- #1
Gabriel Maia
- 72
- 1
Hi. I am reading a paper about gaussian beams and the author says that gaussian beams have simultaneously minimal divergence and minimal transversal extension. In order to prove it, the author states that
[itex]\mathrm{divergenece} \propto \int_{-\infty}^{+\infty} \frac{d\,k_{x}}{2\pi} \int_{-\infty}^{+\infty} \frac{d\,k_{y}}{2\pi} (k_{x}^{2}+k_{y}^{2})|A(k_{x},k_{y})|^2[/itex]
and
[itex]\mathrm{transversal \, extension} \propto \int_{-\infty}^{+\infty} d\,x \int_{-\infty}^{+\infty} d\,y (x^{2}+y^{2})|E|^2[/itex],
where
[itex]E = \int \frac{d\,k_{x}}{2\pi} \int \frac{d\,k_{y}}{2\pi} A(k_{x},k_{y}) \exp[i\,k_{x}\,x + i\,k_{y}\,y+i \, \sqrt{k^2-k_{x}^2-k_{y}^2} \, z] [/itex]
is the electric field amplitude and [itex]A(k_{x},k_{y})[/itex] is the amplitude distribution.
I look it up and I couldn't find these integral definitions of divergence and transversal extension. Does anybody have a source on these? Or could at least give me an idea about how they came to be defined this way?
Thank you very much.
[itex]\mathrm{divergenece} \propto \int_{-\infty}^{+\infty} \frac{d\,k_{x}}{2\pi} \int_{-\infty}^{+\infty} \frac{d\,k_{y}}{2\pi} (k_{x}^{2}+k_{y}^{2})|A(k_{x},k_{y})|^2[/itex]
and
[itex]\mathrm{transversal \, extension} \propto \int_{-\infty}^{+\infty} d\,x \int_{-\infty}^{+\infty} d\,y (x^{2}+y^{2})|E|^2[/itex],
where
[itex]E = \int \frac{d\,k_{x}}{2\pi} \int \frac{d\,k_{y}}{2\pi} A(k_{x},k_{y}) \exp[i\,k_{x}\,x + i\,k_{y}\,y+i \, \sqrt{k^2-k_{x}^2-k_{y}^2} \, z] [/itex]
is the electric field amplitude and [itex]A(k_{x},k_{y})[/itex] is the amplitude distribution.
I look it up and I couldn't find these integral definitions of divergence and transversal extension. Does anybody have a source on these? Or could at least give me an idea about how they came to be defined this way?
Thank you very much.