Can we distinguish if a general wave or disturbance $f$ (not necessarily electromagnetic) which satisfies the wave equation \begin{equation} \frac{1}{v^2}\frac{\partial^2 f}{\partial t^2} = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2} \equiv \nabla^2 f \end{equation} is transverse or longitudinal?
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asked Mar 6, 2016 at 5:11
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$\begingroup$ What do you mean by "in general"? Do you want the identification to be based entirely on the nature of the wave function? $\endgroup$– SchrodingersCatCommented Mar 6, 2016 at 6:15
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$\begingroup$ @SchrodingersCat Yes. Maybe I myself am not very sure at this point, but is there a way to extract information about the mode of propagation by looking at the wave equation? If not, how else could we distinguish them? $\endgroup$– Evangeline A. K. McDowellCommented Mar 6, 2016 at 7:05
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1$\begingroup$ I think that you have to consider a wave equation in dimension>1 since the notions of longitudinal and transverse usually refer to vectors. $\endgroup$– UrgjeCommented Mar 6, 2016 at 10:43
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$\begingroup$ @Urgje yes, the full equation is actually three dimensional. If there is three dimensional, how would I distinguish it then? $\endgroup$– Evangeline A. K. McDowellCommented Mar 6, 2016 at 20:41
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3$\begingroup$ No, because the wave equation applies to both longitudinal and transverse waves. Physical context should tell you what $f$ is supposed to be. $\endgroup$– JavierCommented Mar 7, 2016 at 1:47
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1 Answer
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According to Ibach and Lüth (8th ed.), section 11.4:
Transverse waves obey $\nabla\times f≠0$ and $\nabla f=0$.
Longitudinal waves obey $\nabla\times f=0$ and $\nabla f≠0$.
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2$\begingroup$ This implies that $f$ is a vector. That might not be the case. $\endgroup$ Commented Apr 8, 2020 at 15:34