fred3142
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I'm not sure if this is the appropriate forum for my question as I actually am studying this as part of electrical engineering and I don't actually study physics. Nonetheless, I shall ask and if need be, move my question to another venue.
My question is with regard to how complex permittivity is defined. According to my book
$$
\begin{align*}
\nabla \times \mathbf{\tilde{H}} &= \sigma \mathbf{\tilde{E}} + \jmath\omega\varepsilon \mathbf{\tilde{E}} \\
&= (\sigma + \jmath\omega\varepsilon)\mathbf{\tilde{E}} \\
&= \jmath\omega\underbrace{\left(\varepsilon - \jmath\frac{\sigma}{\omega}\right)}_{\varepsilon_c}\mathbf{\tilde{E}} \\
&= \jmath\omega\varepsilon_c\mathbf{\tilde{E}}
\end{align*}
$$
(\mathbf{\tilde{E}} and \mathbf{\tilde{H}} are phasors.)
I really do not understand why \varepsilon_c \equiv \varepsilon - \jmath\frac{\sigma}{\omega} and not \varepsilon_c \equiv \sigma + \jmath\omega\varepsilon. What is the sense in creating a complex value, \varepsilon_c, and then multiplying by \jmath\omega when you could just modify the definition of \varepsilon_c such that \nabla \times \mathbf{\tilde{H}} = \varepsilon_c \mathbf{\tilde{E}}?
I also have a conceptual question: From what I understand, \varepsilon determines the phase delay between the H and E fields. This phase delay, as far as I know, comes from the finite speed involved in 'rotating' the dipoles in the medium. When these dipoles are 'rotated' though, since they take a finite time to rotate, that implies to me that there are some sort of losses involved in rotating the dipoles. These losses, though, as far as I can tell, are not accounted for in \varepsilon_c \equiv \varepsilon - \jmath\frac{\sigma}{\omega} (I figure the loss due to rotating the dipoles should be part of \Im{\{\varepsilon_c\}} (from what I can tell, \Im{\{\varepsilon_c\}} accounts for the loss and \Re{\{\varepsilon_c\}} accounts for the phase delay); however, \Im{\{\varepsilon_c\}} only seems to take into account frequency and loss from electorns crashing into atoms).
Similarily, I would've thought that the loss that comes from electrons crashing into atoms (which \sigma looks after), would also have the affect of at least somewhat slowing down the wave and causing lag.
Basically, what I'm saying is, why aren't \varepsilon and \sigma also complex numbers? Or maybe they are...
Thank you.
My question is with regard to how complex permittivity is defined. According to my book
$$
\begin{align*}
\nabla \times \mathbf{\tilde{H}} &= \sigma \mathbf{\tilde{E}} + \jmath\omega\varepsilon \mathbf{\tilde{E}} \\
&= (\sigma + \jmath\omega\varepsilon)\mathbf{\tilde{E}} \\
&= \jmath\omega\underbrace{\left(\varepsilon - \jmath\frac{\sigma}{\omega}\right)}_{\varepsilon_c}\mathbf{\tilde{E}} \\
&= \jmath\omega\varepsilon_c\mathbf{\tilde{E}}
\end{align*}
$$
(\mathbf{\tilde{E}} and \mathbf{\tilde{H}} are phasors.)
I really do not understand why \varepsilon_c \equiv \varepsilon - \jmath\frac{\sigma}{\omega} and not \varepsilon_c \equiv \sigma + \jmath\omega\varepsilon. What is the sense in creating a complex value, \varepsilon_c, and then multiplying by \jmath\omega when you could just modify the definition of \varepsilon_c such that \nabla \times \mathbf{\tilde{H}} = \varepsilon_c \mathbf{\tilde{E}}?
I also have a conceptual question: From what I understand, \varepsilon determines the phase delay between the H and E fields. This phase delay, as far as I know, comes from the finite speed involved in 'rotating' the dipoles in the medium. When these dipoles are 'rotated' though, since they take a finite time to rotate, that implies to me that there are some sort of losses involved in rotating the dipoles. These losses, though, as far as I can tell, are not accounted for in \varepsilon_c \equiv \varepsilon - \jmath\frac{\sigma}{\omega} (I figure the loss due to rotating the dipoles should be part of \Im{\{\varepsilon_c\}} (from what I can tell, \Im{\{\varepsilon_c\}} accounts for the loss and \Re{\{\varepsilon_c\}} accounts for the phase delay); however, \Im{\{\varepsilon_c\}} only seems to take into account frequency and loss from electorns crashing into atoms).
Similarily, I would've thought that the loss that comes from electrons crashing into atoms (which \sigma looks after), would also have the affect of at least somewhat slowing down the wave and causing lag.
Basically, what I'm saying is, why aren't \varepsilon and \sigma also complex numbers? Or maybe they are...
Thank you.