4 2009 Fermi Symposium, Washington, D.C., Nov. 2-5 Appendix: synchrotron source model used in pixel-based approach Following [10], we consider a uniform cloud of rel- ativistic electrons in the external uniform magnetic field with component B$_{⊥}$ perpendicular to the line of sight. We assume that the electron energy distribu- tion is described by a power law: where N$_{0}$ and p are constants. For this case, we can write the radiation transfer equation where I$_{ν}$ is the specific intensity at the frequency ν , ds is the differential element of path length, j$_{ν}$ is the emission coefficient (emissivity), κ$_{ν}$ is the absorption (extinction) coefficient. If j$_{ν}$ and κ$_{ν}$ are constant along the line of sight, the solution of the radiation transfer equation is where L is the size of the electron cloud along the line of sight, and κ$_{ν}$L = τ$_{ν}$ is the optical depth. Following [10], this solution can be written in the form: where where e is the electron charge, m$_{e}$ is the electron mass, and c is the speed of light in vacuum. The power law index p of the electron energy dis- tribution may be found from the spectral index α ( S$_{ν}$ ∼ ν $^{α}$) in the optically thin part of the spectrum: p = 1 − 2 α . The optical depth τ$_{m}$ corresponding to the observed spectral peak frequency ν$_{m}$ may be found from equation exp[ τ$_{m}$ ] = 1 + p +4 5 τ$_{m}$ . The frequency ν$_{1}$ corresponding to τ$_{ν}$$_{1}$ = 1 may be found from ν$_{m}$ , p and τ$_{m}$ : ν$_{1}$ = τ 2 / ( p +4) m ν$_{m}$ . From equations (1), (2) and (3), we can derive the perpendicular component of the magnetic field: Note, that if the size of an emitting region is smaller then the angular resolution, only a lower limit on the intensity and, correspondingly, an upper limit on the magnetic field B$_{⊥}$ can be derived. B$_{⊥}$ = ( c$_{5}$ ( p ) c$_{6}$ ( p ) ) 2 S ( ν$_{1}$ ) − 2 ( ν$_{1}$ 2 c$_{1}$ ) 5 . (4) I$_{ν}$ = S ( ν$_{1}$ ) J ( ν ν$_{1}$ , p ), (1) ν$_{1}$ = 2 c$_{1}$ ( N$_{0}$Lc$_{6}$ ( p )) 2 / ( p $^{+4)}$B ( p +2) / ( p +4) ⊥ , (2) S ( ν$_{1}$ ) = c$_{5}$ ( p ) c$_{6}$ ( p ) B − 1 / 2 ⊥ ν$_{1}$ 2 c$_{1}$$^{5}$$^{/}$$^{2}$ , (3) I$_{ν}$ = L ∫ 0 j$_{ν}$e − s ∫ 0 κ ν $^{ds}$ds = j$_{ν}$ κ$_{ν}$ (1 − e − κ ν $^{L}$), dI$_{ν}$ ds = − κ$_{ν}$I$_{ν}$ + j$_{ν}$ , N ( E ) = N$_{0}$E − $^{p}$, J ( z, p ) = z 5 / $^{2}$(1 − $^{exp[}$− z − ( p +4) / $^{2}$]), c$_{1}$ = 3 e 4 πm 3 $_{e}$c 5 = 6 . $^{27}$× 10 $^{18}$, c$_{3}$ = √$_{3}$ 4 π e 3 m$_{e}$c 2 = 1 . $^{87}$× 10 − $^{23}$, c$_{5}$ ( p ) = 1 4 c$_{3}$ Γ ( 3 p − 1 12 ) Γ ( 3 p + 7 12 )( p + 7 / 3 p + 1 ) , c$_{6}$ ( p ) = 1 32 ( c c$_{1}$ ) 2 c$_{3}$ ( p + 10 3 ) Γ ( 3 p + 2 12 ) Γ ( 3 p + 10 12 ) ,