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

2.   =E mc 2 Efficiency

3. 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 1.8x10 -10 =

4. mgh mc 2 == 980x10 4 (h/100 m) 9x10 20 erg 9x10 20 erg 10 -14 ~

5.  mv 2 2 mc 2 == (v/100 km/h) 2 4x10 -15 (v/c) 2 2 =

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

7. = 0.008 x 0.1 ~ 8x10 -4 Fusion

8. = 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

9. Annihilation  = 1

11. Nuclear Fusion/Accreti on Accretion ?? Annihilation

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

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

14. Jets?

15. Power of jets in blazars X-ray cavities

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

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

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

19. 1992: launch of EGRET on the Compton Gamma-ray Observatory

20. Power of jets in blazars X-ray cavities

21.  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

22. 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)

23. 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.

24. 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.

25. GG+ Nature, 2014

26. 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

27. 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

28. disk synchro SSC EC Low energy e- !

29. disk synchro SSC EC “X-ray valley”

30. disk synchro SSC EC “X-ray valley”

31. 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

32. GG+ Nature 2014 If one p per e-

33. 1 proton per electron  no pairs GG+ Nature, 2014

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

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

36. P BZ ~ a 2 B 2 M 2 Blandford Znajek

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

38. 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

39. 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

40. 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

41. 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

42. 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.