Heat dissipated from electric field

[capandbells]

This isn't a homework problem, but I'm having trouble understanding the following equality in a paper I'm reading

Homework Statement

Homework Equations

$$\left<\mathbf{j}(\mathbf{r},t)\cdot\mathbf{E}(\mathbf{r},t)\right>_t = - \frac{1}{2}\mathrm{Re}\left[i\omega\frac{\epsilon(\mathbf{r}) - 1}{4\pi}\mathbf{E_0}(\mathbf{r})\mathbf{E_0^*}(\mathbf{r}) \right]$$
where
$$\mathbf{E}(\mathbf{r},t) = \mathrm{Re}\left[\mathbf{E_0}(\mathbf{r})e^{-i\omega t}\right]$$

The Attempt at a Solution

I understand that
$$\mathbf{j}(\mathbf{r},t) = \sigma \mathbf{E}(\mathbf{r},t) = i\omega\frac{\epsilon(\mathbf{r}) - 1}{4\pi}\mathbf{E}(\mathbf{r},t)$$
but I'm 100% sure how to get from there to the other expression. Mostly I'm confused about how the conductivity gets inside the Re operator. I'm pretty confused in general about how to take the dot product of such a complicated vector expression.

 

[Antiphon]

You're looking at an expression for the amount of power that is heating a dielectric. Since the power is lost to heat, you can define an equivalent ohmic loss, hence the sigma. It's important to note that the permittivity has to be complex as well, and it's the imaginary component that will end up representing a loss.

 

[capandbells]

You're looking at an expression for the amount of power that is heating a dielectric. Since the power is lost to heat, you can define an equivalent ohmic loss, hence the sigma. It's important to note that the permittivity has to be complex as well, and it's the imaginary component that will end up representing a loss.

I understand that, but I don't know how to get from the LHS in that first equation to the RHS.

 

[Antiphon]

The j is the time derivative of the polarization vector dP/dt.

If I have time I'll try to do a derivation.