I was thinking about the classical equations for gravity. I got stuck on two equations: $$\vec{\nabla}.\vec{g}= 0$$ and $$\vec{\nabla} \times \vec{g}= 0$$ The first equation is Gauss law of gravitation in vacuum and the second equation comes from the fact that gravitational forces are conservative. If I take the curl of the second equation what I end up with is $$\nabla^2 \vec{g} =0$$ Whose solution in two dimensions can be written as $$\vec{g}=\vec{A}\sin(kx)(e^{ky}+e^{-ky})$$ My question is: can we interpret this as a wave equation, where the wave is propagating with infinite speed? I know that gravitational waves are predicted by GR and they travel at the speed of light but I wanted to know can the above equation be interpreted as a classical gravitational wave equation? Edit: A general wave equation can be written as $$\nabla^2y-\frac{1}{v²}\frac{\partial^2 y}{\partial t^2}=0$$ when the velocity tends to infinity the second term becomes 0. So Laplace's equation is a wave equation with infinite speed. But there is no physical difference between wave moving infinitely fast or wave not moving at all. This is what I mean when I say a wave with infinite speed.
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asked Jul 24, 2019 at 16:08
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1$\begingroup$ You've ended up with Laplace's equation and the solutions are the harmonic functions. These are not waves in any usual sense of the word. $\endgroup$– John RennieCommented Jul 24, 2019 at 16:33
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1$\begingroup$ I don't understand what a "wave with infinite speed" is supposed to be and how it is distinct from a "wave with zero speed" or "no wave at all". If you explain where you see the difference you might receive more useful answers then just a plain "No" (or "Yes"...). $\endgroup$– ACuriousMind ♦Commented Jul 24, 2019 at 16:35
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3$\begingroup$ Since your equations have no time variable, how would you be able to think of something propagating? $\endgroup$– puppetsockCommented Jul 24, 2019 at 16:39
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3$\begingroup$ @ACuriousMind A general wave equation can be written as $$\nabla^2y-\frac{1}{v²}\frac{\partial^2 y}{\partial t^2}=0$$ when the velocity tends to infinity the second term becomes 0. So Laplace's equation is a wave equation with infinite speed. But as you said there is no physical difference between wave moving infinitely fast or wave not moving at all. $\endgroup$– Manvendra SomvanshiCommented Jul 24, 2019 at 16:50
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2$\begingroup$ It should be noted that one usually imposes the condition that $g(x)\to 0$ as $|x|\to\infty$, in which case one has $g\equiv 0$ by Liouville's theorem. If one drops this, there is an infinite-dimensional space of solutions to the Laplace equation. $\endgroup$– Ryan UngerCommented Jul 24, 2019 at 17:04
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I would say no, since the solution that you found is stationary, it cannot be propagating. For example, in one dimension, a propagating wave needs to have the functional form $\Psi = \Psi (x-ct)$, where $c$ is the (finite) velocity of the wave. In general, a propagating wave in the vacuum constitutes the solution of the equation $\Box \Psi =0$, where $\Box=\nabla^2 - \frac{1}{c^2} \frac{\partial^2}{\partial t^2}$ is the D'alembertian operator in a flat spacetime and we didn't consider the presence of sources in the right-hand side. Actually, the equation that you found for $\vec{g}$ is called Laplace's equation, which also arises when considering stationary and vacuum solutions in Electromagnetism.
answered Jul 24, 2019 at 17:05