The most powerful persistent engine of Nature Gabriele Ghisellini INAF-Osservatorio di Brera Fabrizio Tavecchio, Laura Maraschi, Annalisa Celotti, Tullia.

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Presentation transcript:

The most powerful persistent engine of Nature Gabriele Ghisellini INAF-Osservatorio di Brera Fabrizio Tavecchio, Laura Maraschi, Annalisa Celotti, Tullia Sbarrato With:

  =E mc 2 Efficiency

Sugar saccharose C 12 H 22 O 11 1g  4 kcal= 16.2 kJ = 1e23 eV = 4.7 eV per molecule E mc 2 == 1.6x10 11 erg 9x10 20 erg 9x10 20 erg 1.8x =

mgh mc 2 == 980x10 4 (h/100 m) 9x10 20 erg 9x10 20 erg ~

 mv 2 2 mc 2 == (v/100 km/h) 2 4x (v/c) 2 2 =

Nuclear fission E mc 2 == 0.2x10 9 eV 235x9.4x10 8 eV 235x9.4x10 8 eV 9x10 -4 ~

= x 0.1 ~ 8x10 -4 Fusion

= 0.1 up to 0.3 for accreting Kerr (Thorne 1974) = GM m GM m R mc 2 R mc 2= 1 2 RgRgRgRg 2R

Annihilation  = 1

Nuclear Fusion/Accreti on Accretion ?? Annihilation

M BH = 5x10 -3 M bulge (Kormendy & Ho 2013 ARAA) Accretion ??  BH = 0.1;  = 0.008x0.1=8x10 -4

E BH = 5/8 E = 5/13 E tot Accretion ??

Jets?

Power of jets in blazars X-ray cavities

Power of jets in blazars X-ray cavities Minimum energy t lobe t lobe

Power of jets in blazars X-ray cavities PV work / t sound

Power of jets in blazars X-ray cavities Measure the size, apply SSC model, find  and # of e-

1992: launch of EGRET on the Compton Gamma-ray Observatory

Power of jets in blazars X-ray cavities

 P r = radiation ~ L obs /  2 P e = relat. electrons P p = protons P B = B-field R jet power and accretion luminosity ~10 17 cm Shakura-Sunyaev disk: L d P jet = P e +P p +P B

If you want to compare disk luminosity and jet power, the best sample is: Blazars detected by Fermi  L obs, jet with broad emission lines  L disk Shaw+ 2012: FSRQs (~220 sources) Shaw+ 2013: BL Lacs (26 with BLR)

The jet cannot have less power than what required to produce the observed luminosity: P jet L obs 2222 > If P jet is twice as much,  halves. We can take P rad as the minimum P jet. This limit is model-independent.

The jet cannot have less power than what required to produce the observed luminosity: P jet L obs 2222 > If P jet is twice as much,  halves. We can take P rad as the minimum P jet. This limit is model-independent.

GG+ Nature, 2014

From P rad to P jet  min  total number of e- (and of protons, if no pairs…) Presence of electron positron pairs Radiative model: SSC vs EC vs baryons

From P rad to P jet  min  total number of e- (and of protons, if no pairs…) Presence of electron positron pairs Radiative model: SSC vs EC vs baryons doesn’t fit even more power

disk synchro SSC EC Low energy e- !

disk synchro SSC EC “X-ray valley”

disk synchro SSC EC “X-ray valley”

Nemmen+ Science 2012 blazars GRBs log P rad P jet =10 P rad P jet =10 P rad P jet = P rad P jet = P rad

GG+ Nature 2014 If one p per e-

1 proton per electron  no pairs GG+ Nature, 2014

If one p per e- Relat. electrons Magnetic Field Radiation Disk

Apparent paradox: Jet power proportional to accretion And yet it is greater….

P BZ ~ a 2 B 2 M 2 Blandford Znajek

P BZ ~ a 2 B 2 M 2 P BZ ~ a 2 B 2 R g 2 c Poynting Flux Blandford Znajek

P BZ ~ a 2 B 2 M 2 P BZ ~ a 2 B 2 R g 2 c Poynting Flux P BZ ~ a 2  c 2 R g 2 c  c 2 ~B 2 /8  P BZ ~ a 2 Mc 2 Blandford Znajek

Rotation > Accretion B-field amplified by accretion can tap the spin energy of the hole This process is very efficient The B-field “does not work”, the jet power comes from the BH spin Yet it is the catalyst for the process. No B no jet. But B is linked to accretion, that’s why P jet is propto L d

Shows results from the fiducial GRMHD simulation A0.99fc for a BH with spin parameter a = 0.99; see Supporting Information for the movie. Tchekhovskoy A et al. MNRAS 2011;418:L79-L83 © 2011 The Authors Monthly Notices of the Royal Astronomical Society © 2011 RAS P jet Mc 2

Shows results from the fiducial GRMHD simulation A0.99fc for a BH with spin parameter a = 0.99; see Supporting Information for the movie. Tchekhovskoy A et al. MNRAS 2011;418:L79-L83 © 2011 The Authors Monthly Notices of the Royal Astronomical Society © 2011 RAS P jet Mc 2 On average: 1 g in  1.5 c 2 erg out

Conclusion P jet ~ Mc 2, larger than L d The jet uses to energy stored in the rotation of the hole, that was provided by accretion. But it takes it out faster than when it was put in.