Maxwell Equations Phasor Form Derivation

Question

We know that one of Maxwell's equation in Time-Domain is: $$\nabla \times \vec{E} = - \partial \vec{B}/\partial t$$ The typical steps to convert into Phasor-Domain starts with:

Let $$E(x,y,z,t) = Re[E_s(x,y,z)e^{j\omega t}]$$ And after some substitutions we arrive at: $$Re[\nabla \times \vec{E_s}e^{j\omega t}] = -Re[j \omega \vec{B_s}e^{j\omega t}]$$

Now in most references they just equate the inside parts arriving at the answer correctly.

What I don't understand, in mathematics, when Re[z1] = Re[z2], this does not imply that z1 = z2. But in the above equation this is essentially what we have done.

Can someone explain it?

{I know it has an answer here: Phasor form of Maxwell's Equations, but I did not understand the answer, so if someone can explain it more simply, I would be grateful.}

A side question: What is the transformation used in phasor form? Is it related to Fourier Transform?

Answer 1

Phasors are typically used in situations described by (systems of) ordinary or partial linear differential equations in time with real constant coefficients where you assume that the real time dependent variables have a steady state sinusoidal time dependence with angular frequency $\omega$. These time dependent variables can also be vectors. In general, any real sinusoidal variable $A$ with real amplitude $A_0$ and phase $\delta$ can be written as $$A(t)=A_0 \cos(\omega t+\delta)=A_0 \frac{e^{j\omega t+\delta} +e^{-j\omega t -\delta}}{2}=\bar{A}e^{j\omega t}+\bar{A}^*e^{-j\omega t} \tag{1}$$ ($j$ is the imaginary unit), where $\bar{A}=(A_0/2)$ and $\bar{A^*}=(A_0/2)e^{-j\delta}$ are conjugate complex amplitudes with phase angles $\delta$ depending on the variable. Usually you call the complex amplitude $\bar{A}$ of the $e^{j\omega t}$ time dependence the "phasor", but it can also be the conjugate complex amplitude and its $e^{-j\omega t}$ time dependence.

The first step to arrive at the phasor form of the partial differential equation, and the example of the question's Maxwell equation, is to insert for the sinusoidally time dependent vectors $$\mathbf{E}=\bar{\mathbf{E}}e^{j\omega t}+\bar{\mathbf{E}}^*e^{-j\omega t} \tag{2}$$ and $$\mathbf{B}=\bar{\mathbf{B}}e^{j\omega t}+\bar{\mathbf{B}}^*e^{-j\omega t} \tag{3}$$ into $$\nabla \times \mathbf{E} = - \partial \mathbf{B}/\partial t \tag{4}$$ You differentiate the complex exponentials with respect to time (multiplying by $j\omega$ and $-j\omega$ respectively) and then split the terms with time dependences $e^{j\omega t}$ and $e^{-j\omega t}$ into two equations and divide by the respective exponential time factors. This yields two conjugate complex equation for the conjugate complex amplitudes $$\nabla \times \bar{\mathbf{E}} = - j\omega \bar{\mathbf{B}} \tag{5}$$ $$\nabla \times \bar{\mathbf{E}}^* = j\omega \bar{\mathbf{B}}^* \tag{6}$$ The equation for the conjugate complex amplitude doesn't give any new information. To arrive at a solution for the phasor variables you need to split all of Maxwell's equations into their complex conjugate forms and use only the phasor forms corresponding to the $e^{j\omega t}$ time dependence.

Once you have the complex phasor solution $\bar{A}$ for a real sinusoidal variable A(t) for the angular frequency $\omega$, you obtain the real solution by using the RHS of equation (1).

The same real solution can be obtained by taking the real part of only the first phasor term of the RHS of equation (1): $$A(t)= 2Re{(\bar{A}e^{j\omega t}})=2Re{(\frac{A_0}{2}e^{j\omega t+\delta}})=A_0\cos(\omega t+\delta) \tag{7}$$