Polarisation Some emission processes are intrinsically polarised e.g. synchrotron radiation. B e linearly polarised emission 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.
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. Beam under investigation (Flux = 2 S ) 0 Filter transmits indicated polarisation S S S S 0 1 2 3 Transmitted fluxes the total flux resemblance to horizontal linear polarisation. resemblance to linear polarisation orientated in direction of!" #. resemblance to right-handed circular polarisation. for completely polarised radiation. $ for completely unpolarised radiation. % & ' ( *),+ is the degree of polarisation.
Summary of Key Points - Flux density,.0/, is power per unit area per unit frequency, measured in132 4 5768:9<;>=? 8:9A@ BC8ED - Specific intensity,f#/ is flux density per unit solid angle. It is independent - of distance. For blackbody sourcesf#/ G H / - (Planck function).. / G F /3I I I JLKNMPO>QSRUTWV For uniform I sources JYXT[Z<\ V 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.
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 exerted by radiation isa b cd>ef,g b hif,g isjkdlfm ; the photon flux hndofpm incident on a small area, so the energy flux is just. 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 q transmits only a narrow range of radiation at frequency. The flux of radiation leaking out of A into B via the tube is thenh:rs dutwv t q fm, whereh:rs 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 thath rs b hix s. This quantifyh s must also be isotropic and independent of volume.
If we now consider squeezing an isothermal piston containing ~ y z {} radiation, y z ~ ƒ ~ of total energy ~ {i, we z can { ~ apply the first law to obtain so that { z ˆ} ˆ Š ƒ z ˆ ƒ ˆ {Œ which integrates to give{ z Ž. The radiative flux is then z w Ž z Ž
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 š œ ž Ÿ * i ª ««± Ÿ ³² * µ [ with integer values ofü¹ ¹. º» ¼½ The º ² ½ ¾ ² density À * ³ of states ¼ ¾ Á per ²ÃÂÄ volume ¼ of k-space is thus Á for a total cavity volume. This can written in terms of the density of states in a given direction of a given wavenumber½ : º ² ½» ¼ ½ ¾ º ² ½u¹ÆÅ ½Ç» ½» Å ¾ ½ Ç Á ÂÄ ¼ Assuming two independent polarisation states for the photons (either orthogonal plane polarisations or opposite circular polarisations) we obtain the spectra density of available photon states: º ²Èµ ¹ÆŠɾ º ² ½u¹ÆÅ» ½» µ Ê ¾ Á µ Ç Ë ¼ We multiply this by the mean energy per state and divide by the volume Á to obtain the spectral energy density per unit
solid angle introduced earlier. Remember that the occupancy number for photons is ÌSÍ Î ÏÑÐÓÒ Ô ÕCÖ ÈØÚÙ Û*ÜÆÝ Þ, so that ß:à ÌÈá Ü â ã Ôä å æ Ò Ô ÍçΠϳÌÈÒ ÔÕCÖ èñé Ü³Ù Û Thus the brightnessê à of blackbody radiation is given by ê à â ã Ò Ô æ å ä Û Í Î Ï Ì Ò Ô ÕCÖ è é Ü0Ù Û The total flux from the surface of a blackbody is then ë ÌNé Üìâ í ï î ê àñð Ô â ã íkñ#ö òè Ûó å ä Ò æ é ò ô â ó õ öwø ù The Û7ú Ý:û>ü prefactor ý Ý:ä}þ is the Ý ò Stefan-Boltzman constant, ß}à. We can also now find the energy density per unit frequency and hence the total energy density by using the relationshipÿ âwô Õ å : ß â í ñ Ö òè Ûó å æ Ò æ é ò
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.
Stokes Parameters We conventionally use the four Stokes Parameters to describe the polarisation:!#" %'& $ " $)( $ *!+" % $, " $( $ -!+./" % " $10 234+5 %, 56$87 ( 9!+.:" % " $;3=<?>@4+5 %, 56$A7 ( It follows from these definitions that $ * $ & - $ & 9 $. Thus there are still only three free parameters here. Stokes is proportional to the total flux in the wave. Stokes 9 measures circularity, setting the axis ratio of the ellipse. 9 B for linear polarisation, 9 C B for right handed elliptical polarisation. The remaining parameters Stokes * or - measure the orientation of the ellipse. * - B for circular polarisation.
N Quasi-monochromatic waves D In general the amplitudes E F and E G, and phases H1F and H G, 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. D A completely unpolarised wave will have I J K J L J M since the time averages in the definitions of K and L go to zero, and the field amplitudes in each direction must be on average equal. D One can usefully decompose a wave NPORQSI QTKUQVL W into two components, one completely polarised component I GYX K GUX L G QSI QZK[QVL W and one completely unpolarised: N\O ] I G X K G X L G Q M Q M Q M W D The degree of polarisation ^ is the ratio of the polarised to total intensity: ^ J I G X K G X L G O
a 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 two modes experience different refractive indeces, ` given by g h `1a b ced b t ~} uwvyx{z vsp z/ƒ w c[m f+ghijglk f+gn:ijgrkpo q:rs a is the plasma frequency. g n b u z the gyro-frequency. ˆ 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, Š between the two modes. _ This is equivalent to linearly polarised light with plane of polarisation rotated by s b Š ij. is
Œ lž : š œ Zž ŸS Œ in radians, in metres, œ in particles per cubic metre, ž Ÿ in tesla, inparsecs. Œ Ž is know as the rotation measure.
Antenna Temperature The flux density observed by a telescope is often expressed ª in terms of an Antenna Temperature,. The 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, course) independent of observing frequency. is (of observing a thermal source with the same telescope at different frequencies, the received flux density is the same, because the flux density is proportional to both the intensity («) and the beam solid angle (± ³²³ ). (This is only strictly true if the the aperture efficiencies are the same at the two frequencies).
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: the antenna temperature observed is smaller than the source s actual brightness temperature by a µ ¹ ºV» ¼)µe½ ¾wÀÂÁÄÃÅ factor equal to the ratio. Æ ÈÇ É Ê Ë for a source with a thermal spectrum,.
ï ì ë ë ë Effective area of a Gaussian beam The response, or primary beam, Ì of a telescope falls off as a Gaussian as one moves away from its centre. ÌÎÍPÏ ÐÒÑ ÓZÔ Õ Ö Ïj Ø The Full Width Half Max (FWHM), ÏÚÙÛ ÜÞÝ is the width of the beam when the amplitude is half the maximum value. Ì[Í\ÏßÙÛ ÜàÝ ÐáÑ âã Ø ä ÏßÙÛ ÜÞÝ Ñ Ø The effective area of the Gaussian beam is è é êê. è é êâê Ñ Øí ÏAÌÎÍPÏ ÐÅî)Ï è é êê Ñ ØÂí ÓZÔ ÕæÍñðòÐÅî ð Ñ ØÂí ØÂí è é êê Ñ Í\ÏßÙÛ ÜàÝ Ð ó Ø Í Øæå?ç Ø Ð ô í Ñ Í\ÏßÙÛ ÜàÝ Ð õ å?ç Ø Í Øæå?ç Ø Ð
Brightness Question A Red Giant lies at 20 light years from the Earth. It has half the Sun s surface temperature and 40 times its radius. A White Dwarf lies at 10 light years from the Earth. 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 ö luminosity? ö surface brightness at 30 GHz? ö flux density at 30 GHz at the Earth?
Answer àø ù ú û üeý þ ÿ þ ; þ þ The Red Giant is most luminous ù ü (we are in the R-J region) The White Dwarf has the highest surface brightness.!! $ % #" The Sun has the highest flux density. ( & ')( * #+-,/. & 01 2 4365)( * #+87 9 & :;1 at 30 GHz) 2 436<=(>*