Electromagnetic radiation is created by the varying/accelerating of a system of charges and currents. Suppose that the time dependence of the charges and currents are $\rho(x,t)$ and $J(x,t)$. Then the subsequent radiation will have the same time dependence. In Jackson it is stated that we can assume $\rho$ and $J$ have harmonic time dependence because we can build up any "arbitrary" function as a superposition of sinusoidal functions via Fourier analysis. My understanding of Fourier series is that we can only do this for periodic functions. We always refer to electromagnetic radiation as a wave because of the harmonic time dependence but the radiation has the same time dependence as $\rho(x,t)$ and $J(x,t)$. So what do we do, if the charges and currents are accelerating but not periodically? Then the electromagnetic radiation would not be a wave I think. So why does Jackson state that we can assume harmonic time dependence without losing any generality?

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    $\begingroup$ It would be a wave, and satisfying the wave equation moving with velocity c, it just wouldn't be sinusoidal. $\endgroup$ Commented Jan 31, 2023 at 23:00
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    $\begingroup$ @JosephSanders The solutions of the wave equation aren't required to be periodic. $\endgroup$
    – Triatticus
    Commented Jan 31, 2023 at 23:05
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    $\begingroup$ If the acceleration is not periodic, you can always do the analog of making a Fourier series, which is making a Fourier integral, i.e., a Fourier transform. That is, instead of summing over a set of discrete frequencies, you sum over a all possible frequencies. $\endgroup$
    – march
    Commented Jan 31, 2023 at 23:12
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    $\begingroup$ From the mathematical point of view, searching for solutions in harmonic form means that we apply a Fourier transform to the equation and then we get a solution in Fourier domain. Basically if $\rho$ and $J$ are the function of t in a general sense, you can apply a Fourier Transform to get $\rho(x, \omega)$ and $J(x,\omega)$ and then to multiply it by the transfer function of the system (the one you get when calculating with harmonic functions) and finally reapplying inverse Fourier transform. $\endgroup$ Commented Feb 1, 2023 at 10:26
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    $\begingroup$ To sum up, you can start from the "harmonic solution" and then recalculate a solution for any time dependence $\endgroup$ Commented Feb 1, 2023 at 10:26

2 Answers 2


Fourier analysis does not require periodicity. This was one of the things about Fourier's work that shocked mathematicians. Essentially, any function you can graph may be decomposed into sinusoidal waves.

Imagine electric field lines converging at a charge in a particular position. Now, move that charge to a new position. The electric field lines will converge at the new position. But the news that you've moved the charge can't travel faster than the speed of light. So, there's a kink in the lines joining the old lines with the new.

It turns out that that kink moves at the speed of light. It's a non-periodic electromagnetic wave.


The confusion is between Fourier series, which is expansion for periodic functions, and Fourier transform - which is an expansion for arbitrary functions (satisfying certain mathematical conditions.)

Fourier transform can be though of as a generalization of Fourier series. Some would probably even say that Fourier series simply a particular case of the Fourier transform, although certain care is required when switching from one to another.

Finally, it is mentioning a useful trick: a function defined in an interval, can be extended periodically beyond this interval and expanded in Fourier series.

Since Maxwell equations are linear equations, we can Fourier transform the sources (currents and charges) and the fields, and solve the equation in Fourier space. Thus, even non-periodic fields can be represented in terms of periodic waves.

A related topic is the electromagnetic field radiated by an accelerated point charge: see Liénard-Wiechert potential and Larmor formula.


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