3 synchrotron radiation i - anugeoff/hea/3_synchrotron_radiation_i.pdf · high energy astrophysics:...

Synchrotron Radiation I

1 Examples of synchrotron emitting plasma

Following are some examples of astrophysical objects that are emit-ting synchrotron radiation. These include radio galaxies, quasars andsupernova remnants.

High Energy Astrophysics: Synchrotron Radiation I 1/106

A 20 cm radio image from theVLA of the radio galaxy IC 4296.See Killeen, Bicknell & Ekers,ApJ 325, 180.

This images shows the jets andlobes of the radio emission corre-sponding to a relatively nearby gi-ant elliptical galaxy.


The inner part of the radio galaxy M87 from a 2cm VLA image by Biretta, Zhou & Owen.


VLA 4.9 GHz image of the quasar 3C204.

The linear size is Note the periodic knottystructure in jets that may be

the result of internal shocks and the very bright core. See http://www.cv.nrao.edu/~abridle/images.htm

z 1.112=159 h kpc


The Supernova Remnant Cassio-peia A

Cassiopeia A gets its name from radio astronomers, who “rediscovered” it in 1948 as the strongest radio source in the constellation of Cassiopeia. About 5 years later optical astronomers found the faint wisps, and it was determined that Cas A is the remnant of an explosion that occurred about 300 years ago. The radio emission is synchrotron emission.


The X-ray image of the Cassiopeia A su-pernova remnant on the left was obtained by the Chandra X-ray Observatory using the Advanced CCD Imaging Spectrome-ter (ACIS). Two shock waves are visible: a fast outer shock and a slower inner shock. The inner wave is believed to be due to the collision of the ejecta from the supernova explosion with a shell of ma-terial, heating it to a temperature of ten million degrees. The outer shock wave is a blast wave resulting from the explo-sion.


2 What can we learn from synchrotron emission from an astro-physical object?

• Estimates of total radiative luminosity• Total minimum energy (particles plus field) in the radio emitting particles

• Direction of the magnetic field• Constraints on energy density of emitting particles and estimates of magnetic field

• Total amount of matter converted into energy (relevant for radio galaxies and quasars)

• Accretion rate onto central black hole (radio galaxies and quasars)• Constraints on the mass of the black hole (radio galaxies and qua-sars)

• Information relating to source dynamics, e.g. strengths of shock


waves; velocities of jets

3 Background theory

3.1 Electric fieldWe have the expression for the Fourier component of electric fieldof the pulse from a moving electron derived from the Lienard-Weichert potentials:

(1)

rE iq–4c0---------------e

ir cn n

–

=

i t n X t c

--------------------– dtexp


The Fourier components of a transverse radiation field can be ex-pressed in terms of unit vectors and perpendicular to the wave

direction such that

(2)

In any particular application we choose and so as to simplify

the expression for the resultant electric field and Stokes parameters,defined by:

e1 e2

E E1 e1 E2 e2+=

e1 e2


(3)

I

c0T-------- E1 E1

* E2 E2* + =

Q

c0T-------- E1 E1

* E2 E2* – =

U

c0T-------- E1

* E2 E1 E2* + =

V1i---

c0T-------- E1

* E2 E1 E2* – =


Interpretation of T

Another way to interpret the parameter which is appropriate whenconsidering radiation from a distribution of particles is:

(4)

or

(5)

In the latter case each pulse originates from a different electron in thedistribution.

Nomenclature

We refer to the and components of the electric field as the two

modes of polarisation.

T

T time between pulses=

T 1– No of pulses per unit time=

e1 e2


3.2 EmissivitiesThe basic relation is:

(6)dW

dddt---------------------

c0T--------r2E E

* =


In terms of

(7)

I Q U V

jI dW11

dddt---------------------

dW22dddt---------------------+=

jQ dW11

dddt---------------------

dW22dddt---------------------–=

jU dW12

dddt---------------------

dW12*

dddt---------------------+=

jV idW12

dddt---------------------

dW12*

dddt---------------------–=


3.3 Helical motion of a relativistic particleFor synchrotron emission, we evaluate the above integrals using theexpression for helical motion:

B

B1

v

Pitch angle


(8)

(9)

The derivation of the equations for the helical trajectory of a chargedparticle in a magnetic field are given in the background notes.

v c

Bt 0+ cossin

Bt 0+ sinsin–

cos

=

B Gyrofrequencyq Bm--------- 1– q B

m--------- 1– 0= = = =

0 Non-relativistic gyrofrequency=

qq----- sign of charge= =

pitch angle=


Usually we are concerned with electrons ( ) so that:

(10)

Note that the motion of the velocity vector is anti-clockwise for anelectron.

1–=

v c

Bt 0+ cossin

Bt 0+ sinsin

cos

=


Coordinates of particle

Integrating the velocity gives for the position vector:

(11)X t c

sinB

----------- Bt 0+ sin

sinB

----------- Bt 0+ cos

t cos

X0+=


This consists of a linear motion in the -direction superimposedupon a circular motion in the plane. The radius of the circularmotion is the gyroradius

(12)

zx y–

rGc sinB

------------------mc sin

q B-------------------------= =


3.4 Approximations for ultrarelativistic particles

• . Sometimes is appropriate.

• Most of the radiation beamed into a cone of half-angle . This means that we can expand functions in terms of small powers of the angle between the particle velocity and the direction of emis-sion.

We set up the calculation in the following way (following Rybicki &Lightman):

1» 11

22--------– 1

1


• The origin of time is defined as the instant where the parti-

x

y

z

v

vt a

e1

e2

e3

B

To observer

n

Normal totrajectory

Tangent to particle motion at .t 0=

t 0=


cle’s velocity comes closest to the direction of the observer.• The –axis represents the instantaneous direction of motion at

.• The –axis represents the direction of the centre of curvature.• The –axis completes the system of coordinates (i.e.

).

• is the direction of the observer and makes an angle to the direction of the electron at . We take small.

• Since represents the instant at which the angle between the velocity and the observer’s direction is a minimum, the component of in the direction of the normal, is zero.

• The parameter is the instantaneous radius of curvature of the particle

xt 0=

yz

e3 e1 e2=

n t 0=

t 0=

n ma


• Formally, the pulse of radiation reaching the observer originates from the entire trajectory of the particle. However, most of this radiation originates from a very small region of the particle’s orbit near the origin of the above coordinate system.

3.5 Radius of curvatureSummary of the theory of 3 dimensional curves:

(13)

x x t = Velocity vector vdxdt------= =

Unit tangent vector tvv--= =

Arc length:dsdt----- v=


The curvature ( ), radius of curvature ( ) and curvature normal ( )are given by:

(14)

Using the previous expression for velocity:

(15)

a m

dtds----- m

1a---m= =

dtds-----

dtdt-----

dtds-----

1v---

dtdt-----= =

v c Bt cossin Bt sinsin cos=

tvv-- Bt cossin Bt sinsin cos= =


Since the particle passes through the origin at the positionvector is:

(16)

Remember that the gyroradius

(17)

t 0=

xc sinB

------------------ Bt sinc sinB

------------------ 1 Bt cos– ct cos=

rG Bt sin rG 1 Bt cos– ct cos=

rGc sinB

------------------=


The curvature is determined from:

(18)

This implies

(19)

NB. The radius of curvature is not the same as the radius of the pro-jection of the orbit onto the plane perpendicular to . As one ex-pects, when .

1v---

dtdt----- m

B sin

c-------------------- Bt sin– Bt cos 0= =

m Bt sin– Bt cos 0=

B sin

c--------------------= a

cB sin--------------------

rG

sin2--------------= =

Ba rG= 2=


3.6 The magnetic field in the particle-based system of coordinates

• Since the normal vector is per-

pendicular to the magnetic field, then the magnetic field lies in the plane of and . i.e. in the same plane as .

• Also since the velocity makes an angle of with the magnetic field, the direction of is as shown in the diagram.

m Bt sin– Bt cos 0=

e1 e3 n

B


4 Semi-quantitative treatment of synchrotron emission

4.1 Small angle approximationRecall the expression for the electric field of a radiating charge:

(20)

For a relativistically moving particle, the electric field is highlypeaked in the direction of motion, i.e. when

Eradq

4c0r--------------------

n – · n · 1 n– – 1 n– 3

---------------------------------------------------------------------------------=


is close to zero. Since is the angle between and

(21)

So is a minimum at and increases rapidly for, i.e. for . Since , then the angles we are dealing

with are very small.

The expression for introduces the variable

(22)

This variable occurs frequently in the following theory.

1 n– n

1 n– 1 cos– 1 11

22--------–

112---2–

–=

1

22--------

12---2+

1

22-------- 1 22+ = =

1 n– 0= 1 1 1»

1 n–

1 22+ 1 2/=


4.2 Estimate of critical frequency

Observer

1 1

2

Centre of curvature

P1 P2


The angle turned through and the arc length of the particle orbit arerelated by:

(23)

where is the previously determined radius of curvature.

From the above geometry

(24)

Time taken for trajectory to sweep through relevant angle

(25)

dds-------

1a---=

a

2---= s a 2

a---= =

t2 t1–sc-----

2ac------= =


This is not the period for which the pulse is observed

Partticle moves distance in time .

Front of pulse emitted from here at t1

s D distance to observer=

P1 P2

Back of pulse emitted from here at t2

s t2 t1–


Times of arrival of front and back of pulse:

(26)

i.e. the time between pulses is reduced because the trailing part of thepulse has less distance to travel. This is one example of the effect oftime retardation.

t1 t1D s+

c----------------+= t2 t2

Dc----+=

t2 t1– t2 t1–sc-----– t2 t1– v

c-- t2 t1– –= =

1vc--–

t2 t1– =


The factor can be expanded in powers of as follows:

(27)

Since

(28)

then

(29)

1vc--– 1 –=

11

22--------–

1 –1

22--------


The observed duration of the pulse is therefore

(30)

Fourier theory tells us that if the length of a pulse is then the typ-

ical frequencies present in such a pulse are . Hence thetypical frequencies expected in a synchrotron pulse are given by:

(31)

1 – t2 t1– 1

22-------- t2 t1– a

c3-------- c

B sin--------------------

1

c3--------=

1

B3 sin

--------------------------=

t

t 1–

B3 sin q B

m--------- 3 sin q B

m---------2 sin= =


The critical frequency resulting from the exact synchrotron theory(to follow) is close to this estimate:

(32)

The critical frequency which characterises synchrotron emission is a

factor of higher than the cyclotron frequency and a factor of

higher than the gyrofrequency at which particles gyrate aroundthe magnetic field lines.

c32---B 3sin

32---

q Bm---------3 sin

32---

q Bm

---------2 sin= = =

2 q Bm

---------

3


4.3 Typical example - relativistic electron in a typical interstellar magnetic field

(33)

Note that the critical frequency in this example is about – amicrowave radio frequency.

B 10 5– G 1nT= = 104=

BeBme--------- 1.8

2–10 Hz= =

c32---

eBm------2 sin 1.3

1010 Hzsin= =

rGc sinB

------------------ 1.71010 msin= =

10 GHz


4.4 Relation between frequency and Lorentz factor

(34)

2 32---

eBm------2 sin= =

43

------me----

1 2/B sin 1 2/– 1 2/=

4.9310

B sinnT

--------------- 1 2/–

GHz----------- 1 2/

=

4.9310

B sin10G--------------- 1 2/–

GHz----------- 1 2/

=


5 Detailed development of synchrotron emission

Angle turned through in time :

(35)

t

dds-------

1a---=

1v---

ddt-------

1a---= vt

a------=


In the new set of axes, the motion is circular with no component inthe –direction. The velocity in these axes is therefore:

(36)

z

v c vta

------cos vta

------sin 0=

i.e. vta

------cos vta

------sin 0=


5.1 Axes based upon instantaneous orbital plane

x

y

z

vvt a

e1

e3

B

To observer

n

Normal totrajectory

Tangent to particle motion at .t 0=

e||

eCentre of motion

e2


Rotate axes by angle about - gives new unit vectors

5.2 Significance of and

The axes and are respectively parallel and perpendicular to the

projection of the magnetic field on the plane perpendicular to the lineof sight given by the vector .This is the plane of propagation of theradiation.

e2 e e|| n

e|| e

e|| e

n


5.3 Integral to evaluateWe have our expression for the Fourier component of the electricfield of a pulse:

(37)

5.4 Determination of

We first calculate a simple expression for the quantity .

rE iq–4c0---------------e

ir cn n

–

=

i t n X t c

--------------------– dtexp

n n

n n


Use:

(38)

This gives:

(39)

n e1cos e3sin+=

vta

------ cos e1 vt

a------ sin e2+=

n vta

------ n e1cos vt

a------ n e2sin+=

vta

------ esincos vt

a------ e||sin+=

n n vta

------ e||cossin –

vta

------ esin=


We make the following approximations:

(40)

so that

(41)

Note:

In order to determine terms that are relevant to the small amount ofcircular polarisation that emerges from a synchrotron source, wewould have to expand to the next order in .

1 vta

------ cos 1 vt

a------ vt

a------sin

n n e||vta

------ e– e||

cta

------e–=

1


The term

This term in the argument of the exponential involves . Since

(42)

i t n X t c

--------------------–

X t

dXdt------- v c vt

a------cos vt

a------sin 0= =

and X 0 0=


then

(43)

giving

(44)

X t cav

--------- vta

------sin 1 vta

------cos– 0=

a vta

------sin 1 vta

------cos– 0=

t n X t c

--------------------– t cos 0 sinac--- vt

a------sin 1 vt

a------cos– 0–=

t ac--- vt

a------sincos–=


5.5 Expansion of in small powers of and :

(45)

then

(46)

t n X t c

--------------------– t

cos 112---2–

vta

------ vt

a------

16---

v3t3

a3------------–sin

t ac--- vt

a------ sincos– t 1

vc--–

12---

vc--t2 1

6---

v3t3

a2c------------+ +=


In the first term, we use and put in the remain-

der, giving:

(47)

The characteristic angle enters into the expression.

1vc--–

1

22--------= v c=

t ac--- vt

a------ 1

22--------t 1

2---2t 1

6---

c2t3

a2------------+ +sincos–

t n X t c

--------------------–1

22-------- t 1 22+ 1

3---

c22t3

a2------------------+=

1


Summary so far

(48)

New variables

We put these equations into a dimensionless form by defining newvariables

n n e||vta

------ e– e||

cta

------e–=

i tn X t

c--------------------–

i22-------- 1 22+ t c22

3a2-----------t3+=


(49)

• The purpose of introducing these variables is to make quantities in the integrand of the expression for the Fourier transform of order unity.

• In so doing it helps to represent the new dimensionless variables in a way that makes their physical significance apparent.

2 1 22+= y

ca--------t= a

33c-----------

3=


Significance of

The quantity is the time taken for the particle’s velocity toswing through an angle . Hence

(50)

Note:

• is a dimensionless variable• One expects the major contribution to come from the region of the integrand wherein .

• Integration over will be replaced by integration over .

y

a c 1

yt

Characteristic time ---------------------------------------------------------=

y

y 1t y


Significance of

(51)

Since the radiation is concentrated at the critical frequency, the dom-inant contribution to the result will be around .

a

33c-----------

3 13---

3c a

--------------------3= =

ac---

1c---

cB sin--------------------

1B sin--------------------=

13---

3B sin--------------------------

3 12---

c------

3= =

1


Rearrangement of integrand

For :

(52)

n n

n n e||cta

------e– e||ca---

ac

--------ye–=

e||

-----ye–=


For

(53)

i tn X t

c--------------------–

i tn X t

c--------------------–

i22-------- 1 22+ t c22

3a2-----------t3+=

i22--------

2ac

--------yc22

3a2-----------

a33

c33------------y3+=

i22--------

a3

c---------y

13---

a3

c---------y3+=


(54)

For – variable of integration:

(55)

i tn X t

c--------------------–

ia3

2c3---------------- y

13---y3+=

i32---a

3

3c3--------------

y13---y3+=

32---i y

13---y3+=

t

tac

--------y= dtac

--------dy=


6 Evaluation of Fourier components of electric field

rE iq–4c0---------------e

ir cn n

–

=

i t n X t c

--------------------– dtexp

n n e||

-----ye–

i tn X t

c--------------------–

32---i y

13---y3+ dt

ac

--------dy=


Therefore we have a parallel component and a perpendicular compo-nent of the electric field:

(56)

rE|| iq4c0---------------

ac

------------ eir c 3

2---i y

13---y3+

exp yd–

–=

rE iq4c0---------------

a2

c2---------

eir c y–

32---i y

13---y3+

exp yd=


The integrals over have real and imaginary parts:

(57)

Parallel component

In the integral for the real part only contributes since the im-

aginary part is odd and the different contributions over and cancel.

Therefore, we evaluate:

y

32---i y

13---y3+

exp32--- y

13---y3+

cos=

i32--- y

13---y3+

sin+

rE||

0– 0


Perpendicular component

In the integral for the imaginary part only contributes since

the exponential term is multiplied by . We therefore evaluate:

(58)

32---i y

13---y3+

exp yd–

32--- y

13---y3+

cos yd–

=

rE

y

y32---i y

13---y3+

exp yd–

i y

32--- y

13---y3+

sin yd–

=


These integrals are expressed in terms of Bessel functions:

(59)

where and are modified Bessel functions of or-

der and respectively. These integrals are evaluated in theAppendix.

32--- y

13---y3+

cos yd–

2

3-------K1 3 =

y32--- y

13---y3+

sin–

2

3-------– K2 3 =

K1 3 K2 3

1 3 2 3


Final expressions

(60)

rE|| iq

2 3c0

-----------------------a

c------------ e

irc

--------K1 3 –=

rE q

2 3c0

-----------------------a

2

c2---------

e

irc

--------K2 3 =


7 Ellipticity of single electron emission

The waves corresponding to the above Fourier components are, forthe parallel component:

(61)

(using the reality condition of the Fourier transform).

E|| r t 12------ E|| e it– E|| – eit+ =

12------= E|| e it– E||

* eit+


and for the perpendicular component

(62)

E r t 12------ E e it– E – eit+ =

12------ E e it– E

* eit+ =


Using the expressions for the Fourier components:

(63)

E|| r t q

4 32c0r----------------------------

ac

------------ K1 3 =

ie–i r

c-- t–

iei r

c-- t– –

+

q

2 32c0r----------------------------–

ac

------------ K1 3 t

rc--–

sin=


(64)

E r t q

4 32c0r----------------------------

a2

c2---------

K2 3 =

ei r

c-- t–

ei r

c-- t– –

+

q

2 32c0r----------------------------

a2

c2---------

K2 3 trc--–

cos=


Now consider the path traced out with time by the –vector at afixed . Write

(65)

This describes an ellipse in the coordinate system. We shall

see that but that can be positive or negative.

Er

E|| r t a|| trc--–

sin=

E r t a trc--–

cos=

e e||

a 0 a||


E|| E||

E E

trc--–

E|| r t a|| trc--–

sin= E r t a trc--–

cos=

a|| 0 a|| 0


(66)

• The semi-major and semi-minor axes of the electric vector depend upon (through the variable ) and we have to bear in mind the

a||q

2 32c0r----------------------------–

ac

------------ K1 3 =

q

2 32c0r----------------------------

a

c2-------- – K1 3 =

aq

2 32c0r----------------------------

a2

c2---------

K2 3 =

q

2 32c0r----------------------------

a

c2--------

2 K2 3 =


fact that

(67)

• The rotation of the electric vector and the shape of the polarisation ellipse depends upon

• The angular dependence of the radiation field can be understood from the plots of the functions

(68)

12---c------ 1 22+ 3 2/=

f|| x – K1 3 =

f x 2K2 3 =


where

(69)

These are plotted below for . The variation is similar for all .

xc------=

x 1= x


-2.5 -1.5 -0.5 0.5 1.5 2.5-0.50

-0.25

0.00

0.25

0.50

0.75

1.00

1.25

f||

f

f ||f


As can be seen both the parallel and perpendicular components of theelectric field vanish rapidly for .

Axes of ellipse

This plot shows the variation of the ratio of

for .

Since this ratio is always less than 1 for all the parallel axis is always the minor axis and the perpendicular axis is always the major axis.

1

-2.5 -1.5 -0.5 0.5 1.5 2.5-1.00

-0.75

-0.50

-0.25

0.00

0.25

0.50

0.75

1.00

a|| a f|| f= x 1=


Summary of polarisation properties of single electron emission

From the above:

• is positive for and negative for

• Therefore the radiation is always left polarised for and right polarised for

• The ratio so that the major axis is always .

• That is the major axis of the polarisation ellipse is perpendicular to the projection of the magnetic field on the plane of propagation.

The following figure summarises the above. Note that corre-sponds to looking exactly along the direction of the electron veloci-ty.

a|| 0 0

0 0

a|| a 1 e

0=


8 Integrated single electron emissivity

• This is a good example of how the integrated emission from a source can exhibit polarisation properties that are different from the individual components.

• A qualitative feature that one can note from the above is that the

e

e|| 0

0

Right polarised

Left polarised

V 0

V 0

B

n

v

vn


integrated emission from a single electron will have a low circular polarisation because of the positive and negative contributions to

from negative and positive • Note that because the axis of the ellipse is in the direc-

tion.

Recall our expression for the energy of the pulse corresponding tothe various components of the polarisation tensor:

(70)

V U 0 e

dWdd---------------

c0

--------r2E E* =


n

vB

Illustrating the calculationof total emission from asingle electron. We inte-grate over all directions .n

d d=


The total spectral energy of the pulse, per unit solid angle is:

(71)

The two terms correspond to emission into the two modes. We canuse this expression to determine the total energy emitted by the elec-tron in one gyroperiod and hence the total spectral power.

We first consider the total energy per unit solid angle per unit circu-lar frequency emitted over one entire gyration of the particle. Beginwith the energy radiated per unit solid angle, at a particular point onthe orbit, that we have just calculated:

dW11dd---------------

dW22dd---------------+

dW||dd---------------

dWdd---------------+=


(72)

dW||dd---------------

c0

--------r2E|| E||* =

2q2

123c0

---------------------a

c2-------- 2

2K1 32 =

dWdd---------------

c0

--------r2E E* =

2q2

123c0

---------------------a

c2-------- 2

4K2 3

2 =


These expressions are integrated over all solid angles associatedwith the directions . Let be the polar angles for the direction

. The solid angle for emission around this direction is:

(73)

We can use the solid angle corresponding to a complete orbit sincethe spectrum of a pulse is independent of .

For a complete orbit:

(74)

We have

(75)

n n

d ddsin=

d 2 dsin=

+=


where is the angle we have used in the previous calculations. Sincethe contribution to the integral over solid angle mainly comes from

, then we can put

(76)

1

d d=

d 2 dsin


Hence, the total energy emitted over one gyrational period into theparallel and perpendicular modes is:

(77)

The integral over is replaced by an integral over because of the concentration of

around and the rapid vanishing of the Bessel functions for.

dW|| d

----------------- 2 dW||,dd--------------- d

0

sin

2 dW||,dd--------------- d

–

sin

0 2 = – = dW||, dd

0= 1»


These expressions give the amount of energy radiated per orbit. Theenergy radiated per unit time is given by dividing by the period ofthe orbit

(78)

Hence, the power (i.e. energy per unit time per unit circular frequen-cy) emitted in the parallel mode is

(79)

T2B--------

2mq B

--------------= =1T--- q B

2m--------------=

P|| 2q2

123c0

---------------------a

c2-------- 2

2 q B2m--------------sin=

22K1 3

2–

d


We change the variable of integration to , thereby introducing an-other factor of into the denominator. We also use, for the local ra-dius of curvature,

(80)

Various factors in the numerical factor in from of the expression for

combine to give in the denominator, where

(81)

ac

B sin--------------------=

a2

c2------ 1

B2 sin2

-----------------------=

P|| c2

c2 9

4---B

2 6 sin2=


All of this combines to give:

(82)

The numerical factor in front of the expression for the other power is identical and

(83)

P|| 3

163c0

---------------------q3B sin

m----------------------

c------ 2

=

22K1 3

2 d–

P

P 3

163c0

---------------------q3B sin

m----------------------

c------ 2

4K2 3

2 d–

=


remembering that

(84)

The two integrals involve integration over the variable and in-volve as parameter

(85)

1 22+ 1 2/= 12---c------ 1 22+ 3 2/=

xc------=


These integrals were evaluated by Westfold (1959) in one of the fun-damental papers on synchrotron emission. The results are:

(86)

where the functions are defined by

(87)

22K1 3

2 d–

3

-------x 2– F x G x – =

4K2 3

2 d–

3

-------x 2– F x G x + =

F x x K5 3 z zdx

= G x xK2 3 x =


Hence the components of the single electron emissivities per unitfrequency are:

(88)

P|| 3

1620c---------------------

q3B sinm

---------------------- F x G x – =

P 3

1620c---------------------

q3B sinm

---------------------- F x G x + =


The total synchrotron power per unit circular frequency from a sin-gle electron is:

(89)

Ptot P|| P +3

820c------------------

q3B sinm

----------------------F x = =

xc------=


The functions and are plotted on linear and logarithmicscales in the following figures. The asymptotic forms are as follows:

(90)

F x G x

Small x

F x 43 1 3 21 3/

--------------------------------------x1 3/

G x 23 1 3 21 3/

--------------------------------------x1 3/

Large xF x x

2------e x–

G x x2

------e x–


Features

• Peak of at

• Emission is fairly broadband

with

0 1 2 3 4 50.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0F

(x),

G(x

)

F(x)

G(x)

x c=

F x x c 0.29=

-------- 1


Logarithmic version of theplot of and , show-ing the power law depend-ence of and forsmall .

9 Synchrotron cooling

9.1 Integration of synchro-tron powerA fairly immediate implication of the power radiated by a singleelectron is the loss of energy by the electron.

1e-04 1e-03 1e-02 1e-01 1e+00 1e+011e-04

1e-03

1e-02

1e-01

1e+00

F(x

), G

(x)

F(x)

G(x)

x c=

F x G x

F x G x x


Since the power per unit frequency is:

(91)

then the total power radiated over all frequencies is

(92)

P 3

820c------------------

q3B sinm

----------------------F x =

P P d0

3

820c------------------

q3B sinm

---------------------- F x d0

= =

3

820c------------------

q3B sinm

----------------------c F x xd0

=


The integral

(93)

and

(94)

Therefore,

(95)

F x xd0

x K5 3 t td

x

xd

0

8 327

--------------= =

c32---

qBm-------2 sin2=

P1

60c---------------

q4B2 sin2

m2---------------------------2=


In terms of energy ( ):

(96)

The power of appearing in this expression is why synchrotronemission from protons is not usually considered.

9.2 Other expressions for total power emitted by electrons

Take so that the total power emitted by an electron is

(97)

E mc2=

P1

60------------

q4B2 sin2

m4c5--------------------------- E2=

m4

q e=

P1

60c---------------

e4B2 sin2

m2---------------------------2=


This is often expressed in terms of the Thomson cross-section for anelectron.

The classical radius of an electron is defined by

(98)

Electrostatic potential energy Rest mass energy=

e2

40r0------------------ mec2=

r0 e2

40mec2-------------------------=

2.81815–10 m=


The Thomson cross-section is

(99)

This cross-section is important, inter alia, when considering thescattering of photons by electrons.

In terms of the power emitted by an electron is:

(100)

T83

------r02 e4

602me

2c4-------------------------- 6.65

29–10 m2= = =

T

P1

60c---------------

e4B2 sin2

m2---------------------------2 cT B2

o------

sin2 2= =


9.3 Synchrotron cooling of electronsBy conservation of energy, the rate of energy lost by the electron is

(101)

Hence, the rate of change of Lorentz factor is:

(102)

dEdt------- P– cT B2

o------

sin2 2–= =

ddt-----

1

mec2------------

dEdt-------

Tmec--------- B2

o------

sin2 2–= =


This equation can be put in the form

(103)

Let be the value of at , then assuming that and remain

constant during the time , then

(104)

1

2-----

ddt-----–

Tmec--------- B2

o------

sin2=

0 t 0= B

t

0-----

1

1 0

Tmec--------- B2

o------

sin2 t+

-----------------------------------------------------------------=


This defines a synchrotron cooling time:

(105)

Features

• decreases with increasing – the higher energy electrons

cool the fastest• decreases with increasing magnetic field.

tsyn

mec

T--------- B2 sin2

0--------------------- 1–

01–=

tsyn

tsyn


Example

Take typical values for the lobes of a radio galaxy:

(106)

(107)

Estimates of the ages of radio galaxies often exceed years.Hence there is a need to re-energise particles in the lobes of radiogalaxies. This introduces the need for particle acceleration.

B 10G 1 nT= = 0 104=

tsyn9.11

31–10 3810

6.6529–10

--------------------------------------------------1

9–10 2

4 10 7–--------------------------

1–

01–=

1.6710 yrs

sin2

104----------------- 1–

=

108


10 Appendix: Evaluation of synchrotron integrals

The evaluation of the synchrotron integrals follows from the result(Abramowitz and Stegun, eqn. 10.4.32)

(108)

where is the Airy function. In our case,

(109)

I a x ay3 xy+ cos yd0

3a 1 3/– Ai 3a 1 3/– x = =

Ai z

x32---= a

12---=


Differentiating, with respect to gives:

(110)

so that

(111)

I a x x

x

I a x y ay3 xy+ sin yd0

–=

3a 2 3/– Ai 3a 1 3/– x –=

y ay3 xy+ sin yd0

3a 2 3/– Ai 3a 1 3/– x =


Furthermore, the Airy function and its derivative are related to themodified Bessel functions of order and via:

(112)

See equations (10.4.26) and (10.4.31) of Abramowitz and Stegun.

For and , the argument of the Airy function andits derivative is

. (113)

1 3 2 3

K1 3 3z--- 1 2/

Ai z = K2 3 31 2/

z-----------Ai z –=

z32--- 2 3/

=

x 3 2= a 2=

z 3a 1 3/– x32

------ 1 3/– 3

2------ 3

2------ 2 3/

= = =


Hence, the argument, , of the corresponding Bessel functions is .Hence,

(114)

ay3 xy+ cos yd0

3a 1 3/– Ai 3a 1 3/– x =

32

------ 1 3/–

Ai32

------ 2 3/

=

32

------ 1 3/–

=

1– 32

------ 1 3/

3 1 2/– K1 3

1

3-------K1 3 =


and

(115)

y ay3 xy+ sin yd0

3a 2 3/– Ai 3a 1 3/– x =

32

------ 2 3/–

Ai 32

------ 2 3/

=

32

------ 2 3/–

3 1 2/– 32

------ 2 3/

K2 3– =

1

3-------K

2 3 –=


Finally, the integrals from to are twice the integrals from to:

(116)

– 0

12---y3 3

2---y+cos yd

–

2

3-------K1 3 =

y12---y3 3

2---y+

sin yd0

1

3-------K

2 3 –=


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