I wasn't very sure whether to ask this in the physics forum or here, but the question regards mathematics much more than it does physics.

The following wave function is given (spherical wave): $\psi(\vec{r},t)=f(|\vec{r}|)e^{i(\omega t-k|\vec{r}|)}$. The objective is to find a function $f$ such that $\psi$ satisfies the classic wave equation, that is: $$\frac{\partial^2\psi}{\partial t^2}=c^2\nabla^2\psi$$ Where $c^2=\frac{\omega^2}{k^2}$. When introducing the wave function into the equation, you are left with the following differential equation (you can try it): $$\bigg(2\,\frac{f'(r)}{r}+f''(r)\bigg)+i\,2k\,\bigg(f'(r)+\frac{f(r)}{r}\bigg)=0$$ Which results in $f(r)=\frac{A}{r}$ with $A\in\mathbb{C}$ for both the real and imaginary parts.

I need help in understanding why the result is the same for both parts. I have an intuition that there has to be a very specific reason of why that is. Is the fact that the function is being multiplied with the complex number critical? If it wasn't because of the second time derivative in the left side of the equation and the fact that $c$ is defined the way it is, then $f$ wouldn't exist, right? I hope that I was clear and precise with my question. Thank you.


1 Answer 1


A complex-valued function satisfies the wave equation if and only if both its real and imaginary parts satisfy it. (This is because the equation is linear, with real coefficients.)

So, your question could be equivalently stated as: find $f$ such that both functions $$f(|\vec{r}|)\cos(\omega t-k|\vec{r}|)\quad\text{and}\quad f(|\vec{r}|)\sin(\omega t-k|\vec{r}|)$$ satisfy the wave equation.

At first this looks scary: two equations with one unknown function. But the second equation turns into first by the time shift $$t\mapsto t+\pi/(2\omega)$$
Since the wave equation has constant coefficients, shifting a solution in time gives another solution. This is why it suffices to take care of one, for example $f(|\vec{r}|)\cos(\omega t-k|\vec{r}|)$. The other follows automatically.


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