Metabolics: feeding the monster

•Energetics•Accretion disks•Emitted spectrum•Relativistic discs

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Fueling AGNsConversion of mass to energy with some efficiency

€

L = η M•

c 2 M•

=L

ηc 2=1.8 ×10−3 L44

ηM⊕yr−1

U =1

2

GMm

r L ≈

dU

dt=

1

2

GM

r

dm

dt=

GM M•

r⇒ η ≈

1

2

GM

rc 2=

1

12= 0.083 if

RS = 2GM

c 2≈ 3 ×1013 M8cm last stable orbit = 3RS

and ignoring relativistic effects.

ΔU =1

2

GM

3RS

energy available from a particle falling to 3RS

η = 0.057 for a Schwarzschild metric and 0.42 for a Kerr metric (l.s.o.=1

2RS )

M•

E =LE

ηc 2 ≡ Critical mass accretion rate ≈ 2.2M8M⊕yr−1

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

m= GMr angular momentum per unit mass

r ≈10kpc M ≈1011 M⊕

move this particle to 0.01pc around a 107 M⊕BH

L

m must decrease by 105 times

Accretion: Basic concepts

Tidal disruption of a star

€

A star of mass density ρ near a BH of radius R can approach no closer

than the Roche limit without being tidally disrupted :

rR = 2.4ρ BH

ρ*

⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 3

R ⇒ rR > RS

rR

RS

= 2.43M

4πRS3ρ*

⎛

⎝ ⎜

⎞

⎠ ⎟>1 ⇒ M < 0.64

c 6

G3ρ*

⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 2

≈ 5 ×108 ρ*−1/ 2M⊕

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 time-scales each gas element looses energy via collision and radiative cooling but retains its angular momentum.

• Circular orbits

€

vφ ≡ circular velocity

vφ = ΩR vφ2 = −R

dφ

dR φ is the gravitational potential

a fluid rotates differentially, the rate of shearing A is :

A = RdΩ

dR≠ 0

gas elements at different radii rotate with different angular velocity

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 = 2R R Angular momentum is L= MR2 = 2R 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

∂t=

∂

∂t(2πR ⋅ΔR ⋅Σ) = vR (R, t) ⋅2πR ⋅Σ(R, t) − vR (R + ΔR, t) ⋅2π (R + ΔR) ⋅Σ(R + ΔR, t)

taking the limit for small ΔR this becomes :

R∂Σ

∂t+

∂

∂R(RΣvR ) = 0

Conservation of angular momentum

€

L = MR2Ω = 2πR ⋅ΔR ⋅Σ ⋅R2Ω

∂L

∂t= Lin − Lout = 2πR ⋅vR ⋅Σ ⋅R

2Ω − 2π (R + ΔR) ⋅vR (R + ΔR) ⋅Σ(R + ΔR) ⋅(R + ΔR)2Ω

R∂

∂t(ΣR2Ω) = −

∂

∂R(RΣvR ⋅R

2Ω)

If viscosity is introduced among contiguous gas annuli the rate of change of L will depend also in the viscous force:

€

R∂

∂t(ΣR2Ω) +

∂

∂R(RΣvR ⋅R

2Ω) =1

2π

∂G

∂Rwhere

G(R, t) = νΣA ⋅2πR ⋅R νΣA is the viscous force per unit length around the circonference

and ν is the kinematic viscosity

G is the viscous torque of an outer annulus acting on a neighboring inner one

€

R∂Σ

∂t+

∂

∂R(RΣvR ) = 0 A = R

∂Ω

∂R

R∂

∂t(ΣR2Ω) +

∂

∂R(RΣvR ⋅R

2Ω) =1

2π

∂G

∂R∂

∂t(ΣR2Ω) +

1

R

∂

∂R(ΣvR ⋅R

3Ω) =1

2πR

∂(νΣR ⋅∂Ω /∂R)

∂R

∂Σ

∂t=

1

R

∂ ∂(R2Ω) /∂R[ ]−1⋅∂(νΣR3(−∂Ω /∂R)

∂R

⎧ ⎨ ⎩

⎫ ⎬ ⎭

∂RIf the potential is due to a central point mass M• then :

Ω =GM

R3

∂Σ

∂t=

3

R

∂ R1/ 2 ∂(νΣR1/ 2)

∂R

⎧ ⎨ ⎩

⎫ ⎬ ⎭

∂R

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 :

∂

∂t

1

2mvφ

2 −GMm

R

⎡ ⎣ ⎢

⎤ ⎦ ⎥ where m = 2πR ⋅Σ ⋅δR

=∂

∂t2πR ⋅Σ ⋅δR

Ω2R2

2−

GM

R

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥ = A1

must be balanced by the differences between the rate of flow of the energy in and out of

the ring

∂

∂R2πRvR Σ

Ω2R2

2−

GM

R

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥δR = A2

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 :

δR∂

∂R2πWRφΩR2

[ ] −δR ⋅2πWRφR2 ∂Ω

∂R= A3 where 2πWRφR2 is the viscous torque(stress)

2πWRφR2 = G = 2πR2νΣA ⇒ WRφ = νΣA = νΣR∂Ω

∂RA1 + A2 + A3 = 0

Energy dissipation in the disk

€

The energy dissipated into heat per unit area is :

Q+ =1

2WRφR

∂Ω

∂R (1/2 because the disk has 2 sides)

−2RQ+ =∂

∂tΣR

Ω2R2

2−

GM

R

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥+

∂

∂RΣvR R

Ω2R2

2−

GM

R

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥+

∂

∂RWRφΩR2

[ ]

=∂

∂RWRφΩR2

[ ] +Ω2R3

2

∂Σ

∂t+

∂(ΣvR )

∂R

⎡ ⎣ ⎢

⎤ ⎦ ⎥+ GM

∂Σ

∂t+

∂(ΣvR )

∂R

⎡ ⎣ ⎢

⎤ ⎦ ⎥+

ΣR3

2

∂Ω2

∂t+

3

2R2ΣvRΩ2

Q+ = −1

2R

∂

∂RWRφΩR2

[ ] −3

4RΣvRΩ2

now, remember that M = 2πR ⋅ΔR ⋅Σ ⇒ M•

= 2πR ⋅vR ⋅Σ : so,

Q+ = −1

2R

∂

∂RWRφΩR2

[ ] +MΩ2

•

8π

€

Q+ = −1

2R

∂

∂RWRφΩR2

[ ] +MΩ2

•

8π

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 equilibriumalong 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).

€

∂p

∂z= −ρ

GM

R3z = −ρΩ2z ⇒ p(z) = pc 1−

z

H

⎛

⎝ ⎜

⎞

⎠ ⎟2 ⎡

⎣ ⎢

⎤

⎦ ⎥

pc =1

2ρΩ2H 2 is the central pressure of the disk

p =1

3ρΩ2H 2 =

1

3ΣΩ2H = average pressure (ρ =

Σ

H)

us =p

ρ=

ΩH

3 average sound speed

Viscosity

€

2πWRφR2 is the viscous torque(stress)

2πWRφR2 = G = 2πR2νΣA ⇒ WRφ = νΣA = νΣR∂Ω

∂R= viscous force per unit length

ν is the kinematic viscosity,

ν ≈ lt ⋅ut where ut is the turbulent velocity and lt is turbulent mixing length

lt < H ut < us (sound speed =pH

Σ=

p

ρ) ⇒ ν = αHus with α < 1

Kepler law : Ω =GM

R3 R

∂Ω

∂R= Ω

2

3

WRφ =2

3νΣΩ =

2

3ΣΩ ⋅αHus =

2

3 3ΣΩ2αH 2 =

2

3αpH

Steady disks

€

∂∂t

= 0 ⇒ continuity equation becomes :

∂Σ

∂t= 0 =

1

2πR

∂ M•

∂R⇒ M

•

= const = -2πRvR Σ

does not depend on radius!

momentum equation becomes :

∂

∂RΣR3vRΩ + WRφR2

( ) = 0 =∂

∂R

M•

2πR2Ω + WRφR2

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟⇒

M•

πRΩ =

∂

∂RWRφR2

( )⇒ WRφR2 =M•

πRΩ =

M•

2πR2Ω∫ + c

valid for R > 3RS . In this region R↓, Ω↑, L ↓

R < 3RS matter falls in the BH and vR increases exponentially.

At R ≈ 3RS viscous stresses must stop having big effects⇒ WRφ = 0

€

c = − M•

Ω0R02 = − M

• GM

R03

⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 2

R02 = − M

•

GM( )1/ 2

R01/ 2 = − M

• GM

R3

⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 2

R01/ 2R3 / 2 =

= − M•

ΩR01/ 2R3 / 2

therefore :

WRφR2 =M•

2πR2Ω + c =

M•

2πR2Ω − M

•

ΩR01/ 2R3 / 2

WRφ =MΩ

•

2π1−

R0

R

⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 2 ⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

Q+ = −1

2R

∂

∂RWRφΩR2

[ ] +MΩ2

•

8π=

3MΩ2•

8π1−

R0

R

⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 2 ⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

Energy does not depend on viscosity! It is only the gravitational potential that pays.

€

L = 4π Q+

R0

∞

∫ RdR = 3 M•

1−R0

R

⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 2 ⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

R0

∞

∫ Ω2RdR = 3 M•

GM 1−R0

R

⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 2 ⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

R0

∞

∫ R−2dR

=M•

GM

R0

= accretion rate times the binding energy of the last stable orbit!

independent on dissipative forces

The heat loss per unit area is given by the transfer equation :

Q− = acTc4 /τ

a = Stefan©s constant, τ = ρHk(ρ,Tc )

if the disk is optically thick, each element radiates as a BB with TS ⇒

Q− = σT 4 where σ is the Stefan - Bolzman constant.

putting Q+ = Q− ⇒

Ts =3GM M

•

8πR3σ1−

R0

R

⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 2 ⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

−1/ 4

At each radius the spectrum is a BB with TS so the total spectrum is :

Sν ∝ BνR0

Rout∫ (TS (R))2πRdR =ν 3

exp(hν /kTS ) −1R0

Rout∫ 2πRdR

Tmax occurres at R =49

36R0 ⇒ Tmax = 0.488

3GM M•

8πR3σ

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

1/ 4

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 diffusionA 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.

€

dP

dΩ=

e2

4πc 3ε • ˙ v

2 Larmour formula for the power irradiated in a polarization status ε

dP

dΩ=

e2

4πc 3˙ v

2sin2 θ where θ is the angle between the emitted radiation ad the acceleration

F = eE = eε0E0 sin(ω0t) = me ˙ v ˙ v = ε0

e

me

E0 sin(ω0t)

dP

dΩ=

e4 E02

8πme2c 3

sin2 θ P = dΩ∫ e4 E02

8πme2c 3

sin2 θ =e4 E0

2

3me2c 3

Scattering cross section

€

dσ

dΩ=

irradiated energy per unit time and solid angle

incident energy flux per unit time and area

incident energy per unit area and time = time average of pointing flux

dσ

dΩ=

dP

dΩ

1

< S > S =

c

4πE • E < S >=

c

8πE0

2

dσ

dΩ=

e4 E02

8πme2c 3

8π

cE02

sin2 θ = re2 sin2 θ re =

e2

mec2

e- classic radius

σ T = re2 sin3 θ

0

2π

∫ dθ =8π

3re

2 = 6.65 ×10−25cm2

P =e4 E0

2

3m2c 3= σ TcUrad Urad = S /c = energy density of the incident radiation

This is formula is valid only if the momentum of the incident photon is negligible :

hν /c << mec hν << mec2 mec

2 = 511keV

Compton effectquantum mechanical particle approach

€

E1 = hν 1

p1 =hν 1

c=

h

λ1

€

E2 = hν 2

p2 =hν 2

c=

h

λ 2

€

pe =1

cE 2 − me

2c 4 mec2 = initial e- energy

E = final e- energy

€

Momentum conservation : p1 = p2 + pe ⇒ pe2 = p1

2 + p22 − 2 p1 • p2 = p1

2 + p22 − 2p1p2 cosθ

Energy conservation : E0 = mec2, E = E0

2 + pe2c 2 ⇒ p1c + mec

2 = p2c + E02 + pe

2c 2

c( p1 − p2) + mec2 = E0

2 + pe2c 2 squaring : c 2(p1 − p2)2 + 2cE0( p1 − p2) = pe

2c 2

pe2 = p1

2 + p22 − 2p1p2 + 2E0( p1 − p2) /c

−p1p2 + E0(p1 − p2) /c = −p1p2 cosθ

p1p2(1− cosθ) = E0( p1 − p2) /c multiply each term by hc / p1p2E0 :

hc

E0

(1− cosθ) =( p1 − p2)h

p1p2

now use λ = h / p

Compton effect

€

hc

mec2

(1− cosθ) =

h

λ1

−h

λ 2

⎛

⎝ ⎜

⎞

⎠ ⎟h

h

λ1

h

λ 2

=λ1λ 2

h

h

λ1

−h

λ 2

⎛

⎝ ⎜

⎞

⎠ ⎟= λ 2 − λ1

or

ν 2 =ν 1

1+hν 1

mec2

(1− cosθ) = ν 1 if hν 1 << mec

2

The introduction of this factor changes the definition of

cross section and gives the Klein- Nishina formula :

dσ

dΩ=

e2

mec2

⎛

⎝ ⎜

⎞

⎠ ⎟

2ν 2

ν 1

⎛

⎝ ⎜

⎞

⎠ ⎟

2

sin2 θ +ν 1 −ν 2

4ν 1ν 2

⎡

⎣ ⎢

⎤

⎦ ⎥

Inverse Compton

€

ε1 = initial photon energy in the lab frame

ε1' = initial photon energy in the e - frame

ε2 = photon energy in the lab frame after the scattering

ε2 ' = photon energy in the e - frame after the scattering

ε1'= ε1γ (1− β cosθ) from the lab frame to the e- frame

if ε1'<< mec2 one can apply the Thompson scattering, i.e. ε2 '= ε1'

let's go back to the lab frame : ε2 = ε2 'γ(1+ β cosθ) = ε1γ2(1− β cosθ)(1+ β cosθ)

The photon energy has been incremented by a factor γ 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(ε) = number of photons with energy ε

Urad = N(ε)ε = N(hν )hν

Inverse Compton

€

Let's begin considering non - relativistic Thomson scattering. If the Poynting flux

(power per unit area) of a plane wave incident on the e - is S =c

4π

r E

2 the

r E will

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 = Sσ T = cσ TUrad Urad = S /c =r E

2/4π

Let's consider now the radiation scattered by a relativistic e-. The Thomson formula

is valid in the primed frame if ε'<< mec2. In this case :

P'= cσ TU 'rad

We need to go back in the lab frame. We know that P is a Lorentz invariant, therefore :

P = P'= cσ TU 'rad

and we only need to transform U 'rad in Urad .

Inverse Compton

€

We know that ε '= εγ(1+ β cosθ). The rate at which successive photons arrive is multiplied by the

same factor, so N '(ε') = N(ε)γ (1+ β cosθ). In the e- frame :

U 'rad = N 'hν '= Nγ(1+ β cosθ)hνγ (1+ β cosθ) = Uradγ 2(1+ β cosθ)2

Thus the transformation between U 'rad and Urad depends on the angle θ 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 = Uradγ 2 dϕ sinθdθ(1+ β cosθ)2

0

π

∫0

2π

∫

where is the azimuthal angle around the x axis.

U 'rad = Urad

4γ 2

3−

1

3γ 2(1− β 2)

⎡

⎣ ⎢

⎤

⎦ ⎥= Urad

4(γ 2 −1/4)

3 [γ 2(1− β 2) =1]

P'= P = cσ TU 'rad =4

3cσ TUrad (γ 2 −1/4) = Total power in the radiation field after IC upscattering.

The initial power of the photons was cσ TUrad , so the net power added to the radiation field is:

PIC =4

3cσ TUrad (γ 2 −1/4) − cσ TUrad =

4

3cσ TUrad (γ 2 −1) =

4

3cσ TUradγ 2β 2 [(γ 2 −1) = γ 2β 2]

Thermal Comptonization

€

average fractional energy

change per scattering

⎡

⎣ ⎢

⎤

⎦ ⎥×

mean number of

scatterings

⎡

⎣ ⎢

⎤

⎦ ⎥≡ Compton y parameter

A =ε f

ε i

≈4

3γ 2 =16

kT

mc 2

⎛

⎝ ⎜

⎞

⎠ ⎟2

mean amplification per scattering

assuming a thermal electron distribution : N(E) ≈ E 2 exp(−E /kT)

after k scatterings : εk = ε iAk

the probability of a photon undergoing k scatterings before escaping is :

pk (τ es) ~ ε iτ esk

The intensity of the emergent radiation at εk is proportional to pk (τ es) :

I(εk ) ~ I(ε i)τ esk ~ I(ε i)

εk

ε i

⎛

⎝ ⎜

⎞

⎠ ⎟

−α

⇒ α =−log(τ 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)η[ ]{ }PG

A =1+f

1− f [1− (1− a)η ]

The outgoing luminosity is given by :

Lout = LS + LCU + Lrfl where :

LCU =(1−η ) f

1− f + fη − faηLS and Lrfl =

aηf

1− f + fη − faηLS

For small f LCU and Lrfl are proportional to LS.

For f~1 LCU and Lrfl are determined by a and

Energy balance

€

A =1+f

1− f [1− (1− a)η ]

α =−log(τ es)

log(A) A =16

kT

mc 2

⎛

⎝ ⎜

⎞

⎠ ⎟2

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

La componente continua di hard X-rays 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.

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)

(*) 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.

Lo spettro di emissione dei metalli è sovrapposto allo spettro continuo

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. C. Fabian, G. Miniutti, astro-ph/0507409 v1 18 Jul 2005]

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 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.

Profilo di riga

• effetti RELATIVISTICI

• effetti di ORIENTAZIONE

• posizione dell’ultima orbita stabile (ISCO)

• effetti di IONIZZAZIONE

• profilo di EMISSIVITÀ del disco

Numerosi effetti modificano il profilo della riga del ferro:

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 EFFETTO DOPPLER

RELATIVITA’ SPECIALE

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:

€

1+ z =Eem

Eobs

=uak a

( )em

uak 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) =Eem

Eobs

=ν 1

ν 0

=uα

' kα( )

obs

uα k a( )

em

where uα' =

dxα'

ds= (-1;0,0,0) and uα =

dxα

ds are the emitted and observed frame

quadri - velocities and k a = (−ω

c;r k ).

r k is the direction of propagation of the wave with angular velocity

ω = 2πν ; r k =

2π

λ=

ω

v p

where v p is the phase - velocity

€

ds is the metric. Let us consider the expression for ds provided by Roberson :

ds2 = −c 2 1−2MG

rc 2+ 2 β − γ( )

M 2G2

c 4r2+ ....

⎛

⎝ ⎜

⎞

⎠ ⎟dt 2 + 1+ 2γ

MG

rc 2

⎛

⎝ ⎜

⎞

⎠ ⎟dr2 + r2 sin2 θdϕ 2 + dθ 2

( )

ds2 = −c 2Βdt 2 + Αdr2 + r2 sin2 θdϕ 2 + dθ 2( )

where Β =1−2MG

rc 2 and Α =1+ 2γ

MG

rc 2

ds = cdt Β 1− β 2 Α

Β where β =

v

c and v is the three dimensional velocity.

uα = (u0;ui) where u0 =cdt

ds=

1

Β 1− β 2 Α

Β

and ui =dx i

ds=

v i

c Β 1− β 2 Α

Β

recall that uα = (−1;0,0,0) and that kα = (−ω

c;r k ) then :

uα kα =ω

c; uα kα = −

ω

c

1

Β 1− β 2 Α

Β

+v i

c•

r k

1

Β 1− β 2 Α

Β

€

1

ν 0

=(ua 'kα

' )em

(uα kα )obs

=−

ω

c

−ω

c

1

Β 1− β 2 Α

Β

+v i

c•

r k

1

Β 1− β 2 Α

Β

=

=1

1

Β 1− β 2 Α

Β

(1−v i

c•

r k

c

ω)

=Β 1− β 2 Α

Β(1− β cosθ)

=

=

1−2GM

rc 2

⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 2

1− β 21+ 2γ

MG

rc 2

1−2GM

rc 2

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

(1− β cosθ)

1/ 2

=

= 1−2GM

rc 2

⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 2

1− β 2 1+ 2γMG

rc 2

⎛

⎝ ⎜

⎞

⎠ ⎟ 1−

2GM

rc 2

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥

1/ 2

(1+ β cosθ)

€

1−2GM

rc 2

⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 2

1− β 2 1+ 2γMG

rc 2

⎛

⎝ ⎜

⎞

⎠ ⎟ 1−

2GM

rc 2

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥

1/ 2

(1+ β cosθ) =

1+v

ccosθ −

1

2

v 2

c 2−

GM

rc 2+ .....

leaving only the contributions up to second order in v

c

−1

2

v 2

c 2= transverse Doppler effect

GM

rc 2= gravitational redshift

if R =r

3Rg

; Rg =2GM

c 2 and cosθ = + /−1 than :

v

c=

GM

rc

=1

6R; and

GM

rc 2=

1

6Rν 1

ν 0

=1+1

6R−

1

12R−

1

6R

Effetto di orientazioneLa 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.

(Metrica di Schwarzschild)

Il contenuto di riga nel blu è una misura dell’angolo di inclinazione del disco.

i [deg]

Posizione della ISCOSchwarzschild 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).

2c

GMRg =

Schwarzschild RISCO = 6Rg

Kerr RISCO = 1.24Rg

a/M

parametro di spin

Effetti di ionizzazione

)(

)(4)(

rn

rFr x⋅

=ξ

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.

)(rFx

)(rn

parametro di ionizazione

Flusso di raggi X incidente per unità di area alla distanza r

Densità di elettroni

Profilo di emissività del discoIl 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:

qrr −=)(εq = indice di emissività

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

CONCLUSIONI

• i = 33° ± 1°

• rin = 1.8 ± 0.1 Rg (ISCO)

• a/M = 0.93 ± 0.01 KERR

• ξ < 30 erg cm s-1

• profilo di emissività

Galassia di tipo Seyfert Iz = 0.00775

XMM – NewtonChandra

Osservazione della riga Fe Kα

rin

rbr

rout

qin = 6.9 ± 0.6

qout = 3.0 ± 0.1standard

steep