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I can technically follow the derivation of the EM energy density expression from Maxwell's equations, as here for example. But it seems really strange that, for waves, it does not depend on frequency. So I'm wondering how we could illustrate that more concretely.

The problem is, what kind of thought experiment can you do to show how much energy is in the field, or at least compare it between two different fields? It seems to me you would have to find a situation where the entire field is converted into motion of charges, and then the field energy would just be the resulting kinetic energy of the charges.

How can we show that, for a fixed amplitude and volume, a high-frequency EM wave has the same potential to generate motion as a low-frequency wave?

I am motivated by trying to understand Rayleigh scattering of sunlight in the atmosphere. Here, different frequencies produce the same amplitude of motion in electrons, so higher frequencies produce higher acceleration, hence greater energy. So that's why the scattered light turns more blue -- but that only makes sense if we can understand how the different colors had the same energy to begin with.

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  • $\begingroup$ E & B in the expression for energy density ($E^2+B^2$) can be sinusoidal (example for plane waves). So the energy density does oscillate with certain frequency (for monochromatic waves). The averaged energy density will be independent of frequency though. $\endgroup$
    – KP99
    Commented Oct 1, 2021 at 4:15
  • $\begingroup$ Two EM waves with identical electric field amplitude but different frequencies contain different numbers of photons. The one with lower frequency will contain more photons than the one with higher frequency. $\endgroup$
    – Vincent Thacker
    Commented Oct 1, 2021 at 6:42

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How can we show that, for a fixed amplitude and volume, a high-frequency EM wave has the same potential to generate motion as a low-frequency wave?

The mechanism through which EM waves generate motion is the Lorentz force, $$\vec{F} = q(\vec{E} + \vec{v}\times\vec{B}).$$

The rate of work done by the Lorentz force on a charged particle is then

$$\vec{F} \cdot \vec{v} = q\vec{E}\cdot\vec{v}$$

It then follows that the total rate work of done by the Lorentz force in a volume with charges is $\iiint_V \vec{E}\cdot\vec{J} dV$. The average value in the time domain of this quantity is then frequency independent, if $\vec{E}$ and $\vec{J}$ are harmonic. Of course the instantaneous power will be different.

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  • $\begingroup$ +1 I think this is definitely helpful, but I'm not convinced -- given two $\vec E$ fields, couldn't we just construct a $\vec J$ field for each one, such that the two volume integrals match? That's a specific instance of power consumption for a certain $\vec J$ field, not necessarily a measure of the entire power available in the $\vec E$ field. $\endgroup$
    – Adam Herbst
    Commented Oct 2, 2021 at 1:15
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A possible intuition is the simple harmonic motion, where for a pendulum with small angle of oscillations $$\omega = \sqrt{\frac{g}{l}}$$ or a spring in the elastic range $$\omega = \sqrt{\frac{k}{m}}$$ the energy of the oscillations is related to how much we deflect the spring or the pedulum from its equilibrium position. The frequency is the same, independent of that initial displacement.

An EM wave of a defined frequency can be thought as an infinite row of electric and magnetic vectors, each one oscillating as a SHM, and with a suitable phase shift, so that after a space equal to the wave length they are again in phase.

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  • $\begingroup$ But in this analogy, if you have different frequencies with the same amplitude, the total energy is different. Whereas for the EM field it is the same... $\endgroup$
    – Adam Herbst
    Commented Oct 2, 2021 at 1:02

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