Why does light continue on forever if it was created from some source whose radiation dwindles at a rate of the inverse square of distance. Clearly light can be viewed as an interdependent phenomena, the E field pulling the B field along with it, but if all light must come from a source, and that source creates a field dependent on distance, why doesn't the light die off?
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You seem to be trying to compare static electric fields with EM waves. To do a fair comparison you should set up to two scenarios as close as possible.
You are talking about the static field dropping with distance proportional to $\frac{1}{r^2}$, which tells me you are thinking about a point charge or some equivalent, spherically symmetric configuration (a sphere of charge perhaps). To do a fair comparison with an EM wave, you should also consider a point (or spherical) EM source.
Note: a(n infinite) plane wave is produced by an infinite plane source and is usually described by a sinusoidal EM wave propagating in a particular direction (perpendicular to the plane source), and thus has the form (for example): $$\vec{E} = E_0 \sin(k(x-ct))~\hat{z}$$ This obviously doesn't decay with distance ($x$). But to make a fair comparison, one should really compare this to an infinite plane of charge. Freshman electrostatics tells us that the associated electric field is $$\vec{E} = E_0~\hat{z}$$ ($E_0$ is dependent on the charge density) a constant as well! But these are not very physical scenarios as the require sources on infinite planes.
Now, back to the spherical source for an EM wave:
We are talking about EM waves, so the problem can be explored mathematically, by analyzing the Maxwell equations in vacuum: $$\left[\vec{\nabla}^2-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right]\vec{E} = 0~~~~~~~~~~~~~~~~~\mbox{(and the same equation for $\vec{B}$)}$$ For a spherically symmetric source, it makes sense to use spherical polar coordinates $(r,\theta,\phi)$, under which $$\vec{\nabla}^2\vec{E} = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial \vec{E}}{\partial r}\right)$$ leaving our equation as $$\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial \vec{E}}{\partial r}\right)-\frac{1}{c^2}\frac{\partial^2 \vec{E}}{\partial t^2} = 0$$ The general solution of which is $$\vec{E}(r) = \frac{\vec{E}_{out}(r-ct)}{r} + \frac{\vec{E}_{in}(r+ct)}{r}$$ where $\vec{E}_{out}(r)$ and $\vec{E}_{in}(r)$ are vector valued functions, specified by the boundary conditions (similar results for $\vec{B}$).
Application to your problem:
Let's assume we only have an outgoing wave, that is, $\vec{E}_{in}(r) = 0$ (this is the situation you are thinking of). Now $\vec{E}_{out}(r)$ specifies the profile of the wave - you can assume is sinusoidal if you like - this shape will propagate out. We see the dying off of the amplitude with distance is $\frac{1}{r}$, thus the energy density per second (which is $\propto |\vec{E}|^2$) of the wave dies of at $\frac{1}{r^2}$ and so we do see a "dying off" of light with distance.
Comparison to the photon approach
It's nice to see that the result using photons gives the same result. If we assume that a point source is (isotropically) emitting $N$ photons per second, each of the same energy $E$. Then at a distance $r$ from the source the density of photons each second is $\frac{N}{4\pi r^2}$ and so the energy density per second is $\frac{E~N}{4\pi r^2}$, which gives the same $\frac{1}{r^2}$ "dying off" of light with distance.
Will
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