6
$\begingroup$

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

$$\nabla\cdot\nabla\phi=0$$

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.

$\endgroup$

1 Answer 1

4
$\begingroup$

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.

Once:

$$ \begin{aligned} \nabla \phi &= \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y} , \frac{\partial}{\partial z}\right) \phi(x, y, z) = \\ % &= Gm \left( \frac{x}{(x^2+y^2+z^2)^{3/2}} , \frac{y}{(x^2+y^2+z^2)^{3/2}} , \frac{z}{(x^2+y^2+z^2)^{3/2}} \right) = \\ % &= -g(x, y, z) \end{aligned} $$

Twice:

$$ \begin{aligned} \nabla^2 \phi &= \nabla \cdot \nabla \phi \\ &= Gm \left( \frac{\partial}{\partial x} , \frac{\partial}{\partial y} , \frac{\partial}{\partial z}\right) \left( \frac{x}{(x^2+y^2+z^2)^{3/2}} , \frac{y}{(x^2+y^2+z^2)^{3/2}} , \frac{z}{(x^2+y^2+z^2)^{3/2}} \right) = \\ &= Gm \left( \frac{2x^2-y^2-z^2}{(x^2+y^2+z^2)^{5/2}} + \frac{2y^2-x^2-z^2}{(x^2+y^2+z^2)^{5/2}} + \frac{2z^2-x^2-y^2}{(x^2+y^2+z^2)^{5/2}} \right) = \\ % &= 0 \end{aligned} $$

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.

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$
    – Qmechanic
    Feb 21, 2012 at 21:31
  • $\begingroup$ @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. $\endgroup$
    – Peter4075
    Feb 22, 2012 at 7:49
  • 1
    $\begingroup$ Added a more explicit calculation for you. $\endgroup$ Feb 22, 2012 at 8:30
  • $\begingroup$ @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. $\endgroup$
    – Peter4075
    Feb 22, 2012 at 8:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.