I've read that in mechanical waves that it's(energy) is proportional to the amplitude squared but in electromagnetic waves it's only proportional to the amplitude, is that really true?

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    $\begingroup$ Where have you read that the energy of an electromagnetic wave is proportional to its amplitude (and not amplitude square)? $\endgroup$ – eranreches Nov 17 '17 at 13:46
  • $\begingroup$ In my physics book. $\endgroup$ – Baban Nov 17 '17 at 13:48
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    $\begingroup$ Please show some research effort before asking here, for instance, search for something like "energy electromagnetic wave" with your favourite search engine. If there's something unclear to you about the results, then you can ask a more specific question about that here. $\endgroup$ – ACuriousMind Nov 17 '17 at 13:58

Seems wrong then. The energy density of an electromagnetic wave is


which is certainty proportional to its amplitude squared. For example, a plane wave of amplitude $\vec{E}_{0}$ also satisfies $|\vec{E}_{0}|=c|\vec{B}_{0}|$ such that


since $c^2=\frac{1}{\varepsilon_{0}\mu_{0}}$.

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  • $\begingroup$ although I gave +1, and it is correct as far as energy density goes, BUT I think that there is a difference between energy density and the energy in the amplitude, see ccrma.stanford.edu/~jos/pasp/Acoustic_Energy_Density.html where "Thus, half of the acoustic intensity $ I$ in a plane wave is kinetic, and the other half is potential". There is not potential as such in the em plane wave . So maybe there is no "wrong" in the book but a different deffinition $\endgroup$ – anna v Nov 17 '17 at 14:21
  • $\begingroup$ @annav I agree with this distinction. It is somewhat interesting to see that if one compares the Lagrangian of an acoustic string $\mathcal{L}=\frac{\rho}{2}\left(\frac{\partial y}{\partial t}\right)^2-\frac{T}{2}\left(\frac{\partial y}{\partial x}\right)^2$ to that of an electromagnetic field in vacuum $\mathcal{L}=\frac{\varepsilon_{0}|E|^2}{2}-\frac{|B|^2}{2\mu_{0}}$ a correspondence between potential energy and the magnetic field can be seen. $\endgroup$ – eranreches Nov 17 '17 at 14:31
  • $\begingroup$ Isn’t $E_{0}$ the intensity of the electric field component rather then a amplitude? $\endgroup$ – HolgerFiedler Nov 17 '17 at 18:39
  • $\begingroup$ @HolgerFiedler The intensity of the electromagnetic field is related to the Poynting vector $|\vec{S}|=|\frac{1}{\mu_0}\vec{E}\times\vec{B}|=\sqrt{\frac{\varepsilon_0}{\mu_0}}|\vec{E}_{0}|^2$ where the last equality follows assuming a plane wave. $\endgroup$ – eranreches Nov 17 '17 at 18:43
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    $\begingroup$ @HolgerFiedler the question doesn't ask about photons. The amplitude of the wave can be readily confirmed by for example measuring the accelerating effect it has on electrons. I.e. Thomson scattering. Similarly, interference patterns result from the adding of amplitudes, not intensities. Your interjections on the subject of classical electromagnetism are increasingly bizarre. $\endgroup$ – Rob Jeffries Nov 18 '17 at 0:28

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