Mathematics of phasor representations of sinusoidal variations of fields

The real part of ${\mathbf e}({\mathbf r})={\mathbf E}({\mathbf r})e^{j\omega t}$ is ${\mathbf E}({\mathbf r})\cos (\omega t)$, assuming, as is conventional, that ${\mathbf E}({\mathbf r})$ is real.

The partial time derivative $$\frac{\partial {\mathbf e}}{\partial t} = \omega{\mathbf E}({\mathbf r})\left[ -\sin(\omega t) + j\cos(\omega t)\right] $$

So whether you differentiate the real part of ${\bf e}$ or take the real part of $\partial {\bf e}/\partial t$, makes no difference.

So would the real part just be $\mathbf{e}(\mathbf{r}, t) = \mathbf{E}(\mathbf{r})$?

• $\mathbf{e}(\mathbf{r}, t) = Re[\mathbf{E}(\mathbf{r})e^{j \omega t}] = \mathbf{E}(\mathbf{r}) \cos(\omega t)$ is already real.

How do we deal with the partial derivative $\dfrac{\partial{\mathbf{e}}}{\partial{t}}$?

• $\dfrac{\partial \mathbf e}{\partial t} = \dfrac{\partial \ (\mathbf{E}(\mathbf{r}) \cos(\omega t))}{\partial t} = - \omega \mathbf{E}(\mathbf{r}) \sin(\omega t)$

If you have not taken the real part yet, $$\dfrac{\partial \mathbf e}{\partial t} = \dfrac{\partial \ (\mathbf{E}(\mathbf{r}) \exp(j \omega t))}{\partial t} = j \omega \mathbf{E}(\mathbf{r}) \exp(j \omega t) = \omega \mathbf{E}(\mathbf{r})[-\sin(\omega t) + j \cos(\omega t)]$$ whose real part will still be $-\omega \mathbf E(\mathbf r) \sin(\omega t)$. The order really does not matter.

• In fact, this trick is used many times while solving coupled oscillators. It is much easier to deal with $\exp(j\omega t)$ than $\sin$ and $\cos$ while differentiating.