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

$\displaystyle \frac{m}{r} = \frac{m}{\sqrt{(x-\xi)^2+(y-\eta)^2+(z-\zeta)^2}}.$

The relevance of this concept appears from the fact that its partial derivatives

$\displaystyle \frac{\partial}{\partial x}\!\left(\frac{m}{r}\right) = -\frac{m(...
...\frac{\partial}{\partial z}\!\left(\frac{m}{r}\right) = -\frac{m(z-\zeta)}{r^3}$

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

$\displaystyle x = r\cos\varphi\cos\psi,\quad y = r\cos\varphi\sin\psi,\quad z = r\sin\varphi;$

for attaining all points we set

$\displaystyle R_0 \le r \le R,\quad -\frac{\pi}{2} \le \varphi \le \frac{\pi}{2},\quad
0 \le \psi < 2\pi.$

The cosines law implies that $ PA = \sqrt{r^2-2ar\sin\varphi+a^2}$. Thus the potential is the triple integral

$\displaystyle V(a) = \int_{R_0}^R \int_{-\frac{\pi}{2}}^\frac{\pi}{2} \int_0^{2...
...2}}^\frac{\pi}{2} \frac{r\cos\varphi\,d\varphi}{\sqrt{r^2-2ar\sin\varphi+a^2}},$ (1)

where the factor $ r^2\cos\varphi$ is the coefficient for the coordinate changing

$\displaystyle \left\vert\frac{\partial(x,\,y,\,z)}{\partial(r,\,\varphi,\,\psi)...
...i \\
-r\cos\varphi\sin\psi & r\cos\varphi\cos\psi & 0
\end{matrix}\right\vert.$

We get from the latter integral

$\displaystyle \int_{-\frac{\pi}{2}}^\frac{\pi}{2} \frac{r\cos\varphi\,d\varphi}...
...frac{\pi}{2}}\sqrt{r^2-2ar\sin\varphi+a^2} = \frac{1}{a}[(r+a)-\vert r-a\vert].$ (2)

Accordingly we have the two cases:

$ 1^\circ$. The point $ A$ is outwards the hollow ball, i.e. $ a > R$. Then we have $ \vert r-a\vert = a-r$ for all $ r\in[R_0,\,R]$. The value of the integral (2) is $ \frac{2r}{a}$, and (1) gets the form

$\displaystyle V(a) = \frac{4\pi}{a}\int_{R_0}^R \varrho(r)\,r^2\,dr = \frac{M}{a},$

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

$\displaystyle V'(a) = -\frac{M}{a^2}.$

$ 2^\circ$. The point $ A$ is in the cavity of the hollow ball, i.e. $ a < R_0$ . Then $ \vert r-a\vert = r-a$ on the interval of integration of (2). The value of (2) is equal to 2, and (1) yields

$\displaystyle V(a) = 4\pi\int_{R_0}^R \varrho(r)\,r\,dr,$

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.

Bibliography

1
ERNST LINDELÖF: Differentiali- ja integralilasku ja sen sovellutukset II. Mercatorin Kirjapaino Osakeyhtiö, Helsinki (1932).



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