Metabolics: feeding the
monster
•Energetics
•Accretion
disks
•Emitted
spectrum
•Relativistic
discs
QuickTime™ e un
decompressore Codec YUV420
sono necessari per visualizzare quest'immagine.
Fueling AGNs
Conversion of mass to energy with some efficiency 

L Mc
2

M
L
3 L44
1

1.8
10
M
yr

c 2


1 GMm
dU 1 GM dm GM M
1 GM 1
L


 

= 0.083 if
2
2 r
dt 2 r dt
r
2 rc
12
GM
RS  2 2  3 1013 M 8cm last stable orbit = 3RS
c
and ignoring relativistic effects.
1 GM
U 
energy available from a particle falling to 3R
S
2 3RS
1
  0.057 for a Schwarzschild metric and 0.42 for a Kerr metric (l.s.o. = RS )
2

L
M E = E2  Critical mass accretion rate  2.2M 8 M  yr 1
c
U
The major problem for fueling an AGN is not the energy
requirement but angular momentum considerations.
Consider the angular momentum of a particle in the solar
circle around the Galactic centre:
L
 GMr angular momentum per unit mass
m
r  10kpc M  1011 M 
move this particle to 0.01pc around a 10 7 M  BH
L
must decrease by 10
m

5
times
Accretion: Basic concepts

Tidal disruption of a star
A star of mass density  near a BH of radius R can approach no closer
than the Roche limit without being tidally disrupted
:
 BH 
rR  2.4
 R
 * 
1/ 3
 rR  RS
1/ 2
6 



rR
3M
c
1/ 2
8
 2.4

1

M

0.64

5
10

M



*
3
3
RS
4RS * 
G * 
Accretion disk structure
• Rotating mass of gas in a cylindrical potential well
• Time-scale for processes to redistribute angular
momentum >> dynamical time-scales or radiative timescales  each gas element looses energy via collision and
radiative cooling but retains its angular momentum.
• Circular orbits
v  circular velocity
d
 is the gravitational potential
dR
a fluid rotates differentially, the rate of shearing A is :
v  R
v  R
2
d
0
dR
gas elements at different radii rotate with different angular velocity
AR
Viscosity present among gas anuli rotating with different
 tends to reduce the difference in velocity, dumping
out the shearing motion, and therefore tends to
dissipate energy as heat and radiation. Thus viscosity
converts gravitational potential energy into radiation in
an efficient manner.
Equation of motion of the gas.
Disk has surface density (R,t) and radial velocity vR(R,t).
Consider motion of an annulus of gas with inner radius R and extent R
Mass is: M = 2R R 
Angular momentum is L= MR2  = 2R R  R2
Equation of continuity
The variation of the mass of the annulus with the time must be equal
to the difference between the mass entering the radius R and the
mass going out.
M 
 (2R  R  )  v R (R,t)  2R  (R,t)  v R (R  R,t)  2 (R  R)  (R  R,t)
t t
taking the limit for small
 
R 
(Rv R )  0
t R
R this becomes :
Conservation of angular momentum
L  MR 2  2R  R    R 2
L
 Lin  Lout  2R  v R    R 2  2 (R  R)  v R (R  R)  (R  R)  (R  R) 2 
t
R


(R 2)   (Rv R  R 2)
t
R
If viscosity is introduced among contiguous gas annuli the rate of change
of L will depend also in the viscous force:


1 G
(R 2)  (Rv R  R 2) 
t
R
2 R
where
G(R,t)  A  2R  R A is the viscous force per unit length around the circonference
and  is the kinematic viscosity
G is the viscous torque of an outer annulus acting on a neighboring inner one
R
 

R 
(Rv R )  0
A=R
t R
R


1 G
2
2
R (R ) 
(Rv R  R ) 
t
R
2 R

1 
1  (R   /R)
2
3
(R ) 
(v R  R ) 
t
R R
2R
R
3

1

(

R
( /R) 
2
  (R ) /R 

R
 1 


t R
R
If the potential is due to a central point mass M
GM

R3
 1/ 2  (R1/ 2 ) 
 R

R 
 3 

t R
R

then :
If the rate of change of L would not depend on G then the disk
surface density time derivative would be zero and = ( R).
Conversely,  is a compicate function of R and t.
In general = (R,t, ). If = const. the previous eq. can be
integrated analitically.
If we assume an initial
configuration with all mass
confined at R0 at t0
(t0)=(R- R0), then while time
passes  broadens until all the
mass is distributed toward the
centre and all the angular
momentum is flown toward the
outer disk.
Dissipative processes act to
smooth the differences and
bring the mass toward the
inner disk.
Energy dissipation in the disk
The time variation of the total kinetic
+ potential energy :
 1 2 GMm 
mv 
where m = 2R    R




t 2
R
2 R 2 GM 
 
= 2R    R

 = A1
t 
R 
 2
must be balanced by the differences between the rate of flow of the energy in and out of
the ring
2 R 2 GM 
 

2Rv R 
R  A2
R 
2
R


and also by the difference bewteen the work done by viscous stresses at the side of the ring
and the energy losses into heat due to friction and differential rotation
:


2W R R 2  R  2W R R 2
= A3 where 2W R R 2 is the viscous torque(stress)

R
R

2W R R 2  G  2R 2A  W R  A  R
R
R
A1  A2  A3  0
Energy dissipation in the disk
The energy dissipated into heat per unit area is
:
1

Q  W R R
(1/2 because the disk has 2 sides)
2
R
2 R 2 GM  
  2 R 2 GM   

2
2RQ  R


W
R
 v R R

 R 
t   2
R  R 
R  R
 2
  (v R )  R 3 2 3 2

2 R 3   (v R ) 
2
2

W
R



GM



R
v


 2 t R  t R  2 t 2 R
R R
1 
3
2
Q  
W
R

Rv R 2


R
2R R
4
now, remember that M  2R  R  

1 
M2
2
Q 
W R R 

2R R
8



M  2R  v R   : so,

1 
M2
2
Q 
W R R 

2R R
8

During radial motion half of the liberated potential

energy goes into increasing the kinetic energy and half
goes into heat.
Viscous stresses transfer mechanical energy besides
momentum.
Hydrostatic equilibrium
along z-direction
If the motion in the disk in the z-direction is subsonic, then the
disk is in hydrostatic equilibrium. Gas and radiation pressure
gradient is balanced by the z component of the grav. attraction.
Assuming that the disk is homogeneus (=const along z).
GM
p
   3 z    2 z
R
z
  z 2 
 p(z) = pc 1   
 H  
1
pc  2 H 2 is the central pressure of the disk
2

1
1
2
2 2
p   H   H = average pressure (   )
H
3
3
p H
average sound speed

us 

3
Viscosity
2W R R 2 is the viscous torque(stress)
2W R R 2  G  2R 2A
 W R  A  R
 is the kinematic viscosity,
  lt  ut where ut is the turbulent velocity and
lt  H
ut  us (sound speed =
Kepler law :  
GM
R3
R
pH
p
=
)



 viscous force per unit length
R
lt is turbulent mixing length


2

R
3
2
2
2
2
W R     Hu s 
2H 2 
pH
3
3
3 3
3
  Hus with  < 1
Steady disks

 0  continuity equation becomes :
t



1 M
0
 M  const = -2Rv R 
t
2R R
does not depend on radius!
momentum equation becomes :
 



M
R 3v R   W R R 2  0   R 2  W R R 2 


R
R 2



M

W R R 2  W R R 2 

R


M 2
R  c


2
valid for R > 3R S . In this region R , , L 
R < 3R S matter falls in the BH and v R increases exponentially.
R 
M

R 
At R  3R S viscous stresses must stop having big effects
 WR  0
1/ 2
1/ 2

 GM 


GM
1/
2
c   M 0 R02   M  3  R02   M GM  R1/0 2   M  3  R1/0 2 R 3 / 2 
 R 
 R0 



  M R1/0 2 R 3 / 2
therefore :



M 2
M 2
W R R 
R  c 
R   M R1/0 2 R 3 / 2
2
2

1/ 2
M  R0  
W R 
1   


2   R  

2


1/ 2 
2 


1

M
3M
R
0
Q  
W R R 2 

1


 


2R R
8
8   R  

2
Energy does not depend on viscosity! It is only the
 gravitational potential that pays.
1/ 2
 
 R 1/ 2 

R0   2
2
0
L  4  Q RdR  3 M  
 
 
1 
 RdR  3 M GM  
1 
R dR




R
R


R0
R 0 
R 0 





M GM

 accretion rate times the binding energy of the last stable orbit!
R0
independent on dissipative forces
The heat loss per unit area is given by the transfer equation
:
Q  acTc4 / 
a = Stefan© s constant,   Hk( ,Tc )
if the disk is optically thick, each element radiates as a BB with T
Q  T 4
S

where  is the Stefan - Bolzman constant.
putting Q  Q



1/ 4
1/ 2 

R0  
3GM M
Ts  
1

  
 8R 3 
 R  




At each radius the spectrum is a BB with T
S 

R out
R0
B (TS (R))2RdR 
Tmax occurres at R =

R out
R0
49
R0  Tmax
36
S
so the total spectrum is :
3
2RdR
exp( h /kTS ) 1
1/ 4
 

3GM M 
 0.488
 8R 3 


X-ray spectrum
•
•
Dissipate energy in optically thick disk – cool, no hard X-rays
MUST dissipate in optically thin material so that E >> kT (Compton)
Optically thin accretion flow –
low L/LEdd only!
Magnetic reconnection above
disk – no known alternatives at
high L/LEdd!
Inverse Compton scattering of lower energy photons by
energetic electrons in a corona surrounding the disk
Thomson diffusion
A monochromatic wave interacts with an
e-. The e- will be accelerated and will emit
radiation. The direction of the emitted
radiation will be in general different from
the direction of the incident radiation. If the
particle is not relativistic the frequency of
the emitted radiation will be the same of
the incident radiation.
e2
2
Ý


v
Larmour formula for the power irradiated in a polarization status

4c 3
e2
2
Ý
v
sin 2  where  is the angle between the emitted radiation ad the acceleration
3
4c
e
F  eE  e0 E 0 sin(  0 t)  me vÝ vÝ 0
E 0 sin(  0 t)
me
dP

d
dP

d
dP
e 4 E 02

sin 2 
2 3
d 8me c
e 4 E 02
e 4 E 02
2
P   d
sin  
8me2c 3
3me2c 3
Scattering cross section
d irradiated energy per unit time and solid angle

d
incident energy flux per unit time and area
incident energy per unit area and time = time average of pointing flux
d dP 1
c
c 2

S
EE
 S 
E0
d d  S 
4
8
d
e 4 E 02 8
e2
2
2
2

sin


r
sin

r

e
classic radius
e
e
2 3
2
2
d 8m e c cE 0
me c
2 2
8 2
 T   0 re sin 3 d 
re  6.65 1025 cm 2
3
e 4 E 02
P
  T cU rad U rad  S /c = energy density of the incident radiation
2 3
3m c
This is formula is valid only if the momentum of the incident photon is negligible
h /c  m e c
h  m e c 2
m e c 2  511keV
:
Compton effect
E1  h1
p1 
quantum mechanical particle approach
h1 h

c
1
E 2  h 2
p2 
pe 

h 2 h

c
2
1
2
E 2  me c 4 mec 2  initial e - energy
c
E = final e - energy
Momentum conservation : p1  p2  pe  pe2  p12  p22  2 p1  p2  p12  p22  2 p1 p2 cos 
Energy conservation : E 0  
me c 2, E  E 02  pe2c 2  p1c  me c 2  p2c  E 02  pe2c 2
c( p1  p2 )  me c 2  E 02  pe2c 2 squaring : c 2 ( p1  p2 ) 2  2cE 0 ( p1  p2 )  pe2c 2
pe2  p12  p22  2 p1 p2  2E 0 ( p1  p2 ) /c
 p1 p2  E 0 ( p1  p2 ) /c   p1 p2 cos 
p1 p2 (1 cos )  E 0 ( p1  p2 ) /c
multiply each term by hc / p1 p2 E 0 :
hc
( p  p2 )h
(1 cos  )  1
now use   h / p
E0
p1 p2
Compton effect
 h h 
  h
hc
1 2  12  h h 
(1 cos  ) 

   2  1
h h
mec 2
h 1 2 
1 2
or
1
2 
 1
if h1  me c 2
h 1
(1 cos  )
2
mec
The introduction of this factor changes the definition of
cross section and gives the Klein - Nishina formula :
1
2
2
d  e 2   2   2
1   2 
 
   sin  

d me c 2  1  
41 2 
Inverse Compton
1 = initial photon energy in the lab frame
1' = initial photon energy in the e - frame
2 = photon energy in the lab frame after the scattering
2 ' = photon energy in the e
-
frame after the scattering
1' 1 (1  cos  ) from the lab frame to the e
-
frame
if 1' me c 2 one can apply the Thompson scattering,
i.e. 2 ' 1 '
let' s go back to the lab frame : 2  2 '  (1  cos )  1 2 (1  cos )(1  cos  )
The photon energy has been incremented by a factor  2 . This is for 1 photon and 1 e - .
Now we want to find the energy emitted per unit time by an isotropic distribution of
photons scattered by a isotropic distribution of e
N() = number of photons with energy 
U rad  N()  N(h )h
-
.
Inverse Compton
Let' s begin considering non - relativistic Thomson scattering. If the Poynting flux
2
c
(power per unit area) of a plane wave incident on the e
is S 
E the E will
4
accelerate the e -, and the accelerated e - will in turn emit radiation according to the
Larmour formula. The net result is to scatter a portion of the incoming radiation with
no net tranfer of energy. The scattered radiation had power :
P  S  T  c T U rad
2
U rad  S /c  E /4 
Let' s consider now the radiation scattered by a relativistic e
is valid in the primed frame if
-
. The Thomson formula
' me c 2 . In this case :
P' c T U'rad
We need to go back in the lab frame. We know that P is a Lorentz invariant,
P  P' c T U'rad
and we only need to transform U'rad in U rad .
therefore :
Inverse Compton
We know that '  (1  cos ). The rate at which successive photons arrive is multiplied by the
same factor, so N'(')  N() (1  cos  ). In the e - frame :
U'rad  N' h ' N (1  cos )h (1  cos  )  U rad  2 (1  cos  ) 2
Thus the transformation between U'rad and U rad depends on the angle  between the direction of
the photons and the e - motion. The total energy density in the e - frame of a radiation field that is
isotropic in the observer frame is obtained by integrating over all directions
:
U'rad  U rad 
2
2

 d  sin d (1  cos )
0
2
0
where is the azimuthal angle around the x axis.
4  2 1 2

4( 2 1/4)
2
U'rad  U rad 
  (1  ) U rad
3
3
 3

[ 2 (1  2 )  1]
4
c T U rad ( 2 1/4) = Total power in the radiation field after IC upscattering.
3
The initial power of the photons was c T U rad , so the net power added to the radiation field is :
4
4
4
PIC  c T U rad ( 2 1/4)  c T U rad  c T U rad ( 2 1)  c T U rad  2 2 [( 2 1)   2 2 ]
3
3
3
P' P  c T U'rad 
Thermal Comptonization
average fractional energy  mean number of

 
change per scattering
 scatterings

 Compton y parameter

f 4 2
 kT 2
A     16 2  mean amplification per scattering
mc 
i 3
assuming a thermal electron distribution
: N(E)  E 2 exp(E /kT)
after k scatterings : k  i A k
the probability of a photon undergoing k scatterings before escaping is
pk ( es ) ~ i esk
The intensity of the emergent radiation at




I(k ) ~ I(i ) esk ~ I(i ) k 
i 
 
k is proportional to pk ( es ) :
log(  es )
log( A)
:
Thermal Comptonization
•  depends on 2 parameters: the optical
depth of the medium and the
temperature. Any spectral shape can be
produced with ad hoc choices of these
parameters.
• problem: in AGN  ~ 1 [0.5-1.5]
Two phases disc
(Haardt & Maraschi 90’)
Optically thick emission from the cool layer provides soft
photons input for Comptonization and the hard Comptonized
photons contribute to the heating of the thick phase. The
feedback between the two phases determines the fraction of
power emitted in three main components: a BB from the thick
phase, a power law from Comptonization in the hot layer and
a reflection component. The resulting spectrum is
~independent of the coronal parameters.
Two phase disc
• A fraction f of the gravitational power PG is
dissipated in the hot layer of optical dept <1,
while (1-f)PG is dissipated in the optically thick
phase.
• The total luminosity of the hot phase is
LT=ALS, where LS is the luminosity of the thick
phase. The luminosity added by the hot
phase is LC=(A-1)LS
• LC=LCU+LCD with LCD=LC ~0.5
• Photons directed downward are partly
absorbed and partly reflected:
– Lrfl=aLCD a~0.1-0.2
– Labs=(1-a)LCD will contribute to LS
Energy balance
LS  (1 f )PG  (1 a)LCD for phase 1
ALS  fPG  LS for phase 2
solving for A and L
S
LS  1 f 1 (1 a)PG
f
1 f [1 (1 a)]
The outgoing luminosity is given by
Lout  LS  LCU  Lrfl where :
A  1
LCU 
:
(1 ) f
af
L S and Lrfl 
LS
1 f  f  fa
1 f  f  fa
For small f LCU and Lrfl are proportional to LS.
For f~1 LCU and Lrfl are determined by a and 
Energy balance
f
A  1
1 f [1 (1 a)]
log(  es )

log( A)
 kT 2
A  16 2 
mc 
The energy balance in the first
ew. Implies a relationship
between optical depth and
temperature, and therefore the
spectral shape of the
Comptonized component!
Emitted spectra
Spettro dei raggi X
Il disco di accrescimento che circonda il buco nero è una sorgente di radiazione UV
e X di bassa energia (soft X-rays). Lo spettro di emissione è di tipo termico
(radiazione di corpo nero). La “Comptonizzazione”(*) della componente soft X-rays
in una corona che circonda il disco di accrescimento è una possibile causa dello
spettro a legge di potenza per la componente di raggi X di energia 1-100 KeV (hard
X-rays)
La componente continua di hard Xrays incide sul disco di accrescimento
e produce uno spettro riflesso,
caratterizzato
dai
fenomeni
di
scattering
Compton
e
di
assorbimento fotoelettrico, seguito da
emissione di righe di flourescenza o
diseccitazione di tipo Auger.
Lo spettro di emissione dei metalli è
sovrapposto allo spettro continuo
(*) Si tratta del fenomeno di ICS (Inverse Compton Scattering) per cui un fotone aumenta la propria energia a
seguito di un processo di diffusione su un elettrone.
La riga di emissione Fe Kα
Quali sono le evidenze osservative del BH-Paradigma?
Quali informazioni possono essere estratte dalle misure astronomiche
sulle proprietà fisiche e geometriche del buco nero e del disco di accrescimento?
A causa dell’elevato valore dell’abbondanza cosmica del ferro, la riga Fe Kα è la
componente principale dello spettro di emissione.
Per assorbimento fotoelettrico uno dei due elettroni della
shell K (la shell più interna, con n=1) viene etratto dall’atomo,
lasciando una lacuna. Un elettrone della shell L (n=2) occupa
il “posto vacante”, rilasciando 6.4 KeV di energia.
Lo studio del profilo di riga è un importante
strumento di diagnostica delle proprietà fisiche e
geometriche del buco nero e del disco di
accrescimento
[A. C. Fabian, G. Miniutti, astro-ph/0507409 v1 18 Jul 2005]
Profilo di riga
Numerosi effetti modificano il profilo della riga del ferro:
• effetti RELATIVISTICI
• effetti di ORIENTAZIONE
• posizione dell’ultima orbita stabile (ISCO)
• effetti di IONIZZAZIONE
• profilo di EMISSIVITÀ del disco
La simulazione dei diversi effetti permette di calcolare la forma di riga in
funzione dei parametri fisici e geometrici del sistema costituito dal buco nero
centrale e dal disco di accrescimento. Il confronto dei dati sperimentali con le
previsioni teoriche fornisce importanti informazioni sulla natura degli oggetti
astrofisici.
Effetti relativistici
EFFETTI RELATIVISTICI producono l’allargamento del profilo di riga e lo
spostamento verso il rosso (red-shift) del picco di emissione.
NEWTONIANO
RELATIVITA’
SPECIALE
EFFETTO DOPPLER
EFFETTO DOPPLER
TRASVERSO
BEAMING RELATIVISTICO
RELATIVITA’
GENERALE
REDSHIFT
GRAVITAZIONALE
PROFILO DI RIGA
CONVOLUZIONE DEI
DIFFERENTI EFFETTI
Broad lines from relativistic discs
Axisymmetric disc orbiting a
Schwarzschild BH. Disk
extend from r0 to r1 and it is
observed at inclination i.
The ratio of the emitted
energies of the photons from
a point in the disk to the
observed energies is given
by:
a
u
k
a 
E em
em
1 z 

E obs ua k a 
obs
Second order contri butions to Doppler effect
The expression of the Doppler effect in General Relativity is the
ratio between the observer frame and emitted frame product between
the quadri - velocity
and the quadri - wave vector.
(1 z) 
E em 1

E obs  0
u k 


u k 
'



obs
a
em
dx'
dx
where u =
= (-1;0,0,0) and u =
are the emitted and observed frame
ds
ds
'

quadri - velocities and k  ( ;k ).
c
a
k is the direction of propagation of the wave with angular velocity
 = 2 ; k 
2



vp
where v p is the phase - velocity
ds is the metric. Let us consider the expression for ds provided by Roberson
:
 2MG
 2 
M 2G 2
MG  2 2
2
2
2
2
2
ds  c 1

2




....
d
t

1
2

d
r

r
sin

d


d








2
4 2
2


rc
c
r
rc


ds2  c 2dt 2  dr 2  r 2 sin 2 d 2  d 2 
2MG
and   1 2
2
rc

v
ds  cdt  1  2 where  

c
cdt
u  (u 0 ;u i ) where u 0 

ds

where   1
MG
rc 2
and v is the three dimensional velocity.
1
1  2



and u i 
dx i

ds
recall that u  (1;0,0,0) and that k  ( ;k ) then :
c


1
v
1
u k  ; u k  
 i k
c
c
 c

 1  2
 1  2


vi
c  1  2


1 (u a' k' ) em
 

 0 (u k ) obs  
c



c

1
vi
1
 k
 c

 1  2
 1  2



1
 

1
v
c
(1  cos  )
(1 i  k )
c


 1  2

 1  2

MG 
1/ 2
1
2

2 
 2GM  
2
rc
1 
1
2  
2GM 

rc  

1
2

rc  

(1  cos  )
1/ 2
 2GM 
 1
2 

rc 
1/ 2

MG  2GM 
2 
(1  cos )
1  1 2 2 1
2 

rc 
rc 

1/ 2
 2GM 1/ 2 
MG  2GM 
2 
1  1 2 2 1
(1  cos ) 
1
2  
2 


rc  
rc 
rc 
1/ 2
v
1 v 2 GM
1 cos 

 .....
c
2 c 2 rc 2
leaving only the contributions up to second order in
1 v2
 2  transverse Doppler effect
2c
GM
 gravitational redshift
2
rc
r
2GM
if R 
; Rg = 2 and cos   /1 than :
3Rg
c
GM
v
r  1 ; and GM  1

c
c
rc 2 6R
6R
1
1
1
1
 1


0
6R 12R 6R
v
c
Effetto di orientazione
La forma della riga dipende dall’angolo di inclinazione dell’asse del disco rispetto alla
linea di vista. Al crescere dell’angolo di inclinazione, l’effetto principale è
l’allargamento della riga che si estende verso le più alte energie.
Il contenuto di riga nel blu è una misura dell’angolo di inclinazione del disco.
i [deg]
(Metrica di Schwarzschild)
Posizione della ISCO
Schwarzschild vs Kerr
Il contenuto di riga nel rosso è una stima del raggio della ISCO (Innermost Stable
Circular Orbit) e permette di distinguire un buco nero di tipo Schwarzschild
(statico) da un buco nero di tipo Kerr (rotante).
Rg 
GM
c2
Schwarzschild  RISCO = 6Rg
Kerr  RISCO = 1.24Rg
parametro di spin
a/M
Effetti di ionizzazione
4  Fx (r )
 (r ) 
n( r )
parametro di ionizazione
Fx (r )
n(r )
Flusso di raggi X incidente per unità di area alla distanza r
Densità di elettroni
Se la materia del disco è
fortemente ionizzata, aumenta
l’energia di soglia del processo
di assorbimento fotoelettrico, e
di conseguenza si riduce
l’efficienza per l’emissione
della riga di fluorescenza.
Profilo di emissività del disco
Il profilo di emissività del disco definisce l’efficienza con cui la luce è emessa in
funzione della coordinata radiale del disco. Si assume una legge di potenza:
q = indice di emissività
 (r )  r  q
emissività uniforme
emissività standard
emissività “steep”
Il caso “steep” implica
un’illuminazione più efficiente
a piccoli raggi, cioè nelle
regioni più interne del disco
di accrescimento.
In questo caso, il profilo di
linea si allarga e si estende
verso il rosso: maggiore peso
è dato, infatti, alle zone
centrali del disco, dove
dominano gli effetti di redshift
gravitazionale.
MGC – 6 - 30 -15
Galassia di tipo Seyfert I
z = 0.00775
Osservazione della riga Fe Kα
CONCLUSIONI
• i = 33° ± 1°
XMM – Newton
Chandra
• rin = 1.8 ± 0.1 Rg (ISCO)
• a/M = 0.93 ± 0.01 
KERR
• ξ < 30 erg cm s-1
•r profilo di emissività
out
rbr
rin
qout = 3.0 ± 0.1
standard
qin = 6.9 ± 0.6
steep