I am wondering if any wave packet of the form
$$\psi=\int g(\boldsymbol{k}) e^{i(\boldsymbol{k}\boldsymbol{r}-\omega t)}$$
is always a solution to the classical wave equation? In my understanding, this would raise some condition to the amplitude distribution $g(\boldsymbol{k})$. Inserting this into the classical wave equation, we get:
$\frac{\partial^2\psi}{\partial t^2}=|\boldsymbol{v}|^2\nabla^2 \psi$
$\frac{\frac{\partial^2\psi}{\partial t^2}}{\nabla^2\psi}-|\boldsymbol{v}|^2=0$
$\int g(\boldsymbol{k}) \left(\frac{\omega^2}{|\boldsymbol{k}|^2}-|\boldsymbol{v}|^2\right) e^{i(\boldsymbol{k}\boldsymbol{r}-\omega t)}=0$
In my understanding, this means that only such $g(\boldsymbol{k})$ are allowed for which $g(\boldsymbol{k}) \left(\frac{\omega^2}{|\boldsymbol{k}|^2}-|\boldsymbol{v}|^2\right)$ vanishes for any time and space.
Is this right? And what does this mean about possible solutions?
Best Answer
A general solution can be presented as \begin{equation} \int d^3 \vec k \Big[ g_1(\vec k)e^{i|v||\vec k|t-i\vec k\cdot \vec x}+g_2(\vec k)e^{-i|v||\vec k|t+i\vec k\cdot \vec x}\Big] \qquad (1) \end{equation} with arbitrary $g_1(k)$ and $g_2(k)$. It follows from your formulas, after you express $\omega$ in terms of $k$. Two terms correspond to two ways of taking the square root. If you additionally require that the solution is real, then you need to impose \begin{equation} \Big(g_1(\vec k) \Big)^*=g_2(\vec k). \end{equation}
In (1) the integral goes over spatial $k$ only. In principle, you can make it over all $k$ in space-time (I mean, including $\omega$). Then, as it is clear form what you wrote, you need to solve \begin{equation} g(\vec k,\omega)(\omega^2 - |v|^2 |\vec k|^2)=0. \end{equation} Its general solution is \begin{equation} g(\vec k,\omega) = f(\vec k,\omega) \delta(\omega^2 - |v|^2 |\vec k|^2), \qquad (2) \end{equation} where $\delta$ is the Dirac delta. Note that in (2) only values of $f(\vec k, \omega)$ at $\omega^2 - |v|^2 |\vec k|^2=0$ are relevant. Indeed, if we replace \begin{equation} f(\vec k,\omega)\to f(\vec k,\omega)+(\omega^2 - |v|^2 |\vec k|^2)\alpha (\vec k,\omega) \end{equation} with any $\alpha$, then (2) remains unchanged due to $\delta(x)x=0$. The above replacement allows to change values of $f$ arbitrarily everywhere away from $\omega^2 - |v|^2 |\vec k|^2=0$. Thus, only values of $f$ on $\omega^2 - |v|^2 |\vec k|^2=0$ actually contribute to (2).
The general solution than acquires the form \begin{equation} \int d^3\vec k d\omega f(\vec k,\omega) \delta(\omega^2 - |v|^2 |\vec k|^2) e^{-i\omega t + i\vec k\vec x}. \qquad (3) \end{equation} If you need a real solution, you need to impose \begin{equation} f(-\vec k,-\omega) = \Big( f(\vec k,\omega)\Big)^*. \end{equation} The solution in this form can be easily connected to (1) if one eliminates the $\omega$ integral using the delta function.