Benvenuti al Mera-TeV! Ottobre 2011 Sala “POE” di OAB a Merate Sala “POE” di OAB a Merate

Lorenzo Sironi

FATE DOMANDE! Le domande sono gradite, anche prima della fine dei contributi. L'incontro è volutamente informale, e lo spirito è orientato alla comprensione delle tematiche ed alla interazione tra i partecipanti.

SOCIAL EVENT w Visita alle Cupole Zeiss e Ruths: Mercoledì 5, ore Mercoledì 5, ore 18.15

SOCIAL DINNER Taverna dei Cacciatori – Imbersago Ore Mercoledì Partenza da Osservatorio ore 19.45

Pranzi e Pause Caffè Pranzi: alle nel parco, di fronte alla Cupola Ruths Coffe Breaks: nella Biblioteca, piano seminterrato edificio principale.

Buon Mera-TeV!!

Beaming

Radio-loud AGNs Gamma Ray Bursts ~ 0.1 M o yr -1  ~20 ~ M o in a few sec  ~300

Lorentz transformations: v along x x’ =  (x – vt) y’ = y z’ = z t’ =  (t – v x/c 2 ) for  t = 0   x =  x’/  Contraction for  x’ = 0   t =  t’ time dilation Text book special relativity x =  (x’ + vt’) y = y’ z = z’ t =  (t’ + v x’/c 2 ) To remember: mesons created at a height of ~15 km can reach the earth, even if their lifetime is a few microsec  ct’ life =hundreds of meters.

v=0  =1 v=0.866c  =2 v Can we see contracted spheres? Einstein: Yes!

James Terrel 1959 Roger Penrose 1959 v=0  =1 v NO! v=0.866c  =2 Rotation, not contraction!

Relativity with photons From rulers and clocks to photographs and frequencies Or: from elementary particles to extended objects

The moving square  =0  =0.5 Your camera, very far away

The moving square t=l’/c vt=  l’ l’/  l tot = l’ (  +1/  ) max:2 1/2 l’ (diag) min: l’ (for  =0)

l’ l’cos  =  l’  cos  =  cos  sin  

 ) 



Time CD = c  t e – c  t e  cos    t A =  t e (1-  cos  )   t A =  t e ’  (1-  cos  )  t e = emission time in lab frame  t e ’ = emission time in comov. frame  t e =  t e ’ 

Relativistic Doppler factor   t A =  t e ’  (1-  cos  )  t A =  t e ’  (1-  cos  ) = ’ /  (1-  cos  ) = ’ /  (1-  cos  )  =  =1  (1-  cos  )  (1-  cos  ) Standard relativity Doppler effect You change frame You remain in lab frame

Relativistic Doppler factor  = = = =1  (1-  cos  )  (1-  cos  ) 2  for  =0 o  for  =1/   for   =  At small angles, Doppler wins over Spec. Relat.

2 5 l i g h t y e a r s i n 3 y e a r s … t h e v e l o c i t y i s 8. 3 c

Nucleo v=0.99c

 app =  sin  1-  cos   =  v app = v   t e  sin   t e (1-  cos  )  s app tAtAtAtA  =0 o   app =0 cos  =  ; sin  =1/    app =   =90 o   app =  There is no  Correct?

Aberration of light

sin  = sin  ’/  d  =  d  ’/  2

sin  = sin  ’/  Aberration of light K’ d  = d  ’/  2 K v

Observed vs intrinsic Intensity  3 I’( ’) I( )  I’( ’) ’  ’  = =invariant I( ) = cm 2 s Hz sterad =erg= dA dt d d  E

Observed vs intrinsic Intensity  3 I’( ’) I( )  I’( ’) ’  ’  = =invariant I( ) = cm 2 s Hz sterad =erg= dA dt d d  E

Observed vs intrinsic Intensity  3 I’( ’) I( )  I’( ’) ’  ’  = =invariant I( ) = cm 2 s Hz sterad =erg= dA’ d  ’/  2 E’   3 I’( ’) = I  4 I’ = F  4 F’ =  blueshift  time  2 aberration

v=0 L=100 W

v= c  =10 L=16MW L=10mW L=0.6mW

v= c  =10 blazars radiogalaxies …….?

v= c  =10 blazars radiogalaxies blazars!

jet counterjet (invisible) v v

Radiation processes

 Line emission and radiative transitions in atoms and molecules  Breemstrahlung/Blackbody  Curvature radiation  Cherenkov  Annihilation  Unruh radiation  Hawking radiation  Synchrotron  Inverse Compton

V=0 V=0 E

() V (  =2 ) Charge at time 9.00 Contracted sphere… E-field lines at time 9.00 point to… where the charge is at 9.00 E Breaking news: what happens with the gravitational field?

dP = e 2 a 2 sin 2  d  4  c 3 V Stop at 8:00

dP = e 2 a 2 sin 2  d  4  c 3 P = 2 e 2 a 2 P = 2 e 2 a 2  3  c 3

Synchrotron

Synchrotron  Ingredients: Magnetic field and relativistic charges  Responsible: Lorentz force  Curiously, the Lorentz force doesn’t work. FL =FL =FL =FL =ddt (  mv) =ec v x B 

Total losses P e = P’ e Please, P e is not P received !! P=E/t and E and t Lorentz transform in the same way

Total losses 2e 2 = 3c 3 a’ 2 = 2e 2 3c 3 (a’ 2 + a’ 2 ) P e = P’ e

Total losses P e = P’ e = 2e 2 3c 3 a2a2a2a2 2222 2e 2 = 3c 3 a’ 2 = 2e 2 3c 3 (a’ 2 + a’ 2 ) P e = P’ e a’ =  2 a a ’ || = 0 a = e v B sin   mc

Total losses P e = P’ e = 2e 2 3c 3 a2a2a2a2 2222 2e 2 = 3c 3 a’ 2 = 2e 2 3c 3 (a’ 2 + a’ 2 ) P e = P’ e a’ =  2 a a ’ || = 0 a = e v B sin   mc P S (  ) = 2e 4 3m 2 c 3  B 2   2  2 sin 2 

Total losses P e = P’ e = 2e 2 3c 3 a2a2a2a2 2222 2e 2 = 3c 3 a’ 2 = 2e 2 3c 3 (a’ 2 + a’ 2 ) P e = P’ e a’ =  2 a a ’ || = 0 a = e v B sin   mc P S (  ) = 2e 4 3m 2 c 3  B 2   2  2 sin 2  P S (  ) = 2  T  cU B   2  2 sin 2 

Total losses P e = P’ e = 2e 2 3c 3 a2a2a2a2 2222 2e 2 = 3c 3 a’ 2 = 2e 2 3c 3 (a’ 2 + a’ 2 ) P e = P’ e a’ =  2 a a ’ || = 0 a = e v B sin   mc P S (  ) = 2e 4 3m 2 c 3  B 2   2  2 sin 2  P S (  ) = 2  T  cU B   2  2 sin 2  = = 4  T cU B   2  2 3 If pitch angles are isotropic

Total losses P e = P’ e = 2e 2 3c 3 a2a2a2a2 2222 2e 2 = 3c 3 a’ 2 = 2e 2 3c 3 (a’ 2 + a’ 2 ) P e = P’ e a’ =  2 a a ’ || = 0 a = e v B sin   mc P S (  ) = 2e 4 3m 2 c 3  B 2   2  2 sin 2  P S (  ) = 2  T  cU B   2  2 sin 2  = = 4  T cU B   2  2 3 If pitch angles are isotropic

Log E Log P S v 2 ~ E   ~ E 2 Why  2 ?? P S (  ) = 2  T  U B   2  2 sin 2  What happens when   0 ? Sure, but what happens to the received power if you are in the beam of the particles?

 mc 2  sin  eB rLrLrLrL = v2v2v2v2a = e B  mc = B = 1/T T = 2  r L /v = 1/T T = 2  r L /v Synchrotron Spectrum Characteristic frequency This is not the characteristic frequency e v B sin   mc a =

v<<c v ~ c

 t A = ?

S = S = 1 tAtAtAtA = 2222eB 2  mc  Compare with B. S = B  3

The real stuff x= x 1/3

The real stuff x=

Max synchro frequency Guilbert Fabian Rees 1983  shock t syn = T  6   m e c 2 TB22TB22TB22TB22 = 2   m e c eB  max ~ B 1/2 B 1/2 1 h S,max ~ B  max = m e c 2 /  F = 70 MeV. 2 (+ beaming)

Emission from many particles N(  ) = K  -p The queen of relativistic distributions Log N(  ) Log  Log Log   )  ( ) d  = 1 4444 N(  ) P S d 

Emission from many particles N(  ) = K  -p The queen of relativistic distributions Log N(  ) Log  Log Log   )  ( )  ~ 1 4444 K  -p B 2  2 d  d Emission is peaked!  Emission is peaked!  S= 2222eB 2  mc dddd d

Emission from many particles N(  ) = K  -p The queen of relativistic distributions Log N(  ) Log  Log Log   )  ( )  ~ 1 4444 K B (1+p)/2 (1-p)/2

Emission from many particles N(  ) = K  -p The queen of relativistic distributions Log N(  ) Log  Log Log   )  ( )  ~ 1 4444 K B  +1 -  ====p-12 power law

 ( )  ~ 1 4444 K B  +1 -  So, what? 4  Vol  ( ) ~  s 2 R K B  +1 -  F( ) ~ 4d24d24d24d2 Log Log F  ) K B  +1 If you know  s and R Two unknowns, one equation… we need another one

Synchrotron self-absorption If you can emit you can also absorbIf you can emit you can also absorb Synchrotron is no exceptionSynchrotron is no exception With Maxwellians it would be easy (Kirchhoff law) to get the absorption coefficientWith Maxwellians it would be easy (Kirchhoff law) to get the absorption coefficient But with power laws?But with power laws? Help: electrons able to emit are also the ones that can absorbHelp: electrons able to emit are also the ones that can absorb

A useful trick  -p Many Maxwellians with kT=  mc 2 I( ) = 2 kT 2 /c 2 = 2  mc 2 2 /c 2 Log  Log N(  = 2222eB 2  mc  B) 1/2  ~ (  B) 1/2 5/2 5/2 B 1/2 ~ There is no K !

From data to physical parameters get B insert B and get K t belongs to thick and thin part. Then in principle one observation is enough

Inverse Compton

 Scattering is one the basic interactions between matter and radiation.  At low photon frequencies it is a classical process (i.e. e.m. waves)  At low frequencies the cross section is called the Thomson cross section, and it is a peanut.  At high energies the electron recoils, and the cross section is the Klein-Nishina one.

 = scattering angle 0 1 Thomson scattering hv 0 << m e c 2 hv 0 << m e c 2 tennis ball against a wall tennis ball against a wall The wall doesn’t move The wall doesn’t move The ball bounces back with the same speed (if it is elastic) The ball bounces back with the same speed (if it is elastic) 1 = 0 1 = 0

Thomson cross section dTdTdTdT dddd = r0r0r0r022 (1+cos 2  ) TTTT = r0r0r0r 888 = r0r0r0r0 mec2mec2mec2mec2 e2e2e2e2 a peanut Hurry up!

Why a peanut?

E B

d dd ddP e2a2e2a2e2a2e2a2 4  c 3 sin 2  = Remember:

E B

dTdTdTdT dddd = r0r0r0r022 (1+cos 2  ) % Pol no Pol

Direct Compton x1 =x1 =x1 =x1 = x0x0x0x0 1+x 0 (1-cos  ) x = h mec2mec2mec2mec2  x0x0x0x0 x1x1x1x1 Klein-Nishina cross section

~ E -1 Klein-Nishina cross section

Inverse Compton: typical frequencies Thomson regime Rest frame K’ x’ 1 =x’ x x’ x1x1x1x1 Lab frame K

Min and max frequencies  = 180 o  1 =0 o x 1 =4  2 x  = 0 o  1 =180 o x 1 =x/4  2

Total loss rate vt TTTT Everything in the lab frame n(  ) = density of seed photons of energy  =h n(  ) = density of seed photons of energy  =h v rel = “relative velocity” between photon and electron v rel = c-vcos  c(1-  cos  ) Hurry up!

Total loss rate vt TTTT There are many  1, because there are many  1.. We must average the term 1-  cos  1, getting Hurry up!

Total loss rate There are many  1, because there are many  1.. We must average the term 1-  cos  1, getting U rad { Hurry up!

Total loss rate If seed are isotropic, average over  and take out the power of the incoming radiation, to get the net electron losses: U rad { = = 4  T cU rad   2  2 3 = = 4  T cU B   2  2 3 Compare with synchrotron losses: Hurry up!

Inverse Compton spectrum The typical frequency is:          Going to the rest frame of the e- we see  0  There the scattered radiation is isotropized  Going back to lab we add another  -factor.

The real stuff downupscattering

downupscattering 75%

Emission from many particles N(  ) = K  -p The queen of relativistic distributions Log N(  ) Log  Log Log   )  ( ) d  = 1 4444 N(  ) P C d  Hurry up!

Emission from many particles N(  ) = K  -p The queen of relativistic distributions Log N(  ) Log  Log Log   )  ( )  ~ 1 4444 K  -p U rad  2 d  d Emission is peaked!  Emission is peaked!  dddd d4 =  Hurry up!

Emission from many particles N(  ) = K  -p The queen of relativistic distributions Log N(  ) Log  Log Log   )  ( )  ~ 1 4444 KU rad (2-p)/2 -1/2 Hurry up!

Emission from many particles N(  ) = K  -p The queen of relativistic distributions Log N(  ) Log  Log Log   )  ( )  ~ 1 4444 KU rad -  ====p-12 power law Hurry up!

Synchrotron Self Compton: SSC Due to synchro, then proportional to:  c B  +1 -   c ( ) ~  2 c B  +1 c -   2 c B  +1 c -  Electrons work twice

End

The moving bar

=0=0=0=0

 app ~ 30

Gravity bends space

There is max frequency of synchro radiation produced by shock-accelerated electrons. Even if we have relativistic shocks, so that  can be ~1 for each passage through the shock, there is a max energy attainable which corresponds to a  e for which t syn [propto 1/(  e B 2 )] is comparable to the gyroperiod (propto  e /B). t syn [propto 1/(  e B 2 )] is comparable to the gyroperiod (propto  e /B). This gives a max  e scaling as 1/B 1/2, This gives a max  e scaling as 1/B 1/2, so that S becomes independent of B and which corresponds to a wavelength e 2 /m e c 2 =classical electron radius: i.e. a photon of energy h S,max = m e c 2 /  F = 70 MeV. h S,max = m e c 2 /  F = 70 MeV. Max synchro frequency Guilbert Fabian Rees 1983

FL =FL =FL =FL =ddt (  mv) =ec v x B P S (  ) = 2e 4 3m 2 c 3  B 2   2  2 sin 2  P S (  ) = 2  T  cU B   2  2 sin 2  r 0 =e 2 /m e c 2  T = 8  r 0 /3 2 = = 4  T cU B   2  2 3 If pitch angles are isotropic  =pitch angle  ~constant, at least for one gyroradius a || = 0 a || = 0 a = a = e v B sin   mc

FL =FL =FL =FL =ddt (  mv) =ec v x B P S (  ) = 2e 4 3m 2 c 3  B 2   2  2 sin 2  P S (  ) = 2  T  cU B   2  2 sin 2  r 0 =e 2 /m e c 2  T = 8  r 0 /3 2 = = 4  T cU B   2  2 3 If pitch angles are isotropic  =pitch angle  ~constant, at least for one gyroradius a || = 0 a || = 0 a = a = e v B sin   mc

FL =FL =FL =FL =ddt (  mv) =ec v x B P S (  ) = 2e 4 3m 2 c 3  B 2   2  2 sin 2  P S (  ) = 2  T  cU B   2  2 sin 2  r 0 =e 2 /m e c 2  T = 8  r 0 /3 2 = = 4  T cU B   2  2 3 If pitch angles are isotropic a = a = e v B sin   mc

Emission from many particles N(  ) = K  -p The queen of relativistic distributions Log N(  ) Log  Log Log   )  ( )  ~ 1 4444 K B 2 (2-p)/2 -1/2 B 1/2 B (2-p)/2

Core