I've tried to understand the decomposition of an HF electrical field in a series of plane waves. $$\vec{E}(\vec{r}, t) = \int\int\int \hat{\vec{E}}(\vec{k}) \cdot\mathrm{e}^{\mathrm{i}(\vec{k}.\vec{r}-\omega t)} \mathrm{d}^3\vec k$$

The fourier component corresponding to $\vec{k} = 0$ does not represent any plane wave. But it oscillates with $\omega$. How can I imagine this component, since from Maxwell's equations there is no homogeneous and oscillating field possible?


1 Answer 1


Your expression is a solution of the wave equation $$\vec\Delta\vec E-{1\over c^2}{\partial^2\vec E\over\partial t^2}=0$$ of the electric field only if the pulsation $\omega$ is not a free parameter but a function of $\vec k$, i.e. $\omega=\omega(\vec k)$. When plugging the plane wave $e^{i(\vec k.\vec r-\omega t)}$ into the wave equation, one gets the so-called dispersion law $$\omega^2=||\vec k||^2c^2$$ Therefore, the homogeneous solution $\vec k=0$ leads to $\omega(0)=0$ and, as you correctly stated, does not oscillate.

To answer your comment, introduce the 3D Fourier transform $$\vec E(\vec r,t)=\int \hat{\vec E}(\vec k,t)e^{i\vec k.\vec r}d^3\vec k$$ Note that the Fourier coefficients are time-dependent. When pluging into the wave equation, one gets the differential equation $$-k^2 \hat{\vec E}(\vec k,t)-{1\over c^2}{\partial^2\hat{\vec E} \over\partial t^2}=0$$ whose solutions are $$\hat{\vec E}_\pm(\vec k,t)=\hat{\vec E}(\vec k,0)e^{\pm i\omega t}$$ where $\omega^2=k^2c^2$. $\vec E(\vec r,t)$ is recovered by an inverse Fourier transform. Alternatively, one can introduce the 4D Fourier transform $$\vec E(\vec r,t)=\int\hat{\vec E}(\vec k,\omega)e^{i(\vec k.\vec r-\omega t)}d^3\vec k d\omega$$ In this case, the wave equation implies that the Fourier coefficients vanish unless the dispersion law is satisfied.

  • $\begingroup$ Thank you! So do I understand it correctly, that you can not decompose a field distribution oscillating with a single frequency (e.g. a Hertzian Dipol) into a set of plane waves all having the same frequency? $\endgroup$ Sep 10, 2020 at 13:47
  • $\begingroup$ Exactly! Otherwise, it would not satisfy the wave equation (and therefore Maxwell equations). $\endgroup$
    – Christophe
    Sep 10, 2020 at 14:00
  • $\begingroup$ But one question still remains: What is it what I get if I do the decomposition as above? Since mathematically, the Fourier decomposition is still possible. If I modify the frequency according to the dispersion law, the back-transformation does not yield the original field. $\endgroup$ Sep 11, 2020 at 12:22
  • $\begingroup$ I tried to answer your question above. $\endgroup$
    – Christophe
    Sep 11, 2020 at 13:11
  • $\begingroup$ Sorry for asking again but I struggle with an example: Let's imagine a Hertzian Dipole as source, radiating at one single frequency $\omega$. Now you pic a box in space and fourier transform the field in this box. Then you'll get results for a lot of $\vec{k}$-Vectors that do not obey the dispersion law (I tried in Matlab). But I know that the field should obey the wave equation since it's origin is the Hertzian Dipole. Do the modes not corresponding to the original radiation frequency now correspond to different ones or do I need to set them to zero? $\endgroup$ Sep 11, 2020 at 14:22

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