# I want to use the complex susceptibility to prove that E dot J is independent of the real part

As stated above, I want to use $\chi$=$\chi'$-i$\chi''$ to prove that E$\cdot$J is related to $\chi''$ and independent of $\chi'$.

I need to do this using a monochromatic field E(t)=E($\omega$)*exp(i$\omega$t) and then taking the absorption time-average <E$\cdot$J> (Note when using phasor notation temporal averaging is given by =$\frac{1}{2}$Re[A$\cdot$B$^*$]).

Also using the relation $J_p$=$\partial_t$P (the polarization) and assuming no external currents.

I plugged in both E and Jp and tried using things like the definition of susceptibility which is given by P(t)=$\epsilon_0$$\int$$\chi$(t-$\tau$)$\cdot$E(t)d$\tau$ where the integral goes from $-\infty$ to $\infty$ or the expression in the frequency domain P($\omega$)=$\epsilon_0$$\chi$($\omega$)E($\omega$) but it just got more complicated.

I'd really appreciate some help here.

• Technically, you shouldn't be using a time-average on any oscillating function that has not asymptotic mean. You could use a spatial or temporal ensemble average, which is not the same as a time-average in this case. Apr 12, 2016 at 19:53

You just need the phasor representation of a time derivative.

$P e^{i \omega t} \rightarrow \vec{P}$

$\partial_{t}(P e^{i \omega t}) \rightarrow i \omega \tilde{P}$

Now, since as you said $J=\partial_{t}P$, we have:

$\tilde{J}=i\omega \chi \tilde{E}$

$\langle \vec{E} \cdot \vec{J} \rangle = \frac{1}{2}\text{Re}(i \omega (\chi' + i\chi'')\tilde{E}\tilde{E}^{*}) = -\frac{1}{2}\text{Re}(\omega \chi'' |\tilde{E}|^{2})$.

Physically, this is power dissipation, the same as $\vec{F} \cdot \vec{v}$. The charge movement can be described by either $\vec{P}$ or $\vec{J}$, but either way it's the component of velocity (current) that's in phase with the field that corresponds to power expended. Because $J$ is 90 degrees out of phase with $P$, and $\chi$ describes polarization response:

$\cdot \chi'$ results in polarization in phase with $E$ (current 90 degrees out of phase)

$\cdot \chi''$ results in current in phase with $E$ (polarization 90 degrees out of phase)