= Solution
Write $N=N^*$, $N_i=N_i^*$ and $N_f=N_f^*$ during this derivation. Eliminating the stimulated term between the two <laser rate equations> gives the <photon> balance
$$
\dot n=-\frac{f_c}{\beta}\dot N-\frac n{\tau_c}.
$$
Integrate over the pulse, taking negligible initial and final stored <photon> densities:
$$
\int_0^\infty\frac{n(t)}{\tau_c}\,dt
=\frac{f_c}{\beta}(N_i-N_f).
$$
The <photons> lost from the <optical cavity>[cavity] carry total <energy> $\hbar\omega_LV_c\int n/\tau_c\,dt$. Since $f_cV_c=V_g$,
$$
\boxed{E=\frac{(N_i^*-N_f^*)V_g\hbar\omega_L}{\beta}
=\eta N_i^*V_g\hbar\omega_L.}
$$
This identifies the supplied <energy> formula under the ideal extraction convention that all <photon> loss is useful output. If only a fraction $\zeta=\kappa_{\rm out}/\kappa_{\rm total}$ of the loss exits the output coupler, the measured output <energy> is $\zeta E$. The rate equations by themselves do not distinguish useful output from parasitic loss.
Back to article page
