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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?

Baban
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    Where have you read that the energy of an electromagnetic wave is proportional to its amplitude (and not amplitude square)? – eranreches Nov 17 '17 at 13:46
  • In my physics book. – Baban Nov 17 '17 at 13:48
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    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. – ACuriousMind Nov 17 '17 at 13:58

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

$$u=\frac{\varepsilon_{0}|E|^2}{2}+\frac{|B|^2}{2\mu_{0}}$$

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

$$u=\frac{\varepsilon_{0}|\vec{E}_{0}|^2}{2}+\frac{|\vec{E}_{0}|^2}{2\mu_{0}c^2}=\varepsilon_{0}|\vec{E}_{0}|^2$$

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

eranreches
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    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 https://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 – anna v Nov 17 '17 at 14:21
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    @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. – eranreches Nov 17 '17 at 14:31
  • Isn’t $E_{0}$ the intensity of the electric field component rather then a amplitude? – HolgerFiedler Nov 17 '17 at 18:39
  • @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. – eranreches Nov 17 '17 at 18:43
  • An amplitude is the furthest point of a wave. For EM field this seems not to be practicable. – HolgerFiedler Nov 17 '17 at 19:01
  • Why not? Consider for example the following electric field $\vec{E}=E_{0}\cos(kz-\omega t)\hat{x}$. – eranreches Nov 17 '17 at 19:05
  • https://physics.stackexchange.com/questions/287394/amplitude-of-light-waves – HolgerFiedler Nov 17 '17 at 19:39
  • @HolgerFiedler Can you elaborate more on what you're referring to? – eranreches Nov 17 '17 at 19:42
  • Eranreches, What is the amplitude in mm for EM radiation of say 0,8 mm wavelength? Amplitude is the incorrect wording for intensity. – HolgerFiedler Nov 17 '17 at 19:50
  • Let us continue this discussion in chat. – eranreches Nov 17 '17 at 19:50
  • @HolgerFiedler Your last comments indicate a deep misunderstanding of electromagnetic waves. The amplitude of an electromagnetic wave is well defined and the relationship with the energy flux of the wave is also well defined (it goes as the square of the amplitude). – ProfRob Nov 17 '17 at 21:04
  • @RobJeffries Give please an example of making an EM wave and measuring its amplitude. – HolgerFiedler Nov 17 '17 at 21:35
  • @RobJeffries What I can observe is EM radiation from a thermic source. The intensity of this radiation is related to the sum of the energy of the involved photons (related to their frequency). Give me an example of an EM wave. – HolgerFiedler Nov 17 '17 at 21:48
  • A CW Laser source. – eranreches Nov 17 '17 at 21:49
  • CW Laser emits a continuous stream of monochromatic photons. Where is the wave? Remember a wave has to have a periodicity. CWLaser produces EM radiation. A radio wave has a periodicity but this is due to the periodical acceleration of electrons inside the antenna rod. The emitted EM radiation is pulsed by the antenna generator. But even for such a periodical emission you couldn’t measure an amplitude. You could measure the intensity. – HolgerFiedler Nov 17 '17 at 22:01
  • You are overusing photons. A CW Laser does emit a periodic EM field. As for the measurements, you can't meausre intensity either. You meausre the resistance of a bolometer which depends on its temperature. So amplitude and intensity are similar in this sense, and both exist. – eranreches Nov 17 '17 at 22:18
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    @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. – ProfRob Nov 18 '17 at 00:28