it's known that typical expressions for plane, spherical and cylindrical waves are (for instance in terms of electric field uniformly propagating along r axis, in frequency domain):

  • Plane Wave: $E(r) = E_0 e^{-jkr} $

  • Spherical Wave: $E(r) = \frac{E_0}{r} e^{-jkr} $

  • Cylindrical Wave: $E(r) = \frac{E_0}{\sqrt r} e^{-jkr} $

Which is the formal difference between these expressions?

  • The plane wave is not attenuated with the distance r

  • The spherical wave is attenuated with distance as $\frac{1}{r}$

  • The cylindrical wave is attenuated with distance as $\frac{1}{\sqrt r}$

So, the difference is in how amplitude depends on the distance between the source and the observation point.

But, "plane, spherical, cylindrical" refers to a wavefront's property. A wavefront is a set of points where the wave has the same phase. What does it has to do with it the wave amplitude? I can't see the physical link bewtween the physical meaning of wavefront and the amplitude dependence on the distance.

  • 1
    $\begingroup$ The definition of $r$ (at least in relation to other coordinate variables) and direction of $\mathbf{E}_0$ is different for each of these three cases. Choosing a different coordinate system without altering the expression implicitly changes what the expression means. $\endgroup$ Feb 7, 2021 at 20:16

1 Answer 1


To know what the phase is you need to know what the wave vector is. In the case of the plane wave the wave vector is a vector normal to the planes and the phase is the same over those planes. For the spherical waves the wave vector is radial and that means that the surface over which the phase is constant is a sphere. For the cylindrical it's the same as the spherical but in 2D.

On top of that, for the plane waves the magnitude is the same over all planes. For spherical waves you know that the magnitude is the same over any sphere and the same goes for cylindrical waves but with circles instead of spheres.


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