A week or so back I asked a question about the gravitational potential field $$\phi=\frac{-Gm}{r}, \qquad r\neq 0, $$ and how to show the Laplacian of $\phi$ equals zero for $r\neq 0$? Eventually, (it took a while) I was able to understand that

$$\nabla\cdot\nabla\phi=Gm\left(\frac{2x^{2}-y^{2}-z^{2}+2y^{2}-x^{2}-z^{2}+2z^{2}-x^{2}-y^{2}}{\left(x^{2}+y^{2}+z^{2}\right)^{5/2}}\right)~=~0, \qquad r\neq 0,$$ which was a revelation. But now I'm wondering why Poisson's equation $$\nabla\cdot\nabla\phi=\nabla^{2}\phi=4\pi G\rho$$ doesn't always equal zero as well? Obviously it doesn't, so I'm assuming that inside a mass the gravitational potential field cannot be given by $$\phi=\frac{-Gm}{r}, \qquad r\neq 0.$$ Is that correct? Also, is there a comparably easy formula for gravitational potential inside a mass or does it vary (horribly?) depending on the shape and density of the mass?


3 Answers 3


Yup. Inside the (uniform spherical) mass, IIRC $\phi=-\frac{GM}{2R^3}\left(3R^2-r^2\right)$. Or something like that. So, $$\phi=\begin{cases} -\frac{GM}{r}, & r>R \\ -\frac{GM}{2R^3}\left(3R^2-r^2\right), & r<R \end{cases}$$ The laplacian $\nabla^2\phi$ should be $$4\pi G\rho=\nabla^2\phi=\begin{cases} 0, &r>R \\ 4\pi G\rho_0 &r<R \end{cases}$$ Where $\rho$ is the scalar density field, and $\rho_0=\frac{M}{\frac{4}{3}\pi R^3}$ is the density of the ball.

So this makes sense. Remember that mass density is a field as well as gravitational potential, so calculating it at one point in space doesn't mean that you've done it at all points in space. The laplacian of a discontinuous $\phi$ will give $\rho$ only within the limits of continuity. You have to break the function up.

Just a note: Even for a point particle, $\rho$ is not identically zero everywhere. It's infinity at the origin (check your formula again), so you basically get a dirac $\delta$ function.

For a nonspherical/nonuniform mass there's no such formula. You have to integrate it yourself.

  • $\begingroup$ Thanks. What does IIRC mean? I've just found this in Wikipedia for inside a uniform spherical body$$\phi=\frac{2}{3}\pi G\rho\left(r^{2}-3R^{2}\right)$$which (yippee) I managed to change into your equation. Does all this mean that the equation for the gravitational field $$\mathbf{g}=-\mathbf{\nabla\phi}$$ is only true for points outside of the mass, ie for $r>R$? $\endgroup$
    – Peter4075
    Commented Mar 5, 2012 at 19:43
  • $\begingroup$ @Peter4075 IIRC=If I Recall Correctly. I wasn't too sure if the equation was correct and I didn't want to rederive it. $\mathbf{g}=-\nabla\phi$ is fine. You forgot that $g$ is also a field, and you must apply $\nabla$ in every separate region of continuity. As such, inside a uniform-density-sphere, $\mathbf{g}=\frac{GMr}{R^3}$ (This comes from the Shell theorem--so ar a point inside the sphere we can safely remove the shell outside it). If you apply $\nabla$ to the $r<R$ expression for $\phi$, it should work fine. $\endgroup$ Commented Mar 6, 2012 at 2:58
  • 1
    $\begingroup$ Thanks, that makes sense now. I did try to look up IIRC but for some reason I searched for LLRC and ended up with Low Level Radiation Campaign, Lessons Learnt and Reconciliation Committee or Lincolnshire Land Rover Club. D'oh! $\endgroup$
    – Peter4075
    Commented Mar 6, 2012 at 8:24
  • 1
    $\begingroup$ @peter4075 :p Though I find it strange that you know Laplace's/Poisson's equation and the use of $\nabla$ without knowing about fields inside a body and the shell theorem.. I was taught all this first (probably will get taught about $\nabla$ &co much later). Or are you self-taught? $\endgroup$ Commented Mar 6, 2012 at 11:43
  • $\begingroup$ self taught I'm afraid, and I know it shows. Last time I studied physics and maths was many years ago at school when I took my A levels (here in the UK). I've been trying to teach myself GTR for the last year or so, starting with the free online Leonard Susskind lectures. It's hard but fun. $\endgroup$
    – Peter4075
    Commented Mar 6, 2012 at 12:26

The question has been answered, so just as a remark (which is too long for a comment):

In spherical coordinates, the Laplace operator looks like

$$\Delta f = {1 \over r^2} {\partial \over \partial r} \left( r^2 {\partial f \over \partial r} \right) + {1 \over r^2 \sin \varphi} {\partial \over \partial \varphi} \left( \sin \varphi {\partial f \over \partial \varphi} \right) + {1 \over r^2 \sin^2 \varphi} {\partial^2 f \over \partial \theta^2},$$

so the calculation for $\phi=\frac{-Gm}{r}, \ r\neq 0,$ which only depends on $r$ reduces to calculating

$$-G\ m\ {1 \over r^2} {\partial \over \partial r} \left( r^2 {\partial \over \partial r} \frac{1}{r}\right) .$$

  • $\begingroup$ Thanks. Bear with me if I do this one step at a time. So$$r^{2}\left(\frac{\partial}{\partial r}\left(\frac{1}{r}\right)\right)=-1$$ $$\frac{\partial}{\partial r}\left(-1\right)=0$$ Therefore $\nabla^{2}\phi=0$ which is what it should be. Finally, did you get rid of the second and third terms of your first equation by assuming $\theta=\phi=0$? $\endgroup$
    – Peter4075
    Commented Mar 5, 2012 at 20:04
  • $\begingroup$ @Peter4075: Yes. And if you have a function $x^2\sin(z)$, then $\frac{\partial}{\partial y}(x^2\sin(z))=0$. The function $\frac{1}{r}$ only depends on the radial coordinate, so the angular derivatives are zero. $\endgroup$
    – Nikolaj-K
    Commented Mar 6, 2012 at 8:20
  • $\begingroup$ @Peter4075: The message to be learn is to express the problem so that you use the symmetries. The transformation of tensors (including vectors) and differential operators (including the Laplacian) to a coordinate system at hand can be automatized, with Mathematica, say. This page contains the gradiant in often used systems and you'll also find other lists on wikipedia. For example this, especially if you go into general relativity at one point.. $\endgroup$
    – Nikolaj-K
    Commented Mar 6, 2012 at 10:02

On one hand, for a finite number of point charges, the charge distribution $\rho$ is a finite linear combination of 3d Dirac delta distributions. On the other hand, the Laplacian of a $1/r$ potential is really not identically zero, but also proportional to a 3d Dirac delta distribution. So there is no inconsistency.

  • $\begingroup$ thanks, but this is way over my head. I confess I have no idea what a 3d Dirac delta distribution is. $\endgroup$
    – Peter4075
    Commented Mar 5, 2012 at 20:08
  • 1
    $\begingroup$ I will try to come back later and explain further if I get the time. $\endgroup$
    – Qmechanic
    Commented Mar 5, 2012 at 20:19

Your Answer

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

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