Why EM waves lose energy via the square of distance but the light will not? Does that mean that EM propagate forever in vacuum? 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.
 A: "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.
A: 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.
A: 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.
A: 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.
A: 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.
Meaning,
$\vec{S} \propto \frac{1}{r^2}$
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.
A: 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.
