Self-focusing of multi-Gaussian laser beams in nonlinear optical media as a Kepler’s central force problem

Abstract

In this paper self-focusing of multi-Gaussian laser beam in nonlinear optical media has been investigated theoretically. Saturation of the optical nonlinearity has been incorporated through cubic-quintic model. Moment theory approach in W.K.B approximation has been invoked to find the numerical solution of nonlinear wave equation for the field of incident laser beam. Particularly the dynamical evolutions of the beam width of the laser beam with distance of propagation have been investigated in detail. It has been shown that the self-focusing of the laser beam resembles to Kepler’s central force problem.

Appendices

Appendix 1

Using Eq. (5) in

$$\begin{aligned} I_0=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }|\psi |^2dxdy, \end{aligned}$$

we get

$$\begin{aligned} I_0=\pi E_{00}^2r_0^2(1+3e^{-\frac{x_0^2}{2r_0^2}}). \end{aligned}$$

(21)

In deriving Eq. (21) we have made use of following standard integrals

$$\begin{aligned} \int _0^{\infty }e^{-\frac{(x\mp x_0f)}{r_0^2f^2}}dx & = \frac{\sqrt{\pi }}{2}r_0f\left( 1\pm erf\left( \frac{x_0}{r_0}\right) \right) , \\ \int _0^{\infty }e^{-\frac{x^2}{r_0^2f^2}}dx & = \frac{\sqrt{\pi }}{2}r_0f,\\ \int _0^{\infty }e^{-\frac{((x-x_0f)^2+(x+x_0f)^2)}{2r_0^2f^2}}dx & = \frac{\sqrt{\pi }}{2}r_0fe^{-\frac{x_0^2}{r_0^2}}. \end{aligned}$$

Now, using Eqs. (5) and (21) in

$$\begin{aligned} \langle x^2\rangle & = \frac{1}{I_0}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }x^2|\psi |^2dxdy,\\ \langle y^2\rangle & = \frac{1}{I_0}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }y^2|\psi |^2dxdy, \end{aligned}$$

we get

$$\begin{aligned} \langle x^2\rangle =\langle y^2\rangle =\frac{r_0^2f^2}{\left(1+3e^{-\frac{x_0^2}{2r_0^2}}\right)} \left\{ 1+2e^{-\frac{x_0^2}{r_0^2}}+\left( 2\frac{x_0^2}{r_0^2}+1\right) +\left( 2+\frac{x_0^2}{r_0^2}\right) e^{-\frac{x_0^2}{2r_0^2}}\right\} . \end{aligned}$$

(22)

In deriving Eq. (22) we have used following integral

$$\begin{aligned} \int _0^{\infty }x^2e^{-\frac{(x\mp x_0f)^2}{r_0^2f^2}}dx=\frac{1}{4}r_0fe^{-\frac{x_0^2}{r_0^2}}\left\{ \pm 2r_0x_0f^2+(2x_0^2f^2+r_0^2f^2)e^{-\frac{x_0^2}{r_0^2}}\sqrt{\pi }\left( 1\pm erf\left( \frac{x_0}{r_0}\right) \right) \right\} . \end{aligned}$$

Using Eq. (22) in (6) we get Eq. (7) i.e.,

$$\begin{aligned} \sigma _{M.G}(0)=\frac{\sqrt{2}r_0}{\left(1+3e^{-\frac{x_0^2}{2r_0^2}}\right)^\frac{1}{2}} \left\{ \left( 2+2\frac{x_0^2}{r_0^2}\right) +\left( 2+\frac{x_0^2}{r_0^2}\right) e^{-\frac{x_0^2}{2r_0^2}} +2e^{-\frac{x_0^2}{2r_0^2}}\right\} ^\frac{1}{2}. \end{aligned}$$

Now, differentiating Eq. (6) with respect to ‘z’ we get

$$\begin{aligned} \frac{d\sigma ^2}{dz}=\frac{1}{I_0}\int _{-\infty }^{\infty } \int _{-\infty }^{\infty }(x^2+y^2)\left( \psi ^{\star }\frac{\partial \psi }{\partial z}+\psi \frac{\partial \psi ^{\star }}{\partial z}\right) dxdy. \end{aligned}$$

(23)

From Eq. (3) we can write

$$\begin{aligned} \frac{\partial \psi }{\partial z} & = -\frac{\iota }{2k_0}\left( \frac{\partial ^2\psi }{\partial x^2}+\frac{\partial ^2\psi }{\partial y^2}\right) -\iota \frac{k_0}{2n_0^2}\phi (|\mathbf {E}|^2)\psi , \end{aligned}$$

(24)

$$\begin{aligned} \frac{\partial \psi ^{\star }}{\partial z} & = \frac{\iota }{2k_0}\left( \frac{\partial ^2\psi ^{\star }}{\partial x^2}+\frac{\partial ^2\psi ^{\star }}{\partial y^2}\right) +\iota \frac{k_0}{2n_0^2}\phi (|{\mathbf{E}}|^{\mathbf{2}})\psi ^{\star }. \end{aligned}$$

(25)

Using Eqs. (24), (25) in (23) we get

$$\begin{aligned} \frac{d\sigma ^2}{dz}=\frac{\iota }{2I_0k_0} \int _{-\infty }^{\infty }\int _{-\infty }^{\infty }(x^2+y^2) \left\{ \psi \left( \frac{\partial ^2\psi ^{\star }}{\partial x^2}+\frac{\partial ^2\psi ^{\star }}{\partial y^2}\right) -\psi ^{\star }\left( \frac{\partial ^2\psi }{\partial x^2}+\frac{\partial ^2\psi }{\partial y^2}\right) \right\} dxdy. \end{aligned}$$

(26)

Differentiating Eq. (26) again with respect to ‘z’ we get

$$\begin{aligned} \frac{d^2\sigma ^2}{dz^2} & = \frac{\iota }{2I_0k_0}\int _{-\infty }^{\infty } \int _{-\infty }^{\infty }(x^2+y^2)\left\{ \frac{\partial ^2}{\partial x^2}\left( \frac{\partial \psi ^{\star }}{\partial z}\right) +\left( \frac{\partial ^2}{\partial y^2}\left( \frac{\partial \psi ^{\star }}{\partial z}\right) \right) \psi +\left( \frac{\partial ^2\psi ^{\star }}{\partial x^2}+\frac{\partial ^2\psi ^{\star }}{\partial y^2}\right) \frac{\partial \psi }{\partial z}\right. \nonumber \\&\quad \left. -\,\left( \frac{\partial ^2}{\partial x^2}\left( \frac{\partial \psi }{\partial z}\right) +\frac{\partial ^2}{\partial y^2}\left( \frac{\partial \psi }{\partial z}\right) \right) \psi ^{\star }-\left( \frac{\partial \partial ^2\psi }{\partial x^2}+\frac{\partial ^2\psi }{\partial y^2}\right) \frac{\partial \psi ^{\star }}{\partial z}\right\} dxdy. \end{aligned}$$

(27)

By making use of Eqs. (24) and (25) in above equation one can get Eq. (9) i.e.,

$$\begin{aligned} \frac{d^2}{dz^2}\sigma ^2_{M.G}=\frac{1}{I_0k_0^2}\int \int \left( \left| \frac{\partial \psi }{\partial x}\right| ^2+\left| \frac{\partial \psi }{\partial y}\right| ^2\right) dxdy+\frac{1}{2}\frac{1}{\epsilon _0I_0}\int \int \left( x\frac{\partial \phi }{\partial x}+y\frac{\partial \phi }{\partial y}\right) \psi \psi ^{\star }dxdy. \end{aligned}$$

Appendix 2

Using Eq. (5) we get

$$\begin{aligned} \frac{1}{I_0}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty } \left( \left| \frac{\partial \psi }{\partial x}\right| ^2+\left| \frac{\partial \psi }{\partial y}\right| ^2\right) dxdy=\frac{1}{r_0^2f^2}\frac{\left\{ 1+e^{-\frac{x_0^2}{r_0^2}} \left( 1-\frac{x_0^2}{r_0^2}\right) +e^{-\frac{x_0^2}{2r_0^2}} \left( 2-\frac{x_0^2}{r_0^2}\right) \right\} }{(1+3e^{-\frac{x_0^2}{2r_0^2}})}. \end{aligned}$$

(28)

In deriving Eq. (28) we have used the standard integral

$$\begin{aligned} \int _0^{\infty }(x\mp x_0f)^2e^{-\frac{(x\mp x_0f)^2}{r_0^2f^2}}dx =\frac{1}{4}r_0^2f^2\left\{ \mp 2x_0fe^{-\frac{x_0^2}{r_0^2}}+\sqrt{\pi }r_0f\pm \sqrt{\pi }r_0ferf\left( \frac{x_0}{r_0}\right) \right\} . \end{aligned}$$

Using Eqs. (4), (7), (21) and (28) in Eq. (9) we get Eq. (10) with

$$\begin{aligned} I_n & = \int _0^{\infty }\int _0^{\infty }t_1G^{2n+1}(t_1,t_2)e^{-\frac{t_2^2}{2}}\left[ \Sigma _{j=1}^2\left( t_1+(-1)^j\frac{x_0}{r_0}\right) e^{-\frac{(t_1+(-1)^j\frac{x_0}{r_0})^2}{2}}\right] dt_1dt_2,\\ K_n & = \int _0^{\infty }\int _0^{\infty }t_1^2G^{2n+1}(t_1,t_2)e^{-\frac{t_1^2}{2}}\left[ \Sigma _{j=1}^2e^{-\frac{(t_1+(-1)^j\frac{x_0}{r_0})^2}{2}}\right] dt_1dt_2.\\ G^{2n+1}(t_1,t_2) & = \left[ e^{-\frac{(t_1-\frac{x_0}{r_0})^2+t_2^2}{2}}+e^{-\frac{(t_1+\frac{x_0}{r_0})^2+t_2^2}{2}}+e^{-\frac{t_1^2+(t_2-\frac{x_0}{r_0})^2}{2}}+e^{-\frac{t_1^2+(t_2+\frac{x_0}{r_0})^2}{2}}\right] ^{2n+1},\\ n & = 1,2 \\ t_1 & = \frac{x}{r_0f}, \\ t_2 & = \frac{y}{r_0f}. \end{aligned}$$

The function $G^3(t_1,t_2)$ appears due to cubic nonlinearity of the medium whereas the function $G^5(t_1,t_2)$ appears due to quintic nonlinearity.