with

${\text{\gamma}}^{2}({r}_{1},{t}_{1})=\frac{{b}_{1}({r}_{1},{t}_{1})}{{b}_{2}({r}_{1},{t}_{1})}$ |
(4.120) |

and

$\begin{array}{l}{a}_{1}({r}_{1},{t}_{1})=\frac{\delta}{\delta -1}{K}_{1}{e}^{-{r}_{1}^{2}}{e}^{-{t}_{1}^{2}}+\delta {K}_{1}{p}_{1}{e}^{-{\varrho}^{2}{r}_{1}^{2}}{e}^{-{\tau}^{2}{t}_{1}^{2}}+\frac{{\sigma}^{2}}{4}\\ {a}_{2}({r}_{1},{t}_{1})=\frac{{\delta}^{2}}{\delta -1}{p}_{1}{K}_{1}^{2}{e}^{-{r}_{1}^{2}(1+{\varrho}^{2})}{e}^{-{\tau}^{2}{t}_{1}^{2}}\\ {a}_{3}({r}_{1},{t}_{1})=\sqrt{{a}_{1}{({r}_{1},{t}_{1})}^{2}-4{a}_{2}({r}_{1},{t}_{1}).}\end{array}$ |
(4.121) |

Diffraction - No Beam Walk-off - No Absorption - No Pump Depletion - No Phase Mismatch - Yes

In the nondepleted pump case, the irradiance at the sum frequency given in Eqn. 4.82 can be integrated over the beam area and time to provide the power and energy outputs. Assuming that the incident beams are both circular Gaussian in cross-section and are constant in time, we obtain from Eqns. 4.82 and 4.102

$\begin{array}{l}{P}_{3}(\mathit{\ell})=2\pi {r}_{01}^{2}{\displaystyle \underset{0}{\overset{\infty}{\int}}{I}_{3}(\mathit{\ell}){r}_{1}d{r}_{1}}\\ \text{\hspace{0.17em}}\end{array}$ |

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