# Why does gravitational Laplace's equation equal zero?

1. Feb 22, 2012

### peter46464

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

2. Feb 22, 2012

### peter46464

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!)$$\nabla\cdot\nabla\phi=Gm\left(\frac{2x^{2}-y^{2}-z^{2}+2y^{2}-x^{2}-z^{2}+2z^{2}-x^{2}-y^{2}}{\left(x^{2}+y^{2}+z^{2}\right)^{5/2}}\right)=0$$

3. Feb 22, 2012

### mathfeel

It is much easier to demonstrate this using the representation of the Laplace operator in spherical polar coordinate:
$$\Delta = \frac{1}{r^2} \frac{\partial}{\partial r} r^2 \frac{\partial}{\partial r}$$

4. Feb 22, 2012

### peter46464

How does that work?