0
$\begingroup$

Newton's law of gravitation describes the gravitational potential produced by a mass $m$ as : $G(r)=-k\frac{m}{r}$.However if you solve Laplace's equation for the gravitational potential in polar coordinates at the limit $r\rightarrow\infty$.The solution produces more powers than that of $r^{-1}$,you have odd power of the distance $r^{-(2n+1)},n = 0,1,2,...$ and despite their decreased amplitude they exist.So what is going on here?

$\endgroup$
6
  • $\begingroup$ Hint: what's happening at r = 0 when the potential is not $r^{-1}$. Is the laplace equation the good equation to consider if there is a particle somewhere (creating your field)? $\endgroup$
    – Syrocco
    Jun 12 at 16:35
  • $\begingroup$ @Syrocco the correct equation is the Poisson equation but the solution of the Poisson equation is the linear sum of the solution of the Laplace equation and something else so these terms appear there as well. $\endgroup$
    – user369585
    Jun 12 at 16:46
  • $\begingroup$ if you have a dirac delta mass at 0 (you need it to produce your field), then $1/r$ is indeed a solution. Because, in the distribution sense, $\Delta (1/r)=\delta(r)$. Do you think this equation is still valid if we put a different power of $r$ in the LHS, especially at 0? $\endgroup$
    – Syrocco
    Jun 12 at 16:57
  • $\begingroup$ As said in the answer by Ivan, with multiple sources of gravitational field (or with non natural coordinates), you will see these power appear as a serie! $\endgroup$
    – Syrocco
    Jun 12 at 17:00
  • 1
    $\begingroup$ Try computing the potential of a cubical mass. It won’t be simply $1/r$. $\endgroup$
    – Ghoster
    Jun 12 at 19:57

2 Answers 2

3
$\begingroup$

You need to solve the Poisson equation $\Delta V =-4 \pi G \rho$ with $ \rho = q \delta(\vec{x})$.

In $\vec{x}=0$ it differs from the Laplace equation. You can verify that the other terms in $1/r^l$ satisfy neither the Laplace equation at the origin nor the Poisson equation $\Delta V =-4 \pi G q \delta(\vec{x })$.

One can show that the solution which tends to $0$ at infinity is unique (even with a charge distribution which is not a function but a distribution). Once the uniqueness is admitted, the solution for the distribution $ \rho = q \delta(\vec{x})$ is necessarily spherically symmetric because otherwise we could generate other solutions by simple rotations.

Hope it can help and sorry for my poor english.

$\endgroup$
2
$\begingroup$

Each power term of the solution of Laplace's equation represents a different multipolar term (monopole, dipole, quadrupole …).

For a point mass, you are only interested in the monopole term, i.e., $r^{-1}$.

Such multipolar expansion is quite useful in the context of electromagnetism, but it is also useful in the context of gravitational. For example, gravitational wave radiations are usually associated with oscillations of the quadrupole term.

$\endgroup$

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.