I have read that expression of radiated electric field of oscillating charge is inversely proportional to $r$ (distance from charge).

But the most common diagram used in our books is

enter image description here

But why not here magnitude of crest is decreasing?

I think correct diagram should be

enter image description here

Correct me if I am wrong?

  • 4
    $\begingroup$ The direction of propagation in the top diagram is wrong. The wave crests should be shown moving to the left. For some reason, getting this wrong is not uncommon. $\endgroup$
    – garyp
    Jul 25, 2023 at 11:31
  • $\begingroup$ @garyp How did you determine direction $\endgroup$
    – user373016
    Jul 26, 2023 at 6:19
  • 3
    $\begingroup$ The direction of an EM wave is the direction of the propagation vector $\vec{k} = \vec{E}\times\vec{B}$ $\endgroup$
    – garyp
    Jul 26, 2023 at 11:22
  • $\begingroup$ What does the picture actually represent? The EM field of a single emitted photon or the density of the field for all the photons. The inverse square law is about spreading the photons over an ever increasing area (the density of photons per unit area) not about the field from individual photons decreasing. $\endgroup$
    – matt_black
    Jul 27, 2023 at 13:08
  • $\begingroup$ This diagram doesn't indicate a wave propagating through space. It shows the waveform(s) as a function of time, presumably at the same point in space from a constant source. $\endgroup$ Jul 28, 2023 at 16:13

4 Answers 4


When the oscillations are shown with constant amplitude, it is assumed that the distance from the source is much greater than the wavelength, so that the decrease in amplitude is negligible over the scale of the diagram.

What is misleading about your top diagram is the inclusion of the oscillating source charge. This should be far away to the left of the oscillations that you have shown.

  • 2
    $\begingroup$ To be fair, the OP's diagram is also misleading inasmuch as there would be near-field contributions to $\vec{E}$ and $\vec{B}$ in close proximity to the charge; $\vec{E}$ and $\vec{B}$ don't have a simple plane-wave form there either. $\endgroup$ Jul 25, 2023 at 20:04
  • 1
    $\begingroup$ Please tell me I am correct or not physics.stackexchange.com/a/773472/349020 $\endgroup$ Jul 26, 2023 at 1:58

Of course you are correct that there should be a reduction in amplitude but can you draw it?

When an electric charge oscillates in a source of dimension $d$ and produces an electromagnetic of wavelength $\lambda$ there are three regions to consider.

Near field region, $d\ll r\ll \lambda$ where $r$ is the distance from the source, and is a region where the fields respond almost instantaneously to the source.

Intermediate field region, $r\approx \lambda$, in which analysis is complex.

Far field region, $r\gg \lambda$, in which traveling electromagnetic waves propagate spherically outwards and this is the region where the amplitude is proportional to $1/r$.

So in this region assume that the wavelength of light is $5\times 10^{-7}\, \rm m$ with amplitude $A$, ie $A = \frac kr$ where $r$ is a constant.

Consider the em wave with amplitude $A$ at a distance of $1\,\rm m$ from the source and of amplitude $A'$ when a further two wavelengths away.

$A=\frac k1$ and $A' = \frac {k}{1+10^{-6}} \Rightarrow \frac {A'}{A} =\frac {1}{1+10^{-6}}\approx 1-10^{-6}$, which is an insignificant reduction in amplitude?

Thus, you should not be showing an oscillating charge at the origin which you have used to draw an electromagnetic wave without qualification.


An oscillating electron emits photons. These photons have an E and a B field component and their field components converge into each other and propagate with c. (Whoever has something against photons, please note that photons are part of the standard model of physics and real entities that cause point-like reactions on fluorescent screens). A photon neither spreads out like a volume nor does it divide.

EM radiation consists of zillions of phtons from the many excited electrons. For thermal radiation, where the electrons are chaotically excited and chaotically emit photons, no wave properties of the radiation are directly detectable. There is only the phenomenon of diffraction at edges, as a result of which intensity distributions are visible on an observation screen.

But of course the intensity of the radiation decreases with greater distance from the radiation source. The photon density simply decreases with increasing distance.

Photons are emitted during the synchronous and rectified acceleration of electrons in a conductor. Their wavelength depends on the mean free path length of the electrons, the ohmic resistance, the electronegativity of the conductor material etc. In the case of antenna rods, which do not form a closed circuit, it also depends on the relation of the frequency of the wave generator to the length of the antenna rod and the absorption capacity of electrons at the end of the rod (see Mast Head).

The number of polarised emitted photons are produced - by the current changes of the radio wave generator - with changing intensity. Here, too, the number of photons per area decreases with increasing distance from the source.

Now it also becomes clear why the receiver conductor can be smaller by dimensions than the transmitting antenna. The electronics are able to tune the accelerations of the electrons in the receiving conductor that are excited by a small number of periodically arriving photons and to filter out the modulated information.

Your sketch is correct when you note that

  1. the maxima move forward with c and
  2. the E-maxima are arranged in a ring around the source (as in the case of a ring-shaped water wave of a stone thrown in a pond)
  3. the maxima represent the intensity of the radiation and these decrease with increasing distance from the source.

Everything ends up with only single photons being received from a very, very distant source. In the best case, you can still see the periodicity of the intensity fluctuations after longer observation.


Good point, Bharat. I feel that Philip answered well, and Farcher expanded with further considerations. I'm just noting the size of the reduction of the amplitude in your drawing. It would take some special circumstances to get a 1/3 reduction over one wavelength, and then 1/2 for the next. Maybe a very low frequency and local conditions.


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