Attenuation and Phase constant values in wave equation

[baby_1]

Homework Statement

Attenuation and Phase constant values in wave equation

Relevant Equations

wave equation

Regarding wave equation we are faced with this form
$$\nabla^2 \vec{E}=j\omega \mu \sigma \vec{E}-j\omega \mu\varepsilon \vec{E}=\gamma ^2\vec{E}$$
where
$$\gamma ^2=j\omega \mu \sigma -j\omega \mu\varepsilon $$
$$\gamma =\alpha +j\beta $$
where alpha and beta are attenuation and phase constants respectively. If we have a lossless media(where sigma =0), I need to obtain alpha and beta values:
My attempt:
$$\gamma ^2 =-j\omega \mu\varepsilon =>\gamma ^2=\omega \mu\varepsilon e^{\frac{-j\pi }{2}}=>\gamma=\sqrt{\omega \mu\varepsilon} e^{\frac{-j\pi }{4}}=>\gamma=\sqrt{\omega \mu\varepsilon}(C os(\frac{\pi }{4})+jSin(\frac{\pi }{4}))=>\alpha =\beta => \alpha \neq 0 $$

it means we have a loss value in the equation, However, if we set alpha=0 in the gamma variable all equations are correct. I need to prove alpha is equal to zero when sigma =0.

 

[TSny]

Regarding wave equation we are faced with this form
$$\nabla^2 \vec{E}=j\omega \mu \sigma \vec{E}-j\omega \mu\varepsilon \vec{E}$$

Are you sure that you wrote this correctly? I checked a couple of standard textbooks which write the second term on the right as $-\omega^2 \mu \varepsilon \vec E$ (no factor of $j$ and $\omega$ is squared).

 

[Delta2]

Yes i agree with @TSny, that term comes from $$\mu\epsilon\frac{\partial^2 \vec{E}(r,t)}{\partial t^2}$$ and assuming harmonic time dependence that is $$\vec{E}(r,t)=\vec{E}(r)e^{-j\omega t}$$ we get that it is equal to $$j^2\omega^2\mu\epsilon\vec{E}(r)e^{-j\omega t}$$