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It's a very well-known fact that plane waves can be represented in the complex form: \begin{equation} \mathbf{F}(\mathbf{x},t)=\mathbf{F}_0e^{i(kx-\omega t)} \end{equation} However, I've been struggling to mathematically understand why is that the case. It's a quite important fact to understand, since it has some basic applications in Electromagnetism. Usually the sinuisodal plane wave is used, where the real part of the given equation is taken.

Anyway, if anyone can explain, rigorously, how we can express plane waves in that simple form, I'll be grateful.

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  • $\begingroup$ So what exaclty is your problem, do you want to know why a plane wave can be represented by $e^{i(kx-\omega t)}$ or do you want to know why $e^{i(kx-\omega t)}$ is oscillating similar to a sine and a cosine function? So is your problem in understanding why and how we can use the exponential or what the exponential of a complex number actually means? $\endgroup$
    – Luis
    Commented Dec 17, 2021 at 22:59
  • $\begingroup$ I have no trouble in understanding the meaning of the complex exponential, what I want to know is contained exactly on your first phrase: "So what exaclty is your problem, do you want to know why a plane wave can be represented by $e^{i(kx−ω\omega t)}$ ". $\endgroup$ Commented Dec 17, 2021 at 23:10
  • $\begingroup$ well are you aware of the fomula $e^{ix}=i\sin(x)+\cos(x)$ $\endgroup$
    – physshyp
    Commented Dec 18, 2021 at 3:38

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The basic idea is to leverage the observation that derivatives of exponentials are easier to manipulate than derivatives of trig functions.

The full expression is in fact $e^{i(kx-\omega t+\varphi_0)}$ and is such that, by definition, its real part is the physical quantity: $$ \hbox{Re}(e^{i(kx-\omega t+\varphi_0)}):=\cos(kx-\omega t+\varphi_0)\, . $$ By expanding $\cos(A+B)$ one can get the full linear combination of sine and cosine functions.

The phasor form goes further and eliminates the $e^{-i\omega t}$ part and is very useful since differentiation w/r to $t$ of the physical quantity is just multiplication of the phasor by $-i\omega$, and differentiation w/r to $x$ of the physical quantity is just multiplication of the phasor by $ik$.

Mathematically, the use of phasors transforms some differential equations for physical quantities into algebraic equations for the corresponding phasor forms, which are easier to manipulate than the corresponding equations in terms of sine and cosine.

Note that several equivalent phasors can be used to represent the same physical quantity: if your physical quantity is $\sin(kx-\omega t)$, possible phasors are $-i e^{ikx}$ or $e^{i(kx-\pi/2)}$. To get the physical quantity, one then multiples the phasor by $e^{-i\omega t}$ and then take the real part, bypassing the clumsy use of sine, cosines throughout.

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The most general plane wave solution to the homogenous wave equation is

$\vec{E_{0} } e^{i(\vec{K} \cdot \vec{r}-\omega t+ \phi_{E})}$

The same for the magnetic wave

This is a complex wave, using eulers formula this can be decomposed into an imaginary and real part

Technically the above equation is a valid solution to the inhomogenous wave equation, however physically we take the REAL part of this equation to find the REAL solutions to the homogenous wave equation

the homogenous wave equation is actually 3 vector equations each for each of the components of the E field

To check if this is a solution, we will use substitution

Substituting the above equation into the homogenous wave equation Will not be a function of R or T, thus we have proved that it IS solution.

This will obtain conditions on this equation, that make it satisfy the homogenous wave equation and we will obtain the dispersion relation that$ \frac{\omega}{|\vec{K}|} = C$

AKA, this is a valid solution provided omega and K satisfy the condition that the phase velocity is the speed of light

The reason we use complex notation is that it makes it easier to work with and obtain the dispersion relation. This is purely a mathematical convenience to find the REAL solutions. As we take the REAL PART. however something overlooked is that even the complex wave IS a perfectly valid solution

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  • $\begingroup$ This explains what I couldn't understand. Here are my remarks for others like me. The author of the post asked to explain why a wave can be described by a complex function. I had the same question. My thinking was along the following lines. A real physical wave (a water wave for instance) is, experimentally, a sin wave, let's assume. How is it sin a complex number I wandered? The answer is that we don't measure the wave as a sin. We derive a math equation for the wave, solve it, get the complex function as the result of the solution and, for real physical wave, we keep only the real part. $\endgroup$ Commented May 14 at 15:16

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