Time independence of light in optical fibers

Physics Asked on January 2, 2022

The propagation of light in optical fibers is governed by the equation

$ifrac{partial A}{partial z} = frac{beta_2}{2}frac{partial^2A}{partial T^2} – gamma|A|^2A$

where $A(z,T)$ represents the amplitude of the field envelope and $beta_2$ and $gamma$ are constants.

My textbook (Nonlinear Fiber Optics by Agarwal) says that the amplitude $A$ is independent of $T$ in case of CW radiation at the input end of the fiber ($z=0$).

Why is $A$ independent of $T$?

3 Answers

The $A(z,t)$ in the fibre optics equation is the $A$ in $$ E(z,t)=A(z,t) e^{i(kz-omega t)}. $$ It does not include the $e^{-iomega t}$ part of the CW time dependence therefore.

Answered by mike stone on January 2, 2022

Note that $A$ is the field envelope, which means that $A$ is the average over some duration of time and interval of space. It is closely related to intensity in that respect. For CW excitation, the envelope is constant. There's no mechanism shown that would cause a time-varying envelope within the fiber. In other words the first term on the right serves as a source term, and since the source is CW, the first term on the right is zero.

Answered by garyp on January 2, 2022

That's the definition of "CW radiation" in this context.

CW stands for "continuous wave", but here it is being used to mean "not varying with time".

Much like in electronics the term "DC" literally means not changing direction, but has been co-opted in circuit theory to mean not changing at all.

Answered by The Photon on January 2, 2022

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