5
$\begingroup$

For example, Faraday's law in the time domain is written as $$\nabla \times \vec{\mathbf{E}} = - \frac{\partial\vec{\mathbf{B}}}{\partial t}$$

When using phasor notation, Faraday's law is written as $$\nabla \times \vec{\mathbf{E}} = -j\omega \mu \vec{\mathbf{B}}$$

Why does the time derivative, $\frac{\partial}{\partial t}$ change to $j\omega$?

$\endgroup$
4
  • 6
    $\begingroup$ Well, the first $\vec{\mathbf{B}}$ and the second $\vec{\mathbf{B}}$ are not the same thing (bad notations, the same goes for $\vec{\mathbf{E}}$). The second equation is obtained by taking the Fourier transform of the first equation with respect to the time variable. $\endgroup$ Commented Aug 23, 2017 at 16:22
  • $\begingroup$ @Batominovski I think I figured it out in the answer below. However, the first $\vec{\mathbf{B}}$ and the second $\vec{\mathbf{B}}$ are the same thing; they are just written using different forms. $\endgroup$ Commented Aug 24, 2017 at 16:55
  • $\begingroup$ They are not the same thing. There is no reason why $\vec{\mathbf{B}}$ must be sinusoidal. $\endgroup$ Commented Aug 24, 2017 at 17:15
  • $\begingroup$ Things are explained in this way to the students if the general idea is that they won't understand anything anyway. It's sickening. $\endgroup$ Commented Aug 24, 2017 at 17:30

2 Answers 2

4
$\begingroup$

If $f(t)$ is a sinusoidal function of time, i.e. $$ f(t)=F_0\cos(\omega t +\varphi)=\Re\left\{F_0\mathrm e^{i(\omega t +\varphi)}\right\}=\Re\left\{\underbrace{F_0\mathrm e^{i\varphi}}_{F}\,\mathrm e^{i\omega t }\right\}=\Re\left\{F\,\mathrm e^{i\omega t}\right\} $$ so we can represent the function $f(t)$ through the phasor $F\,\mathrm e^{i\omega t}$.

Taking the first derivative we have $$ f'(t)=\frac{\mathrm d}{\mathrm dt}F_0\cos(\omega t +\varphi)=\frac{\mathrm d}{\mathrm dt}\Re\left\{F\,\mathrm e^{i\omega t}\right\}=\Re\left\{\frac{\mathrm d}{\mathrm dt}F\,\mathrm e^{i\omega t}\right\}=\Re\left\{i\omega F\,\mathrm e^{i\omega t}\right\} $$ so we can represent the derivative $f'(t)$ through the phasor $i\omega F\,\mathrm e^{i\omega t}$.

Let be $\mathcal{\pmb E}(\pmb r,t)$ and $\mathcal{\pmb B}(\pmb r,t)$ sinusoidal fields: $$ \begin{align} \mathcal{\pmb B}(\pmb r,t)&=\pmb B_0(\pmb r)\cos(\omega t +\varphi)=\Re\left\{\pmb B_0(\pmb r)\mathrm e^{i(\omega t +\varphi)}\right\}=\Re\left\{\pmb B(\pmb r)\mathrm e^{i\omega t}\right\}\\ \mathcal{\pmb E}(\pmb r,t)&=\pmb E_0(\pmb r)\cos(\omega t +\vartheta)=\Re\left\{\pmb E_0(\pmb r)\mathrm e^{i(\omega t +\vartheta)}\right\}=\Re\left\{\pmb E(\pmb r)\mathrm e^{i\omega t}\right\} \end{align} $$ where $\pmb B(\pmb r)=\pmb B_0(\pmb r)\mathrm e^{i\varphi}$ and $\pmb E(\pmb r)=\pmb E_0(\pmb r)\mathrm e^{i\vartheta}$. We have $$ \nabla\times \mathcal{\pmb E}(\pmb r,t)=\nabla\times \Re\left\{\pmb E(\pmb r)\mathrm e^{i\omega t}\right\}=\Re\left\{\nabla\times \pmb E(\pmb r)\mathrm e^{i\omega t}\right\} $$ and $$ \frac{\partial}{\partial t}\mathcal{\pmb B}(\pmb r,t)=\frac{\partial}{\partial t}\Re\left\{\pmb B(\pmb r)\mathrm e^{i\omega t}\right\}=\Re\left\{\frac{\partial}{\partial t}\pmb B(\pmb r)\mathrm e^{i\omega t}\right\}=\Re\left\{i\omega \pmb B(\pmb r)\mathrm e^{i\omega t}\right\} $$

Hence the Faraday's law $$ \nabla\times \mathcal{\pmb E}(\pmb r,t)=-\frac{\partial \mathcal{\pmb B}(\pmb r,t)}{\partial t} $$ using phasors becomes $$ \nabla\times \pmb E(\pmb r)=-i\omega\pmb B(\pmb r) $$

In general for a "suitable" function $f(t)$ we use the Fourier transform defined as $$F(\omega) = \int_{-\infty}^{\infty} f(t) \, \mathrm e^{-i \omega t} \mathrm dt$$ and the inverse Fourier transform is $$f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) \, \mathrm e^{i \omega t} \mathrm d\omega$$ So we have $$f'(t) = \frac{\mathrm d}{\mathrm dt}\!\left( \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) \, \mathrm e^{i \omega t} \mathrm d\omega \right)= \frac{1}{2\pi} \int_{-\infty}^{\infty} i \omega \, F(\omega) \,\mathrm e^{i \omega t} \mathrm d\omega$$

Hence the Fourier transform of $f'(t)$ is $ i \omega \, F(\omega)$.

Let be $\mathcal{\pmb E}(\pmb r,t)$ and $\mathcal{\pmb B}(\pmb r,t)$ time varying fields satisfying $$ \nabla\times \mathcal{\pmb E}(\pmb r,t)=-\frac{\partial \mathcal{\pmb B}(\pmb r,t)}{\partial t} $$ Let's take the time Fourier transform $$ \begin{align} \mathcal{F}\{\mathcal{\pmb B}(\pmb r,t)\}&=\pmb B(\pmb r,\omega)=\int_{-\infty}^{\infty}\mathcal{\pmb B}(\pmb r,t)\,\mathrm e^{-i\omega t}\,\mathrm d t\\ \mathcal{F}\{\mathcal{\pmb E}(\pmb r,t)\}&=\pmb E(\pmb r,\omega)=\int_{-\infty}^{\infty}\mathcal{\pmb E}(\pmb r,t)\,\mathrm e^{-i\omega t}\,\mathrm d t \end{align} $$

Taking the Fourier transform of Faraday's law we have $$ \mathcal{F}\left\{\nabla\times \mathcal{\pmb E}(\pmb r,t)\right\}=\nabla\times \mathcal{F}\left\{\mathcal{\pmb E}(\pmb r,t)\right\}=\nabla\times {\pmb E}(\pmb r,\omega) $$ and using the Fourier property for the derivatives $$ \mathcal{F}\left\{-\frac{\partial \mathcal{\pmb B}(\pmb r,t)}{\partial t}\right\}=-i\omega{\pmb B}(\pmb r,\omega) $$ So we find

$$ \nabla\times {\pmb E}(\pmb r,\omega)=-i\omega{\pmb B}(\pmb r,\omega) $$

$\endgroup$
1
$\begingroup$

The $j\omega$ comes from the simple time derivative of a phasor.

Assume $\vec{\mathbf{B}}$ in the time domain is $$\vec{\mathbf{B}} = A\cos{\omega t - \phi}.$$

As a phasor, $\vec{\mathbf{B}}$ is noted as $$\vec{\mathbf{B}} = Ae^{j\omega t - \phi} = Ae^{j\omega t}e^{-\phi}.$$

Taking the time derivative of the phasor $\vec{\mathbf{B}}$, $$\frac{\partial\vec{\mathbf{B}}}{\partial t} = j\omega Ae^{j\omega t}e^{-\phi} = j\omega \vec{\mathbf{B}}.$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .