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The power of an electromagnetic wave is proportional to its amplitude squared. (Watts, usually)

But, it is also proportional to its frequency, correct? The higher the frequency, the greater? ( measured in eV).

Is the TOTAL strength of the wave it's watts and electron-volts combined?

So, what is the difference between a laser, say, with a large number of infrared photons versus an x-ray laser with fewer photons, but higher frequency? They have the same total energy, but do they cut the same? Can a low-frequency laser beam, if intense enough, ionized atoms and molecules like individual photons of x-rays do?

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  • $\begingroup$ I don't quite understand that last line. It's not possible to add two quantities when one is in Watts and the other is in electron volts. $\endgroup$
    – user191954
    Commented Oct 20, 2018 at 3:43
  • $\begingroup$ It is not proportional to the frequency, at least not for classical waves. $\endgroup$
    – ZeroTheHero
    Commented Oct 20, 2018 at 3:49

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The term: electromagnetic wave belongs to classical electrodynamics, where the energy carried by the wave is described by the Poynting vector.

Electromagnetic waves carry energy as they travel through empty space. There is an energy density associated with both the electric field E and the magnetic field B. The rate of energy transport per unit area is described by the vector

$$\vec{S}=\frac{1}{\mu_{0}}\vec{E}\times\vec{B} $$

There is nothing about frequency in this classical definition of the energy carried by electromagnetic waves.

Frequency enters in the quantum mechanical frame, where it is postulated that the classical wave is built up of photons, particles in the standard model of particle physics, of energy $h\times \nu$ where $h$ is the Planck constant and $\nu$ the frequency of the classical wave emergent from zillions of such photons in quantum mechanical superposition.

The higher the frequency the more energy is carried by the photon, and one can get an average number of photons by dividing the power of the beam by $h\times \nu$. Both in the same units, joules preferably as watts are power, i.e. joules per second, one will get the number of photons per second passing the unit area.

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  • $\begingroup$ So, a 'concentrated' wave (laser?) of low frequency has the same energy as a less-concentrated wave of x-ray(s)? Think about a bunch of low-frequency waves constructively interfering, versus a single wave/photon of higher frequency?? $\endgroup$
    – Kurt Hikes
    Commented Oct 20, 2018 at 19:01
  • $\begingroup$ classical waves have extent , the photons are point particles. the word is not concentrated but more or less intensity. an intense low energy laser beam can carry enough energy as a less intense x-ray beam. How this energy is absorbed by a target is a different story , because it depends on the atomic and molecular structures hit. $\endgroup$
    – anna v
    Commented Oct 20, 2018 at 19:11
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For a classical electromagnetic wave of vector potential amplitude $A$ and frequency $\nu$ the energy is proportional to $\omega^2 A^2$, where $\omega = 2\pi \nu$. For a photon the energy is $\hbar \omega $. The difference is due to normalisation of the wave, which is $A^2 V = \hbar / \omega $, $V$ being the volume. Alternatively, if the wave is stated in terms of $E$, the $E$ field amplitude $\omega A$ absorbs the $\omega$ dependency.

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  • $\begingroup$ So, the frequency times two-pi times the amplitude-squared gives the TOTAL energy? $\endgroup$
    – Kurt Hikes
    Commented Oct 20, 2018 at 19:03
  • $\begingroup$ So you are talking about oscillating fields whose amplitudes themselves oscillate? $\endgroup$
    – BioPhysicist
    Commented Oct 20, 2018 at 19:03

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