PlanetPhysics/Potential of Spherical Shell

Let\, $$(\xi,\,\eta,\,\zeta)$$\, be a point bearing a 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 concept 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 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 $$r$$, $$\varphi$$ 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 $$\begin{matrix} V(a) = \int_{R_0}^R \int_{-\frac{\pi}{2}}^\frac{\pi}{2} \int_0^{2\pi}\! \!\frac{\varrho(r)\,r^2\cos\varphi}{\sqrt{r^2-2ar\sin\varphi+a^2}}\,dr\,d\varphi\,d\psi = 2\pi\int_{R_0}^R \varrho(r)\,r\,dr\int_{-\frac{\pi}{2}}^\frac{\pi}{2} \frac{r\cos\varphi\,d\varphi}{\sqrt{r^2-2ar\sin\varphi+a^2}}, \end{matrix}$$ 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 $$\begin{matrix} \int_{-\frac{\pi}{2}}^\frac{\pi}{2} \frac{r\cos\varphi\,d\varphi}{\sqrt{r^2-2ar\sin\varphi+a^2}} = -\frac{1}{a}\sijoitus{\varphi=-\frac{\pi}{2}}{\quad\frac{\pi}{2}}\sqrt{r^2-2ar\sin\varphi+a^2} = \frac{1}{a}[(r+a)-|r-a|]. \end{matrix}$$ 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 $$V(a) = \frac{4\pi}{a}\int_{R_0}^R \varrho(r)\,r^2\,dr = \frac{M}{a},$$ 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 $$V'(a) = -\frac{M}{a^2}.$$

$$2^\circ$$.\, 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, 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.