today at some class at the university we were taught about the propagation of EM (electromagnetic) waves and that they lose energy proportional to the square of distance. Then someone asked: "Why then the light propagate forever in outer space?" And the professor answered: "Light is not electromagnetic wave so it won't lose energy, it's just described as an em." (like a mathematical model)

That answer left me speachless because as far as I know light is em. I would apreciate an answer from someone expert.

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    $\begingroup$ What?! Find another professor. Quickly. $\endgroup$
    – PM 2Ring
    Apr 12, 2022 at 21:46
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    $\begingroup$ Is there an inconsistency with light both "propagating forever" and "lose energy proportional to the square of distance"? $\endgroup$
    – BowlOfRed
    Apr 12, 2022 at 21:48
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    $\begingroup$ I don't see an inconsistency. For any finite distance you can calculate the energy remaining. $\endgroup$
    – BowlOfRed
    Apr 12, 2022 at 21:56
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    $\begingroup$ No energy is lost just spread over a larger area $\endgroup$ Apr 12, 2022 at 22:08
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    $\begingroup$ FYI: The inverse-square law is not a special property of light. It's just simple geometry. If you are receding from a source that sends out rays in all directions, the number of rays that reach you gets fewer and fewer as you get further and further from the source. It works for rays of light, same as it works for anything else that "radiates" outward in straight lines. $\endgroup$ Apr 12, 2022 at 23:55

6 Answers 6


Your professor is wrong! Light is indeed an EM wave and it follows an inverse square law for intensity loss with distance just like all other wavelengths of electromagnetic radiation.

  • $\begingroup$ This seems to me like merely a classical interpretation. Are you proposing that the energy of the light can decay below that carried by a single photon (at the frequency emitted)? $\endgroup$ Apr 13, 2022 at 3:08
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    $\begingroup$ @electronpusher Yes, the expected value of energy of light in a certain solid angle can fall below that carries by a single photon. This means it is likely that you will observe 0 photons, but there is a small chance you can observe 1 or more photons. $\endgroup$
    – Jagerber48
    Apr 13, 2022 at 3:16
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    $\begingroup$ @electronpusher the important point here is that photons are not travelling as classical point particles, instead the wavefunction gives a probability distribution of finding photons in a given area. It is obvious that the expected number of photons in a given region can be less than one (if nothing else, for a finite intensity one can always find a small enough volume that this is true), but when the light is detected it will be as discrete photons $\endgroup$
    – Tristan
    Apr 13, 2022 at 9:35
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    $\begingroup$ @electronpusher: Yes. The same is true of all EM radiation, of course. $\endgroup$ Apr 13, 2022 at 17:31
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    $\begingroup$ @electronpusher That starts to get a little bit philosophical. I can't disagree with what you're saying. I think it's a valid way of stating things. My personal view on quantum mechanics (I'm partial to Everettian interpretation of quantum mechanics) is that the photon is a quantum field that propagates out in all directions and that quantum field does propagate forever. If you perform measurements it's true that there's a chance you will detect no photons... but, at least before the measurement, the quantum field had non-zero support in your vicinity. $\endgroup$
    – Jagerber48
    Apr 14, 2022 at 2:19

By "Light", I assume you mean visible light. visible light is apart of the electromagnetic spectrum, and is itself, an electromagnetic wave.

An electromagnetic wave is a component of the electric and magnetic field, caused by the condition that: $$\frac{\partial \vec{J}}{\partial t} ≠ 0$$

When a charge accelerates, an electromagnetic wave is emmitted.

This wave consists of an electric component and a magnetic component

In the simplest form, for a point source of radiation

$\vec{E} \propto \frac{1}{r} $

$\vec{B} \propto \frac{1}{r} $

Meaning the strength of the Electric and magnetic field components decrease as the wave travels further away.

The poynting vector: $\vec{S} = \frac{1}{\mu_{0}} \vec{E} × \vec{B}$

Denotes the power radiated per unit area. Aka the rate of energy flowing as a result of the EM wave.


$\vec{S} \propto \frac{1}{r^2}$

There is an inverse square law for power radiated.

The rate at which energy flows is inversely proportional to the square of the distance from the source.

The total energy is constant however, as although the energy flow is less the further away you get, the energy is spread over a larger area.

Roughly speaking, calculating the total flow of energy around a spherical surface around the source, the area grows like $r^2$ while the poynting vector grows like $\frac{1}{r^2}$ causing the total rate at which energy flows across the sphere to be constant

$\iint \vec{S} \cdot \vec{da} = $ constant

For all spheres of any radius (growing like ct)

No energy is lost. But the energy flow at any point in space DOES decrease.

Although the flow of energy follows an inverse square law, for any finite distance, you should be able to detect the light.

  • $\begingroup$ I understand your explaination. So no energy is "lost"(converted to some other form) but in contrary it's spread over a wider area so if you are far away from the source the energy you can detect seems reduced. I guess that after a distance: X_max, you can't detect energy at all. So is there a way to calculate this X_max? $\endgroup$ Apr 13, 2022 at 9:45
  • $\begingroup$ That depends on the sensitivity of your instruments. Or perhaps some quantum weirdness, which I am not able to comment on $\endgroup$ Apr 13, 2022 at 11:42
  • $\begingroup$ for example the background cosmic radiation haven't reach that X_max (i guess) because we can still detect it. $\endgroup$ Apr 13, 2022 at 12:33
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    $\begingroup$ @BoliotisManousos There's no X_max. The inverse square relationship asymptotically approaches zero as distance tends to infinity, it never becomes exactly zero. Of course you might not have instruments sensitive enough to detect light weaker than some threshold, but that's just a technical detail, no law of physics. $\endgroup$
    – TooTea
    Apr 13, 2022 at 15:11
  • $\begingroup$ True, I figured that myself later when I was thinking that if energy is proporcional to 1/(distance)^2 then distance must go to infinity so that 1/(distance)^2 go to zero. $\endgroup$ Apr 13, 2022 at 17:31

"Why then the light propagate forever in outer space?" is a very important consideration since, as far as we know, outer space is as much of a vaccuum as we know (Source: NASA Estimations).

We look briefly to the Maxwell Equations (differential form): $$ \begin{align} \nabla \times \vec{\mathbf{B}} -\, \frac1{c^2} \frac{\partial\vec{\mathbf{E}}}{\partial t} & = \mu_0\vec{\mathbf{j}} \\ \nabla \cdot \vec{\mathbf{E}} & = \frac{\rho}{\varepsilon_0} \\ \nabla \times \vec{\mathbf{E}}\, &= -\, \frac{\partial\vec{\mathbf{B}}}{\partial t} \\ \nabla \cdot \vec{\mathbf{B}} & = 0 \end{align} $$

In the vacuum there are no free charges or free currents. More generally (and if you consider the equation for maxwell equations in general media), the current term of the differential equation is expanded to include the polarization and magnetization of the media. If you would like to more closely see the derivation you can see here a short paper by Zhang. Important to note here (as I don't think you will gain much insight from the derivation in the paper), is that when we remove the free charges and currents from the equation, we get to the elegant set of solutions in the vacuum:

$$ \begin{align} \nabla \times \vec{\mathbf{B}} & = \frac1{c^2} \frac{\partial\vec{\mathbf{E}}}{\partial t} \\ \nabla \cdot \vec{\mathbf{E}} & = 0 \\ \nabla \times \vec{\mathbf{E}}\,& = - \frac{\partial\vec{\mathbf{B}}}{\partial t} \\ \nabla \cdot \vec{\mathbf{B}} & = 0 \end{align} $$

With these equations you can insert them in each other and you arrive to the wave equation:

$$ \frac1{c^2} \frac{\partial^2(\vec{\mathbf{E}},\vec{\mathbf{B}})}{\partial t^2} - \nabla^2 (\vec{\mathbf{E}},\vec{\mathbf{B}}) = 0 $$

Very important to note is that this is for monochromatic plane waves (rigorous derivation is here), that means that the analyzed wave only contains one wavelength. It also assumes that the wave is a plane wave, meaning that either the source is far away enough that there are no wavefront changes, or the wave exists in the vacuum without having been generated. The explanation given above the inverse square law does take into account the light-emitting source (therefore the loss of energy is accounted for), and it only applies to single point sources. Stars and other celestial bodies play under different rules because of other circumstances.

That being said, and back to your professor's statement:

"Light is not electromagnetic wave so it won't lose energy, it's just described as an em."

There are three addendums I would make to make it a better statement:

"Light is not (always analyzed as an) electromagnetic wave so (it depends on the regime under which you are approaching it: For plane waves in vacuum) it won't lose energy (because there is no matter interacting with the plane wave, therefore there are no losses to the medium as it is freely propagating in the vacuum), (where) it's just described as an em."

This was also discussed here albeit from a more strict ray-propagation angle, which gives you a further insight into the more rigorous analysis of ray optics and wave optics.

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    $\begingroup$ Downvote: For OP, all electromagnetic waves we see are not plane waves. All real waves obey 1/r^2 law. And I think talking about plane waves in this questions context does a dis-service, as the question is about energy of a wave, and plane waves do not follow the norm. $\endgroup$ Apr 13, 2022 at 16:54
  • $\begingroup$ I provided a generalized explanation for plane monochromatic waves and made it clear that this is for plane waves specifically as a complement to your explanation of waves emanating from a single point source. Both explanations are valid and complementary to each other. All points where rigorous explanations are not carried out are also sourced. $\endgroup$
    – ondas
    Apr 14, 2022 at 8:10
  • $\begingroup$ Although the question directly is about his proffessors comments, the key question that OP said inspired it was "Why then the light propagate forever in outer space?", Talking about plane waves would just confirm the belief that because plane waves don't lose energy, this is why they "travel forever", which is false for real waves. I don't think that OP is at the level to distinguish the difference between plane waves and waves produced by accelerating charges, without getting themselfs confused $\endgroup$ Apr 14, 2022 at 10:34
  • $\begingroup$ Starting your answer off with that quote from one of the students,also certainly doesn't help with OP's question, since plane waves have absolutely nothing to do with that question. In the same paragraph you then say, (as if to answer the question), that space is a vacuum, and then give vacuum solutions to maxwells equations. Which have nothing to do with the question you pose to answer. And then say that in real life 1/r^2 isn't strictly true, leading TO EVEN MORE confusion. $\endgroup$ Apr 14, 2022 at 10:39

Light is simply visible part of the EM spectrum. It obviously follows rules for EM waves, including the 1/r^2 decrease of intensity with propagation distance. This 1/r^2 actually doesn't mean that the wave loses energy. Wave has the same energy, it is just spread over increasingly large area - and sphere surface area grows as r^2. So, the detector which is of fixed size - for example your eye - gets less light.

That the light travels forever in space is true ... but our radio waves travel forever in space too. They just quickly become too faint to be picked up because they didn't start all that strong in the first place.

Our sun at mere few light minutes distance is too bright to look at directly. You wouldn't have any issues looking at it when orbiting Jupiter. Similar star at the distance of several light years is a pale dot on the black night sky. Push it to tens of light years and you wouldn't even see it (without binoculars/telescopes). Push it to millions of light years away and even our biggest telescopes couldn't see it.


Photographers would love it if light did not get weaker at the square of the distance. But it does, leading to things like the "flash guide number" which relates aperture numbers (the inverse of a light-admitting diameter, so needs to be squared to relate to energies) with distance.

Of course light is a mixture of electromagnetic waves. And of course they propagate arbitrarily far in space. But they become weaker (and more spread out) in the process, according to inverse square law.

Now light is quantifiable into individual photons which cannot be subdivided. But as they spread out in space, their density decreases according to inverse square law.


Light is not inherently a particle or a wave, it simply what it is and the best we can do is construct models of it. Sometimes it is best to model it as a wave, sometime it is best model it as a particle. For macroscopic concerns, it is useful to model light as a wave, whose intensity decays with the square of the distance. This classical model is good until the energy in a given direction becomes very small and the "continuum hypothesis" breaks down. Then one must treat light as composed of particles (photons). The energy in a direction cannot decay lower than the energy of a single photon (which depends only on the frequency of the light). This is one of the interesting results of quantum physics.

The professor was somewhat right (though unclear): the reason light travels forever is because it is not really a wave, for such a question light must be modeled as an ensemble of particles and the energy in a direction cannot decay below that carried by a single particle.

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    $\begingroup$ Downvoting without offering a counter-argument doesn't benefit the community. $\endgroup$ Apr 13, 2022 at 2:36
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    $\begingroup$ unfortunately this happens a lot because a large number of people can vote, and some of them think that if they do not like an answer for various reasons, they downvote without commenting. I like you answer because it answers part of the question in the title "Does that mean that EM propagate forever in vacuum" . Maybe if you state that you are answering that part the down votes will stop. $\endgroup$
    – anna v
    Apr 13, 2022 at 3:56
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    $\begingroup$ Downvoted. Even classical light travels forever, the intensity just decays. For quantum light the expected value of intensity (or energy energy flux or whatever) in a given direction can and does decay to arbitrarily small levels, even below that of a single photon. In that case the probability of detecting 0 photons goes very high while the probability of detecting 1 or more photons goes low. $\endgroup$
    – Jagerber48
    Apr 13, 2022 at 4:30
  • $\begingroup$ @electronpusher Maybe don't start off with a tautology: "Light is light." That explains nothing. $\endgroup$
    – hft
    Apr 13, 2022 at 5:41
  • $\begingroup$ +1 for highlighting what the professor (probably) meant, actually. However, I still disagree with the argument: it is perfectly possible to model light as a classical EM wave even at extremely low intensity, as long as you don't care for any interactions with atoms or so. $\endgroup$ Apr 13, 2022 at 13:27

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