Shell theorem

There are three steps to proving Newton's shell theorem. First, the equation for a gravitational field due to a ring of mass will be derived.  Arranging an infinite number of infinitely thin rings to make a disc, this equation involving a ring will be used to find the gravitational field due to a disk.  Finally, arranging an infinite number of infinitely thin discs to make a sphere, this equation involving a disc will be used to find the gravitational field due to a sphere.

The gravitational field ${\displaystyle E}$ at a position called ${\displaystyle P}$ at ${\displaystyle (x,y)=(-p,0)}$ on the x-axis due to a point of mass ${\displaystyle M}$ at the origin is

${\displaystyle E_{\text{point}}={\frac {GM}{p^{2}}}}$

Suppose that this mass is moved upwards along the y-axis to point ${\displaystyle (0,R)}$.  The distance between ${\displaystyle P}$ and the point mass is now longer than before; It becomes the hypotenuse of the right triangle with legs ${\displaystyle p}$ and ${\displaystyle R}$ which is ${\textstyle {\sqrt {p^{2}+R^{2}}}}$. Hence, the gravitational field of the elevated point is:

${\displaystyle E_{\text{elevated point}}={\frac {GM}{p^{2}+R^{2}}}}$

The magnitude of the gravitational field that would pull a particle at point ${\displaystyle P}$ in the x-direction is the gravitational field multiplied by ${\displaystyle \cos(\theta )}$ where ${\displaystyle \theta }$ is the angle adjacent to the x-axis. In this case, ${\displaystyle \cos(\theta )={\frac {p}{\sqrt {p^{2}+R^{2}}}}}$. Hence, the magnitude of the gravitational field in the x-direction, ${\displaystyle E_{x}}$ is:

${\displaystyle E_{x}={\frac {GM\cos {\theta }}{p^{2}+R^{2}}}}$

Substituting in ${\displaystyle \cos(\theta )}$ gives

${\displaystyle E_{x}={\frac {GMp}{(p^{2}+R^{2})^{3/2}}}}$

Suppose that this mass is evenly distributed in a ring centered at the origin and facing point ${\displaystyle P}$ with the same radius ${\displaystyle R}$.  Because all of the mass is located at the same angle with respect to the x-axis, and the distance between the points on the ring is the same distance as before, the gravitational field in the x-direction at point ${\displaystyle P}$ due to the ring is the same as a point mass located at a point ${\displaystyle R}$ units above the y-axis:

${\displaystyle E_{\text{ring}}={\frac {GMp}{(p^{2}+R^{2})^{3/2}}}}$

To find the gravitational field at point ${\displaystyle P}$ due to a disc, an infinite number of infinitely thin rings facing ${\displaystyle P}$, each with a radius ${\displaystyle y}$, width of ${\displaystyle dy}$, and mass of ${\displaystyle dM}$ may be placed inside one another to form a disc.  The mass of any one of the rings ${\displaystyle dM}$ is the mass of the disc multiplied by the ratio of the area of the ring ${\displaystyle 2\pi y\,dy}$ to the total area of the disc ${\displaystyle \pi R^{2}}$.  So, ${\displaystyle dM={\frac {M\cdot 2y\,dy}{R^{2}}}}$. Hence, a small change in the gravitational field, ${\displaystyle E}$ is:

${\displaystyle dE={\frac {Gp\ dM}{(p^{2}+y^{2})^{3/2}}}}$

Substituting in ${\displaystyle dM}$ and integrating both sides gives the gravitational field of the disk:

${\displaystyle E=\int {\frac {GMp\ \cdot {\frac {2y\ dy}{R^{2}}}}{(p^{2}+y^{2})^{3/2}}}}$

Adding up the contribution to the gravitational field from each of these rings will yield the expression for the gravitational field due to a disc.  This is equivalent to integrating this above expression from ${\displaystyle y=0}$ to ${\displaystyle y=R}$ , resulting in:

${\displaystyle E_{\text{disc}}={\frac {2GM}{R^{2}}}\left(1-{\frac {p}{\sqrt {p^{2}+R^{2}}}}\right)}$

To find the gravitational field at point ${\displaystyle P}$ due to a sphere centered at the origin, an infinite amount of infinitely thin discs facing ${\displaystyle P}$, each with a radius ${\displaystyle R}$, width of ${\displaystyle dx}$, and mass of ${\displaystyle dM}$ may be placed together.

These discs' radii ${\displaystyle R}$ follow the height of the cross section of a sphere (with constant radius "${\displaystyle a}$") which is an equation of a semi-circle: ${\displaystyle R={\sqrt {a^{2}-x^{2}}}}$.  ${\displaystyle x}$ varies from ${\displaystyle -a}$ to ${\displaystyle a}$.

The mass of any of the discs ${\displaystyle dM}$ is the mass of the sphere ${\displaystyle M}$ multiplied by the ratio of the volume of an infinitely thin disc divided by the volume of a sphere (with constant radius ${\displaystyle a}$).  The volume of an infinitely thin disc is ${\displaystyle \pi R^{2}\,dx}$, or ${\displaystyle \pi (a^{2}-x^{2})\,dx}$.  So, ${\displaystyle dM={\frac {\pi M(a^{2}-x^{2})\ dx}{{\frac {4}{3}}\pi a^{3}}}}$.  Simplifying gives ${\displaystyle dM={\frac {3M(a^{2}-x^{2})\ dx}{4a^{3}}}}$.  Again, ${\displaystyle x}$ varies from ${\displaystyle -a}$ to ${\displaystyle a}$.

Each discs' position away from ${\displaystyle P}$ will vary with its position within the 'sphere' made of the discs, so ${\displaystyle p}$ must be replaced with ${\displaystyle p+x}$. ${\displaystyle x}$ still varies from ${\displaystyle -a}$ to ${\displaystyle a}$.

Replacing ${\displaystyle M}$ with ${\displaystyle dM}$, ${\displaystyle R}$ with ${\displaystyle {\sqrt {a^{2}-x^{2}}}}$, and ${\displaystyle p}$ with ${\displaystyle p+x}$ in the 'disc' equation yields:

${\displaystyle dE={\frac {\left({\frac {2G[3M(a^{2}-x^{2})]}{4a^{3}}}\right)}{{\sqrt {a^{2}-x^{2}}}^{2}}}\cdot \left(1-{\frac {p+x}{\sqrt {(p+x)^{2}+{\sqrt {a^{2}-x^{2}}}^{2}}}}\right)\,dx}$

Simplifying,

${\displaystyle \int dE=\int _{-a}^{a}{\frac {3GM}{2a^{3}}}\left(1-{\frac {p+x}{\sqrt {p^{2}+a^{2}+2px}}}\right)\,dx}$

Integrating the gravitational field of each thin disc from ${\displaystyle x=-a}$ to ${\displaystyle x=+a}$ with respect to ${\displaystyle x}$, and doing some careful algebra, beautifully yields Newton's shell theorem:

${\displaystyle E={\frac {GM}{p^{2}}}}$

where ${\displaystyle p}$ is the distance between the center of the spherical mass and an arbitrary point ${\displaystyle P}$. The gravitational field of a spherical mass may be calculated by treating all the mass as a point particle at the center of the sphere.

A solid, spherically symmetric body can be modeled as an infinite number of concentric, infinitesimally thin spherical shells. If one of these shells can be treated as a point mass, then a system of shells (i.e. the sphere) can also be treated as a point mass. Consider one such shell (the diagram shows a cross-section):

(Note: the ${\displaystyle d\theta }$ in the diagram refers to the small angle, not the arc length. The arc length is ${\textstyle R\ d\theta }$.)

Applying Newton's Universal Law of Gravitation, the sum of the forces due to the mass elements in the shaded band is

${\displaystyle dF={\frac {Gm\;dM}{s^{2}}}}$

However, since there is partial cancellation due to the vector nature of the force in conjunction with the circular band's symmetry, the leftover component (in the direction pointing towards ${\displaystyle m}$) is given by

${\displaystyle dF_{r}={\frac {Gm\;dM}{s^{2}}}\cos \varphi }$

The total force on ${\displaystyle m}$, then, is simply the sum of the force exerted by all the bands. By shrinking the width of each band, and increasing the number of bands, the sum becomes an integral expression:

${\displaystyle F_{r}=\int dF_{r}}$

Since ${\displaystyle G}$ and ${\displaystyle m}$ are constants, they may be taken out of the integral:

${\displaystyle F_{r}=Gm\int {\frac {\cos \varphi \ dM}{s^{2}}}.}$

To evaluate this integral, one must first express ${\displaystyle dM}$ as a function of ${\displaystyle d\theta }$

The total surface of a spherical shell is

${\displaystyle 4\pi R^{2}}$

while the surface area of the thin slice between ${\displaystyle \theta }$ and ${\displaystyle \theta +d\theta }$ is

${\displaystyle 2\pi R\sin \theta \cdot R\,d\theta =2\pi R^{2}\sin \theta \,d\theta }$

If the mass of the shell is ${\displaystyle M}$, one therefore has that

${\displaystyle dM={\frac {2\pi R^{2}\sin \theta }{4\pi R^{2}}}M\,d\theta =\textstyle {\frac {1}{2}}M\sin \theta \,d\theta }$

and

${\displaystyle F_{r}={\frac {GMm}{2}}\int {\frac {\sin \theta \cos \varphi }{s^{2}}}\,d\theta }$

By the law of cosines,

${\displaystyle \cos \varphi ={\frac {r^{2}+s^{2}-R^{2}}{2rs}}}$

and

${\displaystyle \cos \theta ={\frac {r^{2}+R^{2}-s^{2}}{2rR}}.}$

These two relations link the three parameters ${\displaystyle \theta }$, ${\displaystyle \varphi }$ and ${\displaystyle s}$ that appear in the integral together. As ${\displaystyle \theta }$ increases from ${\displaystyle 0}$ to ${\displaystyle \pi }$ radians, ${\displaystyle \varphi }$ varies from the initial value 0 to a maximal value before finally returning to zero at ${\displaystyle \theta =\pi }$ . At the same time, ${\displaystyle s}$ increases from the initial value ${\displaystyle r-R}$ to the final value ${\displaystyle r+R}$ as ${\displaystyle \theta }$ increases from 0 to ${\displaystyle \pi }$ radians. This is illustrated in the following animation:

(Note: As viewed from ${\displaystyle m}$, the shaded blue band appears as a thin annulus whose inner and outer radii converge to ${\displaystyle R\ \sin \theta }$ as ${\displaystyle d\theta }$ vanishes.)

To find a primitive function to the integrand, one has to make ${\displaystyle s}$ the independent integration variable instead of ${\displaystyle \theta }$.

Performing an implicit differentiation of the second of the "cosine law" expressions above yields

${\displaystyle -\sin \theta \;d\theta ={\frac {-2s}{2rR}}\,ds}$

and thus

${\displaystyle \sin \theta \;d\theta ={\frac {s}{rR}}\,ds.}$

It follows that

${\displaystyle F_{r}={\frac {GMm}{2}}{\frac {1}{rR}}\int {\frac {s\cos \varphi }{s^{2}}}\,ds={\frac {GMm}{2rR}}\int {\frac {\cos \varphi }{s}}\,ds}$

where the new integration variable ${\displaystyle s}$ increases from ${\displaystyle r-R}$ to ${\displaystyle r+R}$ .

Inserting the expression for ${\displaystyle \cos \varphi }$ using the first of the "cosine law" expressions above, one finally gets that

${\displaystyle F_{r}={\frac {GMm}{4r^{2}R}}\int \left(1+{\frac {r^{2}-R^{2}}{s^{2}}}\right)\ ds\ .}$

A primitive function to the integrand is

${\displaystyle s-{\frac {r^{2}-R^{2}}{s}}\ ,}$

and inserting the bounds ${\displaystyle r-R}$ and ${\displaystyle r+R}$ for the integration variable ${\displaystyle s}$ in this primitive function, one gets that

${\displaystyle F_{r}={\frac {GMm}{r^{2}}}}$ ,

saying that the gravitational force is the same as that of a point mass in the centre of the shell with the same mass.

Finally, integrate all infinitesimally thin spherical shell with mass of ${\displaystyle dM}$, and we can obtain the total gravity contribution of a solid ball to the object outside the ball

${\displaystyle F_{total}=\int dF_{r}={\frac {Gm}{r^{2}}}\int dM.}$

Between the radius of ${\displaystyle x}$ to ${\displaystyle x+dx}$, ${\displaystyle dM}$ can be expressed as a function of ${\displaystyle x}$, i.e.,

${\displaystyle dM={\frac {4\pi x^{2}dx}{{\frac {4}{3}}\pi R^{3}}}M={\frac {3Mx^{2}dx}{R^{3}}}}$

Therefore, the total gravity is

${\displaystyle F_{\text{total}}={\frac {3GMm}{r^{2}R^{3}}}\int _{0}^{R}x^{2}\,dx={\frac {GMm}{r^{2}}}}$

which suggests that the gravity of a solid spherical ball to an exterior object can be simplified as that of a point mass in the centre of the ball with the same mass.

For a point inside the shell, the difference is that when θ is equal to zero, ϕ takes the value π radians and s the value R − r. When θ increases from 0 to π radians, ϕ decreases from the initial value π radians to zero and s increases from the initial value R − r to the value R + r.

This can all be seen in the following figure

Inserting these bounds into the primitive function

${\displaystyle s-{\frac {r^{2}-R^{2}}{s}}}$

one gets that, in this case

${\displaystyle F_{r}=0\ ,}$

saying that the net gravitational forces acting on the point mass from the mass elements of the shell, outside the measurement point, cancel out.

Generalization: If ${\displaystyle f={\frac {k}{r^{p}}}}$, the resultant force inside the shell is:

${\displaystyle F(r)={\frac {GMm}{4r^{2}R}}\int _{R-r}^{R+r}\left({\frac {1}{s^{p-2}}}+{\frac {r^{2}-R^{2}}{s^{p}}}\right)\,ds}$

The above results into ${\displaystyle F(r)}$ being identically zero if and only if ${\displaystyle p=2}$

Outside the shell (i.e. ${\displaystyle r>R}$ or ${\displaystyle r<-R}$):

${\displaystyle F(r)={\frac {GMm}{4r^{2}R}}\int _{r-R}^{r+R}\left({\frac {1}{s^{p-2}}}+{\frac {r^{2}-R^{2}}{s^{p}}}\right)\,ds}$

The shell theorem is an immediate consequence of Gauss's law for gravity saying that

${\displaystyle \int _{S}{\mathbf {g} }\cdot \,d{\mathbf {S} }=-4\pi GM}$

where M is the mass of the part of the spherically symmetric mass distribution that is inside the sphere with radius r and

${\displaystyle \int _{S}{\mathbf {g} }\cdot \,d{\mathbf {S} }=\int _{S}{\mathbf {g} }\cdot {\mathbf {\hat {n}} }\,dS}$

is the surface integral of the gravitational field g over any closed surface inside which the total mass is M, the unit vector ${\displaystyle \mathbf {\hat {n}} }$ being the outward normal to the surface.

The gravitational field of a spherically symmetric mass distribution like a mass point, a spherical shell or a homogeneous sphere must also be spherically symmetric. If ${\displaystyle \mathbf {\hat {n}} }$ is a unit vector in the direction from the point of symmetry to another point the gravitational field at this other point must therefore be

${\displaystyle \mathbf {g} =g(r)\mathbf {\hat {n}} }$

where g(r) only depends on the distance r to the point of symmetry

Selecting the closed surface as a sphere with radius r with center at the point of symmetry the outward normal to a point on the surface, ${\displaystyle \mathbf {\hat {n}} }$, is precisely the direction pointing away from the point of symmetry of the mass distribution.

One, therefore, has that

${\displaystyle \mathbf {g} =g(r)\mathbf {\hat {n}} }$

and

${\displaystyle \int _{S}\mathbf {g} \cdot \,d{\mathbf {S} }=g(r)\int _{S}\,dS=g(r)4\pi r^{2}}$

as the area of the sphere is 4πr².

From Gauss's law it then follows that

${\displaystyle g(r)4\pi r^{2}=-4\pi GM}$

i.e. that

${\displaystyle g(r)=-{\frac {GM}{r^{2}}}.}$

It is natural to ask whether the converse of the shell theorem is true, namely whether the result of the theorem implies the law of universal gravitation, or if there is some more general force law for which the theorem holds. More specifically, one may ask the question:

Suppose there is a force ${\displaystyle F}$ between masses M and m, separated by a distance r of the form ${\displaystyle F=Mmf(r)}$ such that any spherically symmetric body affects external bodies as if its mass were concentrated at its centre. Then what form can the function ${\displaystyle f}$ take?

In fact, this allows exactly one more class of force than the (Newtonian) inverse square.[2][3] The most general force as derived in [2] is:

${\displaystyle F=-{\frac {GMm}{r^{2}}}-{\frac {\Lambda Mmr}{3}}}$

where ${\displaystyle G}$ and ${\displaystyle \Lambda }$ can be constants taking any value. The first term is the familiar law of universal gravitation; the second is an additional force, analogous to the cosmological constant term in general relativity.

If we further constrain the force by requiring that the second part of the theorem also holds, namely that there is no force inside a hollow ball, we exclude the possibility of the additional term, and the inverse square law is indeed the unique force law satisfying the theorem.

On the other hand, if we relax the conditions, and require only that the field everywhere outside a spherically symmetric body is the same as the field from some point mass at the centre (of any mass), we allow a new class of solutions given by the Yukawa potential, of which the inverse square law is a special case.

Another generalization can be made for a disc by observing that

${\displaystyle dM={\frac {R^{2}}{2}}{\frac {d\theta \sin ^{2}\theta }{\pi R^{2}}}M={\frac {\sin ^{2}\theta }{2\pi }}M\,d\theta }$

so:

${\displaystyle F_{r}={\frac {GMm}{2\pi }}\int {\frac {\sin ^{2}\theta \cos \varphi }{s^{2}}}\,d\theta ,}$

where ${\displaystyle M=\pi R^{2}\rho }$, and ${\displaystyle \rho }$ is the density of the body.

Doing all the intermediate calculations we get:

${\displaystyle F(r)={\frac {Gm\rho }{8r^{3}}}\int _{R-r}^{R+r}{\frac {(r^{2}+s^{2}-R^{2}){\sqrt {2(r^{2}R^{2}+r^{2}s^{2}+R^{2}s^{2})-s^{4}-r^{4}-R^{4}}}}{s^{2}}}\,ds}$

An analogue for shell theorem exists in general relativity (GR).

Spherical symmetry implies that the metric has time-independent Schwarzschild geometry, even if a central mass is undergoing gravitational collapse (Misner et al. 1973; see Birkhoff's theorem). The metric thus has form

${\displaystyle ds^{2}=-(1-2M/r)\,dt^{2}+(1-2M/r)^{-1}\,dr^{2}+r^{2}\,d\Omega ^{2}}$

(using geometrized units, where ${\displaystyle G=c=1}$). For ${\displaystyle r>R>0}$ (where ${\displaystyle R}$ is the radius of some mass shell), mass acts as a delta function at the origin. For ${\displaystyle r<R}$, shells of mass may exist externally, but for the metric to be non-singular at the origin, ${\displaystyle M}$ must be zero in the metric. This reduces the metric to flat Minkowski space; thus external shells have no gravitational effect.

This result illuminates the gravitational collapse leading to a black hole and its effect on the motion of light-rays and particles outside and inside the event horizon (Hartle 2003, chapter 12).

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