Optical conductivity: Does higher real part mean higher losses/absorption? My question is with regards to the relationship between optical conductivity and optical permittivity. Since the imaginary part of optical permittivity signifies losses/absorption, the real part of optical conductivity signifies losses/absorption.
Does higher real part of optical conductivity mean higher losses/absorption? If so, what is the intuition behind it? Shouldn't higher conductivity mean lower loss? 
Is there something I am missing? 
Any insight is appreciated. 
Thanks!
 A: The Ampere-Maxwell's equation reads $$\nabla \times \vec B=\mu (\vec j+\epsilon \frac {\partial \vec E}{\partial t})=\mu (\sigma \vec E+\epsilon \frac {\partial \vec E}{\partial t}) \tag 1$$ where $\vec j$ the conduction current density, $\sigma$ is the (real) specific conductivity, $\epsilon$ is the (real) absolute permittivity, $\mu$ is the absolute permeability. The expression in the brackets on the RHS is the sum of the conduction and displacement current density, the total current density $\vec j_{tot}$. For sinusoidal time variation  of the fields and currents $\propto \exp(-i\omega t)$ with angular frequency $\omega$, the fields and currents can be expressed as complex amplitudes (phasors) and the total complex current density can be written as $$\vec j_{tot}=\vec j-i\omega \epsilon \vec E=\sigma \vec E-i\omega \epsilon \vec E=(\sigma-i\omega \epsilon) \vec E=\sigma_c \vec E=-i\omega (\epsilon+ i \frac {\sigma}{\omega})\vec E=-i\omega \epsilon_c \vec E \tag 2$$ The last three terms representing the total current show that the total current can be viewed either as being due to a complex conductivity $$\sigma_c=\sigma-i\omega \epsilon=\sigma_r+i\sigma_i \tag 3$$ or due to a complex permittivity $$\epsilon_c=\epsilon+ i \frac {\sigma}{\omega}=\epsilon_r+i\epsilon_i \tag 4$$ were $\sigma=\sigma_r$ and $\epsilon=\epsilon_r$ are the real parts, and $\sigma_i=-\omega\epsilon$ and $\epsilon_i=\frac {\sigma}{\omega}$ the imaginary parts of the  complex conductivity and and permittivity, respectively. Thus both complex conductivity and complex permittivity describe the in-phase and out-of-phase components of the total current density, i.e., the conduction current and displacement current.
According the Poynting Theorem, the power dissipation density of an electromagnetic field (EM wave) is given by the Joule term $$W=\vec j · \vec E\tag 5$$ For a complex current and electric field, this Joule power dissipation density averaged over a cycle becomes  $$W=Re(\frac {\vec E ·\vec j^*}{2})=Re(\frac {\vec E · \sigma_c^* \vec E^*}{2})=Re(\frac {\sigma_c^* |\vec E|^2}{2})=\sigma_r\frac { |\vec E|^2}{2} \tag 6$$ where $Re$ denotes the real part operator and $^*$ the complex conjugate. See, e.g., Simon Ramo et al., Fields and Waves in Modern Communication Electronics, 3rd. ed., 1994, Chapter 3.13 "Poynting's Theorem for Phasors", and Appendix 4. 
Thus a higher real part of the conductivity $\sigma_r$ means, indeed, higher losses and for a wave with the same frequency also a higher absorption. Only the current in phase with the electric field, and thus the (positive) real part of the conductivity, $\sigma_r$, corresponds to a power dissipation and to a wave absorption. Imaginary parts of the permittivity, and thus real parts of the conductivity, can also be caused by in-phase components due to (molecular) damping.
