I am just going over my understanding of the mathematical expression of plane and spherical waves.

For a plane wave we can show that the general mathematical form of such a wave is : $\vec{\Psi}(\vec{r},t) = \vec{A}e^{i(\vec{k} \cdot \vec{r} \pm \omega t)}$. For a spherical waves we can show that the general form of such a wave is of the form: $\Psi(\vec{r},t) = \frac{\vec{A}}{r} e^{i \left(\vec{k}\cdot \vec{r}\pm \omega t\right)} $.

I wanted to understand 2 things from the above, firstly isn’t $\vec{k}$ a function of $\vec{r}$ for spherical waves, since the direction of propagation changes with position, how is this shown in the above equation? The equations for spherical and plane waves look identical except for the $1/r$ factor.

Secondly why is there no attenuation factor for the plane wave equation? I understand the $1/r$ factor for spherical waves as the factor by how much the amplitude must decrease by to conserve energy across the wavefront. Surely the plane waves must have such an expression too?

  • $\begingroup$ Related post by OP: physics.stackexchange.com/q/532302/2451 $\endgroup$
    – Qmechanic
    Commented Feb 22, 2020 at 13:59
  • 2
    $\begingroup$ I believe the $\vec k r$ in your spherical wave expression should be just $kr$. It doesn't make sense to take the exponential of a vector. This should answer your first question. $\endgroup$ Commented Feb 22, 2020 at 15:26
  • $\begingroup$ Does thinking of a plane wave as the limit of a spherical wave as r becomes infinite help with your second question? Increasing r has no effect when the origin is infinitely distant. $\endgroup$
    – DrC
    Commented Feb 22, 2020 at 15:47
  • 1
    $\begingroup$ @AccidentalTaylorExpansion the above is a typo I have now corrected $\endgroup$ Commented Feb 23, 2020 at 12:57

2 Answers 2


Even scalar spherical waves, i.e. solutions of the d'Alembert's equation $$ \frac{1}{v^2}\frac{\partial^2{A}}{\partial{t}^2}-\nabla^2 A=0 $$ in spherical coordinates are less simple as your expression would suggest.

It is not a surprise if one tries to find a solution of the equation by variable separation.

The closest expression to the yours is the asymptotic expression for large r of outgoing waves which is: $$ u(r,\theta,\phi) = \frac{e^{ikr}}{r}f(\theta,\phi)+O\left(\frac{1}{r^2}\right). $$
So, you see that all the information about the directionality of the wave is contained in the angular factor $f(\theta,\phi)$.

An exact (i.e. not asymptotic) general expression for $u(r,\theta,\phi)$ can be obtained in analogy with the plane wave expansion in cartesian coordinates, in terms of a series of spherical Bessel functions multiplied by spherical harmonics.

Notice that the general solution of the vector d'Alambert equation may have a more complicate form than the scalar one. If you are interested, you may have a look at the treatment of the vector equation in the Stratton's textbook Electromagnetic theory.

Summarizing, about your questions:

  1. in the case of spherical waves, the modulus of the wave-vector $\bf k$ enters in the radial components of the solution, while all the information about the direction of wave-packet is left to the coefficients of spherical harmonic components.
  2. As you noticed, a physical explanation for the $1/r$ asymptotic leading behavior is the need of energy conservation along the radial direction. In the case of a plane wave, energy conservation is guaranteed, without need of factors, in any cylindrical volume with the axis parallel to the wave-vector.
  • $\begingroup$ Suppose the E-field of a wave traveling in the z-direction is polarized in the x-direction, $E_y=E_z=0$. Its wave equation is $\nabla^2 E_x=\frac{1}{c^2}\frac{\partial^2E_x}{\partial{t^2}}$ and the solution of which is any function that can be written as $f(z \pm ct)$. I tried $E_x(z,t)=\frac{1}{x-ct} sin(x-ct)$ and it works. So $\psi(r,t)=\frac{A}{r-ct}e^{i(\mathbf k \cdot \mathbf r - \omega t)}$ should satisfy the spherical wave equation with attenuation $\frac{1}{r-ct}$ instead of $\frac{1}{r}$. correct? $\endgroup$
    – Leon Chang
    Commented Oct 21, 2023 at 0:50

firstly isn’t k⃗ a function of r⃗ for spherical waves

The wave vector $\mathbf k$ represents the wave propagation direction and is perpendicular to wavefronts of constant phase, i.e., $\mathbf k \cdot \mathbf r$ = constant. Its magnitude is equal to $\frac{2 \pi}{\lambda}$, not a function of r. For both plane wave and spherical wave, $\mathbf k$ and $\mathbf r$ are parallel, therefore $\mathbf k \cdot \mathbf r$ = kr cos(0) = kr. Specifically, $\mathbf k \cdot \mathbf r$ = kz = $\frac{2 \pi}{\lambda}z$ for the plane wave and $\mathbf k \cdot \mathbf r$ = kr = $\frac{2 \pi}{\lambda}r$ for the spherical wave.

Secondly why is there no attenuation factor for the plane wave equation?

Spherical wave emanates from a point source and spreads out homogeneously in the entire space, so its intensity $I ({Watts}/{m^2}) = \frac{c\varepsilon{E_0}^2}{2}=\frac{P}{4\pi{r^2}}$ whereas plane wave doesn't and its $I = \frac{c\varepsilon{E_0}^2}{2}$ representing a constant energy flux. For the derivation of the dispersion factor $\frac{E_0}{r}$, see Optics, 5-th edition by E. Hecht, 2.9 "Spherical Waves".


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