 2 d -= 4 u ( 0 ) R -- - 1 ­ 02 4
3u (0) = 3 u (0)V -- = R 40 0

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The SSC flux, therefore is: 1 SSC = 3 ------- f ( a ) ( c ) K ( / ) ­ ( a ­ 1 ) / 2 -F Te10 4D2 1 L â ln ( u / l ) u V 0 By comparison, the synchrotron flux is: C2(a) syn F = -------------- D2
a + 1 (a ­ 1) ----------- ­ ---------------e2 2V ------- K e 0 2 0 c (100)

(99)

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6 Models of synchrotron and SSC emission As above let us consider a spherical model in which the radius of the sphere is R . In the expression for the SSC flux 1 SSC = 3 ------ f ( a ) ( c ) K ----- -- F T e 4 D2 1 0 â ln ( u / l ) u ( 0 ) V 0 we have the factor 2 peak u = ----- I c 0
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(a ­ 1) ­ --------------- 1 2 (101)

(102)
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and
peak = 2 j syn R I (103)

Therefore, a model of the combined synchrotron and SSC spectrum involves the synchrotron parameters as well. When constructing models for the entire spectrum, we take this into account. An example is the following spectrum of Markarian 501. In this case, we have also had to allow for the fact that the Klein Nishina cross-section had to be used and the motion of the emitting region is relativistic. Also, the end points of the distribution in have to be included so that this example is quite a comprehensive one.
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CONT: MKN501 IPOL 1634.000 MHZ MRK501,1.4+1.OGEOM.1

5

VLBI Image Markarian 501

of

0

-5

-10

-15

-20 30 15 10 MilliARC SEC Center at RA 16 53 52.21671 DEC 39 45 36.6091 Cont peak flux = 4.2070E-01 JY/BEAM Levs = 1.000E-03 * (-1, 1, 1.500, 3, 5, 10, 20, 30, 50, 100, 200, 300) 25 20 5 0

The X-ray and -ray emission originate from a region about an order of magnitude smaller than the smallest resolvable scale in this image.

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1e+03 2 dN/(dt d dA) (eV cm -2 s -1 )

1e+02

B=0.1 Gauss 1e+01 Ke = 4.3x10 5 cm -3 2 = 3x10 6 R= 5.3x10 15 cm 1e+00 1e+03

1e+05

1e+07

1e+09 (eV)

1e+11

1e+13

X and TeV -ray spectrum of Markarian 501
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7 Inferences from SSC emission on resolvable scales In general, we model synchrotron plus showed for MKN 501. However, when solved it is possible to use both sets of magnetic field. Using the expressions for SSC fluxes: SSC spectra as the emission is data to estimate the synchrotron we rethe and

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C2(a) syn F = -------------D2

a + 1 (a ­ 1) ----------- ­ --------------- e2 2V ------- K e 0 2 0 c (a ­ 1) 1 ­ ---------------2 (104)

) SSC = 0.75 f ( a - ( c ) K ----- --------------------F T e 1 D2 0

ln ( u / l ) u ( 0 ) V 0

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On dividing one equation by the other, a number of factors 2 3 me e 2, V ) and ------------------ = -- ------ . This gives: cancel ( K e, D 2 r0 0 c 2 T
a+1 ----------0 2 =

f (a) ----------------- ------ 2 C 2 ( a ) m e

(a ­ 1) r 0 1 ­ ---------------- ( a ­ 1 ) 2

----- 0

----------------- ln 2

u ---- u ( 0 ) l 0
(105)

syn F â ------------SSC F
1

and, of course, B = ( m e / e ) 0 .
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All of the parameters in the right hand side of the above equation are observable, except for l . We shall see later how we can put constraints on l . The parameter 2 peak u = ----- I c 0
(106)

is also directly estimable from the peak surface brightness of the emitting region.

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8 Example on Cygnus A hot spots X-ray image (contours) of Cygnus A overlaid on the radio image (greyscale) (Wilson, Young and Shopbell, 2000). The bright X-ray regions are coincident with the radio hot spots.
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Spectral break

The radio and X-ray spectrum (in the form of fluence) of the brightest hot spot in Cygnus A. The "bowtie" represents the region of allowable Xray slopes.

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The solid lines show model fits to the data. However, we can estimate the magnetic field using the data from this plot. For example take the point at = 10 8.3 Hz
syn F = 10 10.8 â 10 ­ 26 W m ­ 2

u = 10 10 Hz 1 = 10 17.5 Hz 1 F = 10 9.9 Hz 1 = 0.55

(107)

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The following information on the radio source gives us the value of the energy density of the radio emission in the hot spot. This is obtained from the paper The Structure and Polarisation of Cygnus A at 3.6 cm, R.A. Perley and C.L. Carilli, in Cygnus A ­ Study of a Radio Galaxy, ed. C.L. Carilli & D.E. Harris. Peak flux per beam = 90 mJy Beam size = 0.35 â 0.35 0 = 5 â10 Hz
9

(108)

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The value of the magnetic field, evaluated using equation (105) is B = 1.4 â10
­4

Gauss

(109)

Note that the slope of the spectrum implied by the "bowtie" is not the same as the slope of the radio spectrum. In the model shown in the figure, this is the result of the X-ray emission being affected by the "end effects" referred to earlier. Nevertheless estimates such as these are quite useful for the starting points of detailed models.

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By adopting an angular size for the hot spot we can estimate the minimum energy magnetic field. From the image, one obtains D hot spot 3.5
(110)

The peak surface brightness of the hot spot can be evaluated from the flux per beam and the beam size [see equation (108)] and these parameters can be used to evaluate the minimum energy magnetic field. One obtains for the minimum energy magnetic field in the hot spot B min = 9.2 â10
­5

Gauss

(111)
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This is within a factor of 2 of the inverse Compton value.

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