# The converse of Newton's shell theorem

The shell theorem states that a spherically symmetric body $S$ of mass $m$ has a gravitational field identical to that of a point particle $P$ of mass $m$ located at the center of $S$.

We can ask the converse question: suppose there is a force $F$ between masses $M$ and $m$, separated by a distance $r$ of the form $$F = M m f ( r )$$ such that any spherically symmetric body affects external bodies as if its mass were concentrated at its center. Then what form can the function $f$ take?

According to Wikipedia: https://en.wikipedia.org/wiki/Shell_theorem, the general solution is

$$f(r)=a/r^2+br$$

where $a$ and $b$ are constants.

I attempted to demonstrate this. Replacing the gravitational force with the general force $F=Mmf(r)$ in the calculations to prove the shell theorem, I got that for all distances $r \ge R > 0$, $f$ satisfies:

$$\int_{r-R}^{R+r} (r^2-R^2 +s^2) f(s)\ ds = 4r^2Rf(r).$$

We can take the second derivative of both sides with respect to $r$ to get rid of the integral (we'll need to differentiate inside the integral), and we get the following nasty differential equation:

$$2Rrf''(r)+6Rf'(r)=f'(r+R)(r+R)+2f(r+R)-f'(r-R)(r-R)-2f(r-R).$$

I don't think this really helps. We can easily verify that $f=ar^2+br$ is a solution of this equation, so I didn't make any mistakes thus far (note the equation is linear so it suffices to check it for $f=r$ and $f=1/r^2$).

So how can we demonstrate the result?

Converse shell theorem for an exterior point mass:

1. Assume that the force ${\bf F}_{12}$ between two point masses $m_1$ and $m_2$ is collinear with the difference in positions ${\bf r}_{12}:={\bf r}_1-{\bf r}_2$, is central, and the magnitude $$|{\bf F}_{12}|=m_1m_2 f(|{\bf r}_{12}|)\tag{*}$$ is proportional to both the two point masses. We call the function $f(|{\bf r}_{12}|)$ the specific force.

2. Assume furthermore that the magnitude of the total force between an extended spherically symmetric mass $M$ and an exterior point mass $m$ is of the same form $$|{\bf F}|=mM f(r),\tag{**}$$ where $r\equiv |{\bf r}|$ is the distance between $m$ and the center of $M$.

3. Then the specific force $f$ is a linear combination of

Sketched proof: Let us use the same notation as the Wikipedia page. We consider the outside of a thin shell, i.e. $r\geq R$. Let us for simplicity work in terms of potential energy (rather than force), because it is easier to work with a scalar (rather than a vector) quantity. We may assume that the contribution $dU$ to the potential energy of the mass $m$ and $dM$ with distance $s$ is of the form

$$dU~=~m~dM~u(s),\tag{1.1}$$

where $s\mapsto u(s)$ is an unknown specific potential. Define for later convenience the antiderivative

$$V(s)~:=~\int^s\! ds^{\prime}~s^{\prime}~ u(s^{\prime}), \tag{1.2}$$

so that

$$V^{\prime}(s)~\stackrel{(1.2)}{=}~s~u(s).\tag{1.3}$$

From geometry we have the cosine relation

$$\cos\theta ~=~\frac{r^2+R^2-s^2}{2rR},\tag{1.4}$$

and therefore

$$dM ~=~ \frac{M}{2}\sin\theta~ d\theta ~=~-\frac{M}{2}~d\cos(\theta) ~\stackrel{(1.4)}{=}~\frac{M}{2}\frac{s~ds}{rR}.\tag{1.5}$$

In the last equality of eq.(1.5) we varied $s$ and $\theta$ while holding $r$ and $R$ fixed. Then we have

$$dU~\stackrel{(1.1)+(1.5)}{=}~\frac{mM}{2}\frac{s~ds}{rR}~u(s) ~\stackrel{(1.3)}{=}~\frac{mM}{2rR}ds~V^{\prime}(s).\tag{1.6}$$

So the total potential energy is

$$U(r,R)~=~\int dU ~\stackrel{(1.6)}{=}~mM\frac{V(r+R)-V(r-R)}{2rR}$$ $$~\stackrel{(1.8)}{=}~\frac{mM}{r}\left(V^{\prime}(r)+ \frac{R^2}{6}V^{\prime\prime\prime}(r)+ {\cal O}(R^5)\right).\tag{1.7}$$

In the last equality of eq. (1.7), we Taylor-expanded the potential

$$V(r\pm R)~=~V(r) \pm RV^{\prime}(r) + \frac{R^2}{2}V^{\prime\prime}(r) \pm \frac{R^3}{6}V^{\prime\prime\prime}(r)+ {\cal O}(R^4). \tag{1.8}$$

Let us now implement the main converse shell assumption: We assume that the potential energy $U(r,R)$ up to an ambiguity in the choice of zero-point energy $U_0(R)$ does not depend on the radius $R$ of the thin shell, i.e. $U(r,R)$ is of the form

$$U(r,R)~=~U_1(r)+U_0(R) .\tag{1.9}$$

We see that the only way that eq. (1.7) could be of the form (1.9) is iff

$$V^{\prime\prime\prime}(r)~\stackrel{(1.7)+(1.9)}{\propto}~r,\tag{1.10}$$

i.e. $V^{\prime}$ is a 3rd-order polynomial with a missing 2nd-order term:

$$V^{\prime}(r)~\stackrel{(1.10)}{=}~Ar^3+Br^2+Cr+D\tag{1.11}, \qquad B~=~0.$$

Or in terms of the specific potential

$$u(r)~\stackrel{(1.3)+(1.11)}{=}~Ar^2+C+\frac{D}{r}\tag{1.12}.$$

Or in terms of the specific force

$$-f(r)~=~u^{\prime}(r)~\stackrel{(1.12)}{=}~2Ar-\frac{D}{r^2}\tag{1.13}.$$

$\Box$

• Why does the math work out nicely when you calculate the potential energy but it's a complete mess when you calculate the force (as I tried) – Joshua Benabou Mar 17 '17 at 15:51
• @JoshuaBenabou The potential is easier to work with since it's a scalar, compared to a force, which is a vector. Also the potential satisfies a "cleaner" equation easier to solve by separation of variables since the evaluation of scalar quantities is independent of the choice of coordinate systems. – ZeroTheHero Mar 17 '17 at 16:18
• $\uparrow$ Agree. – Qmechanic Mar 17 '17 at 16:24
• So the integral expression in terms of $f(r)$ can't be worked with at all. You must pass by potential? – Joshua Benabou Mar 17 '17 at 16:31
• @FakeMod: Thanks. The gif is from Wikipedia. – Qmechanic Apr 6 at 11:10

This follows from Gauss's law. The $1/r^2$ part would be for a Gaussian sphere of radius greater the radius if the source, while the $r$ part follows from having a Gaussian sphere with radius inside the source, assuming uniform mass density.

The best way to show this is to start from the differential equation for the potential in spherical coordinates assuming spherical symmetry; this would be the Laplace equation outside and the Poisson equation with a constant term on the right inside the source. Once you have the potential you can get the force through the gradient.

Edit. Start from \begin{align} \nabla^2V(r)&=0\, ,\\ \frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{dV}{dr}\right)&=0\, ,\\ r^2\frac{dV}{dr}&=-a\, , \end{align} for some constant $a$. This is enough to give you the force as $\sim 1/r^2$ outside the source since $F_r=-\frac{dV}{dr}$.

Inside, you need to find a particular solution to $$\nabla^2V=4\pi G\,\rho_m$$ where $\rho_m$ is the mass density. By inspection $dV/dr=4\pi G\rho_m\,r/3$ will work; this entails $F_r=br$ inside the source. The general solution with $\rho_m\ne 0$ is then the sum of the particular solution and the sum of the homogenous problem.

• I guess I'll have to revisit this answer when I've done some higher level physics because I don't understand why the statement that a spherically symmetric body acts as if all its mass is concentrated at its center is equivalent to the condition you gave with the Laplace and Poisson equations. – Joshua Benabou Mar 12 '17 at 17:18
• @JoshuaBenabou The key points are: 1. the spherical symmetry and 2. the potential, which must satisfy $\nabla^2V(r)=4\pi G\rho_m$ as per en.wikipedia.org/wiki/… (I corrected missing $4\pi G$ factors in my answer). – ZeroTheHero Mar 12 '17 at 18:09
• how come the total force is the sum when inside and outside follow different equations? – lucky-guess Mar 17 '17 at 12:53
• @rpfphysics the coefficients $a$ and $b$ that appear in the general solution need to be adjusted for particular solutions. – ZeroTheHero Mar 17 '17 at 13:02

Converse shell theorem for an interior point mass:

1. Assume that the force $${\bf F}_{12}$$ between two point masses $$m_1$$ and $$m_2$$ is collinear with the difference in positions $${\bf r}_{12}:={\bf r}_1-{\bf r}_2$$, is central, and the magnitude $$|{\bf F}_{12}|=m_1m_2 f(|{\bf r}_{12}|)\tag{*}$$ is proportional to both the two point masses. We call the function $$f(|{\bf r}_{12}|)$$ the specific force.

2. Assume furthermore that the total force between an extended spherically symmetric mass $$M$$ and an interior point mass $$m$$ vanish.

3. Then the specific force $$f$$ is an inverse square/Newtonian gravity force.

Sketched proof: We use the same notation as the Wikipedia page and my other Phys.SE answer. We consider the inside of a thin shell, i.e. $$r\leq R$$.

The total radial force should be zero:

\begin{align} 0~=~&F_r(r)\cr ~=~&\int \! dF_r\cr \stackrel{(2.4)}{=}&\frac{mM}{4r^2R}~\int_{R-r}^{R+r} \! ds\left[W^{\prime}(s)+(r^2-R^2)u^{\prime}(s)\right]\cr ~=~&\frac{mM}{4r^2R}\left[W(R\!+\!r)-W(R\!-\!r)+(r^2-R^2)\left(u(R\!+\!r)-u(R\!-\!r)\right)\right] \cr ~=~&\frac{mM}{2rR}\left[W^{\prime}(R)+\frac{r^2}{6}W^{\prime\prime\prime}(R)+(r^2-R^2)\left(u^{\prime}(R)+\frac{r^2}{6}u^{\prime\prime\prime}(R)\right)+{\cal O}(r^4)\right]\cr ~=~&\frac{mM}{2rR}\left[\underbrace{W^{\prime}(R)-R^2u^{\prime}(R)}_{=~0}+r^2 (\ldots)+{\cal O}(r^4)\right].\end{align}\tag{3.1}

In particular the ellipsis $$(\ldots)$$ in eq. (3.1) must vanish:

\begin{align} 0~=~&(\ldots) \cr ~\stackrel{(3.1)}{=}& \frac{W^{\prime\prime\prime}(R)}{6}+ u^{\prime}(R) -R^2\frac{u^{\prime\prime\prime}(R)}{6} \cr ~\stackrel{(2.2)}{=}&-\frac{1}{6}\frac{d^2(R^2f(R))}{dR^2}-f(R)+R^2\frac{f^{\prime\prime}(R)}{6}\cr ~=~&-\frac{2}{3}\left(2f(R) +Rf^{\prime}(R)\right)\cr ~=~&-\frac{2}{3R}\frac{d(R^2f(R))}{dR}\cr ~\stackrel{(2.2)}{=}&\frac{2W^{\prime\prime}(R)}{3R}.\end{align}\tag{3.2}

Hence

$$W^{\prime}(R)~\stackrel{(3.2)}{=}~ {\rm constant},\tag{3.3}$$

and therefore the specific force

$$f(R)~\stackrel{(2.2)}{=}~-\frac{W^{\prime}(R)}{R^2}~\stackrel{(3.3)}{\propto}~ R^{-2}\tag{3.4}$$

is an inverse square force. $$\Box$$

• Thank you Qmechanic for helping me this far. Now how shall we logically prove that $f(R)=kR^{-2}$ is the only solution, i.e. there are no other solutions. Have you proved in your answer. I do not think so. Am I right or wrong? Please have a look at math.stackexchange.com/q/3612111 – Joe Apr 6 at 10:16
• Yes, this answer proves that $f(r)=kr^{-2}$ is the only solution. – Qmechanic Apr 7 at 6:40
• Qmechanic : I think the logic goes as follows: (Please point out if I am wrong) $$\text{Suppose there exists n solutions {f_1(R)=kR^{-2}, f_2(R), f_3(R),......,f_n(R)} }$$ $$\text{You have only proved that f_1(R)=kR^{-2} }$$ $$\text{So next we have to prove n=1, i.e. there exists only one f(R)=kR^{-2}}$$ – Joe Apr 7 at 9:22
• $f(r)=kr^{-2}$ is not an assumption in my proof; it is a conclusion. – Qmechanic Apr 7 at 10:57
• You started from $F=0$ and reached $f(r)=kr^{-2}$. Now what if we start from $F=0$ and by a different approach, we reach $f(r)=f_2(r)$. Then $f(r)=kr^{-2}$ and $f(r)=f_2(r)$ will both be solutions. How can we eliminate the possibility that we may get $f(r)$ other than $kr^{-2}$? – Joe Apr 7 at 11:11

In this answer, we give a proof of the converse shell theorem for an exterior point mass summing over forces rather than potentials, in the spirit of OP's calculation.

Sketched proof: We use the same notation as the Wikipedia page and my other Phys.SE answer. We consider the outside of a thin shell, i.e. $$r\geq R$$. Define for later convenience the antiderivative

$$W(s)~:=~-\int^s\! ds^{\prime}~(s^{\prime})^2~ f(s^{\prime}), \tag{2.1}$$

so that

$$W^{\prime}(s)~\stackrel{(2.1)}{=}~- s^2~f(s).\tag{2.2}$$

The cosine relation yields

$$\cos\phi ~=~\frac{s^2+r^2-R^2}{2sr}.\tag{2.3}$$

The radial contribution $$dF_r$$ to the force on the mass $$m$$ from the ring $$dM$$ with distance $$s$$ is of the form

\begin{align}dF_r~=~& -m~dM~f(s)~\cos\phi\cr ~\stackrel{(1.5)}{=}& -m~\frac{M}{2}\frac{s~ds}{rR}~f(s)~\cos\phi \cr ~\stackrel{(2.3)}{=}&-\frac{mM}{4r^2R}~ds~f(s)~(s^2+r^2-R^2)\cr ~\stackrel{(2.2)}{=}&\frac{mM}{4r^2R}~ds\left[W^{\prime}(s)+(r^2-R^2)u^{\prime}(s)\right].\end{align}\tag{2.4}

The total radial force should be independent of $$R$$:

\begin{align}F_r(r)~=~&\int \! dF_r\cr ~\stackrel{(2.4)}{=}&\frac{mM}{4r^2R}~\int_{r-R}^{r+R} \! ds\left[W^{\prime}(s)+(r^2-R^2)u^{\prime}(s)\right] \cr ~=~&\frac{mM}{4r^2R}\left[W(r\!+\!R)-W(r\!-\!R)+(r^2-R^2)\left(u(r\!+\!R)-u(r\!-\!R)\right)\right] \cr ~=~&\frac{mM}{2r^2}\left[W^{\prime}(r)+\frac{R^2}{6}W^{\prime\prime\prime}(r)+(r^2-R^2)\left(u^{\prime}(r)+\frac{R^2}{6}u^{\prime\prime\prime}(r)\right)+{\cal O}(R^4)\right]\cr ~=~&\frac{mM}{2}\left[\underbrace{r^{-2}W^{\prime}(r)+u^{\prime}(r)}_{=~-2f(r)}+R^2 (\ldots)+{\cal O}(R^4)\right].\end{align} \tag{2.5}

Therefore the ellipses $$(\ldots)$$ in eq. (2.5) must vanish:

\begin{align}0~=~&(\ldots) \cr ~\stackrel{(2.5)}{=}& \frac{W^{\prime\prime\prime}(r)}{6r^2}- \frac{u^{\prime}(r)}{r^2} +\frac{u^{\prime\prime\prime}(r)}{6} \cr ~\stackrel{(2.2)}{=}& -\frac{1}{6r^2}\frac{d^2(r^2f(r))}{dr^2}+ \frac{f(r)}{r^2} -\frac{f^{\prime\prime}(r)}{6} \cr ~=~& \frac{1}{3}\left( \frac{2f(r)}{r^2}-\frac{2f^{\prime}(r)}{r} -f^{\prime\prime}(r)\right)\cr ~=~& -\frac{d}{dr}\left(\frac{1}{3r^2} \frac{d(r^2f(r))}{dr}\right)\cr ~\stackrel{(2.2)}{=}& \frac{d}{dr}\left(\frac{W^{\prime\prime}(r)}{3r^2}\right).\end{align} \tag{2.6}

Hence

$$W^{\prime\prime}(r)~\stackrel{(2.6)}{\propto}~r^2 ,\tag{2.7}$$

so that

$$W^{\prime}(r)~\stackrel{(2.7)}{=}~\alpha r^3+\beta. \tag{2.8}$$

Or in terms of the specific force

$$-f(r)~\stackrel{(2.2)}{=}~\frac{W^{\prime}(r)}{r^2}~\stackrel{(2.8)}{=}~\alpha r+\frac{\beta}{r^2}. \tag{2.9}$$ $$\Box$$

So can we understand this result (2.9) intuitively? Yes:

1. The Coulomb law comes from Gauss' law.

2. The Hooke's law $$\vec{f}(\vec{r})=-k\vec{r}$$ works for every extended mass $$M$$ (not necessarily a spherically symmetric shell) that is invariant under spatial point reflection $$\vec{R}\to -\vec{R}$$. This follows from linearity: Two antipodes $$\vec{f}(\vec{r}\!+\!\vec{R})+\vec{f}(\vec{r}\!-\!\vec{R})=2\vec{f}(\vec{r})$$ contribute as much as twice the midpoint(=center).