(a)

Construct the free-space Green function \(G(x, y; x', y')\) for two-dimensional electrostatics by integrating \(1/R\) with respect to \((z' - z)\) between the limits \(\pm Z\) where \(Z\) is taken to be very large. Show that apart from an inessential constant, the Green function can be written alternately as \begin{align*} G(x, y; x', y') &= - \ln \left [{(x - x')}^2 +{(y - y')}^2 \right ] \\ &= - \ln \left [ \varrho ^2 + \varrho '^2 - 2 \varrho \varrho ' \cos ( \phi - \phi ' )\right ] \end{align*}

Solution. Use the following definitions: \begin{align*} \alpha &\coloneqq \sqrt{{(x - x')}^2 +{(y - y')}^2} \\ u &\coloneqq z - z' \end{align*}

to integrate \begin{align*} \int _{-Z}^{Z} \frac{du}{\sqrt{\alpha ^2 + u^2}} &= \ln \left ( \sqrt{\alpha ^2 + u^2} + u \right ) \biggr \rvert _{-Z}^{Z} \\ &= \ln \frac{\sqrt{Z^2 + \alpha ^2} + Z}{\sqrt{Z^2 + \alpha ^2} - Z} \\ &= \ln \frac{\sqrt{1 + \frac{\alpha ^2}{Z^2}} + 1}{\sqrt{1 + \frac{\alpha ^2}{Z^2}} - 1} \end{align*}

Expanding to first order in \(\frac{\alpha ^2}{Z^2}\) yields \begin{align*} G &= \ln \frac{2 + \frac{\alpha ^2}{2 Z^2}}{\frac{\alpha ^2}{2Z^2}} \\ &= \ln \frac{4Z^2 + \alpha ^2}{\alpha ^2} \\ &= \ln \left (4Z^2 + \alpha ^2\right ) - \ln \alpha ^2 \\ G(Z \gg \alpha ) &= \ln 4Z^2 - \ln \alpha ^2 \end{align*}

\begin{align*} \therefore G(x, y; x', y') &= - \ln \alpha ^2 \\ &= - \ln \left [{(x - x')}^2 +{(y - y')}^2\right ] \end{align*}

Changing to cylindrical coordinates yields \begin{align*} G(x, y;x',y') &= - \ln \left [ \varrho ^2 + \varrho '^2 - 2 \varrho \varrho ' \cos ( \phi - \phi ' )\right ] \end{align*}

(b)

Show explicitly by separation of variables in polar coordinates that the Green function can be expressed as a Fourier series in the azimuthal coordinate, \begin{align*} G = \frac{1}{2 \pi } \sum _{-\infty }^{\infty } e^{i m (\phi - \phi ')}g_m(\varrho , \varrho ') \end{align*}

where the radial Green functions satisfy \begin{align*} \frac{1}{\varrho '}\frac{\partial }{\partial \varrho '} \left ( \varrho ' \frac{\partial g_m}{\partial \varrho '} \right ) - \frac{m^2}{\varrho '^2} g_m = - 4 \pi \frac{\delta (\varrho - \varrho ')}{\varrho } \end{align*}

Note that \(g_m(\varrho , \varrho ')\) for fixed \(\varrho \) is a different linear combination of the solutions of the homogenous radial equation (2.68) for \(\varrho ' < \varrho \) and for \(\varrho ' > \varrho \), with a discontinuity of slope at \(\varrho ' = \varrho \) determined by the source delta function

Solution. The defining equations of the greens function \begin{align*} \int _\Omega \nabla '^2 G(\varrho , \phi ;\varrho ', \phi ') \varrho ' d\varrho ' d\phi ' = -4 \pi \\ \nabla '^2 G(\varrho , \phi ; \varrho ', \phi ') \propto \delta (\varrho - \varrho ')\delta (\phi - \phi ') \end{align*}

can be satisfied if \begin{align*} \nabla '^2 G = -4\pi \frac{\delta (\varrho - \varrho ')\delta (\phi - \phi ')}{\varrho } \end{align*}

Applying the laplacian to the given expansion yields: \begin{align*} \nabla '^2 G &= \left [ \frac{1}{\varrho '} \frac{\partial }{\partial \varrho '} \left (\varrho ' \frac{\partial }{\partial \varrho '}\right ) - \frac{1}{\varrho '^2}\frac{\partial }{\partial \phi '^2} \right ] \left [ \frac{1}{2\pi }\sum _{-\infty }^\infty e^{im(\phi -\phi ')}g_m(\varrho , \varrho ') \right ] \\ &= \frac{1}{2\pi }\sum _{-\infty }^\infty \left [ \frac{1}{\varrho '} \frac{\partial }{\partial \varrho '} \left (\varrho ' \frac{\partial g_m}{\partial \varrho '}\right ) - \frac{m^2 g_m}{\varrho '^2} \right ] e^{im(\phi - \phi ')} \\ &= \frac{1}{2\pi }\sum _{-\infty }^\infty \left [ -4 \pi \frac{\delta (\varrho - \varrho ')}{\varrho } \right ] e^{im(\phi - \phi ')} \\ &= -2 \frac{\delta (\varrho - \varrho ')}{\varrho } \sum _{-\infty }^\infty e^{im(\phi - \phi ')} \\ &= -4 \pi \frac{\delta (\varrho - \varrho ')\delta (\phi - \phi ')}{\varrho } \end{align*}

Showing that the expansion is correct.

(c)

Complete the solution and show that the free-space Green function has the expansion \begin{align*} G(\varrho , \phi ; \varrho ', \phi ') = - \ln (\varrho _>^2) + 2 \sum _{m = 1}^{\infty } \frac{1}{m}{\left ( \frac{\varrho _<}{\varrho _>} \right )}^m \cdot \cos \left [ m (\phi - \phi ') \right ] \end{align*}

where \(\varrho _<\) (\(\varrho _>\)) is the smaller (larger) of \(\varrho \) and \(\varrho '\)

Solution. Since \(g_m(\varrho , \varrho ') = P(\varrho , \varrho ')\) is a solution to the (split) Laplace equation, \(g_m\) can be given by \begin{align*} g_m(\varrho , \varrho ') = \begin{cases} A_m \varrho '^m & \varrho ' < \varrho \\ B_m \varrho '^{-m} & \varrho ' > \varrho \end{cases} \end{align*}

Continuity dictates that \begin{align*} A_m \varrho ^m &= B_m \varrho ^{-m} & \\ \therefore A_m = \alpha _m \varrho ^{-m} &,\;\;\; B_m = \alpha _m \varrho ^m & \end{align*}

Giving \begin{align*} g_m(\varrho , \varrho ') = \begin{cases} \alpha _m{\left (\dfrac{\varrho '}{\varrho }\right )}^m & \varrho ' < \varrho \\ \alpha _m{\left (\dfrac{\varrho }{\varrho '}\right )}^m & \varrho ' > \varrho \end{cases} \end{align*}

The discontinuity in the derivative is determined by the laplace equation for the Green function: \begin{align*} - \frac{4 \pi }{\varrho } &= \frac{d g_m}{d\varrho '} \biggr \rvert _{\varrho _<} - \frac{d g_m}{d \varrho '} \biggr \rvert _{\varrho _>} \\ &= -\frac{2 m \alpha _m}{\varrho } \\ &\therefore \alpha _m = \frac{2 \pi }{m} \end{align*}

\begin{align*} g_m(\varrho , \varrho ') &= \begin{cases} \dfrac{2 \pi }{m}{\left (\dfrac{\varrho '}{\varrho }\right )}^m & \varrho ' < \varrho \\ \dfrac{2 \pi }{m}{\left (\dfrac{\varrho }{\varrho '}\right )}^m & \varrho ' > \varrho \end{cases} \\ &= \frac{2\pi }{m}{\left ( \dfrac{\varrho _<}{\varrho _>}\right )}^m \end{align*}

From part a, \(g_0 = -\ln \varrho _>^2\), so \begin{align*} G(\varrho , \varrho ') &= -\ln ({\varrho _>}^2) + \sum _{m = -\infty , m \ne 0}^{\infty } \dfrac{1}{\lvert m \rvert }{\left (\dfrac{\varrho _<}{\varrho _>}\right )}^{\lvert m \rvert } e^{im (\phi - \phi ')} \\ &= -\ln ({\varrho _>}^2) + \sum _{m = 1}^{\infty } \dfrac{1}{m}{\left (\dfrac{\varrho _<}{\varrho _>}\right )}^{m} e^{-im (\phi - \phi ')} + \sum _{m = 1}^{\infty } \dfrac{1}{m}{\left (\dfrac{\varrho _<}{\varrho _>}\right )}^m e^{im (\phi - \phi ')} \\ &= -\ln ({\varrho _>}^2) + \sum _{m = 1}^{\infty } \dfrac{1}{m}{\left (\dfrac{\varrho _<}{\varrho _>}\right )}^{m} \left ( e^{-im (\phi - \phi ')} + e^{im(\phi - \phi ')} \right ) \\ &= -\ln ({\varrho _>}^2) + 2\sum _{m = 1}^{\infty } \dfrac{1}{m}{\left (\dfrac{\varrho _<}{\varrho _>}\right )}^{m} \cos \left [m (\phi - \phi ')\right ] \end{align*}

Where the absolute values preserve the relations on \(\varrho _>, \varrho _<\)

Two dimensional electric quadrupole focusing fields for particle accelerators can be modeled by a set of four symmetrically placed line charges, with linear charge densities \(\pm \lambda \) as shown in the left hand figure (the right-hand figure shows the electric field lines)

The charge density in two dimensions can be expressed as \begin{align*} \sigma (\varrho , \phi ) = \frac{\lambda }{a} \sum _{n=0}^3{\left (-1\right )}^n \delta (\varrho - a) \delta \left (\phi - \frac{n \pi }{2}\right ) \end{align*}

(a)

Using the Green function expansion from Problem 2.17c, show that the electrostatic potential is \begin{align*} \Phi (\varrho , \phi ) = \frac{\lambda }{\pi \epsilon _0} \sum _{k=0}^\infty \frac{1}{2 k + 1}{\left ( \frac{\varrho _<}{\varrho _>} \right )}^{4k + 2} \cos \left [(4k + 2) \phi \right ] \end{align*}

Solution. \begin{align*} \Phi (\varrho , \phi ) &= \frac{1}{4 \pi \epsilon _0}\int _\Omega \sigma (\varrho ', \phi ')G(\varrho , \phi ; \varrho ', \phi ') \varrho d\varrho d\phi \\ &= \frac{\lambda }{4 \pi \epsilon _0} \sum _{n = 0}^{3}{(-1)}^n \left \{ -\ln (\varrho _>^2) + 2 \sum _{m = 1}^{\infty }\frac{1}{m}{\left ( \frac{\varrho _<}{\varrho _>} \right )}^m \cos \left [m (\phi - \frac{n \pi }{2})\right ] \right \} \\ &= \frac{\lambda }{2 \pi \epsilon _0} \sum _{m = 1}^{\infty }\frac{1}{m}{\left ( \frac{\varrho _<}{\varrho _>} \right )}^m \sum _{n = 0}^{3}{(-1)}^n \cos \left [m (\phi - \frac{n \pi }{2})\right ] \end{align*}

Since the \(\ln \) term cancels after summing over \(n\)
Expanding the second sum yields: \begin{align*} \sum _{n = 0}^{3}{(-1)}^n &\cos \left [m (\phi - \frac{n \pi }{2})\right ] \\ &= \cos (m \phi ) - \cos \left (m \phi - \frac{m \pi }{2}\right ) + \cos \left (m \phi - m \pi \right ) - \cos \left (m \phi - \frac{3 m \pi }{2}\right ) \\ &= \cos (m\phi ) - \cos (m\phi )\cos \left (\frac{m\pi }{2}\right ) + \sin (m \phi )\sin \left (\frac{m \pi }{2}\right ) + \cdots \end{align*}

The first two terms show that \(m\) must be a multiple of 2, and the second two show that it must be an odd multiple of 2 (otherwise the sum is 0): \begin{align*} \sum _{n = 0}^{3}{(-1)}^n \cos \left [m (\phi - \frac{n \pi }{2})\right ] = 2 \cos (m \phi ) \\ \text{for } m = 2 (2k + 1) = 4k + 2 \end{align*}

\begin{align*} \Phi (\varrho , \phi ) &= \frac{\lambda }{\pi \epsilon _0} \sum _{k = 0}^{\infty }\frac{1}{2k + 1}{\left ( \frac{\varrho _<}{\varrho _>} \right )}^{4k + 2} \cos \left [ (4k + 2) \phi \right ] \end{align*}

(b)

Relate the solution of part a to the real part of the complex function \begin{align*} w(z) = \frac{2 \lambda }{4 \pi \epsilon _0} \ln \left [ \frac{(z-ia)(z+ia)}{(z-a)(z+a)} \right ] \end{align*}

where \(z = x + iy = \varrho e^{i \phi }\). Comment on the connection to Problem 2.3

Solution. \begin{align*} \Phi (\varrho , \phi ) &= \frac{\lambda }{\pi \epsilon _0} \sum _{k = 0}^{\infty }\frac{1}{2k + 1}{\left ( \frac{\varrho _<}{\varrho _>} \right )}^{4k + 2} \cos \left [ (4k + 2) \phi \right ] \\ &= \Re \left \{ \frac{2\lambda }{ \pi \epsilon _0} \sum _{k = 0}^{\infty }\frac{1}{4k + 2}{\left ( \frac{\varrho _<}{\varrho _>} e^{i\phi } \right )}^{4k + 2} \right \} \end{align*}

Since \begin{align*} \sum _{n \text{ odd}} \frac{Z^n}{n} = \frac{1}{2}\ln \left (\frac{1 + Z}{1 - Z}\right ) \end{align*}

\begin{align*} \Rightarrow \Phi (\varrho , \phi ) &= \frac{\lambda }{2 \pi \epsilon _0} \Re \left \{ \ln \left [ \frac{ 1 +{\left ( \dfrac{\varrho _>}{\varrho _<}e^{i\phi } \right )}^2 }{ 1 -{\left ( \dfrac{\varrho _>}{\varrho _<}e^{i\phi } \right )}^2 } \right ] \right \} \\ &= \frac{\lambda }{2 \pi \epsilon _0} \Re \left \{ \ln \left [ \frac{ \left (1 + i\dfrac{\varrho _>}{\varrho _<}e^{i\phi }\right ) \left (1 - i\dfrac{\varrho _>}{\varrho _<}e^{i\phi }\right ) }{ \left (1 + \dfrac{\varrho _>}{\varrho _<}e^{i\phi }\right ) \left (1 - \dfrac{\varrho _>}{\varrho _<}e^{i\phi }\right ) } \right ] \right \} \end{align*}

The interior solution has \(\varrho _< = \varrho \) and \(\varrho _> = a\) so the solution becomes \begin{align*} \Phi (\varrho , \phi ) &= \frac{\lambda }{2 \pi \epsilon _0} \Re \left \{ \ln \left [ \frac{ \left (\varrho + iae^{i\phi }\right ) \left (\varrho - iae^{i\phi }\right ) }{ \left (\varrho + ae^{i\phi }\right ) \left (\varrho - ae^{i\phi }\right ) } \right ] \right \} \\ &= \Re \left [ w(\varrho ) \right ] \end{align*}

The exterior solution has \(\varrho _> = \varrho \) and \(\varrho _< = a\) so the solution becomes \begin{align*} \Phi (\varrho , \phi ) &= \frac{\lambda }{2 \pi \epsilon _0} \Re \left \{ \ln \left [ \frac{ \left (i\varrho + ae^{i\phi }\right ) \left (i\varrho - ae^{i\phi }\right ) }{ \left (\varrho + ae^{i\phi }\right ) \left (\varrho - ae^{i\phi }\right ) } \right ] \right \} \end{align*}

Multiply the fraction by \(i^2\) to obtain \(\Phi = \Re \left [w(\varrho )\right ]\)
This is related to problem 2.3 since that problem can be solved with the original line charge and 3 image charges, corresponding to the 4 line charges surrounding the accelerator. Simply take one of the line charges to be at \((x_0, y_0)\) where \(x_0 = y_0\)

(c)

Find expressions for the Cartesian components of the electric field near the origin, expressed in terms of x and y. Keep the \(k = 0\) and \(k = 1\) terms in the expansion. For \(y = 0\) what is the relative magnitude of the \(k = 1\) (\(2^6\)-pole) contribution to \(E_x\) compared to the \(k=0\) (\(2^2\)-pole or quadrupole) term?

Solution. The Cartesian components of the electric field are given by \begin{align*} E_x = -\cos{\theta } \frac{\partial \Phi }{\partial \varrho } + \frac{\sin{\theta }}{\varrho }\frac{\partial \Phi }{\partial \phi } \\ E_y = -\sin{\theta } \frac{\partial \Phi }{\partial \varrho } - \frac{\cos{\theta }}{\varrho }\frac{\partial \Phi }{\partial \phi } \end{align*}

For \(\varrho < a\) we have \begin{align*} \frac{\partial \Phi }{\partial \varrho } &= \frac{2 \lambda }{a \pi \epsilon _0} \sum _{k=0}^{\infty }{\left (\frac{\varrho }{a}\right )}^{4k+1}\cos \left [\phi (4k + 2)\right ] \\ \frac{1}{\varrho }\frac{\partial \Phi }{\partial \phi } &= -\frac{2 \lambda }{a \pi \epsilon _0} \sum _{k=0}^{\infty }{\left (\frac{\varrho }{a}\right )}^{4k+1}\sin \left [\phi (4k + 2)\right ] \end{align*}

Substituting into the original expressions for the components of \(E\): \begin{align*} E_x &= -\frac{2 \lambda }{a \pi \epsilon _0} \sum _{k=0}^{\infty }{\left ( \frac{\varrho }{a} \right )}^{4k + 1} \left \{ \cos \left [\phi \left (4k+2\right )\right ]\cos \phi + \sin \left [\phi \left (4k+2\right )\right ]\sin \phi \right \} \\ &= \frac{2 \lambda }{a \pi \epsilon _0} \sum _{k=0}^{\infty }{\left ( \frac{\varrho }{a} \right )}^{4k + 1} \cos \left [\phi \left (4k+1\right )\right ] \\ E_y &= -\frac{2 \lambda }{a \pi \epsilon _0} \sum _{k=0}^{\infty }{\left ( \frac{\varrho }{a} \right )}^{4k + 1} \left \{ -\sin \phi \cos \left [\phi \left (4k+2\right )\right ] - \cos \phi \sin \left [\phi \left (4k+2\right )\right ] \right \} \\ &= \frac{2 \lambda }{a \pi \epsilon _0} \sum _{k=0}^{\infty }{\left ( \frac{\varrho }{a} \right )}^{4k + 1} \sin \left [\phi \left (4k+3\right )\right ] \end{align*}

Up to k = 1, this yields: \begin{align*} E_x &= \frac{2 \lambda }{a \pi \epsilon _0} \left [ \frac{\varrho }{a}\cos \phi +{\left (\frac{\varrho }{a}\right )}^5\cos 5\phi \right ] \\ E_y &= \frac{2 \lambda }{a \pi \epsilon _0} \left [ \frac{\varrho }{a}\sin 3\phi +{\left (\frac{\varrho }{a}\right )}^5\sin 7\phi \right ] \end{align*}

For \(y = 0\), \(\phi = 0, \pi \): \begin{align*} E_x &= \pm \frac{2 \lambda }{a \pi \epsilon _0} \left [ \frac{\varrho }{a} \pm{\left (\frac{\varrho }{a}\right )}^5 \right ] \end{align*}

The relative strength of the \(k=0\) and \(k=1\) terms is \(\varrho ^4\)