I am trying to understand the derivation of the solution of the wave equation (homogeneous case) while using the Fourier Transformation:

$$\left(\frac 1 {c^2} \frac {\partial^2}{\partial t^2}- \triangle\right)\phi=0$$

The expression for $\phi(\vec r,t)$ is:

$$\phi(\vec r,t)= \iint \tilde \phi(\vec k, \omega)e^{i(\vec k \vec r - \omega t)}d^3kd\omega$$

which I assume is the reverse Fourier transformation, when one knows the $\tilde \phi(\vec k, \omega)$.

Then by substituting the 2nd expression in the first one, we get :$\omega= \pm kc$

Then for $\tilde \phi(\vec k, \omega)$ we have:

$$\tilde \phi(\vec k, \omega)=2\omega f(\vec k, \omega)\delta(\omega^2-c^2k^2)$$

Now, while I am aware that $2\omega$ is a convention, I don't know where does the above expression comes from? I recently learned about the Fourier Transformation and Series, but I didn't come across this expression. Can anyone explain this to me?


1 Answer 1


I'll compare this to a less rigorous way of solving the wave equation that you may be used to.

First let's start by guessing that the solution is a plane wave with $\omega, \vec k$ to be determined. $$\phi(\vec x,t)=A e^{i(\omega t-\vec k\cdot \vec x)}$$ Now plugging this in the wave equation gives $$\left(-\frac{\omega^2}{c^2}+|\vec k|^2\right)Ae^{i(\omega t-\vec k\cdot \vec x)}=0.$$ So this ansatz solves the wave equation provided that $\omega^2=c^2|\vec k|^2\implies\omega=\pm c|\vec k|$. So we are free to choose $\vec k$ and $A$ as long as we replace $\omega$ with $\omega(\vec k)=c|\vec k|$ or $\omega(\vec k)=-c|\vec k|$. We can also take a superposition of these plane waves for different values of $\vec k$: $$\phi(\vec x,t)=\int\mathrm d^3k\, A(\vec k)e^{i(\omega(\vec k) t-\vec k\cdot \vec x)}$$ Your solution is the same as this solution up to some relabelling. Your solution is given by

$$\phi(\vec x,t)=\int\mathrm d^3k\,\mathrm d\omega\, 2\omega f(\vec k,\omega)e^{i(\omega t-\vec k\cdot \vec x)}\delta(\omega^2-c^2k^2)$$

There is an identity for integrating delta functions that have functions in them: $$\int\mathrm d x\, f(x)\delta(g(x))=\sum_i\frac{f(x)}{|g'(x_i)|}$$ where $x_i$ are the solutions to $g(x)=0$. So integrating $\delta(\omega^2-c^2k^2)$ gives $$\int\mathrm d \omega\, f(\omega)\delta(\omega^2-c^2k^2)=\frac{f(ck)}{2ck}+\frac{f(-ck)}{-2ck}$$ since $\omega=\pm ck$ solves $\omega^2-c^2k^2=0$. You can check for yourself that the two solutions now coincide.

  • $\begingroup$ In my eq. I have it $e^{i(kr - \omega t)}$ while you have it $e^{i(\omega t - kr )}$. Is there a difference ? In order to find the solution in the time domain and position space I need to know $\phi(\vec k, \omega)$. But in your solution I couldn't understand the expression : $$\tilde \phi(\vec k, \omega)=2\omega f(\vec k, \omega)\delta(\omega^2-c^2k^2)$$. This is what I initially don't understand. $\endgroup$
    – imbAF
    Dec 29, 2021 at 16:39
  • $\begingroup$ @imbAF Whether you pick $e^{i(kr-\omega t)}$ or$e^{i(\omega t-kr)}$ doesn't matter if you're consistent, sorry I was a bit sloppy here. You need to know $\tilde\phi(\vec k,\omega)$: you already know $\tilde\phi$. If you plug in any function $f(\vec k,\omega)$ you will get a solution that solves the wave equation. You can then plug it in your expression for $\phi(x,t)$ and perform the integral. If you want a specific function for $f$ you need to include boundary conditions. $\endgroup$ Dec 30, 2021 at 11:38

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