Why does gravitational Laplace's equation equal zero?

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SUMMARY

The discussion centers on the application of Laplace's equation in gravitational fields, specifically how it equals zero in empty space. The gravitational potential field is defined as φ = -Gm/r, where φ represents the gravitational potential energy of a unit mass. The divergence of the gravitational field, expressed as ∇·∇φ, leads to Poisson's equation, which simplifies to Laplace's equation (∇²φ = 0) when the density ρ is zero outside a mass. The participant seeks clarification on expressing φ in Cartesian coordinates and calculating the Laplacian correctly, ultimately finding that using spherical coordinates simplifies the process.

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peter46464
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I'm struggling here so please excuse if I'm writing nonsense. I'm trying to understand how, for a gravitational field, Laplace's equation (I think that's the right name) equals zero in empty space.

I understand that the gravitational potential field, a scalar field, is given by \phi=\frac{-Gm}{r} where \phi is the gravitational potential energy of a unit mass in a gravitational field g. The gradient of this is (a vector field) g=-\nabla\phi=-\left(\frac{\partial\phi}{\partial x},\frac{\partial\phi}{\partial y},\frac{\partial\phi}{\partial z}\right) And the divergence of this vector field is \nabla\cdot\nabla\phi=\nabla^{2}\phi=4\pi G\rho and is called Poisson's equation. If the point is outside of the mass, then \rho=0 and Poisson's equation becomes\nabla\cdot\nabla\phi=0 (Laplace's equation). My question is, how do I express \phi=\frac{-Gm}{r} as a function of x,y,z so I can then end up with \nabla\cdot\nabla\phi=0 in empty space? I would have thought that I could write \phi=\frac{-Gm}{r}=\frac{-Gm}{\sqrt{x^{2}+y^{2}+z^{2}}} but when I try to calculate \nabla\cdot\nabla\phi from this, I don't get zero. I do this by assuming (in the simplest case) that both y and z are zero and then taking second derivative of \phi=\frac{-Gm}{r}which should be zero (shouldn't it?) but isn't zero. What am I doing wrong? As simple as possible please.
Thank you
 
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My mistake. Not sure why but I can't simplify like that. The Laplacian of \phi=\frac{-Gm}{r}=\frac{-Gm}{\sqrt{x^{2}+y^{2}+z^{2}}}is (I used the WolframAlpha calculator, which I've only recently discovered - it's very good!)\nabla\cdot\nabla\phi=Gm\left(\frac{2x^{2}-y^{2}-z^{2}+2y^{2}-x^{2}-z^{2}+2z^{2}-x^{2}-y^{2}}{\left(x^{2}+y^{2}+z^{2}\right)^{5/2}}\right)=0
 
It is much easier to demonstrate this using the representation of the Laplace operator in spherical polar coordinate:
\Delta = \frac{1}{r^2} \frac{\partial}{\partial r} r^2 \frac{\partial}{\partial r}
 
How does that work?
 

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