PlanetPhysics/Potential of Spherical Shell

Let\, $(\xi ,\eta ,\zeta )$\, be a point bearing a [[../CosmologicalConstant/|mass\,]] $m$\, and\, $(x,y,z)$\, a variable point. If the distance of these points is $r$, we can define the 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 [[../PreciseIdea/|concept]] appears from the fact that its partial derivatives $\frac{\partial }{\partial x}\left(\frac{m}{r}\right)=-\frac{m(x-\xi )}{r^3},\frac{\partial }{\partial y}\left(\frac{m}{r}\right)=-\frac{m(y-\eta )}{r^3},\frac{\partial }{\partial z}\left(\frac{m}{r}\right)=-\frac{m(z-\zeta )}{r^3}$ 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 [[../Bijective/|function]] \, $\varrho =\varrho (r)$\, 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 $r$, $\phi$ and $\psi$ which are tied to the Cartesian coordinates via $x=r\cos \phi \cos \psi ,y=r\cos \phi \sin \psi ,z=r\sin \phi ;$ for attaining all points we set $R_0\le r\le R,-\frac{\pi }{2}\le \phi \le \frac{\pi }{2},0\le \psi <2\pi .$ The cosines law implies that\, $PA=\sqrt{r^2-2ar\sin \phi +a^2}$. Thus the potential is the triple integral

$\begin{array}{c}V(a)=\underset{R_0}{\overset{R}{\int }}\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\underset{0}{\overset{2\pi }{\int }}\frac{\varrho (r)r^2\cos \phi }{\sqrt{r^2-2ar\sin \phi +a^2}}drd\phi d\psi =2\pi \underset{R_0}{\overset{R}{\int }}\varrho (r)rdr\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{r\cos \phi d\phi }{\sqrt{r^2-2ar\sin \phi +a^2}},\end{array}$

where the factor\, $r^2\cos \phi$\, is the coefficient for the coordinate changing $\left|\frac{\partial (x,y,z)}{\partial (r,\phi ,\psi )}\right|=mod\left|\begin{array}{ccc}\cos \phi \cos \psi & \cos \phi \sin \psi & \sin \phi \\ -r\sin \phi \cos \psi & -r\sin \phi \sin \psi & r\cos \phi \\ -r\cos \phi \sin \psi & r\cos \phi \cos \psi & 0\end{array}\right|.$

We get from the latter integral

Failed to parse (unknown function "\sijoitus"): {\displaystyle \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}}

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 $V(a)=\frac{4\pi }{a}\underset{R_0}{\overset{R}{\int }}\varrho (r)r^{2}dr=\frac{M}{a},$ where $M$ 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 $V^{\prime }(a)=-\frac{M}{a^2}.$

$2^{\circ }$.\, 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 $V(a)=4\pi \underset{R_0}{\overset{R}{\int }}\varrho (r)rdr,$ which is independent on $a$. 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 .