Äîêóìåíò âçÿò èç êýøà ïîèñêîâîé ìàøèíû. Àäðåñ îðèãèíàëüíîãî äîêóìåíòà : http://www.mso.anu.edu.au/~geoff/HEA/8_Radio_Galaxies.pdf Äàòà èçìåíåíèÿ: Tue May 27 07:04:58 2003 Äàòà èíäåêñèðîâàíèÿ: Tue Oct 2 04:01:42 2012 Êîäèðîâêà: Ïîèñêîâûå ñëîâà: spiral galaxy

Radio Galaxies
1 Fanaroff-Riley Class Radio-optical overlay of the prototype FR 1 radio galaxy 3C 31 Credits: Alan Bridle and Robert Laing

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© Geoffrey V. Bicknell


The prototype FR 2 radio galaxy, Cygnus A.Note the difference in the jet structure compared to 3C 31. Dreher, Carilli and Perley, ApJ, 316, 611
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Cygnus A at 850 microns. Only the hot spots and core are visible. Robson et al. MNRAS, 301, 935
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1e+02

Flux density (Jy)

1e+01

Cygnus A hot spots A and D 1e+00

Flux density plot of the hots spots of Cygnus A, utilising microwave to mm data. Data from Robson et al. (1998).

1e-01 1e+03

1e+04

1e+05

1e+06

Frequency (MHz)

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2 Fluence The flux density of a source measures the power per unit area incident on a telescope. Units W m ­ 2 Hz ­ 1 . The power per unit area integrated over a band is
2 dP = F d -----1 dA band

Logarithmic ranges We have also seen that it is often more appropriate to express spectra in logarithmic units. This is especially the case with nonthermal spectra which are often power-laws over a wide
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range of frequency. In order to deal with the large ranges over which spectra are measured, we take the expression for the integrated power and write in the following way:
2 2 d dP = F d = vF ---------1 1 dA band ln 2 = vF d( ln ) ln 1

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The quantity F = Power per unit area per ln is known as the fluence. Plots of fluence give a good way of directly visualising what part of the spectrum dominates the emission from an astrophysical source.

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2.1 Fluence from the hot spots of Cygnus A
1e+06

1e+05 Cygnus A hot spots A and D

This plot of the fluence from the hot spots of Cygnus A shows that the emission is dominated by the region of 10 11 ­ 10 12 Hz

Fluence (Hz Jy)

1e+04 1e+03

1e+04

1e+05

1e+06

Frequency (Mhz)

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The plot of the fluence also shows more readily where the spectrum changes slope ­ or "breaks".

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3 Minimum energy conditions 3.1 Cygnus A example

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Recall the minimum energy parameters for a synchrotron emitting plasma: me a + 1 ­ B min = ------ ----------- ( 1 + c E ) C 2 1 ( a ) ------ ----------- f ( a, 1, 2 ) e 2 me L
2 B min 2 p, min = ----------- ----------a + 1 µ0

I c

2 ----------a+5

where
­1 ­( a ­ 2 ) 1 ­ f ( a, 1, 2 ) = ( a ­ 2 ) 1



2 ­ ( a ­ 2 ) ----- 1
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Inputs required L = Distance through source Angular size corresponding to L H 0 70 km s ­ 1 Mpc = Hubble constant z = Redshift of source I = Surface brightness = spectral index a = 2 + 1 and C 2 ( a ) c E = 0 for electron-positron plasma 1 Mpc = 3.09 â10
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m
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For relatively nearby sources: cz D = -----H0 Cygnus A: z = 0.056075 D = 243 Mpc 22 1 1 = 243 â -------- â ----------- â 3.09 â10 m 180 3600 = 3.64 â10
19

L = D

= angular size

m

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· Assume that the red-circled region is approximately spherical. · For the total energy in this region we ignore the emission I from the small hot spot regions · Assume that the surface brightness from the indicated region is the result of integrating a constant emissivity through a "sphere" of diameter L · The total energy in the sphere is then -- min L 3 . 6
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L

The aim of the following is to obtain a rough estimate of the total energy in the lobe.


3.2 Spreadsheet calculation This sort of calculation can easily be coded in a FORTRAN program. However, it is the sort of calculation that has such a large number of inputs that it lends itself well to a spreadsheet. A template spreadsheet is available on the course web page. The result of that calculation is that the total minimum energy in particles and field is approximately 8.6 â10 Joules . If we allow a factor of 2 for the rest of the lobe (including hot spots) and another factor of 2 for both lobes, then we have a total minimum energy of approximately 3.4 â10
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52

Joules .
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4 Implications 4.1 Mass equivalent 3.4 â10 Joules is a large amount of energy to have accumulated in any part of the Universe and this is a minimum (but probably indicative). The mass equivalent of this amount of energy is
5 E M = ---- 1.9 â10 solar masses c2 52

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Since Mc 2 is the absolute maximum amount of energy that can be extracted from a mass, then this amount of mass represents the absolute minimum mass that is involved in the production of this energy. 4.2 Nucleosynthesis Stars "burn" heavier and heavier elements until they reach Iron at which point the nuclear reactions become endothermic rather than exothermic. The maximum amount of energy that can be obtained from a star is E max = 0.07 % â M * c 2
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Therefore, if all the energy in a radio lobe were to be obtained from stars the mass of stars involved is:
7 3.4 â10 M * --------------------- 2.7 â10 solar masses 0.007 c 2 Supernovae 52

If this were to be the source of energy, one would also have to explain how the energy was channelled into radio emitting particles rather than optical and IR radiation. The standard approach has been to invoke supernovae which are known to produce radio emission.

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If we assume that every supernova produces approximately 10 44 Joules of energy then the number of supernovae required is about
8 3.4 â10 --------------------- 3 â10 10 44 For reasons we shall not go into at this stage, the ages of radio galaxies such as Cygnus A is estimated to be of order 52

10 7 ­ 8 yrs . Hence, the total emission from Cygnus A requires about 3 ­ 30 SNe/yr .

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4.3 The size of the emitting region - Imaging We have seen that the Cygnus A lobes are fed by jets whose origin is an unresolved (at 1 resolution) region in the centre of Cygnus A. This is A VLBI image of Cygnus A at 2 mas resolution. At the distance of Cygnus A VLBI image of Cygnus A (Carilli, Bartel & Linfield, AJ, 1991, 102, 1691) 1 mas = 1.2 pc

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Even at this resolution there is still an unresolved core. So that the 3-30 SNe per year would have to be occurring from within this volume. Compare this to about 1 SNe per 100 years from our entire Galaxy. 4.4 The size of the emitting region - time variability The nuclei of all active galaxies ­ including radio galaxies ­ are observed to vary on timescales as short as 1 day. The variability in the radio region of the spectrum can be explained in terms of interstellar scintillation by the medium of our own galaxy. However, the variability in optical and X-ray emission cannot be explained in this way.
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Naive estimate of size from time variability Observer If an emitting region varies more rapidly than it takes a light ray to travel through it, then the variability would be washed out. L Hence if a source varies on a timescale of t then the size, L , of the emitting region satisfies: 1 day 1 day

13 t ­4 t L < c t = 2.6 â10 ----------- m = 8.4 â10 ----------- pc

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This estimate will be modified later when we incorporate relativistic effects. However, the indication from this estimate is that the timescale is very small compared to the smallest scales that are resolvable with current technology. If we invoke conventional sources of power, then we require either: · A stellar mass density exceeding 4.5 â10 per cubic parsec
16

solar masses

· Or ~10 SNe per year from within a volume of 10 ­ 3 parsec.

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4.5 Black holes The above arguments gradually led to the realisation that black holes provide an appealing mechanism for the production of energy in radio galaxies and active galactic nuclei, generally. Power liberated by a black hole If a mass m spirals into a black hole then an amount of energy E acc 0.3 mc 2

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can be liberated. This is the binding energy of the lowest stable orbit in a maximal Kerr black hole gravitational field. For general accretion onto a black hole, we put for the energy liberated by accretion: E acc = m acc c 2 where m acc = Accreted mass

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In a radio galaxy, this energy appears in the radio lobes after being transported out from the nucleus in jets. Hence, the mass accreted by the black hole is E lobes m acc -------------c2 For Cygnus A
6 ­1 m acc 1.9 â10 -----solar masses 0.1

This represents a lower limit on the mass of the black hole.

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5 Evidence for black holes 5.1 Keplerian velocity of hole If we v (e.g. hole, M R matter orbiting a black have a lump of matter of mass m cold gas, star) orbiting a black then mv 2 GMm --------- = ------------R R2 v = GM --------R
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Numerically, M 1 / 2 R ­1 / 2 ---v = 208 km/s ------------------ pc 10 7 M O This sort of velocity can be detected very easily via optical or radio spectroscopy (depending on the circumstances).

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5.2 The gaseous disk in M87 The gaseous disk in M87 as observed by HST. This press release image is from the work of Holland Ford and collaborators. The gas velocity was measured from observations of the H + [ NII ] lines.

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Geometry i N W Allowing for inclination and Observer orientation on the plane of the sky: i = Angle of inclination Position angle of = major axis Position angle of k = measurement Projected radius of Rk = measurement
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The measured velocities are given by: GM sin2 i cos3 i 2 v k = --------- ------------------------------------------------------------------------------------------- R cos ( + ) [ cos2 i + tan2 ( + ) ] 3 / 2 k k k This model was used by Ford et al. (ApJ, 435, L35) to estimate a black hole mass M bh = ( 2.4 ± 0.7 ) â 10 9 solar masses

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6 Information gained from polarisation 6.1 Example from Pictor A The following image is typical of the sort of image that radio astronomers produce for a polarised source.

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Radio image of Pictor A showing the total intensity contours and the direction and magnitude of the polarisation. (Perley et al, A&A, 328, 12)
The polarization structure in the lobes of Pictor A, at 6 cm wavelength and 1000 resolution. The contours are drawn at 0.4 to 72.4% of the peak brightness of 3.21 Jy/beam with a spacing ratio of 21/2. The dashes show the direction of the observed E-vector, corrected for a foreground RM of 45 rad/ square metre. The projected B-vectors are normal to the dashes. The length of the dash is proportional to the degree of polarization, a length of 1 arcsecond indicates a polarization of 1%. The

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This image of Pictor A shows a typical way of representing polarisation information. The contours represent the total intensity and the lines show the direction and magnitude of the linear polarisation. One of the motivations for polarisation of synchrotron gained from polarisation to utilise the main property of understanding the theory for the radiation is to use the information infer properties of the source. We synchrotron radiation:

Direction of polarisation Projection of B onto the from emissivity plane of propagation Let us look at polarised emission in some detail.
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6.2 The Stokes parameters from a synchrotron emitting region In the coordinate system ^ e || = B p shown we know that for synchrotron radiation e j ( ) 0 j () 0
I Q

jU ( ) = 0

jV ( ) 0

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What do we take for the emissivities in a general coordinate system (e.g. Right ascension and declination). The best way to look at this is to go back and look at the power in a single pulse. c 0 dW * --------------- = ------- r 2 E ( ) E ( ) dd

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After convolving this expression with the particle distribution we obtain the emissivities in the various Stokes parameters so that in an arbitrary coordinate system: 1 * -- ( j I ( ) + j Q ( ) ) E x ( ) E x ( ) 2 1 -- ( j 2 1 -- ( j 2 1 -- ( 2
* ( ) ­ ij V ( ) ) E x ( ) E y ( ) U * ( ) + ij V ( ) ) E x ( ) E y ( ) U * jI ( ) ­ jQ ( ) ) E y ( ) E y ( )

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In the expressions for synchrotron emission, we have calculated the emissivity in a particular coordinate system, in which jI ( ) 0 jQ ( ) 0 jU ( ) = 0 jV ( ) 0

In order to calculate the emissivities in an arbitrary system we need to determine the transformation properties of
* E ( ) E ( ) . We therefore consider an arbitrary coordinate system as shown:

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e ||

ey

The magnetic field coordinate system is related to the arbitrary e system by: e cos sin e x = ex e || ­ sin cos e y with the components of the electric field related by: E cos sin E x = E || ­ sin cos E y

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Let R = cos sin ­ sin cos
t R =

cos ­ sin sin cos

The reverse transformation is Ex E t = R Ey E ||

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Now,
* * Ex E E xE x E xE y t ** ** = E x E y = R E E || R * * Ey E || E x E y E yE y * * E E E E || t = R R * * E E || E || E ||

It just remains to multiply out the matrices. The result is:

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* E xE x E xE * E x E y E yE

* y * y

=

* cos2 E E + sin2 E || E

* ||

* * cos2 E E || ­ sin2 E E

||

* * ­ sin cos ( E E || + E E || ) * cos2 E E || ­ sin2 E E * ||

* * + sin cos ( E E ­ E || E || ) * sin2 E E + cos2 E || E * ||

* * + sin cos ( E E ­ E || E || )

* * + sin cos ( E E || + E E || )

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We then use 1 * -- [ j I ( ) + j Q ( ) ] E x ( ) E x ( ) 2 1 -2 1 -2
* [ j U ( ) ­ ij V ( ) ] E x ( ) E y ( ) * [ j U ( ) + ij V ( ) ] E x ( ) E y ( )

1 * -- [ j I ( ) ­ j Q ( ) ] E y ( ) E y ( ) 2 with primes to denote the emissivities in the magnetic field frame.
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The result of combining the various terms is
j I ( ) + j Q ( ) j U ( ) ­ ij V ( ) j U ( ) + ij V ( ) j I ( ) ­ j Q ( ) jI ( ) + cos 2 j Q ( ) ­ sin 2 j U ( ) sin 2 j Q ( ) + cos 2 j U ( ) + i jV ( ) =

sin 2 j Q ( ) + cos 2 j U ( ) ­ ij V ( ) jI ( ) ­ cos 2 j Q ( ) + sin 2 j U ( )

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The transformation laws for the emissivities are therefore: jI ( ) = jI ( ) j Q ( ) = cos 2 j Q ( ) ­ sin 2 j U ( ) j U ( ) = sin 2 j Q ( ) + cos 2 j U ( ) jV ( ) = jV ( )

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In matrix form 1 0 0 jQ ( ) 0 cos 2 ­ sin 2 = jU ( ) 0 sin 2 cos 2 0 0 0 jV ( ) jI ( ) 0 0 jQ ( ) 0 jU ( ) 1 j () V jI ( )

Note that the transformation for Q and U involves a rotation through 2 .

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6.3 The special case of synchrotron emission We know that for synchrotron emission
U j = 0

Hence, j Q ( ) = cos 2 j Q ( ) ­ sin 2 j U ( ) = sin 2 j Q ( ) + cos 2 For optically thin emission: I Q ( ) = j Q ( ) ds = cos 2 j Q ( ) ds I U ( ) = j U ( ) ds = sin 2 j Q ( ) ds
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j U ( ) = cos 2 j Q ( ) j U ( ) = sin 2 j Q ( )


where j Q ( ) is, for a power-law distribution, the polarised emissivity (See Synchrotron Radiation II) a+1 j Q ( ) = ------------------ j I ( ) a+7/3 3a / 2 = ----------32 3 â 1 a + 19 a ­ ----- -- ----- 4 12 4 12 ------------------------------------------------a+7/3

(a ­ 1) ­ ---------------e2 2 ------- K ( 0 sin ) ----------------- sin 0 c 0
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e || e y e ex

When the angle between the projected magnetic field and the adopted coordinate system is constant, then:

a+1 I Q ( ) = cos 2 j Q ( ) ds = cos 2 ------------------ I ( ) a+7/3 a+1 I U ( ) = sin 2 j Q ( ) ds = sin 2 ------------------ I ( ) a+7/3
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The degree of polarisation can be determined from: [ I Q( ) ]2 + [ I U ( ) ]2 = a+1 2 ------------------ [ I ( ) ] 2 a+7/3

( [ I Q( ) ]2 + [ I U ( ) ]2 )1 / 2 a+1 p = ------------------------------------------------------------------ = -----------------a+7/3 I () The angle is given by: IU () tan 2 = --------------I Q()

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e ||

ey

e ex

The direction determined from gives the direction of the axis perpendicular to the magnetic field. The magnetic field.

If the angle varies through the source, then the degree of polarisation is less than the maximum value given above. Also the direction of the magnetic field is an emission weighted direction. To see how this occurs consider the following expressions for I Q ( ) and I U ( ) .

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a+1 I Q ( ) = cos 2 j Q ( ) ds = cos 2 ( s ) ------------------ j I ( ) ds a+7/3

cos 2 ( s ) j I ( ) ds a+1 = ------------------ I ( ) --------------------------------------------a+7/3 j I ( ) ds
a+1 I U ( ) = sin 2 j Q ( ) ds = sin 2 ( s ) ------------------ j I ( ) ds a+7/3 a+1 = ------------------ I ( a+7/3

sin 2 ( s ) j I ( ) ds ) ------------------------------------------- j I ( ) ds
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Thus, I Q() a+1 --------------- = ------------------ cos 2 a+7/3 I () IU () a+1 --------------- = ------------------ sin 2 a+7/3 I () where the <> indicate an emission-weighted mean. We define the intensity weighted direction of the magnetically oriented coordinates by: IU () tan 2 = --------------I Q()
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The implied direction of the magnetic field in this image is perpendicular to the straight lines. That is, the magnetic field has a circumferential direction with respect to the plasma in the lobe.

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6.4 Polarisation of hot spots
Polarization of the western hot spot of Pictor A, at 3.6 cm wavelength with 400 resolution (left), and at 0.77 by 0.17 resolution (right). The lower resolution map shows the general features of this region, and is contoured at 0.391% and then with a spacing of a factor of 2 between 0.552 and 70.71% of the maximum intensity of 1.55 Jy/beam. The dashed lines again indicate the plane of the electric vector. Their lengths are proportional to the degree of polarization, with 100 equal to 6.67%.

The western hot spot of Pictor A

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The polarisation of the hot spot in Pictor A reveals magnetic field structure that is consistent with that produced by a shock wave.
Perpendicular component of magnetic field enhanced

JET Randomly oriented magnetic field

SHOCK

At the terminal shock of the jet, the component of the magnetic field perpendicular to the shock is amplified, whilst leaving the parallel component unchanged. This lines the field up with the shock as shown
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