Self-Consistent Particle Acceleration in Active Galactic Nuc(5)

Adopting the hypothesis that the nonthermal emission of Active Galactic Nuclei (AGN) is primarily due to the acceleration of protons, we construct a simple model in which the interplay of acceleration and losses can be studied together with the formation o

5 where 2= me=(0:039mp ). Using similar arguments one can calculate the neutrino emissivity resulting from proton-proton collisions. Thus, we write 2 0 (19) Qpp (E; t)= 0:037 ppntarg np ( 3; t); p where 3= E=(0:037mp c2 ). It is easy to show using the above relations that the energy lost per unit time by protons is equal to the amount of energy per unit time given to photons, electrons and neutrinos, i.e., o

ur approximate treatment conserves energy. Finally, it is interesting to note that if we assume the proton losses in Eq. (16) are catastrophic, i.e., if we write3.3. Proton-photon pion production

To take this complicated process into account we assumed that the cross section is given by a -function, i.e., d p (y)=dy= 0 p T (y y0 ), where again y is the photon energy seen from 0 the proton rest frame, p=:25 and y0= 103 . We also take the inelasticity to be kp=:2 and the multiplicity to be 1. These values are found to be in good agreement (better than 10%) with the results of Begelman et al. (1990) who calculate the proton losses due to photopion production using the full cross section. We consider the two basic channels (a) (b)

p+ !p+

0

(26)

Lp (; t)= pp

0 targ ppnp

np (; t)];

(20)

then instead of Eq. (16), we obtain essentially Eq. (6) and can immediately write the analytic solution:

np (; t)= n0 H(

inj

0 1+(tacc=tesc)+ntarg pptacc p

inj)

H(

t=t inje acc )

:

(21)

Comparing the above solution with Eq. (2) we see that the inclusion of proton-proton losses causes the proton distribution to become steeper, but does not a ect the time evolution of the upper cut-o .3.2. Proton-photon pair production

p+ !n++ (27) to be equally probable. Whereas the outgoing protons of channel (a) are e ectively trapped, the neutrons of channel (b) are assumed to escape the source without further attenuation. Thus we treat channel (a) as an energy loss process which preserves proton number in contrast to channel (b) which is treated as a catastrophic proton loss. The photopion collisions of channel (a) move protons of Lorentz factors between and+ d to lower ones and add photons to this range which, before interaction, had Lorentz factors around 0==(1 kp ). Thus, using the -function approximation for the cross section, we nd for channel (a): Lp !p;a (; t)= p 0 1 1 p 0 0 2 0 np (; t)n (y0=; t) np (; t)n (y0=; t) (28)(29)

A photon of (dimensionless) energy x can produce an elec- and for the catastrophic losses of channel (b): 0 tron/positron pair in the Coulomb eld of the proton if the p p Lp !p;b (; t)= 2 1 np (; t)n (y0=; t) threshold conditionpx

2

(22)

is satis ed. Assuming that the resulting electron and positron have the same Lorentz factor as the incoming proton (which is true provided the proton-photon collisions occur predominantly close to threshold), the fractional energy loss of the proton is small. In this case the losses can be considered a continuous process and can be written as

As neutral pions decay essentially instantaneously into rays, channel (a) will provide a source term in the photon equation Eq. (13). Assuming the two -rays have equal energy (see, for example, Stecker 1968), this quantity (x) is related to the incoming proton energy 1 by x= (mp=2me )kp 1 . Thus, we can write the photon source term asp Qp !p (x; t)= np ( 1 )n (x1 ) x 0

(23) where x1= kpmp y0=(2xme ). Three neutrinos an

d a positron are created in the decay chain of a+ from channel (b). Assuming again that these are where produced with equal energies, we can write the electron source Z 1 dxn (x; t) pe (x ) (24) term analogous to the photon term in Eq. (30): p= 2= 0 1 Qe !p (; t)= 2 np ( 2 )n (x2 ) p (31) p is the collision rate and pe(y) is the cross-section of the process in units of the Thomson cross-section as a function of the pho- where= 4 m=(k m ) and x= k m y=(4 m ). 2 e p p 2 p p 0 e ton energy y as seen in the rest frame of the proton. Finally the neutrino emissivity can be written as On the other hand, the rate at which this process injects electrons and positrons, (we make no distinction between 2 3 0 (32) Qp !p (E; t)= 2 np ( 3 )n (x3 ) p mec them) is E Qe !pee(; t)= 2 p np (; t): (25) where 3= 4E=(kp mp c2 ) and x3= kp mpc2 y0=(4E ). p

me Lp !pee(; t)= 2 mp@@ ( p

(30)

p np (

; t))