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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?

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    $\begingroup$ Your wave packet is a linear combination of solutions of the wave equation (i.e. the exponential in the integral, provided that $w(k)=vk$). The wave equation is linear, hence the wave packet is a solution if you substitute $w$ with $vk$ in the integral. $\endgroup$
    – Quillo
    Apr 3, 2023 at 8:27
  • $\begingroup$ Thanks. But does that not mean that every wave packet gives a linear dispersion relation? That would not make sense right? $\endgroup$
    – Guiste
    Apr 3, 2023 at 12:17
  • $\begingroup$ I think what you are referring to is the case of a non-dispersive wave, for which $\omega=vk$. So for a non-dispersive wave, every wave packet is always a solution. However, what in the case of a dispersive wave, for which $\omega\neq vk$? $\endgroup$
    – Guiste
    Apr 3, 2023 at 12:19
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    $\begingroup$ your wave equation is for a non dispersive medium $\endgroup$
    – Quillo
    Apr 3, 2023 at 17:26
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    $\begingroup$ The simple wave equation has always the linear dispersion relations ๐œ” = ±๐‘ฃ๐‘˜ for sinusoidal waves propagating in opposite directions. The ๐‘”(๐‘˜) in your linear superposition integral correspond to the amplitudes of the partial waves which can be arbitrary. $\endgroup$
    – freecharly
    Apr 3, 2023 at 19:10

1 Answer 1

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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.

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  • $\begingroup$ It seems right, but why is the dispersion relation always linear? For a general wave packet, the dispersion relation might become non-linear right? $\endgroup$
    – Guiste
    Apr 3, 2023 at 12:38
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    $\begingroup$ it depends on the equation of motion you solve. If the equation of motion has the mass term, then you get $\omega^2-|v|^2 k^2-m^2=0$. If the equation of motion has higher-derivative terms, you may actually get anything you want. Say, $(\partial_t^2-\nabla^4)\phi=0$ will result in $\omega^2+ k^4=0$ $\endgroup$
    – Dr.Yoma
    Apr 3, 2023 at 13:10
  • $\begingroup$ I think I do not quite get what you mean. Did your derivation not show that $v = \omega/k$, hence a linear dispersion relation for any wave that can be written as a linear superposition of monochromatic waves? $\endgroup$
    – Guiste
    Apr 3, 2023 at 13:40
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    $\begingroup$ Yes. Actually, you get a linear dispersion relation for only one equation of motion, which is $(\partial_t^2-\nabla^2)\phi=0$. Though, it is quite an important one, as it describes all massless particles that we have. However, there are many other equations that you may want to solve. If you have a linear differential equation with the coefficients that do not depend on space-time coordinates, then the easiest way to solve it is by going to the Fourier space. Then, the solution is always a superposition of exponents. The dispersion relation can be anything, depending on the equation. $\endgroup$
    – Dr.Yoma
    Apr 3, 2023 at 15:17
  • $\begingroup$ I got it. I never knew that the classical wave equation always gives a linear dispersion relation, thanks for clarifying this! $\endgroup$
    – Guiste
    Apr 4, 2023 at 4:50

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