Electromagnetic waves are frequently described as "self-propagating", implying a mode of propagation distinct from that of electrostatic fields; but as I understand things, both have strength proportional to the inverse square of the distance from their source. Let me lay out what one ignorant of wave propagation and ignoring the magnetic field expects to see from a moving charge:
- Suppose I am some distance $r$ away from a charged particle moving away from me with constant velocity $v$. Then at time $t$ I will perceive an electric field of strength proportional to $\frac{1}{(r+t\cdot v)^2}$.
- Suppose instead that the charge is oscillating along the vector pointing from it to me, with period $P$ and amplitude $A$. Then I expect to see an electric field of strength proportional to $\frac{1}{(r+A\cdot \sin(t\cdot \frac{2\pi}{P}))^2}$.
- Suppose rather that it oscillates perpendicularly to the vector connecting us. Then I expect to see an electric field whose direction wobbles between right-ish and left-ish with period $P$ and whose magnitude is proportional to $\frac{1}{r^2+A^2\cdot \sin^2(t\cdot \frac{2\pi}{P})}$.
Edit Rephrased the below because I forgot that I was dealing with inverses.
In both situations (2) and (3) the electric field where I stand is the sum of a constant and a periodic function (in case (3) two periodic functions along perpendicular axes), purely as a result of the oscillation of the source charge--no magnetic or special "propagation" effects needed. Obviously I have neglected the finitude of the speed of light in these calculations, which would introduce a tiny bit of distortion.
The periodic component is something like the multiplicative inverse of a squared sine wave, shifted so as to stay finite; some fancy trig likely makes it sinusoidal, since it's pretty dang close. Here are graphs of, respectively, the transverse and longitudinal components of (3), using r=1, P=1, and A=0.1:
Is it the case that the electromagnetic wave produced by Maxwell's equations in (2) and (3) will lose amplitude at precisely the same rate as this "inverse wave" that derives trivially from the inverse square law and the charge's motion? How, then, do we consider the wave "self-propagating" if it has no special powers to resist decay and acts just like the rest of the electric field?
Related desired elaboration: Apparently the Maxwellian wave will have the same frequency as the inverse wave, so how/why do their phases/amplitudes differ? And where do we get the energy for this extra wave?