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Conformal Solution

By letting $w\equiv f(z)$, the Real and Imaginary Parts of $w$ must satisfy the Cauchy-Riemann Equations and Laplace's Equation, so they automatically provide a scalar Potential and a so-called stream function. If a physical problem can be found for which the solution is valid, we obtain a solution--which may have been very difficult to obtain directly--by working backwards. Let

\begin{displaymath}
Az^n = Ar^ne^{in\theta},
\end{displaymath} (1)

the Real and Imaginary Parts then give
$\displaystyle \phi$ $\textstyle =$ $\displaystyle Ar^n\cos(n\theta)$ (2)
$\displaystyle \psi$ $\textstyle =$ $\displaystyle Ar^n\sin(n\theta).$ (3)

For $n=-2$,

$\displaystyle \phi$ $\textstyle =$ $\displaystyle {A\over r^2}\cos(2\theta)$ (4)
$\displaystyle \psi$ $\textstyle =$ $\displaystyle -{A\over r^2}\sin(2\theta),$ (5)

which is a double system of Lemniscates (Lamb 1945, p. 69). For $n=-1$,
$\displaystyle \phi$ $\textstyle =$ $\displaystyle {A\over r}\cos\theta$ (6)
$\displaystyle \psi$ $\textstyle =$ $\displaystyle -{A\over r}\sin\theta.$ (7)

This solution consists of two systems of Circles, and $\phi$ is the Potential Function for two Parallel opposite charged line charges (Feynman et al. 1989, §7-5; Lamb 1945, p. 69). For $n=1/2$,
$\displaystyle \phi$ $\textstyle =$ $\displaystyle Ar^{1/2}\cos\left({\theta\over 2}\right)= A\sqrt{\sqrt{x^2+y^2}+x\over 2}$ (8)
$\displaystyle \psi$ $\textstyle =$ $\displaystyle Ar^{1/2}\sin\left({\theta\over 2}\right)= A\sqrt{\sqrt{x^2+y^2}-x\over 2}\,.$ (9)

$\phi$ gives the field near the edge of a thin plate (Feynman et al. 1989, §7-5). For $n=1$,
$\displaystyle \phi$ $\textstyle =$ $\displaystyle Ar\cos\theta=Ax$ (10)
$\displaystyle \psi$ $\textstyle =$ $\displaystyle Ar\sin\theta=Ay.$ (11)

This is two straight lines (Lamb 1945, p. 68). For $n=3/2$,
\begin{displaymath}
w=Ar^{3/2}e^{3i\theta/2}.
\end{displaymath} (12)

$\phi$ gives the field near the outside of a rectangular corner (Feynman et al. 1989, §7-5). For $n=2$,
\begin{displaymath}
w=A(x+iy)^2=A[(x^2-y^2)+2ixy]
\end{displaymath} (13)


$\displaystyle \phi$ $\textstyle =$ $\displaystyle A(x^2-y^2)=Ar^2\cos(2\theta)$ (14)
$\displaystyle \psi$ $\textstyle =$ $\displaystyle 2Axy=Ar^2\sin(2\theta).$ (15)

These are two Perpendicular Hyperbolas, and $\phi$ is the Potential Function near the middle of two point charges or the field on the opening side of a charged Right Angle conductor (Feynman 1989, §7-3).

See also Cauchy-Riemann Equations, Conformal Map, Laplace's Equation


References

Feynman, R. P.; Leighton, R. B.; and Sands, M. The Feynman Lectures on Physics, Vol. 1. Redwood City, CA: Addison-Wesley, 1989.

Lamb, H. Hydrodynamics, 6th ed. New York: Dover, 1945.



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© 1996-9 Eric W. Weisstein
1999-05-26