- #1
fred3142
- 22
- 0
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
$$
\begin{align*}
\nabla \times \mathbf{\tilde{H}} &= \sigma \mathbf{\tilde{E}} + \jmath\omega\varepsilon \mathbf{\tilde{E}} \\
&= (\sigma + \jmath\omega\varepsilon)\mathbf{\tilde{E}} \\
&= \jmath\omega\underbrace{\left(\varepsilon - \jmath\frac{\sigma}{\omega}\right)}_{\varepsilon_c}\mathbf{\tilde{E}} \\
&= \jmath\omega\varepsilon_c\mathbf{\tilde{E}}
\end{align*}
$$
([itex]\mathbf{\tilde{E}}[/itex] and [itex]\mathbf{\tilde{H}}[/itex] are phasors.)
I really do not understand why [itex]\varepsilon_c \equiv \varepsilon - \jmath\frac{\sigma}{\omega}[/itex] and not [itex]\varepsilon_c \equiv \sigma + \jmath\omega\varepsilon[/itex]. What is the sense in creating a complex value, [itex]\varepsilon_c[/itex], and then multiplying by [itex]\jmath\omega[/itex] when you could just modify the definition of [itex]\varepsilon_c[/itex] such that [itex]\nabla \times \mathbf{\tilde{H}} = \varepsilon_c \mathbf{\tilde{E}}[/itex]?
I also have a conceptual question: From what I understand, [itex]\varepsilon[/itex] 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 [itex]\varepsilon_c \equiv \varepsilon - \jmath\frac{\sigma}{\omega}[/itex] (I figure the loss due to rotating the dipoles should be part of [itex]\Im{\{\varepsilon_c\}}[/itex] (from what I can tell, [itex]\Im{\{\varepsilon_c\}}[/itex] accounts for the loss and [itex]\Re{\{\varepsilon_c\}}[/itex] accounts for the phase delay); however, [itex]\Im{\{\varepsilon_c\}}[/itex] 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 [itex]\sigma[/itex] 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 [itex]\varepsilon[/itex] and [itex]\sigma[/itex] also complex numbers? Or maybe they are...
Thank you.
My question is with regard to how complex permittivity is defined. According to my book
$$
\begin{align*}
\nabla \times \mathbf{\tilde{H}} &= \sigma \mathbf{\tilde{E}} + \jmath\omega\varepsilon \mathbf{\tilde{E}} \\
&= (\sigma + \jmath\omega\varepsilon)\mathbf{\tilde{E}} \\
&= \jmath\omega\underbrace{\left(\varepsilon - \jmath\frac{\sigma}{\omega}\right)}_{\varepsilon_c}\mathbf{\tilde{E}} \\
&= \jmath\omega\varepsilon_c\mathbf{\tilde{E}}
\end{align*}
$$
([itex]\mathbf{\tilde{E}}[/itex] and [itex]\mathbf{\tilde{H}}[/itex] are phasors.)
I really do not understand why [itex]\varepsilon_c \equiv \varepsilon - \jmath\frac{\sigma}{\omega}[/itex] and not [itex]\varepsilon_c \equiv \sigma + \jmath\omega\varepsilon[/itex]. What is the sense in creating a complex value, [itex]\varepsilon_c[/itex], and then multiplying by [itex]\jmath\omega[/itex] when you could just modify the definition of [itex]\varepsilon_c[/itex] such that [itex]\nabla \times \mathbf{\tilde{H}} = \varepsilon_c \mathbf{\tilde{E}}[/itex]?
I also have a conceptual question: From what I understand, [itex]\varepsilon[/itex] 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 [itex]\varepsilon_c \equiv \varepsilon - \jmath\frac{\sigma}{\omega}[/itex] (I figure the loss due to rotating the dipoles should be part of [itex]\Im{\{\varepsilon_c\}}[/itex] (from what I can tell, [itex]\Im{\{\varepsilon_c\}}[/itex] accounts for the loss and [itex]\Re{\{\varepsilon_c\}}[/itex] accounts for the phase delay); however, [itex]\Im{\{\varepsilon_c\}}[/itex] 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 [itex]\sigma[/itex] 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 [itex]\varepsilon[/itex] and [itex]\sigma[/itex] also complex numbers? Or maybe they are...
Thank you.