Where Does the Nonlinear Optics Wave Equation Come From?

[DanSandberg]

From a textbook - The reason why the polarization plays a key role in the description of nonlinear optical phenomena is that a time-varying polarization can act as the source of new components of the electromagnetic field... the wave equation in nonlinear optical media often has the form:

$$\nabla$$ ² E - [tex]\frac{n²}{c²}[/tex] [tex]\frac{d²E}{dt²}[/tex] = [tex]\frac{1}{\epsilon c²}[/tex][tex]\frac{d²P^NL}{dt²}[/tex]

This equation is given with no derivation or justification. Can someone explain where this comes from?

EDIT: I'm having a really hard time getting the equation to come out correctly on the website. Its nabla to the second power operating on the electric field E minus the second time derivative of E times n squared over c squared (where n is the linear refractive index and c is the speed of light) equal to 1 over epsilon c squared times the second time derivative of the polarization. I'll try to uplaod a photo of the equation.

 

[Dr Lots-o'watts]

See Boyd's book, section 2.

 

[Redbelly98]

Can't offer help, but I think this is what the equation in the OP is supposed to be:

$$\nabla^2E - \frac{n^2}{c^2} \ \frac{d^2 E} {dt^2} = \frac{1}{\epsilon c^2} \ \frac{d^2P^{NL}}{dt^2}$$

 

[DanSandberg]

Can't offer help, but I think this is what the equation in the OP is supposed to be:

$$\nabla^2E - \frac{n^2}{c^2} \ \frac{d^2 E} {dt^2} = \frac{1}{\epsilon c^2} \ \frac{d^2P^{NL}}{dt^2}$$

thats exactly it - i think maybe cause I am on a mac? or maybe cause I'm using firefox? I'll see if my linux machine does a better job.

 

[Redbelly98]

You can click on the equation I wrote to see the correct LaTex code. For example, superscripts in LaTex are made using the "^" character, not the [noparse][/noparse] tags.

Other users with macs have been able to write LaTex equations.

 

[Manchot]

It's pretty much the usual derivation of the wave equation, except with a nonlinear polarization term kept along for the ride. That is, take the cross product of Faraday's law, substitute in Ampere's Law, and simplify. You have to also assume that the E field is divergenceless (which is not strictly true here, but is what people do nonetheless).