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I'm struggling here so please excuse if I'm writing nonsense.

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


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 $r\neq 0$? 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 for $r\neq 0$. I'm confused.

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up vote 2 down vote accepted

We want to compute the Laplacian of

$$ \phi=\frac{-Gm}{r}=\frac{-Gm}{\sqrt{x^{2}+y^{2}+z^{2}}} $$

This means applying the $\nabla$ operator twice.


$$ \left(\frac{\partial}{\partial x} , \frac{\partial}{\partial y} , \frac{\partial}{\partial z}\right) \phi(x, y, z) = $$

$$ = Gm \left( \frac{x}{(x^2+y^2+z^2)^{3/2}} , \frac{y}{(x^2+y^2+z^2)^{3/2}} , \frac{z}{(x^2+y^2+z^2)^{3/2}} \right) = $$

$$ = -g(x, y, z) $$


$$ Gm \left( \frac{\partial}{\partial x} , \frac{\partial}{\partial y} , \frac{\partial}{\partial z}\right) \left( \frac{x}{(x^2+y^2+z^2)^{3/2}} , \frac{y}{(x^2+y^2+z^2)^{3/2}} , \frac{z}{(x^2+y^2+z^2)^{3/2}} \right) = $$

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

$$ = 0 $$

because the x's, y's and z's on top cancel out. In reality it's nonzero at $(x,y,z) = 0$ because you can't divide by zero.

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Comment to the answer(v1): The Kronecker delta function should be replaced with the Dirac delta function. The Kronecker delta function is not relevant here. –  Qmechanic Feb 21 '12 at 21:31
Thanks for the bug report, fixed it. –  mtrencseni Feb 21 '12 at 22:23
@mtrencseni - thank you, but that doesn't answer my question. How/why does Laplacian equal zero when the radius doesn't equal zero. I'm thinking that if I differentiate phi twice wrt r I should be getting zero. Is that correct? How does that give zero? I'm a physics novice so don't worry about making your answer too simple. –  Peter4075 Feb 22 '12 at 7:49
Added a more explicit calculation for you. –  mtrencseni Feb 22 '12 at 8:30
@mtrencseni - Now I think I get it. I was making assumptions that I shouldn't (I was assuming I could ignore y and z and find the Laplacian of -Gm/x). I popped the real phi into the WolframAlpha calculator took second partial derivatives, added them all up and there was zero. Brilliant. Thank you very much. –  Peter4075 Feb 22 '12 at 8:59
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