EXTERNAL COMPTON EMISSION IN BLAZARS OF NONLINEAR SYNCHROTRON SELF-COMPTON-COOLED ELECTRONS

We have written the solutions of the differential Equation (6) dependent on the parameter α, which we now intend to discuss in greater detail.

We call it the injection parameter, and it is defined as the square root of the ratio of the nonlinear cooling term to the linear cooling term at the time of injection, i.e.,

$$\begin{eqnarray} \alpha ^2 &=& \frac{|\dot{\gamma }_{{\rm ssc}}(t=0)|}{|\dot{\gamma }_{{\rm syn}}|+|\dot{\gamma }_{{\rm ec}}|} = \frac{A_0q_0\gamma _0^2}{D_0(1+l_{{\rm ec}})} \nonumber \\ &=& 2.13\times 10^{3} \frac{N_{50}\gamma _4^2}{R_{15}^2 (1+l_{{\rm ec}})}, \end{eqnarray} \tag{ 11 }$$

where N denotes the number of radiating particles, and we applied the same scaling law for the parameters as above.

For α ≪ 1 we see that the linear cooling dominates, resulting in the solution (7). If α ≫ 1, the nonlinear cooling at least initially dominates, giving the solutions (8) and (10) for early and late times, respectively. The time t_c marks the transition from nonlinear to linear cooling.

Comparing the α given above to the injection parameter obtained by SBM (where EC losses were neglected), we find that

$$\begin{equation} \alpha = \frac{\alpha _{{\rm SBM}}}{(1+l_{{\rm ec}})^{1/2}}, \end{equation} \tag{ 12 }$$

where α_SBM = A₀q₀γ²₀/D₀.

This implies that the contribution of the external photons lowers the possibility for nonlinear cooling. Solving Equation (11) for the electron density we obtain

$$\begin{equation} q_0 = \frac{D_0(1+l_{{\rm ec}})}{A_0\gamma _0^2}\alpha ^2 > \frac{D_0}{A_0\gamma _0^2}\alpha _{{\rm SBM}}^2 = q_{0,{\rm SBM}}. \end{equation} \tag{ 13 }$$

This also demonstrates the aforementioned fact that with EC losses included it is harder to cool the electrons nonlinearly. Equation (13) shows that for the same value of α, i.e., the relative strength between linear and nonlinear cooling, one needs a higher electron density in the blob compared to the case where the external losses are neglected.

Using Equation (7) in Equation (14) we find for the case of α ≪ 1:

$$\begin{eqnarray} I_{{\rm ec}}(\epsilon _s,\tau) &=& I_0\epsilon _s (1+\tau)^2 G\left(q(\tau),\Gamma (\tau),\epsilon =\epsilon _{{\rm ec}},\gamma =\frac{\gamma _0}{1+\tau } \right) \nonumber \\ && \times H\left[ \tau \right] H\left[ \gamma _0-\gamma _{{\rm min}}(\epsilon =\epsilon _{{\rm ec}})(1+\tau) \right], \end{eqnarray} \tag{ 25 }$$

after performing the simple integrations of the δ-functions and substituting τ = D₀(1 + l_ec)γ₀t. Here we define I₀ = Rcσ_Tq₀Γ²_bu'_ec/(4π²_ecγ²₀),

$$\begin{eqnarray} &&G\left(q(\tau),\Gamma (\tau),\epsilon = \epsilon _{{\rm ec}},\gamma =\frac{\gamma _0}{1+\tau } \right) \nonumber \\ && \quad = G_0(q)+\frac{\epsilon _s^2(1-q)(1+\tau)^2}{2\gamma _0(\gamma _0-\epsilon _s(1+\tau))}, \end{eqnarray} \tag{ 26 }$$

$$\begin{eqnarray} \Gamma (\tau) &=& 4\epsilon _{{\rm ec}}\frac{\gamma _0}{1+\tau }, \end{eqnarray} \tag{ 27 }$$

$$\begin{eqnarray} q(\tau) &=& \frac{\epsilon _s(1+\tau)^2}{4\epsilon _{{\rm ec}}\gamma _0(\gamma _0-\epsilon _s(1+\tau))}, \end{eqnarray} \tag{ 28 }$$

$$\begin{eqnarray} \gamma _{{\rm min}}(\epsilon =\epsilon _{{\rm ec}}) &=& \frac{\epsilon _s}{2}\left[ 1+\sqrt{1+\frac{1}{\epsilon _{{\rm ec}}\epsilon _s}} \right]. \end{eqnarray} \tag{ 29 }$$

Inserting Equation (25) into Equation (23) we obtain after some calculations (see Appendix B) the final expression for the EC fluence in the case of dominant linear cooling:

$$\begin{equation} F(\epsilon _s) = F_0\gamma _0^3\epsilon _{{\rm ec}}^{3/2} \frac{\epsilon _s^{-1/2}}{(1+\epsilon _{{\rm ec}}\epsilon _s)^{3/2}} \left(1-\frac{\epsilon _s}{\epsilon _{{\rm cut}}} \right), \end{equation} \tag{ 30 }$$

where F₀ = I₀/(D₀(1 + l_ec)γ₀) and _cut = 4_ecγ²₀/(1 + 4_ecγ₀).

In the Thomson limit Γ₀ = 4_ecγ₀ ≪ 1 the spectrum cuts off at _s ≈ Γ₀γ₀ ≪ γ₀. The Thomson limit also implies that _ec ≪ (4γ₀)⁻¹. Hence, there is no break in the spectrum, and the denominator in Equation (30) equals unity.

In the Klein–Nishina limit Γ₀ ≫ 1 we find _ec ≫ (4γ₀)⁻¹, and the cutoff at _s ≈ γ₀. Thus, the spectrum breaks at _s = ⁻¹_ec.

3.2.1. Early-time Limit, t ⩽ t_c

For t ⩽ t_c we use Equation (8) in Equation (14), as well as the same substitution for t as above, and obtain after solving the simple integrations

$$\begin{eqnarray} & & I_{{\rm ec}}(\epsilon _s,\tau) = I_0\epsilon _s H\left[ \tau _c-\tau \right](1+3\alpha ^2\tau)^{2/3} \nonumber \\ & &\quad \times G\left(q(\tau),\Gamma (\tau),\epsilon =\epsilon _{{\rm ec}},\gamma =\frac{\gamma _0}{(1+3\alpha ^2\tau)^{1/3}} \right) \nonumber \\ && \quad \times H\left[ \tau \right] H[ \gamma _0-\gamma _{{\rm min}}(\epsilon =\epsilon _{{\rm ec}})(1+3\alpha ^2\tau)^{1/3} ], \end{eqnarray} \tag{ 31 }$$

with τ_c = (α³ − 1)/3α²,

$$\begin{eqnarray} &&G\left(q(\tau),\Gamma (\tau),\epsilon =\epsilon _{{\rm ec}},\gamma =\frac{\gamma _0}{(1+3\alpha ^2\tau)^{1/3}} \right) \nonumber \\ &&= G_0(q)+\frac{\epsilon _s^2(1-q)(1+3\alpha ^2\tau)^{2/3}}{2\gamma _0 (\gamma _0-\epsilon _s(1+3\alpha ^2\tau)^{1/3})}, \end{eqnarray} \tag{ 32 }$$

$$\begin{eqnarray} \Gamma (\tau) &=& 4\epsilon _{{\rm ec}}\frac{\gamma _0}{(1+3\alpha ^2\tau)^{1/3}}, \end{eqnarray} \tag{ 33 }$$

$$\begin{eqnarray} q(\tau) &=& \frac{\epsilon _s(1+3\alpha ^2\tau)^{2/3}}{4\epsilon _{{\rm ec}}\gamma _0 (\gamma _0-\epsilon _s(1+3\alpha ^2\tau)^{1/3})}. \end{eqnarray} \tag{ 34 }$$

Using Equation (23) with Equation (31) the fluence for the early time limit can be calculated. The full details can be found in Appendix C. The solution depends on the external photon energy _ec, giving three different cases.

For $\epsilon _{{\rm ec}}^{-1}<2(\sqrt{2}-1)\gamma _B$ we have

$$\begin{eqnarray} F(\epsilon _s) &=& \frac{F_0\alpha ^3}{5} \frac{\epsilon _s}{\left(1+\frac{\epsilon _s}{\epsilon _{\gamma _B}} \right)\left(1+\frac{\epsilon _s}{2(\sqrt{2}-1)\gamma _B} \right)^{4}} \nonumber \\ && \times \left(1-\frac{\epsilon _s}{\epsilon _{{\rm cut}}} \right), \end{eqnarray} \tag{ 35 }$$

while for $2(\sqrt{2}-1)\gamma _B<\epsilon _{{\rm ec}}^{-1}<2(\sqrt{2}-1)\gamma _0$ we get

$$\begin{eqnarray} F(\epsilon _s) &=& \frac{F_0\alpha ^3}{5} \frac{\epsilon _s}{\left(1+\frac{\epsilon _s}{\epsilon _{\gamma _B}} \right)^{5/2} \left(1+\epsilon _{{\rm ec}}\epsilon _s\right)^{5/2}} \nonumber \\ && \times \left(1-\frac{\epsilon _s}{\epsilon _{{\rm cut}}} \right). \end{eqnarray} \tag{ 36 }$$

The last part is for $2(\sqrt{2}-1)\gamma _0<\epsilon _{{\rm ec}}^{-1}$ and becomes

$$\begin{eqnarray} F(\epsilon _s) &=& \frac{F_0\alpha ^3}{5} \frac{\epsilon _s}{\left(1+\frac{\epsilon _s}{\epsilon _{\gamma _B}} \right)^{5/2} \left(1+\frac{\epsilon _s}{2(\sqrt{2}-1)\gamma _B} \right)^{5/2}} \nonumber \\ && \times \left(1-\frac{\epsilon _s}{\epsilon _{{\rm cut}}} \right). \end{eqnarray} \tag{ 37 }$$

Here we used γ_B = γ₀/α, and $\epsilon _{\gamma _B}= 4\epsilon _{{\rm ec}}\gamma _B^2/(1+4\epsilon _{{\rm ec}}\gamma _B)$.

3.2.2. Late-time Limit, t ⩾ t_c

As before, we can find the intensity for t ⩾ t_c, if we insert Equation (10) into Equation (14). As in the previous cases we exchange t, and obtain

$$\begin{eqnarray} && I_{{\rm ec}}(\epsilon _s,\tau) = I_0\epsilon _s H\left[ \tau -\tau _c \right] \left(\frac{1+2\alpha ^3}{3\alpha ^2}+\tau\right)^2 \nonumber \\ && \quad \times G\left(q(\tau),\Gamma (\tau),\epsilon =\epsilon _{{\rm ec}},\gamma =\frac{\gamma _0}{\frac{1+2\alpha ^3}{3\alpha ^2}+\tau } \right) \nonumber \\ && \quad \times H\left[ \gamma _0-\gamma _{{\rm min}}(\epsilon =\epsilon _{{\rm ec}})(\frac{1+2\alpha ^3}{3\alpha ^2}+\tau) \right], \end{eqnarray} \tag{ 38 }$$

with

$$\begin{eqnarray} && G\left(q(\tau),\Gamma (\tau),\epsilon =\epsilon _{{\rm ec}},\gamma =\frac{\gamma _0}{1+\tau } \right) \nonumber \\ && \quad = G_0(q)+\frac{\epsilon _s^2(1-q) \left(\frac{1+2\alpha ^3}{3\alpha ^2}+\tau\right)^2}{2\gamma _0 \left(\gamma _0-\epsilon _s\left(\frac{1+2\alpha ^3}{3\alpha ^2}+\tau\right) \right)}, \end{eqnarray} \tag{ 39 }$$

$$\begin{eqnarray} \Gamma (\tau) &=& 4\epsilon _{{\rm ec}}\frac{\gamma _0}{\frac{1+2\alpha ^3}{3\alpha ^2}+\tau }, \end{eqnarray} \tag{ 40 }$$

$$\begin{eqnarray} q(\tau) &=& \frac{\epsilon _s(1+\tau)^2}{4\epsilon _{{\rm ec}}\gamma _0(\gamma _0-\epsilon _s(1+\tau))}. \end{eqnarray} \tag{ 41 }$$

Inserting Equation (38) into Equation (23) one can obtain the fluence in the late-time limit. For details we again refer the reader to Appendix C. The results depend also on the external photon energy, giving two cases this time. For $\epsilon _{{\rm ec}}^{-1}<2(\sqrt{2}-1)\gamma _B$ we find

$$\begin{eqnarray} &&F(\epsilon _s<\epsilon _{\gamma _B}) = F_0\gamma _0^3\epsilon _{{\rm ec}}^{3/2} \frac{\epsilon _s^{-1/2}}{(1+\epsilon _{{\rm ec}}\epsilon _s)^{3/2}} \left(1-\frac{\epsilon _s}{\epsilon _{\gamma _B}} \right), \end{eqnarray} \tag{ 42 }$$

while for $\epsilon _{{\rm ec}}^{-1}>2(\sqrt{2}-1)\gamma _B$ we obtain a single power law in the form

$$\begin{eqnarray} &&F(\epsilon _s<\epsilon _{\gamma _B}) = \frac{16}{3}F_0\epsilon _{{\rm ec}}^{3/2}\gamma _0^3\epsilon _s^{-1/2} \left(1-\frac{\epsilon _s}{\epsilon _{\gamma _B}} \right). \end{eqnarray} \tag{ 43 }$$

3.2.3. Total Fluence in the Case α ≫ 1

Combining the results of the previous sections we obtain the total fluence for the IC component due to interactions with the ambient radiation field in the case where the electrons are at first cooled nonlinearly. We have three different cases depending on the value of the normalized external photon energy _ec.

For $\epsilon _{{\rm ec}}^{-1}<2(\sqrt{2}-1)\gamma _B$ we find

$$\begin{eqnarray} F(\epsilon _s) &=& F_0\gamma _0^3\epsilon _{{\rm ec}}^{3/2} \frac{\epsilon _s^{-1/2}}{\left(1+\epsilon _{{\rm ec}}\epsilon _s\right)^{3/2} \left(1+\frac{\epsilon _s}{\epsilon _{\gamma _B}} \right)^2} \nonumber \\ && \times \left(1-\frac{\epsilon _s}{\epsilon _{{\rm cut}}} \right). \end{eqnarray} \tag{ 44 }$$

Secondly, if $2(\sqrt{2}-1)\gamma _B<\epsilon _{{\rm ec}}^{-1}<2(\sqrt{2}-1)\gamma _0$ the fluence becomes

$$\begin{eqnarray} F(\epsilon _s) &=& \frac{16}{3}F_0\gamma _0^3\epsilon _{{\rm ec}}^{3/2} \frac{\epsilon _s^{-1/2}}{\left(1+\frac{\epsilon _s}{\epsilon _{\gamma _B}} \right) \left(1+\epsilon _{{\rm ec}}\epsilon _s\right)^{5/2}} \nonumber \\ && \times \left(1-\frac{\epsilon _s}{\epsilon _{{\rm cut}}} \right). \end{eqnarray} \tag{ 45 }$$

Finally, we obtain for $2(\sqrt{2}-1)\gamma _0<\epsilon _{{\rm ec}}^{-1}$

$$\begin{eqnarray} F(\epsilon _s) &=& \frac{16}{3}F_0\gamma _0^3\epsilon _{{\rm ec}}^{3/2} \frac{\epsilon _s^{-1/2}}{\left(1+\frac{\epsilon _s}{\epsilon _{\gamma _B}} \right) \left(1+\frac{\epsilon _s}{2(\sqrt{2}-1)\gamma _0} \right)^{5/2}} \nonumber \\ && \times \left(1-\frac{\epsilon _s}{\epsilon _{{\rm cut}}} \right). \end{eqnarray} \tag{ 46 }$$

These are all broken power laws, where at least one break ($\epsilon _{\gamma _B}$) depends strongly on the value of α, and therefore on the nonlinear cooling.

We note that the first case corresponds to the extreme Klein–Nishina limit (Γ₀ ≫ 1), the second one to the mild Klein–Nishina limit (Γ₀ > 1), and the last one to the Thomson limit (Γ₀ ≲ 1).

From Equation (ZS-69) we obtain the synchrotron SED for the case α ≪ 1, which is

$$\begin{eqnarray} &&f^{\prime }_s(\nu ^{\prime }) = 5.6 \times 10^{38} R_{15} \delta ^4 \frac{\alpha ^2}{\gamma _4} \left(\frac{\nu ^{\prime }}{\nu _{{\rm syn}}} \right)^{1/2} e^{-\nu ^{\prime }/\nu _{{\rm syn}}} \,{\rm erg\,s^{-1}}, \nonumber \\ \end{eqnarray} \tag{ 47 }$$

with ν_syn = 4.1 × 10¹⁴δbγ²₄ Hz.

The maximum value of the synchrotron SED,

$$\begin{equation} f^{\prime }_{s,{\rm max}}=2.4\times 10^{38} R_{15} \delta ^4 \frac{\alpha ^2}{\gamma _4} \,{\rm erg\,s^{-1}}, \end{equation} \tag{ 48 }$$

is attained at

$$\begin{equation} \nu ^{\prime }_{s,{\rm max}}=\frac{1}{2}\nu _{{\rm syn}}= 2.1\times 10^{14}\delta b\gamma _4^2 \;\hbox{Hz}. \end{equation} \tag{ 49 }$$

Using Equation (ZS-82) we can write the synchrotron SED for α ≫ 1 as

$$\begin{equation} f^{\prime }_{s} = 5.6\times 10^{38} R_{15} \delta ^4 \frac{\alpha ^2}{\gamma _4} \frac{\left(\frac{\nu ^{\prime }}{\nu _{{\rm syn}}} \right)^{1/2}}{1+\frac{\nu ^{\prime }}{\nu _{c}}} e^{-\nu ^{\prime }/\nu _{{\rm syn}}} \,{\rm erg\,s^{-1}}. \end{equation} \tag{ 50 }$$

It peaks at

$$\begin{equation} \nu ^{\prime }_{s,{\rm max}} = \nu _{c}= 2.9\times 10^{14} \delta \frac{b\gamma _4^2}{\alpha ^2} \;\hbox{Hz} \end{equation} \tag{ 51 }$$

with the maximum value

$$\begin{equation} f^{\prime }_{s,{\rm max}} = 2.4\times 10^{38} R_{15} \delta ^4 \frac{\alpha }{\gamma _4} \,{\rm erg\,s^{-1}}. \end{equation} \tag{ 52 }$$

For α ≪ 1, ZS found two different versions of the SSC SED depending on the Klein–Nishina parameter K = 0.136bγ³₄. In the Thomson limit (K ≪ 1) we use Equation (ZS-72) and get

$$\begin{eqnarray} &&f^{\prime }_{{\rm SSC}}(K\ll 1) = 2.8\times 10^{39} R_{15} \delta ^4 (1+l_{{\rm ec}})\frac{\alpha ^4}{\gamma _4} \nonumber \\ &&\times \left(\frac{\nu ^{\prime }}{\nu _{T}} \right)^{3/4} e^{-\nu ^{\prime }/\nu _{T}} \,{\rm erg\,s^{-1}}, \end{eqnarray} \tag{ 53 }$$

while for the Klein–Nishina limit (K ≫ 1) with Equation (ZS-73) we have

$$\begin{eqnarray} f^{\prime }_{{\rm SSC}}(K\gg 1) &=& 2.1\times 10^{40} R_{15} \delta ^4 (1+l_{{\rm ec}})\frac{\alpha ^4}{b\gamma _4^4} \nonumber \\ && \times \frac{\left(\frac{\nu ^{\prime }}{\nu _{B}} \right)^{3/4}}{\left(1+\frac{\nu ^{\prime }}{\nu _{B}} \right)^{7/4}} \left[ 1-\left(\frac{\nu ^{\prime }}{\nu _{\gamma _0}} \right)^{7/12} \right] \,{\rm erg\,s^{-1}}, \nonumber \\ \end{eqnarray} \tag{ 54 }$$

with the constants ν_T = 1.6 × 10²³δbγ⁴₄ Hz, the break frequency ν_B = 2.3 × 10²⁴δb^−1/3 Hz (Schlickeiser & Röken 2008), and $\nu _{\gamma _0}=1.2\times 10^{24}\delta \gamma _4\;\hbox{Hz}$.

The maximum frequencies

$$\begin{equation} \nu ^{\prime }_{{\rm SSC,max}}(K\ll 1) = \frac{3}{4}\nu _{T}= 1.2\times 10^{23} \delta b\gamma _4^4 \;\hbox{Hz} \end{equation} \tag{ 55 }$$

and

$$\begin{equation} \nu ^{\prime }_{{\rm SSC,max}}(K\gg 1) = \frac{3}{4}\nu _{B}= 1.7\times 10^{24} \delta b^{-1/3} \;\hbox{Hz} \end{equation} \tag{ 56 }$$

imply the maximum values

$$\begin{equation} f^{\prime }_{{\rm SSC,max}}(K\ll 1) = 1.0\times 10^{39} R_{15} \delta ^4 (1+l_{{\rm ec}})\frac{\alpha ^4}{\gamma _4} \,{\rm erg\,s^{-1}} \end{equation} \tag{ 57 }$$

and

$$\begin{equation} f^{\prime }_{{\rm SSC,max}}(K\gg 1) = 6.3\times 10^{39} R_{15} \delta ^4 (1+l_{{\rm ec}})\frac{\alpha ^4}{b\gamma _4^4} \,{\rm erg\,s^{-1}}, \end{equation} \tag{ 58 }$$

respectively.

For the SSC SEDs in the case α ≫ 1 we have to discuss three different cases depending on the Klein–Nishina parameter K, and we begin with the Thomson limit (K ≪ 1) using Equation (ZS-85):

$$\begin{eqnarray} f^{\prime }_{{\rm SSC}}(K\ll 1) &=& 2.8\times 10^{39} R_{15} \delta ^{4} (1+l_{{\rm ec}})\frac{\alpha ^4}{\gamma _4} \nonumber \\ &&\times \frac{\left(\frac{\nu ^{\prime }}{\nu _{T}} \right)^{3/4}}{\left(1 + \frac{\alpha ^4\nu ^{\prime }}{\nu _{T}} \right)^{1/2}} e^{-\nu ^{\prime }/\nu _{T}} \,{\rm erg\,s^{-1}}. \end{eqnarray} \tag{ 59 }$$

The maximum value

$$\begin{equation} f^{\prime }_{{\rm SSC,max}}(K\ll 1) = 1.6\times 10^{39} R_{15} \delta ^4 (1+l_{{\rm ec}})\frac{\alpha ^2}{\gamma _4} \,{\rm erg\,s^{-1}} \end{equation} \tag{ 60 }$$

is attained at

$$\begin{equation} \nu ^{\prime }_{{\rm SSC,max}}(K\ll 1) = \frac{1}{4}\nu _{T}= 4.0\times 10^{22} \delta b \gamma _4^4 \;\hbox{Hz}. \end{equation} \tag{ 61 }$$

For the mild Klein–Nishina limit (1 ≪ K ≪ α³) Equation (ZS-88) yields

$$\begin{eqnarray} && f^{\prime }_{{\rm SSC}}(1\ll K\ll \alpha ^3) = 1.5\times 10^{39} R_{15} \delta ^4 (1+l_{{\rm ec}})\frac{\alpha ^4}{\gamma _4} \nonumber \\ && \quad \times \frac{\left(\frac{\nu ^{\prime }}{\nu _{T}} \right)^{3/4}}{\left(1+\frac{\alpha ^4\nu ^{\prime }}{\nu _{T}} \right)^{1/2} \left(1+\frac{\nu ^{\prime }}{\nu _{B}} \right)^{13/4}} \left[ 1-\left(\frac{\nu ^{\prime }}{\nu _{\gamma _0}} \right)^{13/3} \right] \,{\rm erg\,s^{-1}}. \nonumber \\ \end{eqnarray} \tag{ 62 }$$

This triple power law peaks at

$$\begin{eqnarray} &&\nu ^{\prime }_{{\rm SSC,max}}(1\ll K\ll \alpha ^3) \approx \frac{1}{12}\nu _{B}= 1.9\times 10^{23} \delta b^{-1/3} \;\hbox{Hz} \nonumber \\ \end{eqnarray} \tag{ 63 }$$

and reaches a maximum value of

$$\begin{eqnarray} f^{\prime }_{{\rm SSC,max}}(1\ll K\ll \alpha ^3) &\approx& 1.2\times 10^{39} R_{15} \delta ^4 (1+l_{{\rm ec}}) \nonumber \\ && \times \frac{\alpha ^2}{b^{1/3}\gamma _4^2} \,{\rm erg\,s^{-1}}. \end{eqnarray} \tag{ 64 }$$

The last case is the extreme Klein–Nishina limit (1 ≪ α³ ≪ K). We obtain the SED from Equation (ZS-91), resulting in

$$\begin{eqnarray} && f^{\prime }_{{\rm SSC}}(1\ll \alpha ^3 \ll K) = 1.1\times 10^{40} R_{15} \delta ^4 (1+l_{{\rm ec}})\frac{\alpha ^4}{b\gamma _4^4} \nonumber \\ && \quad \times \frac{\left(\frac{\nu ^{\prime }}{\nu _{B}} \right)^{3/4}}{\left(1+\frac{\nu ^{\prime }}{\nu _{B}} \right)^{7/4} \left(1+\frac{\alpha \nu ^{\prime }}{\nu _{\gamma _0}} \right)^{2}} \left[ 1-\left(\frac{\nu ^{\prime }}{\nu _{\gamma _0}} \right)^{13/3} \right] \,{\rm erg\,s^{-1}}. \nonumber \\ \end{eqnarray} \tag{ 65 }$$

The SED peaks with a maximum value

$$\begin{eqnarray} &&f^{\prime }_{{\rm SSC,max}}(1\ll \alpha ^3 \ll K) \approx 3.4\times 10^{39} R_{15} \delta ^4 (1+l_{{\rm ec}})\frac{\alpha ^4}{b \gamma _4^4} \,{\rm erg\,s^{-1}}\nonumber \\ \end{eqnarray} \tag{ 66 }$$

at a peak frequency

$$\begin{eqnarray} &&\nu ^{\prime }_{{\rm SSC,max}}(1\ll \alpha ^3 \ll K) \approx \frac{3}{4}\nu _{B}= 1.7\times 10^{24} \delta b^{-1/3} \;\hbox{Hz}. \nonumber \\ \end{eqnarray} \tag{ 67 }$$

In the case of α ≪ 1 the SED can be calculated from Equation (30), becoming

$$\begin{eqnarray} f^{\prime }_{{\rm ec}}(\nu ^{\prime }) &=& 4.0\times 10^{34} R_{15} l_{{\rm ec}}\frac{\delta ^4\alpha ^2}{\gamma _4^2\epsilon _{{\rm ec}}} \nonumber \\ && \times \frac{\left(\frac{\nu ^{\prime }}{\nu _{{\rm br}}} \right)^{1/2}}{\left(1+\frac{\nu ^{\prime }}{\nu _{{\rm br}}} \right)^{3/2}} \left(1-\frac{\nu ^{\prime }}{\nu _{{\rm cut}}} \right) \,{\rm erg\,s^{-1}}, \end{eqnarray} \tag{ 68 }$$

where ν_br = 1.23 × 10²⁰δ⁻¹_ec Hz and ν_cut = 1.23 × 10²⁰δ_cut Hz. In the Thomson limit (_ec ≪ (4γ₀)⁻¹) the maximum frequency

$$\begin{eqnarray} &&\nu ^{\prime }_{{\rm ec,max}} \approx 1.6\times 10^{20}\delta \epsilon _{{\rm ec}}\gamma _0^2 \;\hbox{Hz} \end{eqnarray} \tag{ 69 }$$

leads to a peak value of

$$\begin{eqnarray} &&f^{\prime }_{{\rm ec,max}} \approx 4.4\times 10^{38} R_{15} l_{{\rm ec}}\frac{\delta ^4\alpha ^2}{\gamma _4} \,{\rm erg\,s^{-1}}. \end{eqnarray} \tag{ 70 }$$

In the Klein–Nishina limit (_ec ≫ (4γ₀)⁻¹) the maximum value

$$\begin{eqnarray} &&f^{\prime }_{{\rm ec,max}} \approx 1.6\times 10^{34} R_{15} l_{{\rm ec}}\frac{\delta ^4\alpha ^2}{\gamma _4^2\epsilon _{{\rm ec}}} \,{\rm erg\,s^{-1}} \end{eqnarray} \tag{ 71 }$$

is attained at

$$\begin{eqnarray} &&\nu ^{\prime }_{{\rm ec,max}} \approx \frac{1}{2}\nu _{{\rm br}}= 6.2\times 10^{19}\frac{\delta }{\epsilon _{{\rm ec}}} \;\hbox{Hz}. \end{eqnarray} \tag{ 72 }$$

Now, we convert the results of Equations (44)–(46) into SEDs, which are the cases for α ≫ 1, yielding for the extreme Klein–Nishina limit $\nu _{{\rm br}}<\nu _{\gamma _{B}r}$

$$\begin{eqnarray} f^{\prime }_{{\rm ec}}(\nu ^{\prime }) &=& 4.0\times 10^{34} R_{15} l_{{\rm ec}}\frac{\delta ^4\alpha ^2}{\gamma _4^2\epsilon _{{\rm ec}}} \nonumber \\ && \times \frac{\left(\frac{\nu ^{\prime }}{\nu _{{\rm br}}} \right)^{1/2}}{\left(1+\frac{\nu ^{\prime }}{\nu _{{\rm br}}} \right)^{3/2} \left(1+\frac{\nu ^{\prime }}{\nu _{\gamma _B}} \right)^{2}} \left(1-\frac{\nu ^{\prime }}{\nu _{{\rm cut}}} \right) \,{\rm erg\,s^{-1}}, \nonumber \\ \end{eqnarray} \tag{ 73 }$$

with $\nu _{\gamma _{B}r}=1.02\times 10^{24}\delta ({\gamma _4}/{\alpha }) \;\hbox{Hz}$ and $\nu _{\gamma _B}=1.23\times 10^{20}\delta \epsilon _{\gamma _B}\;\hbox{Hz}$. It peaks at

$$\begin{equation} \nu ^{\prime }_{{\rm ec,max}} \approx \frac{1}{2}\nu _{{\rm br}}= 6.2\times 10^{19} \delta \epsilon _{{\rm ec}}^{-1} \;\hbox{Hz} \end{equation} \tag{ 74 }$$

with a maximum value of

$$\begin{equation} f^{\prime }_{{\rm ec,max}} \approx 1.6\times 10^{34} R_{15} l_{{\rm ec}} \frac{\delta ^4\alpha ^2}{\gamma _4^2\epsilon _{{\rm ec}}} \,{\rm erg\,s^{-1}}. \end{equation} \tag{ 75 }$$

In the mild Klein–Nishina limit $\nu _{\gamma _{B}r}<\nu _{{\rm br}}<\nu _{\gamma _{0}r}$, with $\nu _{\gamma _{0}r}=1.02\times 10^{24}\delta \gamma _4 \;\hbox{Hz}$, we obtain

$$\begin{eqnarray} f^{\prime }_{{\rm ec}}(\nu ^{\prime }) &=& 4.0\times 10^{34} R_{15} l_{{\rm ec}}\frac{\delta ^4\alpha ^2}{\gamma _4^2\epsilon _{{\rm ec}}} \nonumber \\ && \times \frac{\left(\frac{\nu ^{\prime }}{\nu _{{\rm br}}} \right)^{1/2}}{\left(1+\frac{\nu ^{\prime }}{\nu _{\gamma _B}} \right) \left(1+\frac{\nu ^{\prime }}{\nu _{{\rm br}}} \right)^{5/2}} \left(1-\frac{\nu ^{\prime }}{\nu _{{\rm cut}}} \right) \,{\rm erg\,s^{-1}}. \nonumber \\ \end{eqnarray} \tag{ 76 }$$

The maximum value

$$\begin{equation} f^{\prime }_{{\rm ec,max}} \approx 2.0\times 10^{36} R_{15} l_{{\rm ec}} \frac{\delta ^4\alpha ^{3/2}}{\gamma _4^{3/2}\epsilon _{{\rm ec}}^{1/2}} \,{\rm erg\,s^{-1}} \end{equation} \tag{ 77 }$$

is attained at

$$\begin{equation} \nu ^{\prime }_{{\rm ec,max}} \approx \nu _{\gamma _B}= 1.23\times 10^{24} \delta \frac{\gamma _4}{\alpha } \;\hbox{Hz}. \end{equation} \tag{ 78 }$$

Finally, we have the Thomson limit $\nu _{\gamma _{0}r}<\nu _{{\rm br}}$, where the SED becomes

$$\begin{eqnarray} f^{\prime }_{{\rm ec}}(\nu ^{\prime }) &=& 4.0\times 10^{34} R_{15} l_{{\rm ec}} \frac{\delta ^4\alpha ^2}{\gamma _4^2\epsilon _{{\rm ec}}} \nonumber \\ && \times \frac{\left(\frac{\nu ^{\prime }}{\nu _{{\rm br}}} \right)^{1/2}}{\left(1+\frac{\nu ^{\prime }}{\nu _{\gamma _B}} \right)^{3/2} \left(1+\frac{\nu ^{\prime }}{\nu _{\gamma _{0}r}} \right)^{2}} \left(1-\frac{\nu ^{\prime }}{\nu _{{\rm cut}}} \right) \,{\rm erg\,s^{-1}}. \nonumber \\ \end{eqnarray} \tag{ 79 }$$

It peaks at

$$\begin{equation} \nu ^{\prime }_{{\rm ec,max}} \approx \nu _{\gamma _B}= 4.92\times 10^{28} \delta \frac{\gamma _4^2 \epsilon _{{\rm ec}}}{\alpha ^2} \;\hbox{Hz} \end{equation} \tag{ 80 }$$

(note that we use a different approximation for $\nu _{\gamma _B}$, that is for $\epsilon _{\gamma _B}$, compared to Equation (78)) reaching a maximum value of

$$\begin{equation} f^{\prime }_{{\rm ec,max}} \approx 1.9\times 10^{38} R_{15} l_{{\rm ec}} \frac{\delta ^4\alpha }{\gamma _4} \,{\rm erg\,s^{-1}}. \end{equation} \tag{ 81 }$$

In Figure 1 we present several model SEDs for different sets of parameters. Since the number of free parameters is fairly large, a strict parameter study is beyond the scope of this paper, although we will discuss some aspects below. Therefore, we use some global parameters, which are unchanged through all figures, namely, δ = 10, and Γ_{b, 1} = L'₄₆ = τ₋₂ = R'_pc = R₁₅ = 1. The normalized external photon energy is also fixed at _ec = 10⁻⁴. This leaves us with three free parameters: the injection parameter α, the relative strength of the EC cooling l_ec, and the initial electron Lorentz factor γ₀. These parameters are given at the top of the plots for each case.

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Since we use l_ec as a free parameter and fix almost all parameters it contains, we get the magnetic field strength b as a dependent variable. Increasing l_ec results in a decreasing magnetic field. This choice also demonstrates that the figures given as examples here only cover a very narrow range of possible realizations.

We note that for the top panels of Figure 1 we use Equation (68) for the EC component, while we use Equation (79) in the middle plots and Equations (76) and (73) for the lower plots, respectively. This is due to our choice of parameters.

The first obvious result is that for α ≪ 1 the IC component is significantly influenced, if not dominated, by the EC component. For α ≫ 1 the main contribution comes from the SSC component, at least in the Thomson and mild Klein–Nishina limits. This is expected, since α controls the main cooling mechanism, which should also influence the SED.

Similarly, this is also correct if one considers the linear coolings only, where the relative strength is controlled by l_ec. For example, in the top plots of Figure 1, where we use Equation (68), one can see that for l_ec ≪ 1 the synchrotron component dominates the EC component, while for l_ec ≫ 1 it is the other way round. Since $l_{{\rm ec}}=|\dot{\gamma }_{{\rm ec}}| / |\dot{\gamma }_{{\rm syn}}|$, one could expect that the ratio of the peak values R_{ec, s} = f'_{ec, max}/f'_{s, max}∝l_ec. This is certainly true for the Thomson limit of Equation (68), as is shown in Equation (D2) of Appendix D. However, due to our choice of parameters the top panels of Figure 1 use the Klein–Nishina limit of Equation (68), where Equation (D3) demonstrates that the ratio is more involved and depends also on γ₀ and _ec. Using the parameters of the plots in the top row of Figure 1, we find that the ratios become R_{ec, s} ≈ 0.07 and R_{ec, s} ≈ 66, respectively, consistent with the plots.

In Appendix D we also show the ratios between the maximum values of the SSC and EC components. As one can see, every R_{ssc, ec}∝(1 + l_ec)/l_ec, with the consequence that the ratio is independent of l_ec for l_ec ≫ 1. Although the dependence on other parameters is rather involved and changes from case to case, we can say that a lower value of the external photon energy _ec favors the EC component, while a higher value of the injection parameter α favors the SSC component, as one would expect.

On the other hand, if l_ec ≪ 1, we find R_{ssc, ec}∝l⁻¹_ec. This implies that the EC component is much reduced compared to the SSC component. This is also expected, since in this case the EC cooling is the weakest operating mechanism, if the other parameters are chosen accordingly. It is, of course, still possible that SSC could be lower than EC, since the ratios in Appendix D can always become smaller than unity.

As one can see in the lower panels of Figure 1, we achieve similar results in the mild and extreme Klein–Nishina regimes. In particular, in the latter case, we have an example where the inclusion of the EC cooling has significant effects on the emerging spectrum, giving a different power law and peak value compared to the case where either EC or SSC is neglected.

As stated above, in Figure 1 we just show some model plots with only a very limited range of application, since we fixed most of the free parameters. Some of them, however, might have a significant impact on the resulting SED, such as the radius R of the plasma blob, which we set to R₁₅ = 1. This value corresponds to a light crossing timescale of t_lc = R/(δc) ≈ 0.9h (with δ = 10), which is usually equal to the variability timescale t_var of the source. Naturally, a broad range of variability timescales is observed in blazars, ranging from minutes (as in the cases of PKS 2155−304, Aharonian et al. 2007 and PKS 1222+216, Tavecchio et al. 2011) to months (e.g., Abdo et al. 2010), indicating a large range of possible radii for the emission blobs.

Increasing the radius of our plasma blob would result in a decreasing value of the injection parameter, since α∝R⁻¹ according to Equation (11). This implies that the SSC component would become less important, leading to a dominant EC flux in the high-energy regime.

On the other hand, long variability timescales can also be the result of inefficient cooling or other factors with long timescales, such as a rotating jet (e.g., Agudo et al. 2012). Our simple model assumes an instantaneous and effective cooling of the electrons far from steady state, favoring short variability timescales, at least if one is interested in the effects of nonlinear SSC cooling. Schlickeiser (2009) as well as Zacharias & Schlickeiser (2010) showed that the nonlinear process begins orders of magnitudes earlier to cool the electrons than the standard linear synchrotron cooling, implying shorter variability timescales. However, a true analysis of the variability should be performed with the help of light curves, which we aim to present in another paper.

We should also note that our discussion adopts a point-like emission zone, which means that we neglect additional effects from photon retardation variability processes from the finite size of the emission zone. For an extended analysis of these effects we refer to the discussion in Eichmann et al. (2010, 2012).

Another parameter, which might easily have different values, is the Doppler factor δ. In the literature it usually covers a range 10–30, but extreme values of 50 and higher have been used to model blazar spectra (e.g., Begelman et al. 2008), although the latter are mostly regarded as unrealistic.

For our plots we used δ = 10. Increasing that value would not change the overall appearance of the SED, since in our simple model the dependences on δ are the same in all cases; however the peak luminosities would increase significantly. In the plots in Figure 1 we achieve γ-ray luminosities roughly between 10³⁹ erg s⁻¹ and 10⁴⁶ erg s⁻¹. The former value is rather low for blazars. However, the values of the peak luminosity depend strongly on the parameters, which might differ from the ones used to create the plots.

As an example: with δ = 50, α = 50, and l_ec = 100 one would arrive at an SSC luminosity of roughly 10⁵⁰ erg s⁻¹. This is close to the highest recorded luminosity of blazars, achieved by 3C 454.3 with a peak luminosity of 2.1 × 10⁵⁰ erg s⁻¹ (Abdo et al. 2011b). In principle, the above-stated values for α and l_ec could be achieved by setting N₅₀ ≈ 120 and b = 0.14. The high number of electrons in the source may not be that unrealistic, since a large outburst requires something extreme going on, such as a large mass load into the jet. According to Equation (61) the peak frequency of the SSC component would be at roughly 3 × 10²³ Hz, which is reasonable (Abdo et al. 2011b). Using Equation (51) we obtain a synchrotron peak frequency roughly at 8 × 10¹¹ Hz, which is in the far-infrared regime. Still, a Doppler factor of 50 is quite high, severely questioning the validity of the results of our simple model. Only a detailed spectral analysis, potentially also including a discussion of variability aspects (see above), can decide about the validity. However, such an analysis is beyond the scope of this paper.

Another important aspect is the energy transported and dissipated by the content of the jet. The initial comoving energy density of the electrons is given by u_e = q₀γ₀m_ec². Using Equation (13), we obtain

$$\begin{equation} u_e = 0.09 \frac{(1+l_{{\rm ec}})\alpha ^2}{R_{15}\gamma _4} \,\hbox{erg}\,\hbox{cm}^{-3}. \end{equation} \tag{ 82 }$$

The power transported by these electrons is given by $P_e = \pi R^2 c\beta _{\Gamma _b}\Gamma _b^2 u_e$, yielding with $\beta _{\Gamma _b}\approx 1$

$$\begin{equation} P_e = 8.4\times 10^{41} \Gamma _{b,1}^2 R_{15} \frac{(1+l_{{\rm ec}})\alpha ^2}{\gamma _4} \,{\rm erg\,s^{-1}}. \end{equation} \tag{ 83 }$$

A widely used measure to quantify whether the jet is dominated by particles or the magnetic field is the equipartition parameter

$$\begin{equation} \frac{u_e}{u_B} = 2.2 \frac{(1+l_{{\rm ec}})\alpha ^2}{b^2 R_{15}\gamma _4}, \end{equation} \tag{ 84 }$$

which can be rewritten with the help of Equation (4), giving

$$\begin{equation} \frac{u_e}{u_B} = 24.4 \frac{R_{{\rm pc}}^{\prime 2}}{L_{46}^{\prime }\tau _{-2}R_{15}} \frac{(1+l_{{\rm ec}})l_{{\rm ec}} \alpha ^2}{\Gamma _{b,1}^2 \gamma _4}. \end{equation} \tag{ 85 }$$

Table 1 lists the values of these constraints for the plots given in Figure 1. Depending on the parameters used we obtain a wide range of possible realizations.

A first result is that we are far from equipartition with all our setups, and only the case with the lowest energies involved shows a dominant magnetic field. This is reasonable, since the number of electrons in this case is also at a minimum in our models. This is also the only case where the synchrotron peak dominates the IC peak. However, we should note that this might be an artifact of our parameter settings, i.e., the inverse dependence of b on l_ec, while we left all other parameters in l_ec unchanged. Keeping b at some value and changing some other parameters in l_ec might give a different picture.

The energy density u_e and the transferred power P_e of the electron naturally increase with the number of particles in the blob. Comparing our values of P_e with, e.g., those of Ghisellini et al. (2009), we find that our plots cover a range that is usually below the obtained values for FSRQs, which is again due to our choice of parameters. The case with α = 10, l_ec = 100, and γ₄ = 1 is, however, exactly in the range obtained by these authors. Higher values, as we used for the discussion about 3C 454.3 above, are also possible. So, from the energetics point of view our model is definitely supported.

For t ⩽ t_c we use Equation (8) in Equation (14) and obtain

$$\begin{eqnarray} &&I_{{\rm ec}}(\epsilon _s,t) = \frac{3Rc\sigma _T}{16\pi }\epsilon _s\int \nolimits _0^{\infty } d \epsilon \frac{\frac{4}{3}\Gamma _b^2 u^{\prime }_{{\rm ec}} \delta \left(\epsilon -\epsilon _{{\rm ec}} \right)}{\epsilon ^2} \int \nolimits _{\gamma _{{\rm min}}}^{\infty } d \gamma \frac{q_0}{\gamma ^2} H\left[ t_c-t \right] G(q,\Gamma) \delta \left(\gamma -\frac{\gamma _0}{(1+3D_0(1+l_{{\rm ec}})\gamma _0\alpha ^2t)^{1/3}} \right). \end{eqnarray} \tag{ C1 }$$

Using the same definitions of τ and I₀ one can easily arrive at Equation (31). Then, the fluence integral can be written as

$$\begin{eqnarray} &&F(\epsilon _s) = F_0\epsilon _s\int \nolimits _0^{\infty } d \tau (1+3\alpha ^2\tau)^{2/3} G\left(q(\tau),\Gamma (\tau),\epsilon =\epsilon _{{\rm ec}},\gamma =\frac{\gamma _0}{(1+3\alpha ^2\tau)^{1/3}} \right) H\left[ \tau _c-\tau \right] H\left[ \frac{\gamma _0}{\gamma _{{\rm min}}(\epsilon _{{\rm ec}})}-(1+3\alpha ^2\tau)^{1/3} \right]. \nonumber \\ \end{eqnarray} \tag{ C2 }$$

Examining the Heaviside functions we find that τ_c is the upper limit as long as $\epsilon _s<\epsilon _{\gamma _B}$, while for $\epsilon _s>\epsilon _{\gamma _B}$ the upper limit becomes [(γ₀/γ_min)³ − 1]/3α². As long as they are not really needed we will refer to both of them as the respective variable with the subscript up.

The break energy is given by

$$\begin{equation} \epsilon _{\gamma _B}= \frac{\gamma _B}{1+\frac{1}{4\gamma _B\epsilon _{{\rm ec}}}}, \end{equation} \tag{ C3 }$$

where γ_B = γ₀/α.

Substituting μ = (1 + 3α²τ)^1/3 we obtain

$$\begin{equation} F(\epsilon _s) = \frac{F_0\epsilon _s}{\alpha ^2} \int \nolimits _1^{\mu _{{\rm up}}} d \mu \,\mu ^4 G(q,\Gamma,\mu), \end{equation} \tag{ C4 }$$

where $\mu _{{\rm up}}(\epsilon _s<\epsilon _{\gamma _B}) = \alpha$ and $\mu _{{\rm up}}(\epsilon _s>\epsilon _{\gamma _B}) = \gamma _0/\gamma _{{\rm min}}$.

Inverting q(μ) yields

$$\begin{equation} \mu = 2\epsilon _{{\rm ec}}\gamma _0 q \left(\sqrt{1+\frac{1}{\epsilon _s\epsilon _{{\rm ec}}q}}- 1 \right). \end{equation} \tag{ C5 }$$

Inserting this into Equation (C4) with $q_{{\rm up}}(\epsilon _s<\epsilon _{\gamma _B}) = \epsilon _s\alpha ^2/(4\epsilon _{{\rm ec}}\gamma _0(\gamma _0-\epsilon _s\alpha))$ and $q_{{\rm up}}(\epsilon _s>\epsilon _{\gamma _B}) = 1$, we find

$$\begin{eqnarray} &&F(\epsilon _s) = \frac{16F_0\epsilon _{{\rm ec}}^5\gamma _0^5}{\alpha ^2}\epsilon _s\int \nolimits _{\frac{\epsilon _s}{4\epsilon _{{\rm ec}}\gamma _0(\gamma _0-\epsilon _s)}}^{q_{{\rm up}}} d q \frac{q^4}{\sqrt{1+\frac{1}{\epsilon _s\epsilon _{{\rm ec}}q}}} \left(\sqrt{1+\frac{1}{\epsilon _s\epsilon _{{\rm ec}}q}}- 1 \right)^6 \left[ G_0(q)+2\epsilon _s\epsilon _{{\rm ec}}q(1-q) \right]. \end{eqnarray} \tag{ C6 }$$

Using again the substitution x = (_secq)⁻¹ we eventually obtain

$$\begin{eqnarray} &&F(\epsilon _s) = \frac{16F_0\gamma _0^5}{\alpha ^2}\epsilon _s^{-4} \int \nolimits _{x_{{\rm down}}}^{\frac{4\gamma _0(\gamma _0-\epsilon _s)}{\epsilon _s^2}} d x \frac{\left(\sqrt{1+x} - 1 \right)^6}{x^6 \sqrt{1+x}} \left[ G_0\left(\frac{1}{\epsilon _s\epsilon _{{\rm ec}}x} \right) + \frac{2}{x}\left(1-\frac{1}{\epsilon _s\epsilon _{{\rm ec}}x} \right) \right], \end{eqnarray} \tag{ C7 }$$

with the lower limits $x_{{\rm down}}(\epsilon _s<\epsilon _{\gamma _B}) = 4\gamma _0(\gamma _0-\alpha \epsilon _s)/\epsilon _s^2\alpha ^2$ and $x_{{\rm down}}(\epsilon _s>\epsilon _{\gamma _B}) = (\epsilon _s\epsilon _{{\rm ec}})^{-1}$.

Since this integral cannot be solved in closed form either we will now consider similar approximations as in the linear case. Additionally, we will now consider each case of the lower limit in turn.

Beginning with $\epsilon _{\gamma _B}<\epsilon _s<\epsilon _{{\rm cut}}$ we use the lower limit $x_{{\rm down}}(\epsilon _s>\epsilon _{\gamma _B}) = (\epsilon _s\epsilon _{{\rm ec}})^{-1}$. In case of $\epsilon _{{\rm ec}}>(2(\sqrt{2}-1)\gamma _B)^{-1}$ we find for $2(\sqrt{2}-1)\gamma _0<\epsilon _s<\epsilon _{{\rm cut}}$ that both limits are lower than unity and we approximate the fraction in Equation (C7) with

$$\begin{equation} \frac{(\sqrt{1+x} - 1)^6}{x^6 \sqrt{1+x}} \approx 2^{-6}, \end{equation} \tag{ C8 }$$

and G(x) becomes to leading order G(x) ≈ 2/(_secx²). Hence,

$$\begin{eqnarray} &&F(2(\sqrt{2}-1)\gamma _0<\epsilon _s<\epsilon _{{\rm cut}}) \approx \frac{16F_0\gamma _0^5}{\alpha ^2}\epsilon _s^{-4} \int \nolimits _{(\epsilon _s\epsilon _{{\rm ec}})^{-1}}^{\frac{4\gamma _0(\gamma _0-\epsilon _s)}{\epsilon _s^2}} d x \frac{1}{2^5\epsilon _s\epsilon _{{\rm ec}}x^2} = \frac{F_0\gamma _0^5}{2\alpha ^2}\epsilon _s^{-4} \left[ 1-\frac{\epsilon _s}{\epsilon _{{\rm cut}}} \right]. \end{eqnarray} \tag{ C9 }$$

For $\epsilon _{\gamma _B}<\epsilon _s<2(\sqrt{2}-1)\gamma _0$ we see that the upper limit is larger than unity, while the lower limit is still smaller than unity. We will therefore split the integral. For the part below unity we use the above-mentioned approximation, while for the part larger than unity we approximate the integrand with x^−7/2. Thus,

$$\begin{eqnarray} &&F(\epsilon _{\gamma _B}<\epsilon _s<2(\sqrt{2}-1)\gamma _0) \approx \frac{16F_0\gamma _0^5}{\alpha ^2}\epsilon _s^{-4} \left[ \int \nolimits _{(\epsilon _s\epsilon _{{\rm ec}})^{-1}}^{1} \frac{ d x }{2^5\epsilon _{{\rm ec}}\epsilon _sx^2} + \int \nolimits _{1}^{\frac{4\gamma _0(\gamma _0-\epsilon _s)}{\epsilon _s^2}} d x x^{-7/2} \right] \approx \frac{F_0\gamma _0^5}{2\alpha ^2}\epsilon _s^{-4}. \end{eqnarray} \tag{ C10 }$$

Combining the last two equations we obtain

$$\begin{equation} F(\epsilon _{\gamma _B}<\epsilon _s<\epsilon _{{\rm cut}}) = \frac{F_0\gamma _0^5}{2\alpha ^2} \epsilon _s^{-4} \left(1-\frac{\epsilon _s}{\epsilon _{{\rm cut}}} \right). \end{equation} \tag{ C11 }$$

For $\epsilon _{{\rm ec}}<(2(\sqrt{2}-1)\gamma _B)^{-1}$ we can make use of Equations (C9) and (C10), while x ≪ 1 and x ≈ 1, respectively, yielding

$$\begin{eqnarray} &&F(2(\sqrt{2}-1)\gamma _0<\epsilon _s<\epsilon _{{\rm cut}}) \approx \frac{F_0\gamma _0^5}{2\alpha ^2}\epsilon _s^{-4} \left[ 1-\frac{\epsilon _s}{\epsilon _{{\rm cut}}} \right] \end{eqnarray} \tag{ C12 }$$

and

$$\begin{eqnarray} &&F \big(\epsilon _{{\rm ec}}^{-1}<\epsilon _s<2(\sqrt{2}-1)\gamma _0\big) \approx \frac{F_0\gamma _0^5}{2\alpha ^2}\epsilon _s^{-4}. \end{eqnarray} \tag{ C13 }$$

If both limits are larger than unity we approximate the integrand with x^−7/2 and obtain

$$\begin{eqnarray} &&F \big(\epsilon _{\gamma _B}<\epsilon _s<\epsilon _{{\rm ec}}^{-1}\big) \approx \frac{16F_0\gamma _0^5}{\alpha ^2}\epsilon _s^{-4} \int \nolimits _{(\epsilon _s\epsilon _{{\rm ec}})^{-1}}^{\frac{4\gamma _0(\gamma _0-\epsilon _s)}{\epsilon _s^2}} d x x^{-7/2} = \frac{32F_0\gamma _0^5\epsilon _{{\rm ec}}^{5/2}}{5\alpha ^2}\epsilon _s^{-3/2}\left[ 1-\left(\frac{\epsilon _s}{\epsilon _{{\rm cut}}} \right)^{5/2} \right]. \end{eqnarray} \tag{ C14 }$$

Combining the Equations (C12)–(C14) we can write the final result as

$$\begin{eqnarray} &&F(\epsilon _{\gamma _B}<\epsilon _s<\epsilon _{{\rm cut}}) = \frac{F_0\gamma _0^5\epsilon _{{\rm ec}}^{5/2}}{2\alpha ^2} \frac{\epsilon _s^{-3/2}}{(1+\epsilon _{{\rm ec}}\epsilon _s)^{5/2}} \left(1-\frac{\epsilon _s}{\epsilon _{{\rm cut}}} \right). \end{eqnarray} \tag{ C15 }$$

Now, we turn our attention to the case where $\epsilon _s<\epsilon _{\gamma _B}$, which means that the lower limit in the integral of Equation (C7) becomes x_down = 4γ₀(γ₀ − α_s)/²_sα².

The first thing we notice is that the upper limit is always larger than unity. This is also true for the lower limit, except for $\epsilon _{{\rm ec}}^{-1}<2(\sqrt{2}-1)\gamma _B$. In the latter case, we obtain with the same approximations as above

$$\begin{eqnarray} &&F(2(\sqrt{2}-1)\gamma _B<\epsilon _s<\epsilon _{\gamma _B}) \approx \frac{16F_0\gamma _0^5}{\alpha ^2}\epsilon _s^{-4} \left[ \int \nolimits _{\frac{4\gamma _0(\gamma _0-\alpha \epsilon _s)}{\epsilon _s^2\alpha ^2}}^{1} \frac{ d x }{2^5\epsilon _{{\rm ec}}\epsilon _sx^2} + \int \nolimits _{1}^{\frac{4\gamma _0(\gamma _0-\epsilon _s)}{\epsilon _s^2}} d x x^{-7/2} \right] \approx \frac{F_0\gamma _0^3}{8\epsilon _{{\rm ec}}}\epsilon _s^{-3}. \end{eqnarray} \tag{ C16 }$$

In the case where both limits are larger than unity, we find

$$\begin{eqnarray} &&F(\epsilon _s<\epsilon _{\gamma _B}) \approx \frac{16F_0\gamma _0^5}{\alpha ^2}\epsilon _s^{-4} \int \nolimits _{\frac{4\gamma _0(\gamma _0-\alpha \epsilon _s)}{\epsilon _s^2\alpha ^2}}^{\frac{4\gamma _0(\gamma _0-\epsilon _s)}{\epsilon _s^2}} d x x^{-7/2} \approx \frac{F_0\alpha ^3}{5}\epsilon _s. \end{eqnarray} \tag{ C17 }$$

Finally, we can give the fluence in the early-time limit covering all energies. However, we have to distinguish three cases for the external photon energy.

For $\epsilon _{{\rm ec}}^{-1}<2(\sqrt{2}-1)\gamma _B$ we have

$$\begin{eqnarray} &&F(\epsilon _s) = \frac{F_0\alpha ^3}{5} \frac{\epsilon _s}{\left(1+\frac{\epsilon _s}{\epsilon _{\gamma _B}} \right)\left(1+\frac{\epsilon _s}{2(\sqrt{2}-1)\gamma _B} \right)^{4}} \left(1-\frac{\epsilon _s}{\epsilon _{{\rm cut}}} \right), \end{eqnarray} \tag{ C18 }$$

while for $2(\sqrt{2}-1)\gamma _B<\epsilon _{{\rm ec}}^{-1}<2(\sqrt{2}-1)\gamma _0$ we get

$$\begin{eqnarray} &&F(\epsilon _s) = \frac{F_0\alpha ^3}{5} \frac{\epsilon _s}{\left(1+\frac{\epsilon _s}{\epsilon _{\gamma _B}} \right)^{5/2} \left(1+\epsilon _{{\rm ec}}\epsilon _s\right)^{5/2}} \left(1-\frac{\epsilon _s}{\epsilon _{{\rm cut}}} \right). \end{eqnarray} \tag{ C19 }$$

The last part is for $2(\sqrt{2}-1)\gamma _0<\epsilon _{{\rm ec}}^{-1}$ and becomes

$$\begin{eqnarray} &&F(\epsilon _s) = \frac{F_0\alpha ^3}{5} \frac{\epsilon _s}{\left(1+\frac{\epsilon _s}{\epsilon _{\gamma _B}} \right)^{5/2} \left(1+\frac{\epsilon _s}{2(\sqrt{2}-1)\gamma _B} \right)^{5/2}} \left(1-\frac{\epsilon _s}{\epsilon _{{\rm cut}}} \right). \end{eqnarray} \tag{ C20 }$$

Hence, we obtained Equations (35)–(37).

As before, we can find the intensity for t ⩾ t_c, if we insert Equation (10) into Equation (14):

$$\begin{eqnarray} &&I_{{\rm ec}}(\epsilon _s,t) = \frac{3Rc\sigma _T}{16\pi }\epsilon _s\int \nolimits _0^{\infty } d \epsilon \frac{\frac{4}{3}\Gamma _b^2 u^{\prime }_{{\rm ec}} \delta \left(\epsilon -\epsilon _{{\rm ec}} \right)}{\epsilon ^2} \int \nolimits _{\gamma _{{\rm min}}}^{\infty } d \gamma \frac{q_0}{\gamma ^2} H\left[ t-t_c \right] G(q,\Gamma) \delta \left(\gamma -\frac{\gamma _0}{\frac{1+2\alpha ^3}{3\alpha ^2}+D_0(1+l_{{\rm ec}})\gamma _0t} \right). \end{eqnarray} \tag{ C21 }$$

Equation (38) is recovered if one uses the definitions of τ and I₀, again. The fluence then follows as

$$\begin{eqnarray} &&F(\epsilon _s) = F_0\epsilon _s\int \nolimits _{\tau _c}^{\frac{\gamma _0}{\gamma _{{\rm min}}}-\frac{1+2\alpha ^3}{3\alpha ^2}} d \tau \left(\frac{1+2\alpha ^3}{3\alpha ^2}+\tau \right)^2 G\left(q,\Gamma,\epsilon =\epsilon _{{\rm ec}},\gamma =\frac{\gamma _0}{\frac{1+2\alpha ^3}{3\alpha ^2}+\tau } \right). \end{eqnarray} \tag{ C22 }$$

Substituting y = ((1 + 2α³)/3α²) + τ we obtain

$$\begin{eqnarray} &&F(\epsilon _s) = F_0\epsilon _s\int \nolimits _{\alpha }^{\frac{\gamma _0}{\gamma _{{\rm min}}}} d y y^2 G\left(q,\Gamma,\epsilon =\epsilon _{{\rm ec}},\gamma =\frac{\gamma _0}{y} \right). \end{eqnarray} \tag{ C23 }$$

Inverting q(y) gives $y=2\epsilon _{{\rm ec}}\gamma _0 q (\sqrt{1+({1}/{\epsilon _s\epsilon _{\rm ec}q})}- 1)$, which leads to

$$\begin{eqnarray} &&F(\epsilon _s) = 8F_0\epsilon _{{\rm ec}}^3\gamma _0^3\epsilon _s\int \nolimits _{\frac{\epsilon _s\alpha ^2}{4\epsilon _{{\rm ec}}\gamma _0(\gamma _0-\alpha \epsilon _s)}}^{1} d q \frac{q^2\left(\sqrt{1+\frac{1}{\epsilon _s\epsilon _{{\rm ec}}q}}-1 \right)^4}{\sqrt{1+\frac{1}{\epsilon _s\epsilon _{{\rm ec}}q}}} \left[ G_0(q) + 2\epsilon _s\epsilon _{{\rm ec}}q(1-q) \right]. \end{eqnarray} \tag{ C24 }$$

Finally, we use again x = (_ecsq)⁻¹ and thus the fluence becomes

$$\begin{eqnarray} &&F(\epsilon _s) = 8F_0\gamma _0^3\epsilon _s^{-2} \int \nolimits _{(\epsilon _s\epsilon _{{\rm ec}})^{-1}}^{\frac{4\gamma _0(\gamma _0-\alpha \epsilon _s)}{\epsilon _s^2\alpha ^2}} d x \frac{\left(\sqrt{1+x}-1 \right)^4}{x^4\sqrt{1+x}} \left[ G_0\left(\frac{1}{\epsilon _s\epsilon _{{\rm ec}}x} \right) + \frac{2}{x}\left(1-\frac{1}{\epsilon _s\epsilon _{{\rm ec}}x} \right) \right]. \end{eqnarray} \tag{ C25 }$$

As before, we will consider approximative results. For $\epsilon _{{\rm ec}}^{-1}<2(\sqrt{2}-1)\gamma _B<\epsilon _s<\epsilon _{\gamma _B}$ both limits are smaller than unity, and we approximate the integrand with (2³_secx²)⁻¹ giving

$$\begin{eqnarray} &&F(2(\sqrt{2}-1)\gamma _B<\epsilon _s<\epsilon _{\gamma _B}) \approx 8F_0\gamma _0^3\epsilon _s^{-2} \int \nolimits _{(\epsilon _s\epsilon _{{\rm ec}})^{-1}}^{\frac{4\gamma _0(\gamma _0-\alpha \epsilon _s)}{\epsilon _s^2\alpha ^2}} d x \frac{1}{2^3\epsilon _s\epsilon _{{\rm ec}}x^2} = F_0\gamma _0^3\epsilon _s^{-2} \left(1-\frac{\epsilon _s}{\epsilon _{\gamma _B}} \right). \end{eqnarray} \tag{ C26 }$$

For $\epsilon _s<\epsilon _{{\rm ec}}^{-1}<2(\sqrt{2}-1)\gamma _B<\epsilon _{\gamma _B}$ both limits are larger than unity and the integrand can be written as x^−5/2, yielding

$$\begin{eqnarray} &&F\big(\epsilon _s<\epsilon _{{\rm ec}}^{-1}<2(\sqrt{2}-1)\gamma _B\big) \approx 8F_0\gamma _0^3\epsilon _s^{-2} \int \nolimits _{(\epsilon _s\epsilon _{{\rm ec}})^{-1}}^{\frac{4\gamma _0(\gamma _0-\alpha \epsilon _s)}{\epsilon _s^2\alpha ^2}} d x x^{-5/2} = \frac{16}{3}F_0\epsilon _{{\rm ec}}^{3/2}\gamma _0^3\epsilon _s^{-1/2} \left[ 1-\left(\frac{\epsilon _s}{\epsilon _{\gamma _B}} \right)^{3/2} \right]. \end{eqnarray} \tag{ C27 }$$

In the intermediate part $\epsilon _{{\rm ec}}^{-1}<\epsilon _s<2(\sqrt{2}-1)\gamma _B<\epsilon _{\gamma _B}$ we can use both limits to obtain

$$\begin{eqnarray} &&F\big(\epsilon _{{\rm ec}}^{-1}<\epsilon _s<2(\sqrt{2}-1)\gamma _B\big) \approx 8F_0\gamma _0^3\epsilon _s^{-2} \left[ \int \nolimits _{(\epsilon _s\epsilon _{{\rm ec}})^{-1}}^{1} d x \frac{1}{2^3\epsilon _s\epsilon _{{\rm ec}}x^2} + \int \nolimits _{1}^{\frac{4\gamma _0(\gamma _0-\alpha \epsilon _s)}{\epsilon _s^2\alpha ^2}} d x x^{-5/2} \right] \approx F_0\gamma _0^3 \epsilon _s^{-2}. \end{eqnarray} \tag{ C28 }$$

Collecting terms, we find the fluence in the late-time limit for $\epsilon _{{\rm ec}}^{-1}<2(\sqrt{2}-1)\gamma _B$

$$\begin{eqnarray} &&F(\epsilon _s<\epsilon _{\gamma _B}) = F_0\gamma _0^3\epsilon _{{\rm ec}}^{3/2} \frac{\epsilon _s^{-1/2}}{(1+\epsilon _{{\rm ec}}\epsilon _s)^{3/2}} \left(1-\frac{\epsilon _s}{\epsilon _{\gamma _B}} \right). \end{eqnarray} \tag{ C29 }$$

For $\epsilon _{{\rm ec}}^{-1}>2(\sqrt{2}-1)\gamma _B$ we obtain a single power law in the form

$$\begin{eqnarray} &&F(\epsilon _s<\epsilon _{\gamma _B}) = \frac{16}{3}F_0\epsilon _{{\rm ec}}^{3/2}\gamma _0^3\epsilon _s^{-1/2} \left(1-\frac{\epsilon _s}{\epsilon _{\gamma _B}} \right). \end{eqnarray} \tag{ C30 }$$

Both equations equal Equations (42) and (43), respectively.

In the case α ≪ 1 we use Equations (48) and (70) for the Thomson limit of the EC component and obtain

$$\begin{eqnarray} &&R_{{\rm ec},s} = 1.8\; l_{{\rm ec}}. \end{eqnarray} \tag{ D2 }$$

In the Klein–Nishina limit of the EC component we use Equation (71) and obtain

$$\begin{eqnarray} &&R_{{\rm ec},s} = 6.6\times 10^{-5} \frac{l_{{\rm ec}}}{\gamma _4\epsilon _{{\rm ec}}}. \end{eqnarray} \tag{ D3 }$$

For α ≫ 1 we have to distinguish between the three cases for ν_br. For the extreme Klein–Nishina limit we use Equations (52) and (75), giving

$$\begin{eqnarray} &&R_{{\rm ec},s} = 6.6\times 10^{-5} l_{{\rm ec}} \frac{\alpha }{\gamma _4\epsilon _{{\rm ec}}}. \end{eqnarray} \tag{ D4 }$$

For the mild Klein–Nishina limit we use Equation (77), yielding

$$\begin{eqnarray} &&R_{{\rm ec},s} = 8.3\times 10^{-3} l_{{\rm ec}} \left(\frac{\alpha }{\gamma _4\epsilon _{{\rm ec}}} \right)^{1/2}, \end{eqnarray} \tag{ D5 }$$

while we use Equation (81) for the Thomson limit, resulting in

$$\begin{eqnarray} &&R_{{\rm ec},s} = 0.8\; l_{{\rm ec}}. \end{eqnarray} \tag{ D6 }$$

In this case we have to distinguish not only between the possible values of α, but also between the different cases of the Klein–Nishina parameter K.

For α ≪ 1 we use for the synchrotron peak value Equation (48), again, while for the SSC peak value for K ≪ 1 we use Equation (57). The ratio then becomes

$$\begin{eqnarray} &&R_{{\rm ssc},s} = 4.4 (1+l_{{\rm ec}})\alpha ^2. \end{eqnarray} \tag{ D7 }$$

For K ≫ 1 we use Equation (58), giving

$$\begin{eqnarray} &&R_{{\rm ssc},s} = 26.8 (1+l_{{\rm ec}})\frac{\alpha ^2}{b\gamma _4^3}. \end{eqnarray} \tag{ D8 }$$

Using Equation (52) for the synchrotron peak value in the case α ≫ 1, we find with Equation (60) the ratio in the case³ K ≪ 1:

$$\begin{eqnarray} &&R_{{\rm ssc},s} = 6.8 (1+l_{{\rm ec}})\alpha. \end{eqnarray} \tag{ D9 }$$

The ratio for the case 1 ≪ K ≪ α³ can be found using Equation (64), yielding

$$\begin{eqnarray} &&R_{{\rm ssc},s} = 5.2 (1+l_{{\rm ec}})\frac{\alpha }{b^{1/3} \gamma _4}. \end{eqnarray} \tag{ D10 }$$

Finally, we obtain the ratio for 1 ≪ α³ ≪ K with the help of Equation (66):

$$\begin{eqnarray} &&R_{{\rm ssc},s} = 14.4 (1+l_{{\rm ec}})\frac{\alpha ^3}{b\gamma _4^3}. \end{eqnarray} \tag{ D11 }$$

Compared to ZS all ratios gain a factor (1 + l_ec), raising the ratio, i.e., the Compton dominance, potentially by a large amount.

Although this ratio is not between the two different components of the blazar SED, it still might be useful to give the ratios between the possible realizations of the high-energy component since, depending on the parameters, it might be possible to discriminate between either the SSC or the EC scenario.

D.3.1. Small Injection Parameter, α ≪ 1

Firstly, we use the Thomson limit for the EC SED, where the peak value is given by Equation (70). Equation (57) gives the case K ≪ 1 for the SSC component, yielding the ratio

$$\begin{eqnarray} &&R_{{\rm ssc,ec}} = 2.3 \frac{(1+l_{{\rm ec}})}{l_{{\rm ec}}} \alpha ^2. \end{eqnarray} \tag{ D12 }$$

For K ≫ 1 we obtain with Equation (58)

$$\begin{eqnarray} &&R_{{\rm ssc,ec}} = 14.3 \frac{(1+l_{{\rm ec}})}{l_{{\rm ec}}} \frac{\alpha ^2}{b\gamma _4^3}. \end{eqnarray} \tag{ D13 }$$

Secondly, we use Equation (71) for the peak value of the EC SED in the Klein–Nishina limit. For K ≪ 1 we obtain with Equation (57)

$$\begin{eqnarray} &&R_{{\rm ssc,ec}} = 6.3\times 10^4 \frac{(1+l_{{\rm ec}})}{l_{{\rm ec}}} \alpha ^2 \gamma _4 \epsilon _{{\rm ec}}. \end{eqnarray} \tag{ D14 }$$

For K ≫ 1 we use Equation (58), yielding

$$\begin{eqnarray} &&R_{{\rm ssc,ec}} = 4.0\times 10^5 \frac{(1+l_{{\rm ec}})}{l_{{\rm ec}}} \frac{\alpha ^2\epsilon _{{\rm ec}}}{b\gamma _4^2}. \end{eqnarray} \tag{ D15 }$$

D.3.2. Large Injection Parameter, α ≫ 1

In this case we obtain nine different ratios, which will be given below.

Beginning with the case K ≪ 1 we will use Equation (60) for the SSC peak values. For $\nu _{{\rm br}}<\nu _{\gamma _{B}r}$ we use Equation (75), which becomes

$$\begin{eqnarray} &&R_{{\rm ssc,ec}} = 1.0\times 10^5 \frac{(1+l_{{\rm ec}})}{l_{{\rm ec}}} \gamma _4\epsilon _{{\rm ec}}. \end{eqnarray} \tag{ D16 }$$

In the case $\nu _{\gamma _{B}r}<\nu _{{\rm br}}<\nu _{\gamma _{0}r}$ we take Equation (77), yielding

$$\begin{eqnarray} &&R_{{\rm ssc,ec}} = 8.0\times 10^2 \frac{(1+l_{{\rm ec}})}{l_{{\rm ec}}} \left(\alpha \gamma _4\epsilon _{{\rm ec}}\right)^{1/2}. \end{eqnarray} \tag{ D17 }$$

Thirdly, we have the case $\nu _{\gamma _{0}r}<\nu _{{\rm br}}$, where we use Equation (81) to obtain

$$\begin{eqnarray} &&R_{{\rm ssc,ec}} = 8.4 \frac{(1+l_{{\rm ec}})}{l_{{\rm ec}}} \alpha. \end{eqnarray} \tag{ D18 }$$

Continuing with the case 1 ≪ K ≪ α³ we take Equation (64) for the SSC peak values. With Equation (75) we get for $\nu _{{\rm br}}<\nu _{\gamma _{B}r}$

$$\begin{eqnarray} &&R_{{\rm ssc,ec}} = 7.5\times 10^4 \frac{(1+l_{{\rm ec}})}{l_{{\rm ec}}} \frac{\epsilon _{{\rm ec}}}{b^{1/3}}. \end{eqnarray} \tag{ D19 }$$

For $\nu _{\gamma _{B}r}<\nu _{{\rm br}}<\nu _{\gamma _{0}r}$ we use Equation (77), again, giving

$$\begin{eqnarray} &&R_{{\rm ssc,ec}} = 6.0\times 10^2 \frac{(1+l_{{\rm ec}})}{l_{{\rm ec}}} \frac{\alpha ^{1/2}\epsilon _{{\rm ec}}^{1/2}}{b^{1/3}\gamma _4^{1/2}}. \end{eqnarray} \tag{ D20 }$$

In the case $\nu _{\gamma _{0}r}<\nu _{{\rm br}}$ Equation (81) gives the correct ratio:

$$\begin{eqnarray} &&R_{{\rm ssc,ec}} = 6.3 \frac{(1+l_{{\rm ec}})}{l_{{\rm ec}}} \frac{\alpha }{b^{1/3}\gamma _4}. \end{eqnarray} \tag{ D21 }$$

Finally, we use Equation (66) for the case 1 ≪ α³ ≪ K. With Equation (75) we yield for $\nu _{{\rm br}}<\nu _{\gamma _{B}r}$

$$\begin{eqnarray} &&R_{{\rm ssc,ec}} = 2.1\times 10^5 \frac{(1+l_{{\rm ec}})}{l_{{\rm ec}}} \frac{\alpha ^2\epsilon _{{\rm ec}}}{b\gamma _4^2}. \end{eqnarray} \tag{ D22 }$$

In the case $\nu _{\gamma _{B}r}<\nu _{{\rm br}}<\nu _{\gamma _{0}r}$ we take Equation (77), obtaining

$$\begin{eqnarray} &&R_{{\rm ssc,ec}} = 1.7\times 10^3 \frac{(1+l_{{\rm ec}})}{l_{{\rm ec}}} \frac{\alpha ^{5/2}\epsilon _{{\rm ec}}^{1/2}}{b\gamma _4^{5/2}}. \end{eqnarray} \tag{ D23 }$$

Finally, we have the case $\nu _{\gamma _{0}r}<\nu _{{\rm br}}$, where we find with Equation (81)

$$\begin{eqnarray} &&R_{{\rm ssc,ec}} = 17.9 \frac{(1+l_{{\rm ec}})}{l_{{\rm ec}}} \frac{\alpha ^3}{b\gamma _4^3}. \end{eqnarray} \tag{ D24 }$$

We see in all cases that for l_ec ≫ 1 the ratio is independent of this value, while for l_ec ≪ 1 the ratio becomes R_{ssc, ec}∝l⁻¹_ec. The latter implies a growing dominance of the SSC component over the EC component, which is expected. The former shows that which component might dominate the IC hump critically depends on the other parameters.