First, we consider that the sample is illuminated by the excitation light $S(r,s,t)$, where $r,s$, and $t$ are the position, direction, and the time, respectively. When the most general expression of the photon transport in a strong turbid medium is RTE and the radiance $L(r,s,t)$ can be expressed as Display Formula
$[1c\u2202\u2202t+s\xb7\u2207+\mu a(r)+\mu s(r)]L(r,s,t)=\mu s(r)\u222b4\pi P(s,s\u2032)L(r,s\u2032,t)d\Omega \u2032+S(r,s,t),$(1)
where $\mu s,\mu a$, and $c$ are the scattering and absorption coefficients, and speed of light, respectively. $P(s,s\u2032)$ is the scattering phase function, which is the probability of the scattering with an incident direction $s\u2032$ and the scattering direction $s$. The right-side integral with respect to the solid angle $d\Omega \u2032$ sums up all incoming light at this point. In case of the fluorescence, the source term $S$ becomes Display Formula$S=\gamma \u03f5N(r)4\pi \u222bdt\u2032\u222b4\pi d\Omega F(t\u2212t\u2032)Lex(r,s,t\u2032),$(2)
where $F(t)$, $\gamma ,\u03f5$, and $N(r)$ are the fluorescence decay function, quantum efficiency, molar absorption coefficient, and concentration of the fluorophore, respectively. The suffix “ex” means the excitation light while the “fl” means fluorescence. This source term is neglecting a reabsorption–emission process, where fluorophores are excited by the fluorescence. The convolution integral can be eliminated by introducing $Lfl*$, which satisfies $Lfl=\u222b0tdt\u2032Lfl(t\u2032)*F(t\u2212t\u2032)$. $Lfl*$ is actually IF-TPSF of the fluorescence. Here, the position and direction in L are not shown to reduce the complexity of equations. Then, one can obtain an RTE for $Lfl*$ as Display Formula$[1c\u2202\u2202t+s\xb7\u2207+\mu afl(r)+\mu sfl(r)]Lfl*(t)=\mu sfl(r)\u222b4\pi Pfl(s,s\u2032)Lfl*(t)d\Omega \u2032+\gamma \mu a,flex4\pi \u222bd\Omega Lex(t),$(3)
where $\mu a,fl=eN(r)$, meaning the absorption coefficient of the fluorophore.