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This is from the slide used in class and I have a few questions to ask.

First question: In the first line of equations of the phasor form, why is there a negative sign? Is this a sign error?

How is the effective dielectric constant of metal derived in this case? Specifically, for the third line of equation of the phasor form, why is that we can say the below equation is true?

$$ \nabla \times \vec{H}(\vec{r}) = \epsilon_{eff}(\omega)\frac{d\vec{E}(\vec{r},t)}{dt} $$

Isn't this ignoring the displacement current?


2 Answers 2


For what concerns the first question, recall that there are two conventions on the time dependence associated to phasors, one typically used by engineers and one typically used by physicists.

In the engineer's convention, a sinusoidal quantity $x(t)$ is represented by

$$x(t) = \operatorname{Re}[X\mathrm{e}^{\mathrm{j}\omega t}],$$

such that the phasor associated to the derivative of $x(t)$ is $\mathrm{j}\omega X$.

In the physicist's convention, $x(t)$ is represented by

$$x(t) = \operatorname{Re}[X\mathrm{e}^{-\mathrm{i}\omega t}],$$

such that the phasor associated to the derivative of $x(t)$ is $-\mathrm{i}\omega X$, with the minus sign. This is the convention used in the slides of your class.

About your second question, the equation you reported is NOT true.

In fact, note first that $\vec{H}(\vec{r})$ and $\vec{E}(\vec{r})$ are the phasors associated to $\vec{H}(\vec{r},t)$ and $\vec{E}(\vec{r},t)$, respectively: even though the same symbols are used in the slides, those quantities are not the same!

The last equation in phasor form is

$$\nabla \times \vec{H}(\vec{r}) = -\mathrm{i}\omega\epsilon_\mathrm{eff}(\omega)\vec{E}(\vec{r})$$

To go back to the time domain, phasors are no longer enough and one should use the Fourier transform. From the properties of Fourier transforms, the associated time-domain equation is

$$\nabla \times \vec{H}(\vec{r},t) = \epsilon_\mathrm{eff}(t)*\frac{\mathrm{d}\vec{E}(\vec{r},t)}{\mathrm{d}t}$$

where the asterisk denotes the operation of convolution and $\epsilon_\mathrm{eff}(t)$ is the inverse Fourier transform of $\epsilon_\mathrm{eff}(\omega)$.

For the special case in which $\epsilon_\mathrm{eff}(\omega)$ is independent of $\omega$, that is, $\epsilon_\mathrm{eff}(\omega) = \epsilon_\mathrm{eff} = \mathrm{const.}$, then

$$\nabla \times \vec{H}(\vec{r},t) = \epsilon_\mathrm{eff}\frac{\mathrm{d}\vec{E}(\vec{r},t)}{\mathrm{d}t}$$

Final note: actually, there are even more conventions on phasors, see this other answer of mine.


Q1 Presumably we're using $\vec E(\vec r, t) = \vec E(\vec r) e^{-i\omega t}$.

[We're assuming also that $\vec H(\vec r, t) = \vec H(\vec r) e^{-i\omega t}$ and $\vec J(\vec r, t) = \vec J(\vec r) e^{-i\omega t}$.

Q2 Just substitute the expression for $\epsilon_\text{eff}(\omega)$ (given on the right) into the final version of the phasor form, tidy up, and you'll get the middle version of the phasor form – in which the second term on the right is the 'displacement current' term.


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