I am currently studying Optics, fifth edition, by Hecht. In chapter 2.9 Spherical Waves, the author says the following:

$$\dfrac{\partial^2}{\partial{r}^2}(r \psi) = \dfrac{1}{v^2} \dfrac{\partial^2}{\partial{t}^2} (r \psi) \tag{2.71}$$ Notice that this expression is now just the one-dimensional differential wave equation, Eq. (2.11), where the space variable is $r$ and the wavefunction is the product $(r \psi)$. The solution of Eq. (2.71) is then simply $$r \psi(r, t) = f(r - vt)$$ or $$\psi(r, t) = \dfrac{f(r - vt)}{r} \tag{2.72}$$ This represents a spherical wave progressing radially outward from the origin, at a constant speed $v$, and having an arbitrary functional form $f$. Another solution is given by $$\psi(r, t) = \dfrac{g(r + vt)}{r}$$ and in this case the wave is converging toward the origin. The fact that this expression blows up at $r = 0$ is of little practical concern. A special case of the general solution $$\psi(r, t) = C_1\dfrac{f(r - vt)}{r} + C_2 \dfrac{g(r + vt)}{r} \tag{2.73}$$ is the harmonic spherical wave $$\psi(r, t) = \left( \dfrac{\mathcal{A}}{r} \right) \cos k(r \mp vt) \tag{2.74}$$ or $$\psi(r, t) = \left( \dfrac{\mathcal{A}}{r} \right) e^{ik(r \mp vt)} \tag{2.75}$$ wherein the constant $\mathcal{A}$ is called the source strength.

Now, see my question 1. here. It seems to me that the author has made the same error of writing $\cos k(r \mp vt)$ and $e^{ik(r \mp vt)}$, instead of $\cos (kr \mp vt)$ and $e^{i(kr \mp vt)}$, respectively. But this repeat of the error now makes me wonder: Is this actually an error on the part of the author, or am I misunderstanding something?

I would greatly appreciate it if people would please take the time to clarify this.


1 Answer 1


It's no error. $k$ has dimensions of inverse length, $r$ has dimensions of length, $v$ has dimensions of length per time, and $t$ has dimensions of time.

What you propose is dimensionally incorrect, as $kr$ is dimensionless and $vt$ has dimensions of length. On the other hand, $k(r\mp vt)$ is a valid operation, and it gives us an overall dimensionless quantity that we need for the argument of an exponential function.

You might be getting mixed up with distributing that $k$ to get $kr\mp \omega t$, which is a typical way to denote such an argument. $\omega$ had dimensions of inverse time, and $v=\omega/k$

  • $\begingroup$ Ahh, yes, that's what I the author did in the context of my other question physics.stackexchange.com/q/535702/141502: $$E(x, y, z, t) = E_0 e^{i\hat{\mathbf{k}} \cdot (\vec{\mathbf{r}} - \omega t)}$$ So the user Semoi's answer physics.stackexchange.com/a/535706/141502 was correct in saying that the author made an error with $E(x, y, z, t) = E_0 e^{i\hat{\mathbf{k}} \cdot (\vec{\mathbf{r}} - \omega t)}$, and it should actually be $E(x, y, z, t) = E_0 e^{i(\hat{\mathbf{k}} \cdot \vec{\mathbf{r}} - \omega t)}$? $\endgroup$ Jun 26, 2020 at 13:30
  • $\begingroup$ @ThePointer Yes. Both of your questions are essentially about the same issue. $\endgroup$ Jun 26, 2020 at 13:31
  • $\begingroup$ I think being able to recall the author's last error made me think that this was also an error. But, as you say, the last one actually was an error, whereas this one is not and is actually slightly different, since we have $v$ (the velocity) instead of $\omega$ (the angular frequency), right? $\endgroup$ Jun 26, 2020 at 13:34
  • $\begingroup$ @ThePointer Yes $\endgroup$ Jun 26, 2020 at 13:41

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