Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook