Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

Suppose you are given a sphere of fixed radius and you can design a radial force (and associated radial potential) . Can you

(1) Give example of other potentials that are constant in the interior of the sphere?

(2) Characterize all potentials that are uniform in the sphere's interior?

Literature references are particularly appreciated.

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.

share|improve this question
    
Does the potential you describe need to yield the Newtonian gravitational field associated with the shell? –  Brian Trundy Jul 20 '12 at 19:47
    
No, not necessarily. It would yield a designer force between particles that might be positive at some distances and negative at others. However inside the sphere the cumulative force would be zero. –  shrdlu Jul 20 '12 at 19:57

1 Answer 1

up vote 1 down vote accepted

I'm answering my own question after a night of sleep.


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.

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 $0\le r <1$. 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:

$$\bullet \ F(z)=0 \Rightarrow Q(\ell) = \frac{ \Phi_0}{4 \pi \ell} , \qquad \text{the Newtonian potential},$$ $$\bullet \ F(z)=z \Rightarrow Q(\ell) = \frac{ \Phi_0}{4 \pi},\qquad \text{a constant potential,}$$ $$\bullet \ F(z)=(3z-z^3)/2 \Rightarrow Q(\ell) =\frac{ \Phi_0}{4 \pi} \cdot \frac{3\ell-\ell^2}{2},\qquad \text{a quadratic potential.}$$

share|improve this answer
    
I'm embarrassed - I completely and utterly failed to understand this question. Your solution is quite elegant. –  Brian Trundy Jul 23 '12 at 16:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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