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

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?

• Your question is mixing up two ideas. The solution is actually pretty simple: The Poynting vector measure of the energy of the wave is correct, but inside it is the quantisation. Each photon inside the EM wave would have energy $hf$, which implies a fixed amplitude of the wave. However, at the level at which you would use Poynting vector, there would be huge numbers $n$ of these photons, thus averaging out to classically measurable energies $nhf$; for your two scenarios, the Poynting vector is the same, but different $f$ just means different $n$ to make the overall the same. Nov 6, 2023 at 7:43
• No I do not agree. Also think of it in terms of kinetic energy, let's say you have a rope attached to a rotating wheel. The rotational kinetic energy of the wheel is: K= 1/2 * I * w . I = 1/2 mr^2. w is the angular speed. We know w=2*pi*frequency. While r we could say is the same as the amplitude of the wave of the rope. Well it is pretty simple to see that the energy to make a higher frequency wave, is higher. Naturally also the amplitude has influence over the energy of the associated rope wave. So In electromagnetic waves how can I look for this measure?Which propety? Nov 6, 2023 at 9:44
• There is nothing for you to disagree; I am trying to tell you that new physics, in particular quantum theory, is coming in and changing the understanding. You are not allowed to insist upon a classical understanding of something that is inherently quantum. An understanding of this that is purely stuck at the Poynting vector level, will simply not show that kind of behaviour that we are talking about. Nov 6, 2023 at 9:48
• @DaveTechICX44 Poynting vector measures not energy, but energy flux, incident energy per unit area. This is a different thing than plane wave energy, i.e. in Poynting vector a lot of waves are summed-up in energy flux calculation. Do not mix different quantities. Energy of single plane wave is $h \nu$. Nov 6, 2023 at 9:51
• Nice question. A classical plane electromagnetic wave extending in the whole space is an infinite energy configuration of the field. For fields you have local energy density (or better, the stress energy momentum tensor). You can calculate the tensor for a plane wave configuration and that's the answer to your question (but remember that is more an energy density rather than an energy). The energy is given by integrating over the space region of interest. Nov 7, 2023 at 9:28

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.

• See my answer to this Nov 7, 2023 at 12:59

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.

• How is flux of complex Poynting vector phasor $\hat{S}$ relevant to the question? Nov 6, 2023 at 21:33
• The question is about energy of wave, not about rate of change of energy of wave. Rate of change of the Poynting energy as opposed to Poynting energy does have additional $\omega$ factor, but this is not what OP asks about. Nov 6, 2023 at 21:37
• @JanLalinsky I am not familiar with the concept you call "Poynting energy" but davetechicx44's question explicitly referred to the Poynting vector and its relationship to frequency of oscillation; see above "We know a way of measuring energy of a electromagnetic wave is the Poynting vector, which is independent of the frequency, etc." and I think my question answers that explicitly. And you may also wish to read John Rennie's "handwaving" answer above that effectively explains in words how the $S$ is directly related to that reactive power and frequency. Nov 7, 2023 at 9:44
• Poynting energy is volume integral of the Poynting energy density, and its rate of change is proportional to frequency of the wave. The question asks about energy of EM wave, and makes an incorrect statement about the Poynting vector, which should not be taken as basis for answering. Poynting vector or its surface integral is not a measure of energy of EM wave. Volume integral of Poynting energy density is. Nov 7, 2023 at 12:46
• See my answer to this Nov 7, 2023 at 12:58

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.