How to Measure Energy of a Electromagnetic Wave accounting also for frequency?

Question

We know a way of measuring energy of a electromagnetic wave is the Poynting vector, which is independent of the frequency. But let's say we want to make two different electromagnetic waves, with different wavelengths and so different frequency, but with the same amplitude (so same Poynting vector).

In my opinion the guy in the image is spending more energy for the wave with higher frequency. But the Poynting vector, doesn't care. So is there another property of Electromagnetic Waves that accounts for the energy that the guy used to make the wave?

Answer 1

I'm going to attempt an intuitive explanation of this. It will inevitably be a bit hand-wavy so beware of taking this too literally but I hope it will help clarify the situation.

Suppose our physicist is just moving a ball on a spring (not necessarily at the natural frequency):

The physicist is not doing any work and it doesn't matter with what frequency they move the ball they still aren't doing any work. This is because any energy that needs to be supplied to stretch the spring is returned as the spring contracts again, so the net energy supplied is zero.

This is a little counterintuitive because we all know waving our arm faster is more effort, but that's because our muscles cannot use the energy returned to them. The physicist will certainly get tired faster the faster they move the ball, but that's just their inefficient muscles and not a fundamental feature of the ball spring system.

The only way the physicist would need to put in work is if the energy of the ball and spring was leaking away somewhere e.g. in a mechanical system this could be due to friction. Replace the ball by a charge and now the energy leaks away because it is generating an EM wave that carries away energy as described by the Poynting vector.

The point of this is that while it may seem unintuitive it does not take extra work to move the charge faster because the net work to move the charge is zero and this is independent of the frequency. The only work you need to supply is to compensate for the energy being carried away by the wave.

Answer 2

This is a good question and it should have an explanation within classical EM. There the answer is to be found in Poynting's theorem in which the Poynting vector is just one term.

From Maxwell's equations you can derive by using simple vector analytical identities that $$\operatorname{div}(\mathbf{E}\times \mathbf{H}) = -\mathbf{E} \cdot \mathbf{J} - \big( \epsilon \mathbf{E} \cdot \frac{\partial \mathbf{E}}{\partial t} + \mu \mathbf{H} \cdot \frac{\partial \mathbf{H}}{\partial t} \big) \tag{1}\\ $$ or in complex amplitude form

$$\begin{align} \operatorname{div} (\hat{\mathbf{E}}\times \hat{\mathbf{H}}^*) &=-\hat{\mathbf{E}} \cdot \hat{\mathbf{J}}^* +\mathfrak{j}\omega \big(\mu |\hat{\mathbf{H}}|^2 - \epsilon |\hat{\mathbf{E}}|^2 \big).\end{align} \tag{2}$$

Already in classical EM was noted that the vector flux of the power (work rate) cannot be uniquely represented by the Poynting vector as defined by $\mathbf S = \mathbf{E}\times \mathbf{H}$ because one can always add to it an arbitrary solenoidal field, say, $\mathbf s$ such that $\rm{div} \mathbf s = 0$, to $\mathbf S$ to be a new Poynting vector $\mathbf S +\mathbf s$ having all measurable results be unchanged.

In fact, the only thing that is measurable is the net flux through the closed boundary of a volume, say $\mathcal V$, that is $\int_{\mathcal V}\operatorname{div}\mathbf{S}dv=\oint_{\partial \mathcal V} \mathbf S d\mathbf{\sigma}$.

$$\oint_{\partial \mathcal V} \mathbf S d\mathbf{\sigma}= -\int_{\mathcal V}dv{\mathbf{E}} \cdot {\mathbf{J}}- \int_{\mathcal V}dv\big( \epsilon \mathbf{E} \cdot \frac{\partial \mathbf{E}}{\partial t} + \mu \mathbf{H} \cdot \frac{\partial \mathbf{H}}{\partial t} \big) \tag{3}$$ or with $\hat{\mathbf{S}}=\hat{\mathbf{E}}\times \hat{\mathbf{H}}^*$ $$W=\oint_{\partial \mathcal V} \mathbf S d\mathbf{\sigma}=-\int_{\mathcal V}dv\hat{\mathbf{E}} \cdot \hat{\mathbf{J}^*}+ \mathfrak{j}\omega \int_{\mathcal V}dv\big(\mu |\hat{\mathbf{H}}|^2 - \epsilon |\hat{\mathbf{E}}|^2 \big). \tag{4}$$ In Eq (4) on the left you have $W$ representing the net radiated (propagated) power flux going in and out through the boundary $\partial \mathcal V$ of a fixed volume $\mathcal V$, while on the right you have the energy rate of the electric and magnetic fields within that fixed volume. When $\mathbf J=0$ you have a lossless case.

The accumulated energy/work rate in $\mathcal V$ is represented by and depends explicitly on the frequency $\omega$ multiplying factor, exactly as you have expected it. When $\mathbf J=0$ you have the lossless case. But note that the multiplicative dependence is on the imaginary that is reactive part and not on the real dissipative part, exactly as in an LC resonator in which the electric and magnetic energies oscillates between the two. If $\mathcal V$ is a lossless cavity with conducting walls then $W=0$, nothing goes in and nothing goes out.

Answer 3

We know a way of measuring energy of a electromagnetic wave is the [Poynting vector][2], which is independent of the frequency.

Energy of EM wave in some volume $V$ is measured by integral of the Poynting energy density over that volume:

$$ E_{wave} = \int_V \frac{1}{2}\epsilon_0 E^2 + \frac{1}{2\mu_0} B^2 dV. $$

The Poynting vector does not "measure" energy of EM wave, but it is directly related to its intensity, that is, how much energy comes per unit time per unit area perpendicular to direction of EM energy flux in space. So that total energy coming through any simply connected and oriented surface $\boldsymbol {\Sigma}$ per unit time is:

$$ \frac{dE_{income}}{dt} = \int_\Sigma d\boldsymbol{\Sigma} \cdot (\mathbf E\times \mathbf B/\mu_0 ). $$

When the wave is a localized pulse, after it passes entirely through the surface, $E_{income} = E_{wave}$. So the Poynting vector is related to how fast the energy passes through the surface, not to total value of the wave energy.

In my opinion the guy in the image is spending more energy for the wave with higher frequency. But the Poynting vector, doesn't care. So is there another property of Electromagnetic Waves that accounts for the energy that the guy used to make the wave?

The image does not explain what the wave represents. If it is rope oscillating due to forced oscillations of the end, and if we are comparing two oscillations with same amplitude of oscillation (elongation), then yes, the guy is spending more energy per unit time to maintain oscillation at higher frequency. But this is because the same amplitude of oscillation at higher frequency means amplitude of velocity is much higher, and thus both kinetic and potential energy are higher and more gets dissipated per unit time (this requires some calculations to show). Energy of the rope consists of both the potential and the kinetic component, and at higher frequency, both these components are higher, despite the amplitude of motion being the same.

With EM wave, it is similarly sum of two components, electric and magnetic energy, whose values are related to frequency. These energies are proportional to square of electric and magnetic field, and these in turn depend on how fast one oscillates with charges that generate those wave fields. So in a sense, energy of EM wave generated does depend on the frequency of the forced oscillations of the charge, but this dependence is fully taken into account in the Poynting energy formula above, via the amplitudes of electric and magnetic field. The higher the frequency of oscillation, the stronger the electric and magnetic field generated. Electric field strength in the wave zone (far enough from the charge) is proportional to acceleration of the charge, thus to square of frequency of oscillation and to amplitude of charge's oscillations $y_0$:

$$ E \propto \omega^2 y_0. $$ $$ B \propto \omega^2 y_0. $$

Thus the Poynting energy density (energy per unit volume), depends on frequency as $\omega^4$, if we somehow keep the charge oscillation amplitude constant. This is approximatelly valid for Poynting energy density e.g. in case it is due to the Rayleigh scattering in atmosphere or water, where the charge oscillation amplitude is constrained by the size of molecules, or density fluctuations. Hence intensity of the Rayleigh scattering goes as $\omega^4$, and thus intensity of blue scattered light is higher than that of red light, even while both are associated with charge oscillations of similar amplitude.