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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?

Qmechanic
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abouttostart
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  • Well, what about $x=y=z=0$, or equivalently $r=0$? –  Apr 29 '20 at 11:21
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    https://en.wikipedia.org/wiki/Gauss%27s_law – bemjanim Apr 29 '20 at 11:23
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    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 $) – Ofek Gillon Apr 29 '20 at 11:24
  • @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? – abouttostart Apr 29 '20 at 11:48
  • @OfekGillon And also gravitational field clearly has its "source" (mass). Why did I get the result that div F = 0? Thanks. – abouttostart Apr 29 '20 at 11:51
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    @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) – Ofek Gillon Apr 29 '20 at 12:03
  • @OfekGillon Ah, so it is always true that if Laplacian is 0 then the divergence of the force is zero (and that means I am making an assumption that the source of the force is a point mass), however not all potentials of the conservative force fields satisfy laplacian = 0 because in general the source may not be from a point (e.g. a point mass)? Sorry for repeating, I just wanted to make sure I understood it correctly. – abouttostart Apr 29 '20 at 12:10
  • @curious maybe the sources are not points but outside the region of interest (say the gravity outside earth will still have a divergence of 0, but all of the sources are just outside the region of interest. The correct statement is "wherever the laplacian is 0, there is no source there". This is easy to prove just as you did. – Ofek Gillon Apr 29 '20 at 12:17
  • Possible duplicates: https://physics.stackexchange.com/q/488220/2451 , https://physics.stackexchange.com/q/378828/2451 and links therein. – Qmechanic Apr 29 '20 at 13:40

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