Äîêóìåíò âçÿò èç êýøà ïîèñêîâîé ìàøèíû. Àäðåñ îðèãèíàëüíîãî äîêóìåíòà : http://www.mso.anu.edu.au/~geoff/HEA/1_Radiation_field.pdf Äàòà èçìåíåíèÿ: Thu Mar 21 15:11:49 2013 Äàòà èíäåêñèðîâàíèÿ: Fri Feb 28 01:00:52 2014 Êîäèðîâêà: Ïîèñêîâûå ñëîâà: m 101

Description of the Radiation Field
Based on: Chapter 1 of Rybicki & Lightman, Radiative Processes in Astrophysics, and Chapter 12 of Shu, Radiation.

High Energy Astrophysics: Radiation Field

© G.V. Bicknell


1 Introduction We see astrophysical objects in different wavebands from radio through to optical to X-ray and gamma-ray because of the radiation they emit, which propagates, both through the source and the intervening medium. Therefore, it is important to understand the properties of the radiation field and the manner in which it is described. One of the most important diagnostics of radiation from an astrophysical source is that afforded by polarisation.

High Energy Astrophysics: Radiation Field

2/112


For example, the direction of the magnetic field in a synchrotron emitting source is perpendicular to the direction of the E -vector of the predominantly linear polarised radiation. Polarisation is also affected by the medium through which transverse waves travel (Faraday rotation). Hence it is important to have a sound theoretical basis from which to discuss polarisation.

High Energy Astrophysics: Radiation Field

3/112


2 Specific intensity and flux 2.1 Definition of specific intensity n d n d dA dA = dA cos

dA

High Energy Astrophysics: Radiation Field

4/112


Specific intensity is defined by:
Solid angle

Electromagnetic energy passing through surface normal = I d dAdtd to surface within elementary solid angle Specific
intensity

(1)

High Energy Astrophysics: Radiation Field

5/112


In terms of circular frequency : Electromagnetic energy passing through surface normal = I d dAdtd to surface
(2)

We often require the energy flowing through dA at an angle to the normal. We can derive the relevant expression in the following way. Consider the elementary surface dA which is · normal to the ray, and · the projection of dA (see the above figure)
High Energy Astrophysics: Radiation Field 6/112


Then dA = dA cos and Electromagnetic energy passing through surface = I cos d dAdtd (4) at an angle to surface We put I = I
High Energy Astrophysics: Radiation Field

(3)

(5)

7/112


to emphasize the fact that the specific intensity can vary with direction with respect to the normal n . Units of I : We have Joules = I frequency area time solid angle (6) so that the units of I are: Watts m ­ 2 Hz ­ 1 Str ­ 1
(7)

High Energy Astrophysics: Radiation Field

8/112


or ergs s ­ 1 cm ­ 2 Hz ­ 1 Sr ­ 1
(8)

2.2 Flux density The flux density, F , is the power per unit area of the radiation field. It is therefore defined by dF = I cos d
(9)

High Energy Astrophysics: Radiation Field

9/112


The flow of energy per unit area per unit time per unit frequency through a surface with normal n is given by: F Wm ­ 2 Hz ­ 1 = I cos d
(10)

The density in this case refers to the "per unit frequency" part. More about this below.

High Energy Astrophysics: Radiation Field

10/112


2.3 Relationship of flux density to power at telescope
Source of cosmic radiation

n n

I

Bundle of rays from source

High Energy Astrophysics: Radiation Field

11/112


The solid angle of the bundle of rays is usually defined by the resolving power of the telescope. The power per unit area received by the telescope is I cos d . Frequently in radio astronomy, you will hear people refer to the flux density of a source.


High Energy Astrophysics: Radiation Field

12/112


2.4 Surface brightness For a small region of solid angle normal to the direction in which a telescope is pointing, we have: n

dA

F F = I I = --------
High Energy Astrophysics: Radiation Field

(11)

13/112


Hence, the specific intensity is the flux received per unit of telescope area per unit of solid angle. For this reason, some astronomers, and particularly radio astronomers, refer to the specific intensity as surface brightness. This equation is also frequently used to estimate surface brightness of a source from an image when the image is represented in terms of flux density per beam. More about this later.

High Energy Astrophysics: Radiation Field

14/112


Units of flux density Often, especially in radio astronomy, we use the unit of a Jansky (after one of the discoveries of cosmic radio radiation) 1 Jansky (Jy) = 10 ­ 26 Wm ­ 2 Hz ­ 1
(12)

2.5 Momentum flux density Since the energy passing through a surface in a given direction is dE = I cos d dtd dA
High Energy Astrophysics: Radiation Field

(13)

15/112


then the magnitude of momentum passing through the surface in the same direction is I dE dp = --------- = ---- cos d dtd dA c c
(14)

However, momentum is a vector quantity. The component of momentum in the direction of the normal is I dp cos = ---- cos2 d dtd dA c
(15)

High Energy Astrophysics: Radiation Field

16/112


The momentum flux density, , is the momentum per unit time per unit frequency per unit area passing in all directions through the surface, so that 1 = -- I cos2 d c
(16)

High Energy Astrophysics: Radiation Field

17/112


2.6 The cies

Integration over frequency total flux of energy per unit area between two frequen 1 and 2 is just the integral of the flux density be2 Total flux = F Wm ­ 2 = F d 1

tween these limits.
(17)

High Energy Astrophysics: Radiation Field

18/112


Pressure is force per unit area, or equivalently, momentum flux density per unit area, so that the total radiation pressure on the surface can be found by integrating the momentum flux density over frequency.
2 Pressure = p Nm ­ 2 = d 1 (18)

The total intensity is just the integral of the specific intensity over frequency.
2 Intensity = I Wm ­ 2 Str ­ 1 = I d 1
High Energy Astrophysics: Radiation Field

(19)

19/112


In particular the total flux is the frequency-integrated flux density. Sometimes people loosely refer to flux density as flux and I have seen pedantic thesis examiners insist on such carelessness being expunged from a thesis before it can be accepted. 2.7 Radiation energy density Define Energy in volume dV u dVd d = solid angle d and frequency interval dv
High Energy Astrophysics: Radiation Field

(20)

20/112


Consider a cylinder of length cdt , cross-sectional area dA dA d cdt

High Energy Astrophysics: Radiation Field

21/112


The energy of radiation in cylinder within a cone of solid angle d is: dE = u d dVd = u d dAcdtd = cu d dAdtd
(21)

High Energy Astrophysics: Radiation Field

22/112


All of the radiation within the cylinder passes through dA in the time dt . Hence, cu d dAdtd = I d dAdtd cu = I I u = ---c

(22)

High Energy Astrophysics: Radiation Field

23/112


Therefore, the total energy density at a point within the volume is given by: 1 4 u = -- I d = ----- J c c
4

where

1 J = ----- I d = Mean intensity 4
4

(23)

High Energy Astrophysics: Radiation Field

24/112


2.8 Constancy of specific intensity in free space d 1 d 2

Ray
dA 1 dA 2 d 1 = --------R2 R dA 2 dA 1 d 2 = --------R2
(24)
25/112

High Energy Astrophysics: Radiation Field


Consider the energy flux through an elementary surface dA 1 within solid angle d 1 . Consider all of the photons which pass through dA 1 in this direction which then pass through dA 2 . We take the solid angle of these photons to be d 2 . By conservation of energy the energy passing through both surfaces within the corresponding solid angles is identical. Hence,
1 2 dE = I dA 1 dtd 1 d 1 = I dA 2 dtd 2 d 2 (25)

High Energy Astrophysics: Radiation Field

26/112


The solid angles are given by: dA 2 d 1 = --------R2 This implies that: dA 1 dA 2 d 1 dA 1 = d 2 dA 2 = ------------------R2 Since d 1 = d 2 , then
1 2 I = I = constant along ray
High Energy Astrophysics: Radiation Field

dA 1 d 2 = --------R2

(26)

(27)

(28)
27/112


2.9 Spontaneous emission Various emission processes along a ray contribute to the specific intensity. The emissivity, in principle, is angle-dependent, e.g. synchrotron emission depends upon the angle between the emission direction and the magnetic field. The emissivity is defined by: Energy radiated from volume dV (29) in time dt into freqency interval d = dE = j dVdtd d into solid angle d
Emissivity

High Energy Astrophysics: Radiation Field

28/112


If the emissivity is isotropic, then 1 j = ----- P 4
Radiated power per unit volume (30)

The emission may be considered to be isotropic for two reasons: 1. The emission mechanism is independent of direction. 2.The emission may be considered to be the random superposition of a number of anisotropic emitters, e.g. synchrotron emission from a tangled magnetic field.

High Energy Astrophysics: Radiation Field

29/112


Effect on specific intensity ds

dA d

Energy added to beam from emission from within dV = dAds is given by: dE = j dVd dtd = j dAdsd dtd
High Energy Astrophysics: Radiation Field

(31)
30/112


This is radiated into the solid angle d emerging from dA so that the change in specific intensity is given by: dI dAd dtd = j dAdsd dtd dI ------- = j ds
(32)

High Energy Astrophysics: Radiation Field

31/112


2.10 Absorption Often absorption is presented in the following form: dI ------- = ­ I ds
Coefficient of absorption (33)

and we shall see examples of this later on.

High Energy Astrophysics: Radiation Field

32/112


3 The radiative transfer equation 3.1 The fundamental equation Putting emission and absorption into the one equation, we have dI ------- = j ­ I ds Note from This tions
(34)

that the emissivity can include scattering of photons other directions into the direction being considered. is what makes the solution of radiative transfer equaa challenging problem in general.
33/112

High Energy Astrophysics: Radiation Field


3.2 Solution for emission only (optically thin emission) I 0 s=0
Emitting (and absorbing) region

Emergent intensity I Ray Free space s = s1

High Energy Astrophysics: Radiation Field

34/112


dI s ------- = j I s = I 0 + j s ds ds 0
s1 = I 0 + j s ds 0

(35)

In many cases: · I 0 = 0 · We approximate the medium by one with constant properties. This gives, I = j s1
High Energy Astrophysics: Radiation Field

(36)

35/112


Once outside the emitting region, the specific intensity is constant, unless another emitting or absorbing region is encountered. 3.3 Absorption only dI ------- = ­ I ds I s = I s 0 exp ­ s ds 0
s

(37)

High Energy Astrophysics: Radiation Field

36/112


This introduces the optical depth between 0 and s s = s ds 0
­ s I s = I 0 e s (38)

In terms of , the specific intensity along a ray is given by:
(39)

3.4 Both emission and absorption The differential form of the optical depth is d = ds
High Energy Astrophysics: Radiation Field

(40)
37/112


We can write the radiative transfer equation in the form: dI ------------ = S ­ I ds i.e. where j S = ------ = Source function
(42)

dI ------- = S ­ I d

(41)

High Energy Astrophysics: Radiation Field

38/112


Write the transfer equation as dI -------- + I = S d
The integrating factor is e so that d -------- e I = S e e I = I 0 + e S d d 0 ­ ­ ­ I = e I 0 + e S d 0 (44) (43)

High Energy Astrophysics: Radiation Field

39/112


The optical depth delineates the region which contributes most significantly to the intensity of an emerging ray. To see this, consider
s ­ = s ds ­ s ds 0 0 s s

= s ds s = s s

(45)

High Energy Astrophysics: Radiation Field

40/112


I 0 s=0

s s
Free

Emergent inten-

I

Ray

This part gives s s
Emitting (and absorbing) region

It is obvious that the dominant contribution to the integral comes from regions wherein the optical depth, s s 1 .
High Energy Astrophysics: Radiation Field 41/112


3.5 Relationship between flux and luminosity
Telescope



I

The flux density received at the telescope is given by: F =
source



I cos d I d


(46)

(We put 0 because for a distant source, all rays differ very little in their direction.)
High Energy Astrophysics: Radiation Field 42/112


ds dA

dV = dA ds

Let dA be the cross-sectional area of the bundle of light rays at the source. For a distant source (distance R ), the element of solid angle is given by dA d = -----R2
High Energy Astrophysics: Radiation Field

(47)
43/112


where R is approximately the same for all parts of the source and dA is the cross-sectional area of a bundle of light rays as shown. Hence, 1 F ----- I dA R 2 For optically thin radiation I = j ds
ray (48)

1 1 F = ----- j dA ds = ----- j dV R 2 Source R 2 Source
High Energy Astrophysics: Radiation Field

(49)

44/112


where V is the volume (element dV ). The equation 1 1 F = ----- j dV = ----- Volume integrated emissivity (50) R 2 Source R2 shows the origin of the inverse square law for flux density.

High Energy Astrophysics: Radiation Field

45/112


3.6 Isotropic emissivity If the emissivity is isotropic 1 Total power emitted per unit volume j = ----- P = --------------------------------------------------------------------------------------4 4 solid angle 1 F = -----------4R2 where L = Monochromatic luminosity = Luminosity per unit frequency
High Energy Astrophysics: Radiation Field

L P dV = ------------4R2 Source

(51)

(52)

46/112


3.7 Calculation of luminosity Knowing the flux density, one can calculate the monochromatic luminosity, from L = 4 R 2 F
2 F d Total luminosity = L tot = 4 R 0 (53) (54)

3.8 Example: The luminosity of a radio source A typical extragalactic radio source would have a flux density at 1.4 GHz of a Jansky at a redshift of 0.1.
High Energy Astrophysics: Radiation Field 47/112


The distance (for small redshifts) is 300 000 0.1 cz D = ------ = ---------------------------------- Mpc 70 H0 430 Mpc so that the monochromatic luminosity is L = 4 430 3.1 10 10 ­ 26 WHz ­ 1 2.2 10 WHz ­ 1
25 22 2 (56) (55)

High Energy Astrophysics: Radiation Field

48/112


F ­ log F
Typical spectrum of an extragalactic radio source

log

u = 10 ­ 100 GHz
High Energy Astrophysics: Radiation Field 49/112


Typically such a source has a spectral index of = 0.7 between a generally undefined lower frequency, l , and an upper cutoff frequency u = 10 ­ 100 GHz , so that the total luminosity,

High Energy Astrophysics: Radiation Field

50/112


u 0 ­ u ­ ----- d ----- L tot L ----d = L 0 l l 0 0 0 0 0 0

L 0 0 ----- 1 ­ u = -------------1 ­ 0 l L 0 1 ­ 0 u -------------- ----- -1 ­ 0

0 (57) 0

High Energy Astrophysics: Radiation Field

51/112


Thus, for the parameters of our source,
35 2.2 10 1.4 10 10 0.3 -----L tot ----------------------------------------------- 2.7 10 W 1.4 0.3 (58) 25 9

7 10 L sun

8

High Energy Astrophysics: Radiation Field

52/112


4 Polarisation 4.1 Monochromatic plane wave Plane wave solutions of Maxwell's equations: E = E 0 exp i t ­ k x B = B 0 exp i t ­ k x = circular frequency = ck k = wave number = k n E 0 = amplitude of Electric field k E0 B 0 = Amplitude of magnetic field = --------------- = c n E 0 B0 n = E0 n = 0
High Energy Astrophysics: Radiation Field 53/112

(59)


y

B0

The electric vector deterE0 mines all of the parameters of the wave. Since there x Direction are two independent comn of propa- ponents of E 0 there are Plane of electric & In general, we put magnetic field E 0 = E 0 1 e 1 + E 0 2 e 2 two modes of polarisation.
(60)

where e 1 and e 2 are the unit vectors in the x and y directions.
High Energy Astrophysics: Radiation Field 54/112


Electromagnetic waves, far from the point of origin can be considered to be locally plane. Electric vector at a point in space Consider a wave at the location x = x 0 . In component form the electric field of the wave may be written
i i t ­ k x0 i t e i ­ k x0 E = A e e = A e i t ­ = A e (61)

High Energy Astrophysics: Radiation Field

55/112


Hence the real part of this wave may be expressed as E = A cos t ­ = 1 2
(62)

The parameters are the phases of the two modes. They are not both arbitrary since the origin of time is arbitrary. However, the difference = 2 ­ 1 is arbitrary.

High Energy Astrophysics: Radiation Field

56/112


4.2 The polarisation ellipse e2 e2 E e 1 Definition of axes for polarisae1

tion ellipse. is the angle of rotation from the arbitrary axes to the principal axes (primed).

Consider a general elliptically polarised monochromatic wave. The electric vector is given by: E = A 1 cos t ­ 1 e 1 + A 2 cos t ­ 2 e 2
High Energy Astrophysics: Radiation Field

(63)
57/112


N.B. The phases of both component are not free parameters since one phase can be adjusted by a change of the time origin. However, the relative phase 2 ­ 1 is arbitrary. 4.2.1 What we are aiming for Write the electric field in axes corresponding to the principal axes of the ellipse: E = E 1 cos t e 1 + E 2 sin t e 2
(64)

High Energy Astrophysics: Radiation Field

58/112


e2
E 2 = E 0 sin

e2 E

e1

Polarisation ellipse

e1
E 1 = E 0 cos

Relationship between parameters

High Energy Astrophysics: Radiation Field

59/112


The aim of the following is to determine the parameters E 1 and E 2 in terms of A 1 A 2 2 ­ 1 and . To do so, we use the relations between primed and unprimed unit vectors: e1 cos sin e 1 = e2 ­ sin cos e 2
(65)

High Energy Astrophysics: Radiation Field

60/112


Then write the electric field in the principal axis system in terms of the electric field in the arbitrary axes:

High Energy Astrophysics: Radiation Field

61/112


E = E 1 cos t cos e 1 + sin e 2 + E 2 sin t ­ sin e 1 + cos e 2 = E 1 cos cos t ­ E 2 sin sin t e + E 1 sin cos t + E 2 = A 1 cos t ­ 1 e 1 + = A 1 cos t cos 1 + A

1 (66)

cos sin t e 2 A 2 cos t ­ 2 e 2 sin t sin 1 e 1 1

+ A 2 cos t cos 2 + A 2 sin t sin 2 e 2

High Energy Astrophysics: Radiation Field

62/112


Equate the coefficients of e 1 e 2 and sin t cos t within those terms. A 1 cos 1 = E 1 cos A 1 sin 1 = ­ E 2 sin A 2 cos 2 = E 1 sin A 2 sin 2 = E 2 cos

(67)

High Energy Astrophysics: Radiation Field

63/112


So far, so good, but this is not the best form in which to describe the relationship between these coefficients. The following quadratic relationships are easy to verify:
2 2 2 A 1 = E 1 cos2 + E 2 sin2 2 2 2 A 2 = E 1 sin2 + E 2 cos2 (68)

High Energy Astrophysics: Radiation Field

64/112


together with:
2 A 1 A 2 cos 1 cos 2 = E 1 sin cos 2 A 1 A 2 sin 1 sin 2 = ­ E 2 sin cos (69)

A 1 A 2 sin 1 cos 2 = ­ E 1 E 2 sin2 A 1 A 2 cos 1 sin 2 = E 1 E 2 cos2

High Energy Astrophysics: Radiation Field

65/112


We now form the following combinations of the above:
2 2 2 2 A1 + A2 = E1 + E2 2 2 2 2 A 1 ­ A 2 = E 1 ­ E 2 cos2 ­ sin2 2 2 = E 1 ­ E 2 cos 2 (70)

High Energy Astrophysics: Radiation Field

66/112


2 2 A 1 A 2 cos 2 ­ 1 = E 1 ­ E 2 sin cos 2 2 E1 ­ E2 = ----------------------- sin 2 2 A 1 A 2 sin 2 ­ 1 = E 1 E 2 (71)

High Energy Astrophysics: Radiation Field

67/112


Now define the additional angle , the parameter t and the phase difference by: E 1 = E 0 cos E1 t = ----- = cot E2 2 ­ 1 = E 2 = E 0 sin
(72)

= Ratio of semi-major to semi-minor axis

High Energy Astrophysics: Radiation Field

68/112


The above quadratic relations become:
2 2 2 E0 = A1 + A2 2 2 2 E 0 cos 2 cos 2 = A 1 ­ A 2 2 E 0 cos 2 sin 2 = 2 A 1 A 2 cos 2 E 0 sin 2 = 2 A 1 A 2 sin (73)

So the amplitudes A 1 A 2 and the phase difference define the parameters E 0 and .
High Energy Astrophysics: Radiation Field 69/112


e2
E 2 = E 0 sin

e2 E

e1

Polarisation ellipse

e1
E 1 = E 0 cos

Relationship between parameters

High Energy Astrophysics: Radiation Field

70/112


4.3 Interdependence of parameters Note that the 4 above equations are not independent. Take the last 3 equations; the sum of the squares of the left-hand
4 sides is E 0 . The sum of the squares of the right hand sides is 2 2 22 A 1 ­ A 2 2 + 4 A 1 A 2 cos2 + sin2 2 2 22 = A1 ­ A2 2 + 4 A1 A2 2 2 = A1 + A2 2 (74)

High Energy Astrophysics: Radiation Field

71/112


So the sum of the squares of these 3 equations gives:
4 2 2 E0 = A1 + A2 2 2 2 2 E0 = A1 + A2 (75)

which of course is the first equation.

High Energy Astrophysics: Radiation Field

72/112


5 The Stokes parameters ­ definitions c 0 c 0 2 -2 -2 I = ------- E 0 = ------- A 1 + A 2 2 2 c 0 c 0 2 -2 -2 Q = ------- E 0 cos 2 cos 2 = ------- A 1 ­ A 2 2 2 c 0 c 0 U = ------- E 0 cos 2 sin 2 = ------- 2 A 1 A 2 cos -2 2 2 c 0 c 0 V = ------- E 0 sin 2 = ------- 2 A 1 A 2 sin -2 2 2
High Energy Astrophysics: Radiation Field

(76)

73/112


(The reason for the factor of c 0 2 is the relation to the Poynting flux in the following.) These equations can also be expressed in the form: Q = I cos 2 cos 2 U = I cos 2 sin 2 V = I sin 2 Squaring each equation and adding: I2 = Q2 + U2 + V2
(78) (77)

High Energy Astrophysics: Radiation Field

74/112


Note the close correspondence between these equations for I 2 2 and defining these angles as polar coordinates. This correspondence is exploited when we discuss the Poincare sphere.

High Energy Astrophysics: Radiation Field

75/112


5.1 Fractional polarisation We define the fractional polarisation of a wave by: Q q = --- = cos 2 cos 2 I U u = --- = cos 2 sin 2 I V v = -- = sin 2 I

(79)

High Energy Astrophysics: Radiation Field

76/112


The direction of the principal axis is therefore given by: u U tan 2 = -- = --q Q
(80)

High Energy Astrophysics: Radiation Field

77/112


5.2 The Poincare sphere Polarised light can be represented v q P u v 2 u q 2

in terms of the Poincare = cos 2 cos 2 = cos 2 sin 2 (81) = sin 2

High Energy Astrophysics: Radiation Field

78/112


sphere. The Stokes parameters for fractional polarisation can be represented in terms of the parameters 2 and 2 as polar angles. The Poincare sphere makes it easy to determine the relevant ranges of and . From the diagram it is obvious that 0 2 2 ­ -- 2 2 0 -­ -- -2 4 4
(82)

Physically, the reason for this is as follows: 1.Rotation of an ellipse by and + give the same
High Energy Astrophysics: Radiation Field 79/112


ellipse. 2. Recall the definition of E 1 = E 0 cos E1 t = ----- = cot E2 E 2 = E 0 sin E2 t ­ 1 = ----- = tan E1

(83)

When varies between 4 , E 2 varies between E 1 . This is the appropriate range for the semi-minor axis.

High Energy Astrophysics: Radiation Field

80/112


5.3 Linear and circular polarisation 5.3.1 Linear polarisation If the phase difference between the two components in the arbitrary reference system is = 0 , then V = c 0 A 1 A 2 sin = 0 sin 2 = 0 2 = 0 This implies that E 1 = E 0 cos = E 0 E 2 = E 0 sin = 0
(85)
81/112

(84)

The only value of in the appropriate range is = 0 .

High Energy Astrophysics: Radiation Field


Hence the electric field is: E = E 0 cos t e 1
(86)

i.e. the electric vector oscillates in one direction ­ hence the name linear polarisation. 5.3.2 Circular polarisation A purely circularly polarised wave is defined by equal amplitudes of the two components, differing in phase by 2 , i.e. A1 = A2 = 2
(87)

High Energy Astrophysics: Radiation Field

82/112


Since, c 0 2 -2 Q = ------- A 1 ­ A 2 2 c 0 U = ------- 2 A 1 A 2 cos 2 then q=Q=0 u=U=0
(89)

(88)

High Energy Astrophysics: Radiation Field

83/112


The only remaining Stokes parameter in this case is: c 0 V = ------- 2 A 1 A 2 cos 2 c 0 ------- 2 A 1 A 2 sin 2 v = --------------------------------------------- = 1 c 0 2 ------- A 1 + A 2 -2 2
(90)

High Energy Astrophysics: Radiation Field

84/112


The equations defining and are: cos 2 cos 2 = 0 cos 2 sin 2 = 0 sin 2 = 1 The solution for this is 2 = -- = -2 4 and = arbitrary
High Energy Astrophysics: Radiation Field

(91)

(92)

(93)

85/112


Hence E2 = E1 and the two waves are: E = E 0 cos t e 1 + E 0 sin t e 2 E = E 0 cos t e 1 ­ E 0 sin t e 2
(95) (94)

The first solution represents a vector moving anti-clockwise in a circle as seen by an observer facing the wave ­ This is known as left circularly polarised or positive helicity.
High Energy Astrophysics: Radiation Field 86/112


The second represents a vector moving clockwise in a circle as seen by an observer facing the wave ­ This is known as right circularly polarised or negative helicity. 5.3.3 General elliptical polarisation In the general case when q u v 0 E = E 0 cos cos t e 1 + E 0 sin sin t e 2
(96) (97)

When v 0 and consequently 0 then E rotates anticlockwise (since cos 0 and sin 0 ) and the wave is left-polarised.
High Energy Astrophysics: Radiation Field 87/112


When v 0 and consequently 0 , E rotates clockwise (since cos 0 and sin 0 ) and the wave is right-polarised. 5.3.4 Direction of the major axis Take q = cos 2 cos 2 u = cos 2 sin 2 v = sin 2
(98)

High Energy Astrophysics: Radiation Field

88/112


then it is clear that 1 ­1 u u tan 2 = -- = -- tan -2 q q
(99)

In the case of linear polarisation this is the direction of the line of oscillation of the electric field.

High Energy Astrophysics: Radiation Field

89/112


5.4 The Poincare sphere revisited Left-polarised v P 2 q Right-polarised 2 u
The point P on the Poincare sphere represents the fractional polarisation of a monochromatic wave. The North and South poles of the sphere, v = 1 and v = ­ 1 respectively, represent lefthanded and right-handed circular polarisation.
90/112

Pure linear polarisation

High Energy Astrophysics: Radiation Field


5.5 Relationship to the Poynting flux The Poynting flux is EB S = ------------0
(100)

For a transverse wave with wave-vector k and normal k n = -- : k kE EB E2 B = ------------ ------------- = ---------- n 0 0
High Energy Astrophysics: Radiation Field

(101)

91/112


With = ck , E2 2 2 S = -------- n = c 0 E 2 n = c 0 E 1 cos2 t + E 2 sin2 t n (102) 0 c Averaged over a period, the Poynting flux is: S = c 0
2 2 2 E1 E2 c 0 E0 ------ + ------ = --------------2 2 2 (103)

The Stokes parameter I is the Poynting flux of electromagnetic energy.

High Energy Astrophysics: Radiation Field

92/112


6 Polarisation of a quasi-monochromatic wave 6.1 The electric field As before, we write the electric field as E = A 1 cos t ­ 1 e 1 + A 2 cos t ­ 2 e 2
­i 1 t i t = Re A 1 t e e e1 ­i 2 t i t + Re A 2 t e e e2 (104)

High Energy Astrophysics: Radiation Field

93/112


The previous section is concerned with the case in which the waves are purely monochromatic so that A and are constant. In the following a complex notation based on the above is used. We consider quasi-monochromatic waves for which E = E1 t e i t e1 + E2 t e i t e2
­i 1 t E1 t = A1 t e ­i 2 t E2 t = A2 t e (105)

High Energy Astrophysics: Radiation Field

94/112


where the time scale of variation of the waves is much longer than the wave period. This is relevant to the situation where the estimate of the Stokes parameters involves averages over many periods. For example, to consider a radio wave as a monochromatic wave, one would have to sample at the rate of once every 10 ­ 9 s or so. In reality, measurements at a radio telescope require integration times of about 5 minutes.

High Energy Astrophysics: Radiation Field

95/112


6.2 Stokes parameters for quasi-monochromatic waves We define the Stokes parameters as time averages with angular brackets <>:

High Energy Astrophysics: Radiation Field

96/112


c 0 c 0 2 * -2 I = ------- A 1 + A 2 = ------- E 1 E 1 2 2 c 0 c 0 2 * -2 Q = ------- A 1 ­ A 2 = ------- E 1 E 1 2 2 c 0 * U = c 0 A 1 A 2 cos = ------- E 1 2

* + E2 E2 * ­ E2 E2 * E 2 + E 1 E 2

(106)

c 0 * * V = c 0 A 1 A 2 sin = ------- E 1 E 2 ­ E 1 E 2 2i

High Energy Astrophysics: Radiation Field

97/112


In this case I 2 Q 2 + U 2 + V 2 , as we now show. Denote c 0 C = ------- , then 2
* * * * I 2 = C 2 E1 E1 2 + E2 E2 2 + 2 E1 E1 E2 E2 * * * * Q 2 = C 2 E1 E1 2 + E2 E2 2 ­ 2 E1 E1 E2 E2 (107)

* * * * U 2 = C 2 E1 E2 2 + E2 E1 2 + 2 E1 E2 E2 E1

* * * * V 2 = C 2 ­ E1 E2 2 ­ E2 E1 2 + 2 E1 E2 E2 E1

High Energy Astrophysics: Radiation Field

98/112


These imply that
* * I 2 ­ Q 2 + U 2 + V 2 = 4 C 2 [ E1 E1 E2 E2 * * ­ E1 E2 E2 E1 (108)

Since we dealing with time averages, e.g. 1 * = -- T E t E * t dt E1 E1 - 1 T0 1
(109)

High Energy Astrophysics: Radiation Field

99/112


where T is the time of integration, then, using the CauchySchwarz inequality,
* * * * E1 E1 E2 E2 E1 E2 E2 E1 (110)

implying that I2 ­ Q2 + U2 + V2 0
(111)

High Energy Astrophysics: Radiation Field

100/112


7 Superposition of independent waves Another important case to consider is where the radiation received by a detector is composed of a number of independent components - "independent" meaning that the amplitudes and phases of the components are uncorrelated. We put:

High Energy Astrophysics: Radiation Field

101/112


i E = E i
i ­i i i E = A e

= 1 2

(112)

i A = Amplitude of part of i th component i = Phase of part of i th component

High Energy Astrophysics: Radiation Field

102/112


The Stokes parameters consist of terms of the form
* E E which we can write as: * * E E = E i E j ij (113)

Now,
i j ij ­i ­ ­ i i* ij ij E E j = A A e = A A e

ij = Phase difference between the and parts of components i and j

(114)

High Energy Astrophysics: Radiation Field

103/112


The essence of independent waves is that the phase differences between them be randomly distributed over 0 2 . For this reason,
i* E E j = 0 when i j (115)

Hence,
i* i* E E j = E E i i (116)

High Energy Astrophysics: Radiation Field

104/112


and I = Ii
i

Q = Qi
i

U = Ui
i

V = Vi
i

(117)

i.e. the Stokes parameters are the sums of the Stokes parameters of the individual waves.

High Energy Astrophysics: Radiation Field

105/112


8 Partially polarised radiation 8.1 Separation into polarised and unpolarised components Consider the relations for the Stokes parameters for an arbitrary quasi-monochromatic wave:

High Energy Astrophysics: Radiation Field

106/112


* * I = C E1 E1 + E2 E2 * * Q = C E1 E1 ­ E2 E2 * * U = C E1 E2 + E2 E1 (118)

C * * V = --- E 1 E 2 ­ E 2 E 1 i If, on average, the amplitudes of the two parts of the wave are the same, then
* * Q = E1 E1 ­ E2 E2 = 0
High Energy Astrophysics: Radiation Field

(119)

107/112


Consider
* * U = E 1 E 2 + E 2 E 1 = 2 A 1 A 2 cos (120)

If the phase difference between the two components varies in such a way, that it averages to zero, then U = 0 . Similarly for V . Radiation with these properties is called unpolarised and is distinguished by: I0 Q=U=V=0
(121)

High Energy Astrophysics: Radiation Field

108/112


Similarly, if radiation is composed of a number of indeii pendent components and the phase difference 12 is randomly distributed, then we also have

Q = U = V = 0.

(122)

High Energy Astrophysics: Radiation Field

109/112


We separate EM radiation into polarised and unpolarised components as follows: I I ­ Q2 + U2 + V2 1 / 2 Q2 + U2 + V2 1 / 2 Q= 0 Q + (123) U 0 U V 0 V
Unpolarised compoPolarised component

High Energy Astrophysics: Radiation Field

110/112


The fractional polarisation is: I pol Q2 + U2 + V2 1 / 2 r = -------- = ---------------------------------------------I I
(124)

8.2 Polarisation from astrophysical sources In general, radiation from astrophysical sources is only weakly polarised ­ at the level of 1 or 2%. However, radiation from synchrotron sources can be very highly polarised ­ up to 50-70% in some cases and this is often a good indication of the presence of synchrotron emission.

High Energy Astrophysics: Radiation Field

111/112


Nevertheless, even polarisation at the level of 1 or 2% can be extremely important.

High Energy Astrophysics: Radiation Field

112/112