Then the energy distribution of the secondary electrons/positrons in volume $\Delta V (r)$, $$\left({dN(r)\over dE_e}\right)\approx {d^2N\over dVdE_e}\Delta V(r)
\sim {1\over \dot{E}_e}{l_{cur}n_{GJ}\Delta V(r)\over
E_{cur}}\ln\left({E_{cur}\over E_e}\right).$$ The corresponding photon spectrum of the synchrotron radiation is $$F_{syn}(E_{\gamma},r)={3^{1/2}e^3B(r)\sin\beta \over mc^2 h}{1\over
E_{\gamma}}\int^{E_{max}}_{E_{min}}\left({dN(r)\over
dE_e}\right) F(x)dE_e$$ where $x=E_{\gamma}/E_{syn}$. $$E_{syn}(r)={3\over 2}\left({E_e\over mc^2}\right)^2{h e B(r)\sin\beta 
(r)\over mc}$$ is the typical photon energy, and $F(x)=x\int^{\infty}_xK_{5/3}(y)dy$, where $K_{5/3}(y)$ is the modified Bessel function of order 5/3. Similarly, the spectrum of inverse Compton scattered photons in the volume $\Delta V(r)$ is $$F_{ICS}(E_{\gamma}, r)=\int^{E_{max}}_{E_{min}}
\left({dN(r)\over dE_e}\right)\left({d^2N_{ICS}(r)\over
dE_{\gamma}dt}\right)
dE_e,$$ where $${d^2N(r)_{ICS}\over dE_{\gamma}dt}=\int^{\epsilon_2}_{\epsilon_1}
n_{syn}(\epsilon,r)F(\epsilon, E_{\gamma}, E_e)d\epsilon,$$ and $$F(\epsilon, E_{\gamma}, E_e)={3\sigma_Tc\over 4 (E_e/mc^2)^2}{1\over
\epsilon}\left[2q\ln q +(1+2q)(1-q)+{(\Gamma q)^2(1-q)\over 2(1+\Gamma
q)}\right],$$ with $\Gamma=4\epsilon(E_e/mc^2)/mc^2$, $q=E_1/\Gamma (1-E_1)$ with $E_1=E_{\gamma}/E_e$ and $1/4(E_e/mc^2) < q <1$. The number density of the synchrotron photons with energy $\epsilon$ is $$n_{syn}(\epsilon, r)={F_{syn}(\epsilon)\over cr^2\Delta\Omega},$$ where $F_{syn}$ is the calculated synchrotron radiation flux, and $\Delta \Omega$ is the usual beam solid angle.

Since the outer gap of the Crab pulsar is very thin, it is sufficient to use one representative surface to calculate the high-energy radiation. For a given viewing angle not only the light curve can be determined but also the exact emission regions in the outer gap are known. Figure 3 shows the emission 