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Why does gravitational Laplace's equation equal zero?

  1. Feb 22, 2012 #1
    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 [tex]\phi=\frac{-Gm}{r}[/tex] where [itex]\phi[/itex] is the gravitational potential energy of a unit mass in a gravitational field [itex]g[/itex]. The gradient of this is (a vector field) [tex]g=-\nabla\phi=-\left(\frac{\partial\phi}{\partial x},\frac{\partial\phi}{\partial y},\frac{\partial\phi}{\partial z}\right)[/tex] And the divergence of this vector field is [tex]\nabla\cdot\nabla\phi=\nabla^{2}\phi=4\pi G\rho[/tex] and is called Poisson's equation. If the point is outside of the mass, then [itex]\rho=0[/itex] and Poisson's equation becomes[tex]\nabla\cdot\nabla\phi=0[/tex] (Laplace's equation). My question is, how do I express [itex]\phi=\frac{-Gm}{r}[/itex] as a function of [itex]x,y,z[/itex] so I can then end up with [itex]\nabla\cdot\nabla\phi=0[/itex] in empty space? I would have thought that I could write [tex]\phi=\frac{-Gm}{r}=\frac{-Gm}{\sqrt{x^{2}+y^{2}+z^{2}}}[/tex] but when I try to calculate [tex]\nabla\cdot\nabla\phi[/tex] from this, I don't get zero. I do this by assuming (in the simplest case) that both [itex]y[/itex] and [itex]z[/itex] are zero and then taking second derivative of [tex]\phi=\frac{-Gm}{r}[/tex]which should be zero (shouldn't it?) but isn't zero. What am I doing wrong? As simple as possible please.
    Thank you
  2. jcsd
  3. Feb 22, 2012 #2
    My mistake. Not sure why but I can't simplify like that. The Laplacian of [tex]\phi=\frac{-Gm}{r}=\frac{-Gm}{\sqrt{x^{2}+y^{2}+z^{2}}}[/tex]is (I used the WolframAlpha calculator, which I've only recently discovered - it's very good!)[tex]\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[/tex]
  4. Feb 22, 2012 #3
    It is much easier to demonstrate this using the representation of the Laplace operator in spherical polar coordinate:
    [tex]\Delta = \frac{1}{r^2} \frac{\partial}{\partial r} r^2 \frac{\partial}{\partial r}[/tex]
  5. Feb 22, 2012 #4
    How does that work?
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