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Next: Waves Up: Basics Potential Solutions Previous: Cylindrical Coordinates,

Spherical Coordinates - $F(r, \theta )$

For axisymmetric solutions, that is solution independent of $\phi$ Laplace's equation is
\begin{displaymath}
{1\over r^{2}}{\partial \over \partial r}\left (r^{2}{\part...
... (\sin \theta {\partial F\over \partial
\theta}\right ) = 0.
\end{displaymath} (5.27)

As in the case of cylindrical coordinates there are many particular solutions. There are a few basic important solutions and the rest are given in terms of powers of $r$ and Legendre polynomials in $\cos
\theta$.
  1. An isolated point source (point charge, magnetic monopole, point source or sink) at the origin has

    \begin{displaymath}
F = {c\over r}.
\end{displaymath}

  2. If the point source is located at ${\bf r}_{0} = (r_{0},
\theta_{0})$ then the potential at a general point ${\bf r} = (r, \theta )$ is

    \begin{displaymath}
F = {c \over \vert{\bf r} - {\bf r}_{0}\vert}.
\end{displaymath}

  3. A uniform field at large $r$ in the $z$-direction is

    \begin{displaymath}
F = - c r\cos \theta = - cz, \qquad -\nabla F = c{\bf k}.
\end{displaymath}

  4. A dipole of moment $m$, orientated in the ${\bf k}$ direction has a potential

    \begin{displaymath}
F = m{\cos \theta \over r^{2}}.
\end{displaymath}

Be careful of the difference in forms for the point sources in spherical coordinates and the line sources in cylindrical coordinates.

To obtain the general solutions, we look for seperable solutions along the lines of the cylindrical case. Hence, to solve (5.27) we set

\begin{displaymath}
F = f(r) g(\theta).
\end{displaymath}

Thus, we obtain

\begin{displaymath}
{1\over r^{2}}\left (r^{2}f^{\prime}\right )^{\prime}g + {f...
...r d\theta}\left( \sin \theta {dg\over
d\theta} \right ) = 0.
\end{displaymath}

Rearranging by dividing by $fg$ gives,

\begin{displaymath}
{r^{2}f^{\prime\prime} + 2 rf^{\prime}\over f} = -{1\over \...
...theta}\left (\sin \theta {dg\over d\theta}\right ) =
n(n+1).
\end{displaymath}

The separation constant is taken as $n(n+1)$ where $n$ is an integer. Thus, the radial variations satisfy

\begin{displaymath}
r^{2}f^{\prime\prime} + 2r f^{\prime} - n(n+1)f = 0.
\end{displaymath}

It is straightforward to show that solutions to this are given by

\begin{displaymath}
f = r^{n} \hbox{ and } f = r^{-(n+1)}.
\end{displaymath}

The $\theta $ variations are given by solving

\begin{displaymath}
\sin \theta {d^{2}g\over d\theta^{2}} + \cos \theta {dg\over
d\theta} + n(n+1)\cos \theta g = 0.
\end{displaymath}

This equation may be simplified by changing the independent variable from $\theta $ to

\begin{displaymath}
s = \cos \theta, \hbox{ and } \sin \theta = \sqrt{1 - s^{2}}.
\end{displaymath}

Thus,

\begin{displaymath}
{dg\over d\theta} = -\sin \theta {dg\over ds}, \qquad {d^{2...
...g\over ds} = (1 - s^{2}){d^{2}g\over ds^{2}} - s {dg\over ds}.
\end{displaymath}

Hence our equation for $g$ becomes

\begin{displaymath}
(1-s^{2}){d^{2}g\over ds^{2}} - 2s{dg\over ds} + n(n+1) g = 0,
\end{displaymath}

and the solution is given in terms of Legendre polynomials of order $n$,

\begin{displaymath}
g = P_{n}(s).
\end{displaymath}

Finally, the potential may be expressed as either
\begin{displaymath}
F = r^{n}P_{n}(\cos \theta) \hbox{ or }F =
r^{-(n+1)}P_{n}(\cos \theta).
\end{displaymath} (5.28)

Like the trigonometric functions in the cylindrical case, the Legendre polynomials are orthogonal since,

\begin{displaymath}
\int_{0}^{2\pi}\sin \theta P_{n}(\cos \theta)
P_{m}(\cos\theta)d\theta = 0, \hbox{ for } m\ne n.
\end{displaymath}

The first two Legendre polynomials give
  1. $n=0$ so that

    \begin{displaymath}
P_{0}(\cos\theta) = 1, \qquad F = 1 \hbox{ or } F = {1\over r}.
\end{displaymath}

  2. $n=1$ and

    \begin{displaymath}
P_{1}(\cos\theta) = \cos\theta, \qquad F = r\cos\theta \hbox{
or } F = {\cos\theta \over r^{2}}.
\end{displaymath}

Example 5. .6Find the incompressible, irrotational flow of an ideal fluid or plasma, uniform at infinity, past a sphere of radius $a$. The general solution that is even in $\theta $ and uniform at infinity is

\begin{displaymath}
F = {c_{0}\over r} - V_{0}r\cos \theta + c_{1}{\cos\theta \...
...} + \sum_{n=2}^{\infty}{c_{n}\over r^{n+1}}P_{n}(\cos \theta).
\end{displaymath}

On the surface of the sphere we require that the radial velocity component vanishes. Thus,

\begin{displaymath}
v_{r} = - {\partial F \over \partial r} = 0, \hbox{ on } r = a.
\end{displaymath}

This means that

\begin{displaymath}
0 = {c_{0}\over a^{2}} + \left (V_{0} + {2c_{1}\over
a^{3...
...m_{n=2}^{\infty}c_{n}{n(n+1)\over
a^{n+2}}P_{n}(\cos\theta).
\end{displaymath}

From the orthogonality condition on the Legendre polynomials every coefficient must be zero, giving,

\begin{displaymath}
c_{n}=0, \hbox{ for } n\geq 2, \quad c_{0} = 0, \qquad c_{1} =
-{V_{0}a^{3}\over 2}.
\end{displaymath}

Thus,

\begin{displaymath}
F = -\left ( V_{0}r + {V_{0}a^{3}\over r^{2}}\right )\cos \theta.
\end{displaymath}


next up previous
Next: Waves Up: Basics Potential Solutions Previous: Cylindrical Coordinates,
Prof. Alan Hood
2000-11-06