I have recently found out that word "intensity" is somewhat ambiguous for electromagnetic waves. I always thought of it as the magnitude of the Poynting vector (field), but this is not always correct.

When considering the interaction with of EM waves with matter (i.e. electronic transitions), the magnetic part is often completely neglected (as the interaction of matter with electric fields is generally much stronger). In this context, one might consider the amplitude of the electric field to be the intensity and indeed this is the same as the magnitude of the Poynting vector for a single plane wave. However this is not true for a superposition of plane waves.

Now I always thought (without deeper consideration) that the size of the Poynting vector is proportional to the amount of photons in the EM field, but apparently it is more important to consider the size of the electric field (or maybe the size of the magnetic field in different cases) than the amount photons for many applications. Now if the interaction things with light can be understood in terms of the interactions with the fields, what does the size of the Poynting vector actually mean in practice? Are there any processes that depend on this rather than directly on the amplitude of the electric or magnetic field?

EDIT: Although there is some interesting discussion below, my question has been interpreted slightly differently from what I intended (nevertheless learned a lot though!). I am a graduate student, so I am certainly familiar with the basic of electrodynamics. To be very concrete, my research involves making an interference pattern of plane wave coming from different directions. This pattern excited fluorescent beads, which are then measured with a "camera". I expected that resulting image would be related to $\langle|\mathbf{S}(\mathbf{r})|\rangle$, but it turns out to be much closer to $\langle|\mathbf{E}(\mathbf{r})|^2\rangle$. This made me wonder what $\langle|\mathbf{S}(\mathbf{r})|\rangle$ really means and what it would be relevant for.


5 Answers 5


If there exist both an electric field $\vec{E}$ and a magnetic field $\vec{B}$, then the Poynting vector is defined as: \begin{equation} \vec{S}\equiv\frac{1}{\mu_0}(\vec{E}\times\vec{B}). \end{equation}

And it quantifies the energy per unit area, per unit time transported by the electromagnetic fields.

And the actual definition of intensity in this case is - The average power (or energy per unit time) per unit area transported by an electromagnetic wave. So intensity $I\equiv\langle\vec{S}\rangle$ (where $\langle...\rangle$ means averaging).
Now consider our electric and magnetic fields are given as follows: \begin{equation} \vec{E}(\vec{r},t)=E_0\space cos(kz-\omega t + \delta)\hat{x},\\ \vec{B}(\vec{r},t)=\frac{1}{c}E_0\space cos(kz-\omega t + \delta)\hat{y}. \end{equation} Then, $\vec{S}=c\epsilon_0E_0^2cos^2(kz -\omega t+\delta)\hat{z},$ which gives $\langle\vec{S}\rangle=\frac{1}{2}c\epsilon_0E_0^2=I.$
Note that $\vec{B}=0\implies \vec{S}=0.$

I cannot say much about the relation between Poynting vector and light quanta (photon), but to tell you more about the use of Poynting vector see below.

enter image description here

If you send current through a wire you do some work on it. Joule heating of the wire is a measure of it.
As in figure, an electric field is there, with magnitude $E=\frac{V}{L}$ ($V$: potential difference on the wire), and direction as indicated.
Resulting magnetic field is there, magnitude $B=\frac{\mu_0 I}{2\pi a}$, direction: around the curved periphery of the cylinder.
So Poynting vector? $S=\frac{VI}{2\pi a L}$, direction: perpendicular to both $E-$ and $B-$ fields, i.e., radially inward the wire.
To interpret physically - the energy stored in the electric and magnetic field passes in through the surface of the wire; and we just calculated that amount of energy per unit area, per unit time (or, power delivered per unit area).

[Everything as understood from Electrodynamics by D.J. Griffiths.]

  • 1
    $\begingroup$ Thanks for the example! I was stuck thinking about waves... What is interesting to me here is that a current wouldn't give an electric field if there was no resistance in the wire. Hence the Poynting vector would be zero and indeed there is no Joule heating. Thus in a sense there is an electric field because there is energy transfer (resistance). I suppose in the way when an electron "uses" the electric field to get excited, it produces a magnetic field and hence a non-zero Poynting vector. That doesn't necessarily help to understand the Poynting vector of a wave though. $\endgroup$
    – ReinTurtle
    Jan 20, 2023 at 16:39
  • $\begingroup$ Field and Matter are two independent entities. This statement gets stronger when we see that momentum of matter alone is not conserved, but sum of momenta of matter and field (here charged particle and EM fields) is conserved. Field need not be associated with some matter (like electron) to exists. Field also carries energy along with momentum. So a propagating electromagnetic wave can have its own corresponding Poynting vector, no need to find charges. $\endgroup$ Jan 20, 2023 at 17:01

Given the stress-energy tensor of the EM field, the Poynting vector is its $(0,i)$ part, for $i=1,2,3$.

To understand what this means, compare this with the $(0,i) $ part of the stress-energy tensor of a fluid: it is the momentum density of the field! Therefore, its magnitude is just the norm of the momentum density.

This is, at least semantically, simpler than Wikipedia explanation: "The Poynting vector (or Umov–Poynting vector) represents the directional energy flux (the energy transfer per unit area per unit time) or power flow of an electromagnetic field", see also this question.

Note: I do not think that it's very convenient to think in terms of photon number (or photon number density), because the number of photons is not conserved (a photon is its own antiparticle), see also the nice answers to this question. However, I like Gerard's answer, but to have a reasonable interpretation in terms of number of photons, notice that you should have a fixed frequency $\nu$, which is typically not the case (and its expectation value in QED is a more complex expression, see this). However, a simpler formula is reported here, if this is accurate then the simple interpretation in terms of the number operator is OK.


Now if the interaction things with light can be understood in terms of the interactions with the fields, what does the size of the Poynting vector actually mean in practice? Are there any processes that depend on this rather than directly on the amplitude of the electric or magnetic field?

The Poynting vector, a function of total electric and total magnetic field, is interpreted as density of flux of EM energy in macroscopic theory (watts per unit area), due to energy interpretation of the Poynting theorem. The Poynting theorem is the following mathematical relation based solely on Maxwell's equations:

$$ \mathbf j \cdot \mathbf E = -\nabla \cdot \mathbf S - \partial_t w $$ where $w$ is the well-known Poynting energy density expression in terms of $E^2,B^2$.

The energy interpretation of this equation adopts additional assumption: that $\mathbf j \cdot \mathbf E$ is an accurate expression of total work per unit volume per unit time done by EM field on matter, so the terms on the right-hand side of this equation can be interpreted as convergence of EM energy flux density vector and rate of decrease of energy density at the same point of space. Under this assumption, Poynting's theorem becomes a law of local conservation of energy of matter and EM field. One can say work done on matter particle in a region of space by EM forces equals EM energy that came in through the surface of the region from the outside, plus decrease of EM energy in the region.

Such interpretation of the Poynting theorem equation is either incorrect, or correct but more complicated in the microscopic EM theory, where total fields are $\mathbf e, \mathbf b$, for the following reasons:

  • if the charged particle that is experiencing EM force is a point, the term $\mathbf j \cdot \mathbf e$ is too singular and its meaning cannot be consistently defined; total electric field diverges at the particle. So the assumption about $\mathbf j\cdot \mathbf e$ is invalid and Poynting's theorem while true, is useless. We can derive another similar theorem that can be interpreted as local conservation of energy, if we base it on some regular expression for EM work. There is the Frenkel theory which defines its own energy density and energy flux density expressions based on the idea that point particles experience only external field, and their own field does not affect them. This leads to expressions for EM energy and EM energy flux density that are different from the Poynting ones.

  • if the charged particle is extended in space with a finite charge density everywhere, the term $\mathbf j \cdot \mathbf e$ is fine and formula for the Poynting vector is fine and can be interpreted in terms of local conservation of energy, but $\mathbf e$ is usually unknown to us, because near and inside the particles, it is affected by the particle's own field in a non-trivial way:

$$ \mathbf e = \mathbf e_{external} + \mathbf e_{due~to~particle}. $$

People sometimes assume (as you did) they can simply put into the Poynting formula the macroscopic field $\mathbf E$ or total external microscopic field $\mathbf e_{external}$ and obtain the correct microscopic energy flux density. But that is not correct. The electric field due to extended charged particle itself is contributing to the total electric field acting on the particle; particle's own field acts and can do work on the very same particle. This energy exchange is not captured by Poynting's vector based on macroscopic field $\mathbf E,\mathbf B$, or based on external microscopic fields $\mathbf e_{external},\mathbf b_{external}$.

In your example, macroscopic Poynting's vector vanishes at a certain point of space, while macroscopic electric field does not. But that is possible in microscopic theory too only if there is no moving charged particle there. If there is some moving charged particle or group of moving charged particles there, these will affect the total microscopic field so that magnetic field does not vanish. Poynting's vector based on microscopic fields won't vanish there anymore and EM energy flux into or from the charged particle can be non-zero.

When electric field accelerates a charged particle, the particle generates magnetic field component proportional to the velocity everywhere near and inside the particle, and total magnetic field no longer vanishes where the particle is. This way, the particle's own magnetic field due to its motion makes it possible for vector of EM energy flux to not vanish, and for electric energy to be transformed into kinetic energy of the particle. This is the case irrespective of whether the charged particle is a point or extended charge distribution with finite density.

  • $\begingroup$ Wow, interesting stuff. I never knew about this additional assumption to get to the conventional interpretation of the Poynting vector and I never considered that the field of a particle also influences the situation... This certainly explains a lot of my confusion. $\endgroup$
    – ReinTurtle
    Jan 23, 2023 at 14:25
  • $\begingroup$ @Jan , I would like to know more about this "frenkel theory" you keep talking about. Could you please send some material on the topic for a detailed picture of what its saying... $\endgroup$ Sep 28, 2023 at 16:18
  • $\begingroup$ Could you please suggest some sources from which one can learn more about frenkel's theory? $\endgroup$
    – Arjun
    Sep 28, 2023 at 16:19
  • $\begingroup$ @nickbros123 J. Frenkel, Zur Elektrodynamik punktfoermiger Elektronen, Zeits. f. Phys., 32, (1925), p. 518-534. dx.doi.org/10.1007/BF01331692 In English, this article also explains it concisely: R. C. Stabler, A Possible Modification of Classical Electrodynamics, Physics Letters, 8, 3, (1964), p. 185-187. dx.doi.org/10.1016/S0031-9163(64)91989-4 $\endgroup$ Oct 3, 2023 at 19:06

So here I am going to reference Griffith's (4th edition) Introduction to Electrodynamics.

The general solution to Maxwell's equations are called "plane waves". Plane waves are notable for being described by a few equations, namely 9.49, 9.50, and 9.56. These are technically only valid for monochromatic light, but can be generalized to all frequencies with an integral over the frequency range or as a sum over individual frequencies. \begin{align} \tilde{E}&=\tilde{E}_0 e^{i(\vec{k}\ \cdot\ \vec{r} - \omega t)} \hat{n}\\ \tilde{B}&=\tilde{B}_0 e^{i(\vec{k}\ \cdot\ \vec{r} - \omega t)} \hat{k}\times\hat{n}\\ &=\frac{1}{c}\hat{k}\times\tilde{E} \\ \hat{n}\ &\cdot\ \hat{k} = 0 \\ \vec{S} &= \frac{\vec{E}\times\vec{B}}{\mu_0} \\ \end{align}

Where the complex vectors with the tilde relate to the real vectors by the relation $\vec{V} = \Re(\tilde{V})$.

Since $\hat{n}$, $\hat{k}$, and $\hat{n}\times\hat{k}$ are an orthonomal set of basis vectors, the magnitude of the Poynting vector is $$ |S| = \frac{|E||B|}{\mu_0} $$ So clearly the magnetic field is not being neglected from the magnitude of the pointing vector. We can further simplify this expression by replacing the magnitude of the magnetic field with a relevant expression from above. We take the magnitude of the magnetic field to be $|B|=\frac{|E|}{c}=\sqrt{\epsilon_0\mu_0}|E|$. This is also equation 9.54 and can be derived from the above set of equations. So we have $$ |S| = \frac{|E|^2}{\mu_0 c} $$ finally with reference to 9.42, $c=\frac{1}{\sqrt{\mu_0\epsilon_0}}\rightarrow \epsilon_0 c = \sqrt{\frac{\epsilon_0}{\mu_0}}$ $$ |S|=\sqrt{\frac{\epsilon_0}{\mu_0}}|E|^2=c\epsilon_0|E|^2 $$

It is worth noting that this quantity is measured in watts per unit area. We can easily convert to a photon density with the equation $E_\text{single photon energy}=h\nu$. The average energy density is given directly by equation 9.60. The number density of photons is $n$ and may be taken to be $$ n = \frac{\left<u\right>}{h\nu} = \frac{c\epsilon_0 E_0^2}{2h\nu} $$

  • $\begingroup$ Indeed this is true for a single wave, but for example with two identical counterpropagating waves, right in the middle you will have an oscillating electric field and 0 magnetic field. Hence the Poynting vector here is 0. Clearly then $|S|$ is not proportional to $|E|^2$. What about photons in this case? $\endgroup$
    – ReinTurtle
    Jan 20, 2023 at 16:46
  • $\begingroup$ With 2 waves the electric field may be described by the equation $\tilde{E}=\tilde{E}_{1,0} e^{i(\vec{k}_1\ \cdot\ \vec{r} - \omega t)} \hat{k}_1 + \tilde{E}_{2,0} e^{i(\vec{k}_2\ \cdot\ \vec{r} - \omega t)} \hat{k}_2$. In general you will still have the same quantities $c\epsilon_0|E|^2_{1,0}$ and $c\epsilon_0|E|^2_{2,0}$, but a third will be present which will represent the same beating phenomena that two sound waves will cause. $\endgroup$
    – Gerald
    Jan 20, 2023 at 16:57
  • $\begingroup$ Regardless, the greater the energy density, the greater the photon density. When the beating creates a low (quiet) spot the number of photons will be less than when the beating is high. $\endgroup$
    – Gerald
    Jan 20, 2023 at 16:57
  • $\begingroup$ S=0 for a particular place and particular moment, not forever and everywhere. We tend to find the average Poynting vector, time average or spatial average. Also, Poynting vector is a measure of energy carried per unit time per unit area; so at a particular place and particular time the energy can be zero, but other places have the rest of the energy. Although I think detailed analysis needs a well defined question. $\endgroup$ Jan 20, 2023 at 17:09

The Poynting vector describes the magnitude and direction of the flow of energy in electromagnetic waves per unit surface. That's a little too less, but hope it helps:)

  • $\begingroup$ flow of energy in electromagnetic waves per unit $\mathbf{AREA}$ $\endgroup$
    – hyportnex
    Jan 20, 2023 at 15:28

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