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potential of spherical shell (Topic)

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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Cross-references: function, concept, mass

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Physics Classification: 04.20.-q (Classical general relativity )
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