# Wave vector in spherical coordinates

I understand the wave vector for plane waves in Cartesian coordinates. Along the direction of propagation of wave, $$k = \sqrt{k_x^2+k_y^2+k_z^2} = \frac{2\pi}{\lambda}$$ If $$(k_x, k_y, k_z) \neq 0$$, this would mean we can find troughs and crests moving along $$(x,y,z)$$ directions respectively. The dot product in wave equation in Cartesian coordinates, $$\vec{dr}=(dx,dy,dz)$$ would be $$\vec{k}\cdot\vec{dr} = k_x dx + k_y dy + k_z dz$$

What would the interpretation be in case of spherical coordinates? Can we still have wave vector, $$k = (k_{r},k_{\theta},k_{\phi})$$ along $$(r,\theta,\phi)$$ directions such that $$k = \sqrt{k_r^2+k_{\theta}^2+k_{\phi}^2} = \frac{2\pi}{\lambda}$$ If $$(k_{r},k_{\theta},k_{\phi}) \neq 0$$, would this mean we can find troughs and crests moving along $$(r,\theta,\phi)$$ directions respectively. Is the following dot product in wave equation in spherical coordinates, $$\vec{dr}=(dr,d\theta,d\phi)$$ correct? $$\vec{k}\cdot\vec{dr}= k_{r}dr + k_{\theta}rd\theta + k_{\phi}r\sin\theta d\phi$$

• The "dot product" $\vec{k}\cdot\vec{dr}= k_{r}dr + k_{\theta}rd\theta + k_{\phi}r\sin\theta d\phi$ doesn't make much sense. It could be appropriate in a small fraction of circumstances, but you'd have to be careful to explain what's happening around it. Feb 1, 2021 at 16:49
• $\vec{dr}=(dr,d\theta,d\phi)~$ this is not correct
– Eli
Feb 1, 2021 at 17:50
• Be careful with position vectors in polar coordinate systems. The coordinates of a point in spherical for example are (r,th,ph), but the position vector is r alone. The angles are contained within the unit vector for r, because this is a curvilinear system. Look up 'position vector in spherical coordinates' and you should find plenty of information about it. Feb 5, 2021 at 12:23

As a general rule, having a wave vector $$\vec k$$ implies that you have a travelling wave, i.e. one which transports momentum and energy.

This is particularly relevant to part of your question,

would this mean we can find troughs and crests moving along $$(r,\theta,\phi)$$ directions respectively,

because you cannot have a travelling wave along the $$\theta$$ direction $$-$$ the energy would accumulate at the north axis and you'd need a source along the south. A travelling wave along the $$\phi$$ axis is less of a problem (you can have energy circling around indefinitely) but it's still a problem (you would have energy circling around indefinitely).

That said, the general answer to your question is: yes. The way to get at it is to rephrase the Cartesian wavevector into saying that we want to solve the wave equation, $$\nabla^2 f = \frac{1}{c^2}\partial_t^2 f,$$ first by separating out $$f(\mathbf r,t)=f(\mathbf r)e^{-i\omega t}$$ into a space- and time-dependent parts, and then handling the resulting Helmholtz equation for the spatial part by using Cartesian coordinates, $$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2} = k^2 f,$$ and using separation of variables. Here the wavenumber arises as a straight definition $$k^2=\omega^2/c^2$$, and once you solve the ODE of each individual coordinate into individual exponentials, the result can be folded back into $$f(\mathbf r)=e^{i\mathbf k\cdot \mathbf r}$$, with a wavevector $$\mathbf k$$ whose norm coincides with $$k$$.

Now, if you want the wave behaviour of $$f(\mathbf r)$$ to go along the spherical coordinate surfaces, then you do the same, but now you address the Helmholtz equation $$\nabla^2 f = k^2 f$$ by expressing $$\nabla^2$$ in spherical coordinates, and you separate the variables as $$f(\mathbf r)=R(r)\Theta(\theta)\Phi(\phi)$$. The resulting procedure is standard, so I won't reproduce it here (try e.g. Jackson's Classical Electrodynamics for a thorough solution, or any book on advanced mathematical methods for physics). The upshot is that:

• The solution for the angular variables is that $$\Theta(\theta)\Phi(\phi) = Y_{lm}(\theta,\phi)$$ must be one of the family of spherical harmonics, which are functions which "wave" exclusively in the angular directions, with nodes along the $$\theta=\rm const$$ and $$\phi=\rm const$$ directions. However, because of the problems mentioned at the start, particularly in the $$\theta$$ direction, they are not travelling waves $$-$$ they are standing waves.

Moreover, the spherical-harmonics solutions do not have any associated "wave vector". Instead, the description of the wave moves over to the two indices $$l$$ and $$m$$, which describe the total number of nodes and how they're split into the azimuth and altitude directions.

• For the radial variable, the Helmholtz equation becomes the spherical Bessel equation, $$r^{2}{\frac {d^{2}R}{dr^{2}}}+2r{\frac {dR}{dr}}+\left(k^2r^{2}-l(l+1)\right)R=0,$$ and its solutions are the spherical Bessel functions. As a general rule, all of these are spherical waves, with crests and troughs along the radial direction, but there's several different types:

• The spherical Bessel functions of the first kind, $$j_l(kr)$$, describe standing waves, and are regular at the origin.
• The spherical Bessel functions of the second kind, $$y_l(kr)$$, describe standing waves, but they have a singularity at the origin, so they can only be used if the origin is not part of the region of interest.
• The spherical Hankel functions, $$h_l^{\pm}(kr)$$, describe travelling waves, with a source (or sink) at the origin, where they have a singularity.

\begin{align*} &\text{with}\\ \vec{R}&=r\,\left[ \begin {array}{c} \cos \left( \phi \right) \sin \left( \theta \right) \\ \sin \left( \theta \right) \sin \left( \phi \right) \\ \cos \left( \theta \right) \end {array} \right] \\ \vec{dR}&=\left[ \begin {array}{c} \cos \left( \phi \right) \sin \left( \theta \right) \\ \sin \left( \theta \right) \sin \left( \phi \right) \\ \cos \left( \theta \right) \end {array} \right] \,dr\\&+\left[ \begin {array}{c} -r\sin \left( \theta \right) \sin \left( \phi \right) \\ r\cos \left( \phi \right) \sin \left( \theta \right) \\ 0\end {array} \right] \,d\phi\\&+ \left[ \begin {array}{c} r\cos \left( \phi \right) \cos \left( \theta \right) \\ r\cos \left( \theta \right) \sin \left( \phi \right) \\ -r\sin \left( \theta \right) \end {array} \right] \,d\theta \end{align*}

thus $$~\vec{dR}\ne (dr~,d\phi~,d\theta)$$