Äîêóìåíò âçÿò èç êýøà ïîèñêîâîé ìàøèíû. Àäðåñ îðèãèíàëüíîãî äîêóìåíòà : http://www.mrao.cam.ac.uk/~kjbg1/lectures/lect1_2.pdf
Äàòà èçìåíåíèÿ: Fri Oct 3 17:47:33 2003
Äàòà èíäåêñèðîâàíèÿ: Tue Oct 2 03:59:47 2012
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Ïîèñêîâûå ñëîâà: ï ï ï ï ï ï ï ï ï ï ï ï ï ï ï ï ï ï ï ï ï ï ï ï ï ï ï ï ï ï ï ï ï ï ï ð ð ð
Polarisation
Some emission processes are intrinsically polarised e.g. synchrotron radiation.

circularly polarised emission

Scattering processes can either increase or decrease the amount of polarisation. Magnetic fields can rotate the plane of polarisation through the Faraday effect. We can characterise a source's polarisation through its Stoke's Parameters: I, Q, U and V.

?? ?Å?Å???Å

??

? ÅÅ?Å

?



B

e

linearly polarised emission


Stoke's Parameters: I, Q, U and V
Consider 4 different filters, each of which under natural unpolarised illumination will transmit half the incident light.

polarisation. - "resemblance" to linear polarisation orientated in direction of . - "resemblance" to right-handed circular polarisation.

for completely unpolarised radiation.

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Beam under investigation (Flux = 2 S0 )
Filter transmits indicated polarisation

S

0

S

1

S

2

S

3

Transmitted fluxes
- the total flux - "resemblance" to horizontal linear

for completely polarised radiation.

is the degree of polarisation.


Summary of Key Points
ba `Y XW V UT S R QP I HG F ? ` " t s 5G p
Flux density, measured in , is power per unit area per unit frequency,

Specific intensity, is flux density per unit solid angle. It is independent of distance.

For uniform sources where is often either or for compact or extended sources respectively. A source's polarisation can be characterised by its Stoke's parameters. gives the total flux. and give the degree of linear polarisation, the degree of circular polarisation.

s



Qs G p



Gr

q

q 5G p

For blackbody sources

ih gV f ed XW V
(Planck function).

c

GF



b y xw vu t s

E E E E

E


Blackbody Radiation
Thermodynamics
We can characterise the radiation from a blackbody -- one in which matter and radiation are in thermodynamic equilibrium -- from thermodynamic arguments alone. Adkins (Equilibrium Thermodynamics, CUP) gives a thorough treatment. Treating the radiation as a gas of photons, we see that the pressure exer ted by radiation is ; the photon , so the energy flux is just flux incident on a small area is . Consider two perfectly reflecting cavities containing radiation, each separately in equilibrium at temperature T. We make small holes in each cavity and connect them with a tube and a filter which transmits only a narrow range of radiation at frequency . The flux of radiation leaking out of A into B via the tube is then , where is the energy density per unit wavelength in the radiation field in A in the direction of the tube. The second law of thermodynamics prohibits any net energy flow between two bodies at equal temperatures, so we conclude that .

sh

This quantify ume.

must also be isotropic and independent of vol-

Dg if h



"m lf kd j Dg "f be d

Xrs h



"m f

ixs h

q



t

wv ut d Xrs h

rs h

pm of nd h q


If we now consider squeezing an isothermal piston containing radiation, of total energy , we can apply the first law to obtain so that



z

?

w

z

z

{

which integrates to give

. The radiative flux is then

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|

~

{

z

z i| {

~ }| {

z



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} | z {

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z

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Full expressions for the radiant intensity
A perfectly reflecting cubic cavity at temperature and of side contains equilibrium radiation. The wave modes (using travelling wave boundary conditions) must be of the form

Assuming two independent polarisation states for the photons (either or thogonal plane polarisations or opposite circular polarisations) we obtain the spectra density of available photon states:

We multiply this by the mean energy per state and divide by the volume to obtain the spectral energy density per unit

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The density of states

per volume of k-space is thus for a total cavity volume . This can written in terms of the density of states in a given direction of a given wavenumber :

?

É

?

É ?

with integer values of

.

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Î

È

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Ê


solid angle introduced earlier. Remember that the occupancy number for photons is , so that

The total flux from the surface of a blackbody is then

The prefactor is the Stefan-Boltzman constant, . We can also now find the energy density and hence the total energy density by per unit frequency using the relationship :

?

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ò

?

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ð

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Thus the brightness

of blackbody radiation is given by

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ä â

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?ò ì Øñ Ü ï ?í ý ò ð

üò ì xù Ü

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ï ú  û b? X? í U? ë


Describing Polarisation
Monochromatic waves
The most general form for a pure monochromatic wave is one of elliptical polarisation, formed by taking two plane waves at right angles with a phase difference. Thus three parameters specify the wave -- two electric field amplitudes and the phase difference.

Thus a pure monochromatic wave is always polarised.

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)" (%

?? ? ? ?

' & ?? ? % ?

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Stokes Parameters
We conventionally use the four Stokes Parameters to describe the polarisation:

It follows from these definitions that Thus there are still only three free parameters here. Stokes

is propor tional to the total flux in the wave.

Stokes measures circularity, setting the axis ratio of the ellipse. for linear polarisation, for right handed elliptical polarisation. The remaining parameters Stokes or measure the orientation of the ellipse. for circular polarisation.

6W

86 E

86 B

@ VU T6 S

@ eU T6 S

f

E

2 61

g

&@ 6 D7 64 6 C4 3 A@ 6 98 7 64 6 54 3

D 7 CS dR cb a` YQ 6 7 4 X4 CF 3 D 7 CS R Q ?P IH 6 7 4 G4 CF 3

W

B

f

2

E

2

2

21

B

B

2

2

W

E

f

2

W

1

W

0 0 0 0 0

.


Quasi-monochromatic waves
In general the amplitudes and , and phases and , are functions of time. The wave is no longer monochromatic, but if the observing bandwidth is small it makes sense to talk about quasi-monochromatic waves. There is now the possibility that there is an unpolarised component in the wave. A completely unpolarised wave will have since the time averages in the definitions of and go to zero, and the field amplitudes in each direction must be on average equal.

and one completely unpolarised:

qv

qu

y

qs



The degree of polarisation total intensity:

is the ratio of the polarised to

v u s y x

One can usefully decompose a wave components, one completely polarised component

into two

t

v

v

Ip r

t

u

u



t

v u s q v

w



s

w



w



qi q

v

qu

pi

q u

qs

q s

t



y x

x

qr h h h

h

w


Faraday Rotation
Faraday Rotation: Linearly polarised light which passes through a plasma which contains a uniform magnetic field has its position angle rotated. The modes of propogation of radio waves in a magnetised plasma are right- and left-handed elliptically polarised.

the gyro-frequency.

The phase velocities of the two modes are different and one sence of polarisation runs ahead of the other. Linearly polarised light can be considered as a superposition of equal components of right- and left-handed elliptically polarised light. On propagating through the plasma the different phase velocities result in a phase difference, between the two modes. This is equivalent to linearly polarised light with plane of polarisation rotated by .

z

u



is the plasma frequency.



The two modes experience different refractive indeces, given by

ng

s Xr ?q po k jg Xi )n Cg f

lk jg &i h Cg f



j i



m

ed



s



w G z v {z yx wv u ~} | t

I



hg



is


is know as the rotation measure.





in radians, in metres, in tesla, inparsecs.

in par ticles per cubic metre,





X





l Å ?? ?




Antenna Temperature
The flux density observed by a telescope is often ex. The pressed in terms of an Antenna Temperature, indicates that a correction has been applied for atmospheric opacity. is the equivalent temperature of the power received at an antenna from a source. Hence, for a uniform source which fills the telescope beam (i.e. an extended source):

- the antenna temperature is equal to the source's brightness temperature. - for a source with a thermal spectrum, is (of course) independent of observing frequency. - observing a thermal source with the same telescope at different frequencies, the received flux density is the same, because the flux density is propor tional to both the intensity ( ) and the beam solid angle ( ). (This is only strictly true if the the aper ture efficiencies are the same at the two frequencies).

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?? ?

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Ã? ÃÂ ?



Á

?? ?

?

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For a source that is smaller than the telescope beam (often called a point source), antenna temperature is a less useful concept, and the following are true:

ÀÛ Ú

Ù ×Ø Ö

- for a source with a thermal spectrum,

ÕÈ ÔÓ ÒÑ wÐ ÏÎ eÍ AÅ Ì

Ë Ê ÉÈ ÇÆ Å

Ä

- the antenna temperature observed is smaller than the source's actual brightness temperature by a factor equal to the ratio . .


Effective area of a Gaussian beam
The response, or primary beam, of a telescope falls off as a Gaussian as one moves away from its centre.

The Full Width Half Max (FWHM), is the width of the beam when the amplitude is half the maximum value.

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The effective area of the Gaussian beam is

.

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"Brightness" Question
A Red Giant lies at 20 light years from the Ear th. It has half the Sun's surface temperature and 40 times its radius. A White Dwarf lies at 10 light years from the Ear th. It has double the Sun's surface temperature and 1/40 times its radius. Assuming that all three radiate as black bodies, which of the Sun, the Red Giant or the White Dwarf have the highest luminosity?

? ? ?

surface brightness at 30 GHz? flux density at 30 GHz at the Ear th?


bB a@ T` RQ P? ? YI X 8 W VU 5C % ?B A@ TS RQ P? ? &I H 8 GF ED 5C % ?B A@ )9 ?? ? 8 2 "
The Sun has the highest flux density.

0

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? ? ?

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Answer

;



?? ?? ? ? Å
?

e ?? ? ?

( at 30 GHz)

The White Dwarf has the highest surface brightness. The Red Giant is most luminous (we are in the R-J region)

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