Author's derivation of time-independent form of Maxwell's equations

Question

Laser Electronics, 3rd edition, by Joseph T. Verdeyen, gives the following:

To describe an electromagnetic wave, we need two field-intensity vectors, $\mathbf{e}$ and $\mathbf{h}$, which are related to each other by

$$\nabla \times \mathbf{h} = \mathbf{j} + \epsilon_0 \dfrac{\partial{\mathbf{e}}}{\partial{t}} + \dfrac{\partial{\mathbf{p}}}{\partial{t}},$$

$$...$$

where $\mathbf{p}$ is the polarisation current induced by the electric field.

...

Most of the time we will be talking about sinusoidal variations of the field and use the phasor representation

$$\mathbf{j}(\mathbf{r}, t) = \mathcal{R}e[\mathbf{J}(\mathbf{r})e^{j\omega t}],$$

$$\mathbf{e}(\mathbf{r}, t) = \mathcal{R}e[\mathbf{E}(\mathbf{r})e^{j\omega t}],$$

$$\mathbf{p}(\mathbf{r}, t) = \mathcal{R}e[\mathbf{P}(\mathbf{r})e^{j\omega t}],$$

$$...$$

where $\mathcal{R}e$ is the real part, $\mathbf{r} = x \mathbf{a}_x + y \mathbf{a}_y + z \mathbf{a}_z$, $\mathbf{a}_i$ is the unit vector in the $i$th direction, and the capital letters $\mathbf{E}$ and $\mathbf{H}$ are complex vector quantities depending on space coordinates but not on time.

By substitution, we obtain the time-independent form of Maxwell's equations:

$$\nabla \times \mathbf{H} = \mathbf{J} + j \omega \epsilon_0 \mathbf{E} + j \omega \mathbf{P} = \mathbf{J} + j \omega \mathbf{D},$$

$$...$$

where $\mathbf{D} = \epsilon_0 \mathbf{E} + \mathbf{P}$

It seems to me that the author thinks that $\mathcal{R}e[\mathbf{J}(\mathbf{r})e^{j \omega t}] = \mathbf{J}$, when it seems to me that it should be $\mathcal{R}e[\mathbf{J}(\mathbf{r})e^{j \omega t}] = \mathcal{R}e[\mathbf{J}(\cos(\omega t) + j \sin(\omega t))] = \mathbf{J} \cos(\omega t)$?

I would greatly appreciate it if people could please take the time to review this.

Answer 1

So, the original equation is $$\operatorname{rot}(h) = j + \varepsilon_0 \partial_t e + \partial_tp$$ Substituting back everything, we get that $$\operatorname{rot}\left(\Re\left(H\exp(i \omega t)\right)\right) = \Re\left(J\exp(i \omega t)\right) + \varepsilon_0 \partial_t \Re\left(E\exp(i \omega t)\right) + \partial_t\Re\left(P\exp(i \omega t)\right)$$ $$\Re\left(\operatorname{rot}\left(H\exp(i \omega t)\right)\right) = \Re\left(J\exp(i \omega t)\right) + \Re\left(\varepsilon_0 \partial_t E\exp(i \omega t)\right) + \Re\left(\partial_tP\exp(i \omega t)\right)$$ $$\Re\left(\operatorname{rot}\left(H\exp(i \omega t)\right)\right) = \Re\left(J\exp(i \omega t) + \varepsilon_0 \partial_t E\exp(i \omega t)+\partial_tP\exp(i \omega t)\right)$$ $$\Re\left(\operatorname{rot}\left(H\right)\exp(i \omega t)\right) = \Re\left(J\exp(i \omega t) + \varepsilon_0 E\partial_t \exp(i \omega t)+P\partial_t\exp(i \omega t)\right)$$ $$\Re\left(\operatorname{rot}\left(H\right)\exp(i \omega t)\right) = \Re\left(J\exp(i \omega t) + \varepsilon_0 Ei \omega \exp(i \omega t)+Pi \omega\exp(i \omega t)\right)$$ $$0 = \Re\left(J\exp(i \omega t) + \varepsilon_0 Ei \omega \exp(i \omega t)+Pi \omega\exp(i \omega t)-\operatorname{rot}\left(H\right)\exp(i \omega t)\right)$$ Now let $$f=J + \varepsilon_0 Ei \omega +Pi \omega -\operatorname{rot}\left(H\right)$$ Then we have that $$0=\Re(f \exp(i \omega t))$$ $$0=\Re(f)\Re(\exp(i \omega t))-\Im(f)\Im(\exp(i \omega t))$$ And it should be true $\forall t$, so if we let $t=0$, we get that $$0=\Re(f)$$ And if we let $\omega t = \frac{\pi}{2}$, we get that $$0=\Im(f)$$ Which means that $$\Re(f)+i\Im(f)=f=0$$