Bremsstrahlung & Synchotron Radiation

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Question 1 In an optically thick source the observed flux is dependent on the projected area and the excitation temperature with a given velocity Vobs (Shaw et al., 2013). Consequently, optically thin sources are also dependent on projected area and excitation temperature, but their flux is calculated by obtaining an integration of the emissivity line over the volume. Therefore, in this case the observed flux will be given by fv=1-e-t2hv4 c31ehv/kTex-14πr2/d2 EQ 1 Where Tex represents the excitation temperature. In an optically thin source Tv << 1 J(k)T which is the emission coefficient can be given by hvk÷ehvkt-1Therefore, the middle equation in EQ 1 can be represented with Jv assuming it is a free path. With respect to unit mass Jv can be given by Ev⍴/4πUsing the optical length Ev⍴/4πThe equation can be differentiated to dEm=Ev4πdvdmdt dm=⍴dV ………(Equation 2) Where ⍴ is the density. Thus, equation one can be written as fv=1-e-t2hv4 c3Ev⍴4π4πr2/d2 ….(Equation 3) Question 2 The Flux Density Fv=1-e-t2hv4 c31ehv/kTex-14πr2/d2 as in Q1 Whereby, 1ehv/kTex-1 = Ev⍴4πTherefore Fv=1-(1.602*10-19)-12*1*1(3.0*108)3 Ev⍴4π 4πr2/d2Which is Fv=1-(1.602*1019)2(3.0*108)3 ⍴dV 4πr2/d2=-1.068x1043 X1011x100x4π10021002=-1.342088x1030JyB) Question 3 The Lorentz factor ϒ has a velocity of V=ϒv(1+ϒ)In a constant accelerating the relativistic motion is obtained by ω=vr=eBϒMeTherefore, the acceleration ω can be given by ω=ebBϒme Ω with respect to frequency is given by 2πf; Thus, this equates to 2πf=ebBϒme Where f=1Hz, assuming mass of the electron is 1, the magnetic flux will be given

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