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I saw from "Advanced Engineering Mathematics, 10th Edition" by Kreyszig, p. 400, that the solution $V$ of the Laplace's equation,

$$\nabla^2 V = \frac{\partial^2V}{\partial x^2}+\frac{\partial^2V}{\partial y^2}+\frac{\partial^2V}{\partial z^2} = 0,$$

is a potential function.

However I thought I can manipulate the expression as $\nabla^2V=0 \ \rightarrow \ \nabla \cdot (\nabla V)=0$. Because $\vec{F}=-\nabla V$, I can write the expression as $\nabla \cdot \vec{F} = 0$.

I did an example calculation with the gravitational force:

$$\vec{F} = - \frac{GMm}{r^2} \hat{r} = - \frac{GMm}{(x^2+y^2+z^2)^{3/2}} \ (x, y, z)$$

and I got $\text{div} \ \vec{F} = \vec{0}$.

But why is this not introduced in most mathematics / physics books?

Or is there any exception?

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  • $\begingroup$ Well, what about $x=y=z=0$, or equivalently $r=0$? $\endgroup$ – FakeMod Apr 29 at 11:21
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    $\begingroup$ en.wikipedia.org/wiki/Gauss%27s_law $\endgroup$ – bemjanim Apr 29 at 11:23
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    $\begingroup$ For a potential with no "sources" this is true by definition. But there are potentials with sources for which the correct formula is $\nabla^2 V \propto -\rho $. (Then, $\nabla \cdot \vec{F} \neq 0 $) $\endgroup$ – Ofek Gillon Apr 29 at 11:24
  • $\begingroup$ @OfekGillon Where did I make an error (in the derivation from the Laplace's equation)? From that it seems like the formula applies to every conservative force. Is writing $\nabla^2 = \nabla \cdot \nabla$ mathematically invalid? $\endgroup$ – curious Apr 29 at 11:48
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    $\begingroup$ @curious you calculated for a point source - does the divergence in the origin (the place of the source) equal to 0? It isn't. For a spherical planet, you will get a whole region of space (where the planet is) that the divergence won't be 0. You didn't make a mistake in your math: if the Laplacian is 0, the divergence will be 0, but not for every conservative force, the laplacian will be 0 (for example, gravity from a not-point source) $\endgroup$ – Ofek Gillon Apr 29 at 12:03