# Negative real part AC conductivity

I am reading this paper, where the authors are calculating the frequency dependence of the chiral magnetic effect, i.e., $${\bf J} = \sigma^{\text{CME}}(\omega) {\bf B}$$. The authors find, see for instance Figure 1, that the real part of the conductivity shows a resonance behavior of the form $$\sigma \propto [(4- (\omega/\mu)^2]^{-1}$$. Hence, for frequencies $$\omega > \mu$$, the conductivity is negative and goes to zero for large omega.

Now I was wondering what the physical meaning is of a negative (real part of the) conductivity. In the paper, the authors state that

The real part of the conductivity becomes negative above the resonance frequency. This is a typical resonance behavior and implies that when the imaginary part vanishes the response is 180 degrees out of phase with the applied magnetic field.

I do not fully understand this statement. I get the fact that $$e^{i\pi} = -1$$ such that the negativeness implies that there is a phase difference of $$\pi$$ between the conductivity and the perturbation (in this case the magnetic field). What does this physically mean? Can you measure it?

I tried to compare it to the simplest model of a complex conductivity that there is, the Drude model, but there the real part is strictly positive.

• Maybe its related, didnt check out the paper: In NMR (and atomic spectroscopy and such), it is common to have "negative absorbtion", which basically means that the medium amplifies a resonant perturbation (i. e. when resonant light passes through an active medium in a laser you get gain).- the medium is adding energy to the perturbation. "Negative conductivity" could mean something similar, that instead of dissipating the perturabtion, the medium actually makes it stronger, i. e. there's gain. Remember that having a negative conductivity is equivalent to having negative resistivity. – Gyromagnetic Jun 22 '19 at 10:31