Guided Waves - Helmholtz Equation

Physics Asked by javierhersan on December 22, 2020

I am having trouble understanding the Guided Fields derivation process. Firstly, We derive the Helmholtz equation easily:

$$Δ⋅vec{E}+ frac{∂^2vec{E}}{∂z^2} – gamma$$0$$⋅vec{E}= 0$$

Then we decompose the Helmholtz equation into two new equations thanks to decomposing the field in tangential and transversal components ($$vec{E}=vec{E_t}+vec{E_z}$$):

$$Δ_t⋅vec{E_t}+ frac{∂^2vec{E_t}}{∂z^2} – gamma$$0$$⋅vec{E_t}= 0$$

$$Δ_t⋅vec{E_z}+ frac{∂^2vec{E_z}}{∂z^2} – gamma$$0$$⋅vec{E_z}= 0$$

The first thing I do not undestand is why we can apply variable separation to solve this differential equation, are not we leaving some solutions applying this method?

$$E_z(x,y,z) = F_E(x,y)⋅e^{-gamma⋅z}$$

$$H_z(x,y,z) = F_H(x,y)⋅e^{-gamma⋅z}$$

Secondly, the boundary conditions, which are derived from Maxwell’s equation and the boundary with a perfect conductor, are the following:

TE Mode: $$frac{∂F_H}{∂n} = 0$$

TM Mode: $$F_E = 0$$

How do we get the boundary conditions of the TE Mode?

And lastly having all this conditions and the Helmholtz equation, how we can assure that $$k_c^2$$ is real and greater than $$0$$

$$Δ_t⋅F_{E/H}+ k_c^2⋅F_{E/H}= 0$$ , $$frac{∂F_H}{∂n} = 0$$, $$F_E = 0$$

Thank you very much, any help is appreciated.

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