A constant potential due to a designed radial force inside a spherical shell - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T23:46:56Z http://mathoverflow.net/feeds/question/102764 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102764/a-constant-potential-due-to-a-designed-radial-force-inside-a-spherical-shell A constant potential due to a designed radial force inside a spherical shell shrdlu 2012-07-20T19:06:27Z 2012-07-21T14:03:04Z <p>It is well known that the gravitational forces due to a spherical shell of uniform density cancels in the interior of the shell (in three dimensions). Another way to state this is that the gravitational potential is uniform in the interior of the sphere.</p> <p>Suppose you are given a sphere of fixed radius and you can design a radial force (and associated radial potential) . Can you</p> <p>(1) Give example of other potentials that are constant in the interior of the sphere?</p> <p>(2) Characterize all potentials that are uniform in the sphere's interior?</p> <p>Literature references are particularly appreciated. </p> <p>I know that the solutions to (2) are a vector space and are a solution to an integral equation, but it is not clear that this is a fruitful line of attack.</p> http://mathoverflow.net/questions/102764/a-constant-potential-due-to-a-designed-radial-force-inside-a-spherical-shell/102818#102818 Answer by shrdlu for A constant potential due to a designed radial force inside a spherical shell shrdlu 2012-07-21T14:03:04Z 2012-07-21T14:03:04Z <p>I'm answering my own question after a night of sleep.</p> <hr> <p>Without loss of generality we consider the unit sphere centered at the origin. Define the potential between two point masses separated by a distance $\ell$ as $Q(\ell)$. The potential at a point at radius $r$ inside the unit sphere can be found by taking a point at $(x,y,z)=(0,0,r)$ and integrating $Q$ over the unit sphere which yields $$\Phi(r) = 2 \pi \int_0^\pi Q(\ell) \sin \phi \, d \phi$$ where $\ell = \sqrt{1+r^2 - 2 r \cos \phi}$. We want $\Phi(r)=\Phi_0$, a constant.</p> <p>Change variables in the integral from $\phi$ to $\ell$. Note that $\ell^2 = 1+r^2 - 2 r \cos \phi$ so $2 \ell d\ell = 2 r \sin \phi d \phi$. So the integral equation can be rewritten as $$\frac{ \Phi_0}{4 \pi}= \frac{1}{2r} \int_{1-r}^{1+r} \ell Q(\ell) ~ d \ell$$ Now, stare at the righthand side. It is the average value of $\ell Q(\ell)$ over the interval $[1-r,1+r]$, which we want to be constant for $0\le r &lt;1$. The answer is $$Q(\ell) = \frac{ \Phi_0}{4 \pi} \cdot \frac{\left [ 1 +f(\ell-1) \right ]}{\ell} .$$ where $f$ is any odd function. Three examples:</p> <p>$$\bullet \ F(z)=0 \Rightarrow Q(\ell) = \frac{ \Phi_0}{4 \pi \ell} , \qquad \text{the Newtonian potential},$$ $$\bullet \ F(z)=z \Rightarrow Q(\ell) = \frac{ \Phi_0}{4 \pi},\qquad \text{a constant potential,}$$ $$\bullet \ F(z)=(3z-z^3)/2 \Rightarrow Q(\ell) =\frac{ \Phi_0}{4 \pi} \cdot \frac{3\ell-\ell^2}{2},\qquad \text{a quadratic potential.}$$</p>