Talk:PlanetPhysics/Potential of Spherical Shell

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%%% This file is part of PlanetPhysics snapshot of 2011-09-01 %%% Primary Title: potential of spherical shell %%% Primary Category Code: 04.20.-q %%% Filename: PotentialOfSphericalShell.tex %%% Version: 2 %%% Owner: pahio %%% Author(s): pahio %%% PlanetPhysics is released under the GNU Free Documentation License. %%% You should have received a file called fdl.txt along with this file. %%% If not, please write to gnu@gnu.org. \documentclass[12pt]{article} \pagestyle{empty} \setlength{\paperwidth}{8.5in} \setlength{\paperheight}{11in}

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Let\, $(\xi,\,\eta,\,\zeta)$\, be a point bearing a \htmladdnormallink{mass\,}{http://planetphysics.us/encyclopedia/CosmologicalConstant.html} $m$\, and\, $(x,\,y,\,z)$\, a variable point. If the distance of these points is $r$, we can define the {\em potential} of\, $(\xi,\,\eta,\,\zeta)$\, in\, $(x,\,y,\,z)$\, as

$$\frac{m}{r} = \frac{m}{\sqrt{(x-\xi)^2+(y-\eta)^2+(z-\zeta)^2}}.$$ The relevance of this \htmladdnormallink{concept}{http://planetphysics.us/encyclopedia/PreciseIdea.html} appears from the fact that its partial derivatives $$\frac{\partial}{\partial x}\!\left(\frac{m}{r}\right) = -\frac{m(x-\xi)}{r^3},\quad \frac{\partial}{\partial y}\!\left(\frac{m}{r}\right) = -\frac{m(y-\eta)}{r^3},\quad \frac{\partial}{\partial z}\!\left(\frac{m}{r}\right) = -\frac{m(z-\zeta)}{r^3}$$ are the components of the gravitational force with which the material point\, $(\xi,\,\eta,\,\zeta)$\, acts on one mass unit in the point\, $(x,\,y,\,z)$\, (provided that the measure units are chosen suitably).

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

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

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

We get from the latter integral \begin{align} \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{align} Accordingly we have the two cases:

$1^\circ$.\, The point $A$ is outwards the hollow ball, i.e. $a > R$.\, Then we have\, $|r-a| = a-r$\, for all\, $r\in[R_0,\,R]$.\, The value of the integral (2) is $\frac{2r}{a}$, and (1) gets the form $$V(a) = \frac{4\pi}{a}\int_{R_0}^R \varrho(r)\,r^2\,dr = \frac{M}{a},$$ where $M$ is the mass of the hollow ball. Thus {\em 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 $$V'(a) = -\frac{M}{a^2}.$$

$2^\circ$.\, The point $A$ is in the cavity of the hollow ball, i.e. $a < R_0$ .\, Then\, $|r-a| = r-a$\, on the interval of integration of (2). The value of (2) is equal to 2, and (1) yields $$V(a) = 4\pi\int_{R_0}^R \varrho(r)\,r\,dr,$$ which is independent on $a$. That is, {\em 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}.

\begin{thebibliography}{8} \bibitem{lindelof}{\sc Ernst Lindel\"of}: {\em Differentiali- ja integralilasku ja sen sovellutukset II}.\, Mercatorin Kirjapaino Osakeyhti\"o, Helsinki (1932). \end{thebibliography}

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