View source for PlanetPhysics/Potential of Spherical Shell

Let\, <math>(\xi,\,\eta,\,\zeta)</math>\, be a point bearing a [[../CosmologicalConstant/|mass\,]] <math>m</math>\, and\, <math>(x,\,y,\,z)</math>\, a variable point. If the distance of these points is <math>r</math>, we can define the ''potential''  of\, <math>(\xi,\,\eta,\,\zeta)</math>\, in\, <math>(x,\,y,\,z)</math>\, as
<math>\frac{m}{r} = \frac{m}{\sqrt{(x-\xi)^2+(y-\eta)^2+(z-\zeta)^2}}.</math>
The relevance of this [[../PreciseIdea/|concept]] appears from the fact that its partial derivatives
<math>\frac{\partial}{\partial x}\!\left(\frac{m}{r}\right) = -\frac{m(x-\xi)}{r^3},\quad
\frac{\partial}{\partial y}\!\left(\frac{m}{r}\right) = -\frac{m(y-\eta)}{r^3},\quad
\frac{\partial}{\partial z}\!\left(\frac{m}{r}\right) = -\frac{m(z-\zeta)}{r^3}</math>
are the components of the gravitational force with which the material point\, <math>(\xi,\,\eta,\,\zeta)</math>\, acts on one mass unit in the point\, <math>(x,\,y,\,z)</math>\, (provided that the measure units are chosen suitably).

The potential of a set of points\, <math>(\xi,\,\eta,\,\zeta)</math>\, is the sum of the potentials of individual points, i.e. it may lead to an integral.\\

We determine the potential of all points\, <math>(\xi,\,\eta,\,\zeta)</math>\, of a hollow ball, where the matter is located between two concentric spheres with radii <math>R_0</math> and <math>R\, (>R_0)</math>. Here the density of mass is assumed to be presented by a continuous [[../Bijective/|function]] \, <math>\varrho = \varrho(r)</math>\, at the distance <math>r</math> from the centre <math>O</math>. Let <math>a</math> be the distance from <math>O</math> of the point <math>A</math>, where the potential is to be determined. We chose <math>O</math> the origin and the ray <math>OA</math> the positive <math>z</math>-axis.

For obtaining the potential in <math>A</math> we must integrate over the ball shell where <math>R_0 \le r \le R</math>. We use the spherical coordinates <math>r</math>, <math>\varphi</math> and <math>\psi</math> which are tied to the Cartesian coordinates via
<math>x = r\cos\varphi\cos\psi,\quad y = r\cos\varphi\sin\psi,\quad z = r\sin\varphi;</math>
for attaining all points we set
<math>R_0 \le r \le R,\quad -\frac{\pi}{2} \le \varphi \le \frac{\pi}{2},\quad
0 \le \psi < 2\pi.</math>
The cosines law implies that\, <math>PA = \sqrt{r^2-2ar\sin\varphi+a^2}</math>. Thus the potential is the triple integral
<blockquote><math>\begin{matrix}
V(a) =
\int_{R_0}^R \int_{-\frac{\pi}{2}}^\frac{\pi}{2} \int_0^{2\pi}\!
\!\frac{\varrho(r)\,r^2\cos\varphi}{\sqrt{r^2-2ar\sin\varphi+a^2}}\,dr\,d\varphi\,d\psi
= 2\pi\int_{R_0}^R \varrho(r)\,r\,dr\int_{-\frac{\pi}{2}}^\frac{\pi}{2}
\frac{r\cos\varphi\,d\varphi}{\sqrt{r^2-2ar\sin\varphi+a^2}},
\end{matrix}</math></blockquote>
where the factor\, <math>r^2\cos\varphi</math>\, is the coefficient for the coordinate changing
<math>\left|\frac{\partial(x,\,y,\,z)}{\partial(r,\,\varphi,\,\psi)}\right| =
\!\mod\!\left|\begin{matrix}
\cos\varphi\cos\psi & \cos\varphi\sin\psi & \sin\varphi \\
-r\sin\varphi\cos\psi & -r\sin\varphi\sin\psi & r\cos\varphi \\
-r\cos\varphi\sin\psi & r\cos\varphi\cos\psi & 0
\end{matrix}\right|.</math>

We get from the latter integral
<blockquote><math>\begin{matrix}
\int_{-\frac{\pi}{2}}^\frac{\pi}{2}
\frac{r\cos\varphi\,d\varphi}{\sqrt{r^2-2ar\sin\varphi+a^2}}
= -\frac{1}{a}\sijoitus{\varphi=-\frac{\pi}{2}}{\quad\frac{\pi}{2}}\sqrt{r^2-2ar\sin\varphi+a^2}
= \frac{1}{a}[(r+a)-|r-a|].
\end{matrix}</math></blockquote>
Accordingly we have the two cases:

<math>1^\circ</math>.\, The point <math>A</math> is outwards the hollow ball, i.e. <math>a > R</math>.\, Then we have\, <math>|r-a| = a-r</math>\, for all\,
<math>r\in[R_0,\,R]</math>.\, The value of the integral (2) is <math>\frac{2r}{a}</math>, and (1) gets the form
<math>V(a) = \frac{4\pi}{a}\int_{R_0}^R \varrho(r)\,r^2\,dr = \frac{M}{a},</math>
where <math>M</math> is the mass of the hollow ball. Thus ''the potential outwards the hollow ball is exactly the same as in the case that all mass were concentrated to the centre'' . A correspondent statement concerns the attractive force
<math>V'(a) = -\frac{M}{a^2}.</math>

<math>2^\circ</math>.\, The point <math>A</math> is in the cavity of the hollow ball, i.e. <math>a < R_0</math> .\, Then\, <math>|r-a| = r-a</math>\, on the interval of integration of (2). The value of (2) is equal to 2, and (1) yields
<math>V(a) = 4\pi\int_{R_0}^R \varrho(r)\,r\,dr,</math>
which is independent on <math>a</math>. That is, ''the potential of the hollow ball, when the density of mass depends only on the distance from the centre, has in the cavity a constant value, and the hollow ball influences in no way on a mass inside it'' .

==All Sources==
<ref name='lindelof'/>

==References==
<references>
<ref name='lindelof'>{\sc Ernst Lindel\"of}: {\em Differentiali- ja integralilasku
ja sen sovellutukset II}.\, Mercatorin Kirjapaino Osakeyhti\"o, Helsinki (1932).
</ref></references>

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