Ellipsoid Space Charge Model for Electron Beam Dynamics Simulations

Abstract

The pulsed beam current in linear accelerators can approach significant values up to tens of Amperes. Such high currents cause many specific adverse effects in the accelerators. One of such effects is the repelling forces of the space charge as they become comparable to the forces of the electromagnetic accelerating fields, and can influence the stability of phase and radial particle motion. In the numerical analysis of the beam dynamics in linear accelerators, it is necessary to choose one of the different space charge models depending on the desired accuracy, complexity, speed and computer resources availability. The most accurate results are achieved by the numerical solution of the Poisson equation. However, this method may require significant resources and time and may not be suitable when fast analysis is required in the linac design stages. Analytical methods, on the other hands, base on the analytical solution of Poisson equations for the pre-defined shape of the particles distribution inside the beam. One of the most popular analytical space charge models is the ellipsoidal beam approximations. Despite being a well-developed model, many published approaches lack some important features as fully three-dimensional ellipsoid asymmetry and multi-bunch model. In this paper, we will derive the equations of the space charge field for the full-3D non-relativistic ellipsoid bunch step by step, starting from the Poisson equation, and compare this model with the other known models.

Appendices

APPENDIX A

1.1 INTEGRALS FOR THE “COMPRESSED” ELLIPSOID

The integral I₀ can be calculated by the following way:

$$\begin{gathered} {{I}_{0}} = \int\limits_\lambda ^\infty {\frac{{d\xi }}{{({{a}^{2}} + \xi )\sqrt {{{c}^{2}} + \xi } }}} \\ = c\int\limits_\lambda ^\infty {\frac{{d({\xi \mathord{\left/ {\vphantom {\xi {{{c}^{2}}}}} \right. \kern-0em} {{{c}^{2}}}})}}{{({{a}^{2}} + {{{{c}^{2}}\xi } \mathord{\left/ {\vphantom {{{{c}^{2}}\xi } {{{c}^{2}}}}} \right. \kern-0em} {{{c}^{2}}}})\sqrt {1 + {\xi \mathord{\left/ {\vphantom {\xi {{{c}^{2}}}}} \right. \kern-0em} {{{c}^{2}}}}} }}} \\ = c\int\limits_{\lambda /{{c}^{2}}}^\infty {\frac{{dt}}{{({{a}^{2}} + {{c}^{2}}t)\sqrt {1 + t} }}.} \\ \end{gathered} $$

(A1)

After the variable substitution

$$y = \sqrt {1 + t} .$$

(A2)

We get

$$\begin{gathered} {{I}_{0}} = c\int\limits_{\sqrt {1 + \lambda /{{c}^{2}}} }^\infty {\frac{{2ydy}}{{[{{a}^{2}} + {{c}^{2}}({{y}^{2}} - 1)]y}}} \\ = \frac{{2c}}{{{{a}^{2}} - {{c}^{2}}}}\int\limits_{\sqrt {1 + \lambda /{{c}^{2}}} }^\infty {\frac{{dy}}{{1 + \frac{{{{c}^{2}}}}{{{{a}^{2}} - {{c}^{2}}}}{{y}^{2}}}}} . \\ \end{gathered} $$

(A3)

Next, the result depends on the fact whether the ellipsoid is “compressed” (a > c) or “stretched” along the rotation axis. In case of the “compressed ellipsoid, the multiplier before y² in the integral function is positive and thus:

$$\begin{gathered} {{I}_{0}} = \frac{{2c}}{{{{a}^{2}} - {{c}^{2}}}}\sqrt {\frac{{{{a}^{2}} - {{c}^{2}}}}{{{{c}^{2}}}}} \int\limits_{\sqrt {1 + \lambda /{{c}^{2}}} }^\infty {\frac{{d\left( {\sqrt {\frac{{{{c}^{2}}}}{{{{a}^{2}} - {{c}^{2}}}}} y} \right)}}{{1 + {{{\left( {\sqrt {\frac{{{{c}^{2}}}}{{{{a}^{2}} - {{c}^{2}}}}} y} \right)}}^{2}}}}} \\ = \frac{2}{{\sqrt {{{a}^{2}} - {{c}^{2}}} }}\left. {\arctan \frac{c}{{\sqrt {{{a}^{2}} - {{c}^{2}}} }}y} \right|_{{\sqrt {1 + \lambda /{{c}^{2}}} }}^{\infty } \\ = \frac{2}{{\sqrt {{{a}^{2}} - {{c}^{2}}} }}\left( {\frac{\pi }{2} - \arctan \sqrt {\frac{{{{c}^{2}} + \lambda }}{{{{a}^{2}} - {{c}^{2}}}}} } \right) \\ = \frac{2}{{\sqrt {{{a}^{2}} - {{c}^{2}}} }}\arccos \sqrt {{{\frac{{{{c}^{2}} + \lambda }}{{{{a}^{2}} - {{c}^{2}}}}} \mathord{\left/ {\vphantom {{\frac{{{{c}^{2}} + \lambda }}{{{{a}^{2}} - {{c}^{2}}}}} {\left( {1 + \frac{{{{c}^{2}} + \lambda }}{{{{a}^{2}} - {{c}^{2}}}}} \right)}}} \right. \kern-0em} {\left( {1 + \frac{{{{c}^{2}} + \lambda }}{{{{a}^{2}} - {{c}^{2}}}}} \right)}}} . \\ \end{gathered} $$

(A4)

Or

$${{I}_{0}} = \frac{2}{{\sqrt {{{a}^{2}} - {{c}^{2}}} }}\arccos \sqrt {\frac{{{{c}^{2}} + \lambda }}{{{{a}^{2}} + \lambda }}} .$$

(A5)

Then, we’ll find I_xy for the compressed ellipsoid by applying the variables substation used for I₀

$$\begin{gathered} {{I}_{{xy}}} = \int\limits_\lambda ^\infty {\frac{{d\xi }}{{{{{({{a}^{2}} + \xi )}}^{2}}\sqrt {{{c}^{2}} + \xi } }}} \\ = c\int\limits_\lambda ^\infty {\frac{{d({\xi \mathord{\left/ {\vphantom {\xi {{{c}^{2}}}}} \right. \kern-0em} {{{c}^{2}}}})}}{{{{{({{a}^{2}} + {{c}^{2}}{\xi \mathord{\left/ {\vphantom {\xi {{{c}^{2}}}}} \right. \kern-0em} {{{c}^{2}}}})}}^{2}}\sqrt {1 + {\xi \mathord{\left/ {\vphantom {\xi {{{c}^{2}}}}} \right. \kern-0em} {{{c}^{2}}}}} }}} \\ = c\int\limits_{\lambda /{{c}^{2}}}^\infty {\frac{{dt}}{{{{{({{a}^{2}} + {{c}^{2}}t)}}^{2}}\sqrt {1 + t} }}} \\ = \frac{{2c}}{{{{{({{a}^{2}} - {{c}^{2}})}}^{2}}}}\int\limits_{\sqrt {1 + \lambda /{{c}^{2}}} }^\infty {\frac{{dy}}{{{{{\left( {1 + \frac{{{{c}^{2}}}}{{{{a}^{2}} - {{c}^{2}}}}{{y}^{2}}} \right)}}^{2}}}}} . \\ \end{gathered} $$

(A6)

Let’s introduce the following parameters for the compressed ellipsoid:

$$\alpha = {{{{c}^{2}}} \mathord{\left/ {\vphantom {{{{c}^{2}}} {({{a}^{2}} - {{c}^{2}})}}} \right. \kern-0em} {({{a}^{2}} - {{c}^{2}})}} > 0.$$

(A7)

And the integral I₁:

$$\begin{gathered} {{I}_{1}}(\alpha ) = \int {\frac{{dy}}{{1 + \alpha {{y}^{2}}}}} = \frac{1}{{\sqrt \alpha }}\int {\frac{{d(\sqrt \alpha y)}}{{1 + {{{(\sqrt \alpha y)}}^{2}}}}} \\ = \frac{1}{{\sqrt \alpha }}\arctan (\sqrt \alpha y). \\ \end{gathered} $$

(A8)

Then

$$\begin{gathered} - \frac{{\partial {{I}_{1}}(\alpha )}}{{\partial \alpha }} = \int {\frac{{{{y}^{2}}dy}}{{{{{(1 + \alpha {{y}^{2}})}}^{2}}}}} = \frac{1}{\alpha }\int {\frac{{[ - 1 + (1 + {{y}^{2}})]dy}}{{{{{(1 + \alpha {{y}^{2}})}}^{2}}}}} \\ = - \frac{1}{\alpha }\int {\frac{{dy}}{{{{{(1 + \alpha {{y}^{2}})}}^{2}}}} + \frac{1}{\alpha }\int {\frac{{dy}}{{1 + \alpha {{y}^{2}}}}} } , \\ \end{gathered} $$

(A9)

or in the other words,

$$\begin{gathered} \int {\frac{{dy}}{{{{{(1 + \alpha {{y}^{2}})}}^{2}}}}} = {{I}_{1}} + \alpha \frac{{\partial {{I}_{1}}(\alpha )}}{{\partial \alpha }} = \frac{1}{{\sqrt \alpha }}\arctan (\sqrt \alpha y) \\ + \,\,\alpha \frac{\partial }{{\partial \alpha }}\left[ {\frac{1}{{\sqrt \alpha }}\arctan (\sqrt \alpha y)} \right] = \frac{1}{{\sqrt \alpha }}\arctan (\sqrt \alpha y) \\ - \,\,\frac{1}{{2\sqrt \alpha }}\arctan (\sqrt \alpha y) + \frac{\alpha }{{\sqrt \alpha }}\frac{1}{{1 + \alpha {{y}^{2}}}}\frac{y}{{2\sqrt \alpha }} \\ = \frac{1}{{2\sqrt \alpha }}\left[ {\arctan (\sqrt \alpha y) + \frac{{\sqrt \alpha y}}{{1 + \alpha {{y}^{2}}}}} \right] \\ \end{gathered} $$

(A10)

and thus,

$$\begin{gathered} {{I}_{{xy}}} = \frac{{2c}}{{{{{({{a}^{2}} - {{c}^{2}})}}^{2}}}}\int\limits_{\sqrt {1 + \lambda /{{c}^{2}}} }^\infty {\frac{{dy}}{{{{{\left( {1 + \alpha {{y}^{2}}} \right)}}^{2}}}}} \\ = \frac{1}{{{{{({{a}^{2}} - {{c}^{2}})}}^{{3/2}}}}}\left[ {\underbrace {\frac{\pi }{2} - \arctan \sqrt {\frac{{{{c}^{2}} + \lambda }}{{{{a}^{2}} - {{c}^{2}}}}} }_{\arccos \sqrt {({{c}^{2}} + \lambda )/({{a}^{2}} + \lambda )} } - \frac{{\sqrt {\frac{{{{c}^{2}} + \lambda }}{{{{a}^{2}} - {{c}^{2}}}}} }}{{1 + \frac{{{{c}^{2}} + \lambda }}{{{{a}^{2}} - {{c}^{2}}}}}}} \right] \\ \end{gathered} $$

(A11)

or

$${{I}_{{xy}}} = \frac{1}{{{{{({{a}^{2}} - {{c}^{2}})}}^{{3/2}}}}}\left[ {\arccos \sqrt {\frac{{{{c}^{2}} + \lambda }}{{{{a}^{2}} + \lambda }}} - \frac{{\sqrt {({{c}^{2}} + \lambda )({{a}^{2}} - {{c}^{2}})} }}{{{{a}^{2}} + \lambda }}} \right].$$

(A12)

For the integral I_z, we will repeat the similar steps:

$$\begin{gathered} {{I}_{z}} = \int\limits_\lambda ^\infty {\frac{{d\xi }}{{({{a}^{2}} + \xi ){{{({{c}^{2}} + \xi )}}^{{3/2}}}}}} = \frac{1}{c}\int\limits_\lambda ^\infty {\frac{{d({\xi \mathord{\left/ {\vphantom {\xi {{{c}^{2}}}}} \right. \kern-0em} {{{c}^{2}}}})}}{{({{a}^{2}} + {{c}^{2}}{\xi \mathord{\left/ {\vphantom {\xi {{{c}^{2}}}}} \right. \kern-0em} {{{c}^{2}}}}){{{(1 + {\xi \mathord{\left/ {\vphantom {\xi {{{c}^{2}}}}} \right. \kern-0em} {{{c}^{2}}}})}}^{{3/2}}}}}} {\text{ }} = \frac{1}{c}\int\limits_{\lambda /{{c}^{2}}}^\infty {\frac{{dt}}{{({{a}^{2}} + {{c}^{2}}t){{{(1 + t)}}^{{3/2}}}}}} \\ = \frac{2}{c}\int\limits_{\sqrt {1 + \lambda /{{c}^{2}}} }^\infty {\frac{{dy}}{{\left[ {{{a}^{2}} + {{c}^{2}}({{y}^{2}} - 1)} \right]{{y}^{2}}}}} = \frac{2}{c}\int\limits_{\sqrt {1 + \lambda /{{c}^{2}}} }^\infty {\left( {\frac{1}{{{{y}^{2}}}} - \frac{{{{c}^{2}}}}{{{{a}^{2}} - {{c}^{2}} + {{c}^{2}}{{y}^{2}}}}} \right)} \frac{{dy}}{{{{a}^{2}} - {{c}^{2}}}} \\ = \frac{2}{{c({{a}^{2}} - {{c}^{2}})}}\left[ {\left. {\frac{1}{y}} \right|_{\infty }^{{\sqrt {1 + \lambda /{{c}^{2}}} }} - \frac{{{{c}^{2}}}}{{{{a}^{2}} - {{c}^{2}}}}\int\limits_{\sqrt {1 + \lambda /{{c}^{2}}} }^\infty {\frac{{dy}}{{1 + \frac{{{{c}^{2}}}}{{{{a}^{2}} - {{c}^{2}}}}{{y}^{2}}}}} } \right] = \frac{2}{{c({{a}^{2}} - {{c}^{2}})}}\left[ {\frac{c}{{\sqrt {{{c}^{2}} + \lambda } }} - \frac{{{{c}^{2}}}}{{{{a}^{2}} - {{c}^{2}}}}\frac{{\sqrt {{{a}^{2}} - {{c}^{2}}} }}{c}\frac{{^{{^{{^{{^{{}}}}}}}}}}{{_{{_{{_{{_{{_{{}}}}}}}}}}}}} \right. \\ \times \,\,\left. {\int\limits_{\sqrt {({{c}^{2}} + \lambda )/({{a}^{2}} - {{c}^{2}})} }^\infty {\frac{{dz}}{{1 + {{z}^{2}}}}} } \right] = \frac{{2c}}{{c({{a}^{2}} - {{c}^{2}})}}\left[ {\frac{1}{{\sqrt {{{c}^{2}} + \lambda } }} - \frac{1}{{\sqrt {{{a}^{2}} - {{c}^{2}}} }}\left( {\underbrace {\frac{\pi }{2} - \arctan \sqrt {\frac{{{{c}^{2}} + \lambda }}{{{{a}^{2}} - {{c}^{2}}}}} }_{\arccos \sqrt {({{c}^{2}} + \lambda )/({{a}^{2}} + \lambda )} }} \right)} \right] \\ \end{gathered} $$

(A13)

finally,

$${{I}_{z}} = \frac{2}{{{{{({{a}^{2}} - {{c}^{2}})}}^{{3/2}}}}}\left( {\sqrt {\frac{{{{a}^{2}} - {{c}^{2}}}}{{{{c}^{2}} + \lambda }}} - \arccos \sqrt {\frac{{{{c}^{2}} + \lambda }}{{{{a}^{2}} + \lambda }}} } \right).$$

(A14)

APPENDIX B

1.1 INTEGRALS FOR THE “STRETCHED” ELLIPSOID

The integral I₀ as show in in Appendix A (A3), is equal to:

$${{I}_{0}} = \frac{{2c}}{{{{a}^{2}} - {{c}^{2}}}}\int\limits_{\sqrt {1 + \lambda /{{c}^{2}}} }^\infty {\frac{{dy}}{{1 + \frac{{{{c}^{2}}}}{{{{a}^{2}} - {{c}^{2}}}}{{y}^{2}}}}} .$$

(B1)

For the stretched ellipsoid, c > a, so by introducing the parameter

$$\alpha = {{{{c}^{2}}} \mathord{\left/ {\vphantom {{{{c}^{2}}} {({{c}^{2}} - {{a}^{2}}) > 1}}} \right. \kern-0em} {({{c}^{2}} - {{a}^{2}}) > 1}}$$

(B2)

we have:

$$\begin{gathered} {{I}_{0}} = \frac{{2c}}{{{{a}^{2}} - {{c}^{2}}}}\int\limits_{\sqrt {1 + \lambda /{{c}^{2}}} }^\infty {\frac{{dy}}{{1 - {{{(\sqrt \alpha y)}}^{2}}}} = \frac{{2{{c}^{2}}}}{{c({{c}^{2}} - {{a}^{2}})}}} \int\limits_{\sqrt {1 + \lambda /{{c}^{2}}} }^\infty {\frac{{dy}}{{{{{(\sqrt \alpha y)}}^{2}} - 1}}} = \frac{{2\alpha }}{{c\sqrt \alpha }}\int\limits_{\sqrt {1 + \lambda /{{c}^{2}}} }^\infty {\frac{{d(\sqrt \alpha y)}}{2}} \left( {\frac{1}{{\sqrt \alpha y - 1}} - \frac{1}{{\sqrt \alpha y + 1}}} \right) \\ = \frac{1}{{\sqrt {{{c}^{2}} - {{a}^{2}}} }}\left. {\log \frac{{\sqrt \alpha y - 1}}{{\sqrt \alpha y + 1}}} \right|_{{\sqrt {1 + \lambda /{{c}^{2}}} }}^{\infty } = \frac{1}{{\sqrt {{{c}^{2}} - {{a}^{2}}} }}\log \frac{{c\sqrt {{{(1 + {\lambda \mathord{\left/ {\vphantom {\lambda {{{c}^{2}}}}} \right. \kern-0em} {{{c}^{2}}}})} \mathord{\left/ {\vphantom {{(1 + {\lambda \mathord{\left/ {\vphantom {\lambda {{{c}^{2}}}}} \right. \kern-0em} {{{c}^{2}}}})} {({{c}^{2}} - {{a}^{2}})}}} \right. \kern-0em} {({{c}^{2}} - {{a}^{2}})}}} + 1}}{{c\sqrt {{{(1 + {\lambda \mathord{\left/ {\vphantom {\lambda {{{c}^{2}}}}} \right. \kern-0em} {{{c}^{2}}}})} \mathord{\left/ {\vphantom {{(1 + {\lambda \mathord{\left/ {\vphantom {\lambda {{{c}^{2}}}}} \right. \kern-0em} {{{c}^{2}}}})} {({{c}^{2}} - {{a}^{2}})}}} \right. \kern-0em} {({{c}^{2}} - {{a}^{2}})}}} - 1}} \\ = \frac{1}{{\sqrt {{{c}^{2}} - {{a}^{2}}} }}\log \frac{{\sqrt {{{({{c}^{2}} + \lambda )} \mathord{\left/ {\vphantom {{({{c}^{2}} + \lambda )} {({{c}^{2}} - {{a}^{2}})}}} \right. \kern-0em} {({{c}^{2}} - {{a}^{2}})}}} + 1}}{{\sqrt {{{({{c}^{2}} + \lambda )} \mathord{\left/ {\vphantom {{({{c}^{2}} + \lambda )} {({{c}^{2}} - {{a}^{2}})}}} \right. \kern-0em} {({{c}^{2}} - {{a}^{2}})}}} - 1}} = \frac{1}{{\sqrt {{{c}^{2}} - {{a}^{2}}} }}\log \frac{{1 + \sqrt {{{({{c}^{2}} - {{a}^{2}})} \mathord{\left/ {\vphantom {{({{c}^{2}} - {{a}^{2}})} {({{c}^{2}} + \lambda )}}} \right. \kern-0em} {({{c}^{2}} + \lambda )}}} }}{{1 - \sqrt {{{({{c}^{2}} - {{a}^{2}})} \mathord{\left/ {\vphantom {{({{c}^{2}} - {{a}^{2}})} {({{c}^{2}} + \lambda )}}} \right. \kern-0em} {({{c}^{2}} + \lambda )}}} }}. \\ \end{gathered} $$

(B3)

Thus,

$${{I}_{0}} = \frac{1}{{\sqrt {{{c}^{2}} - {{a}^{2}}} }}\log \frac{{1 + \sqrt {{{({{c}^{2}} - {{a}^{2}})} \mathord{\left/ {\vphantom {{({{c}^{2}} - {{a}^{2}})} {({{c}^{2}} + \lambda )}}} \right. \kern-0em} {({{c}^{2}} + \lambda )}}} }}{{1 - \sqrt {{{({{c}^{2}} - {{a}^{2}})} \mathord{\left/ {\vphantom {{({{c}^{2}} - {{a}^{2}})} {({{c}^{2}} + \lambda )}}} \right. \kern-0em} {({{c}^{2}} + \lambda )}}} }}.$$

(B4)

Then, by using the expression (A5), we find

$${{I}_{{xy}}} = \int\limits_\lambda ^\infty {\frac{{d\xi }}{{{{{({{a}^{2}} + \xi )}}^{2}}\sqrt {{{c}^{2}} + \xi } }}} = \frac{{2c}}{{{{{({{a}^{2}} - {{c}^{2}})}}^{2}}}}\int\limits_{\sqrt {1 + \lambda /{{c}^{2}}} }^\infty {\frac{{dy}}{{{{{\left( {1 + \frac{{{{c}^{2}}}}{{{{a}^{2}} - {{c}^{2}}}}{{y}^{2}}} \right)}}^{2}}}}} $$

(B5)

and for the stretched ellipsoid:

$${{I}_{{xy}}} = \frac{{2c}}{{{{{({{c}^{2}} - {{a}^{2}})}}^{2}}}}\int\limits_{\sqrt {1 + \lambda /{{c}^{2}}} }^\infty {\frac{{dy}}{{{{{\left( {\frac{{{{c}^{2}}}}{{{{a}^{2}} - {{c}^{2}}}}{{y}^{2}} - 1} \right)}}^{2}}}}} = \frac{{2{{\alpha }^{2}}}}{{{{c}^{3}}\sqrt \alpha }}\int\limits_{\sqrt {1 + \lambda /{{c}^{2}}} }^\infty {\frac{{d(\sqrt \alpha y)}}{{{{{[{{{(\underbrace {\sqrt \alpha y}_x)}}^{2}} - 1]}}^{2}}}}} = \frac{2}{{{{{({{c}^{2}} - {{a}^{2}})}}^{{3/2}}}}}\int\limits_{\sqrt {({{c}^{2}} + \lambda )/({{c}^{2}} - {{a}^{2}})} }^\infty {\frac{{dx}}{{{{{({{x}^{2}} - 1)}}^{2}}}}} .$$

(B6)

To calculate this integral, let’s introduce the other integral:

$${{I}_{1}}(\beta ) = \int {\frac{{dx}}{{{{x}^{2}} - {{\beta }^{2}}}} = \frac{1}{{2\beta }}\log \frac{{x + \beta }}{{x - \beta }}} .$$

(B7)

Then

$$\begin{gathered} \int {\frac{{dx}}{{{{{({{x}^{2}} - {{\beta }^{2}})}}^{2}}}} = \frac{\partial }{{\partial {{\beta }^{2}}}}} \int {\frac{{dx}}{{{{x}^{2}} - {{\beta }^{2}}}}} = \frac{1}{{2\beta }}\frac{\partial }{{\partial \beta }}\left( {\frac{1}{{2\beta }}\log \frac{{x + \beta }}{{x - \beta }}} \right) \\ = - \frac{1}{{4{{\beta }^{2}}}}\log \frac{{x + \beta }}{{x - \beta }} + \frac{1}{{4{{\beta }^{2}}}}\frac{{x - \beta }}{{x + \beta }}\frac{{(x - \beta ) - (x + \beta )( - 1)}}{{{{{(x - \beta )}}^{2}}}} = \frac{1}{{4{{\beta }^{2}}}}\left( {\log \frac{{x - \beta }}{{x + \beta }} - \frac{{2x}}{{{{x}^{2}} - {{\beta }^{2}}}}} \right), \\ \end{gathered} $$

(B8)

thus,

$$\begin{gathered} {{I}_{{xy}}} = \frac{2}{{{{{({{c}^{2}} - {{a}^{2}})}}^{{3/2}}}}}\int\limits_{\sqrt {({{c}^{2}} + \lambda )/({{c}^{2}} - {{a}^{2}})} }^\infty {\frac{{dx}}{{{{{({{x}^{2}} - 1)}}^{2}}}}} = \frac{1}{{2{{{({{c}^{2}} - {{a}^{2}})}}^{{3/2}}}}}\left. {\left( {\log \frac{{x - 1}}{{x + 1}} - \frac{2}{{{{x}^{2}} - 1}}} \right)} \right|_{{\sqrt {({{c}^{2}} + \lambda )/({{c}^{2}} - {{a}^{2}})} }}^{\infty } \\ = \frac{1}{{{{{({{c}^{2}} - {{a}^{2}})}}^{{3/2}}}}}\left( {\frac{{\sqrt {{{({{c}^{2}} + \lambda )} \mathord{\left/ {\vphantom {{({{c}^{2}} + \lambda )} {({{c}^{2}} - {{a}^{2}})}}} \right. \kern-0em} {({{c}^{2}} - {{a}^{2}})}}} }}{{{{({{c}^{2}} + \lambda )} \mathord{\left/ {\vphantom {{({{c}^{2}} + \lambda )} {({{c}^{2}} - {{a}^{2}})}}} \right. \kern-0em} {({{c}^{2}} - {{a}^{2}})}} - 1}}} \right.\left. { + \,\,\frac{1}{2}\log \frac{{\sqrt {{{({{c}^{2}} + \lambda )} \mathord{\left/ {\vphantom {{({{c}^{2}} + \lambda )} {({{c}^{2}} - {{a}^{2}})}}} \right. \kern-0em} {({{c}^{2}} - {{a}^{2}})}}} - 1}}{{\sqrt {{{({{c}^{2}} + \lambda )} \mathord{\left/ {\vphantom {{({{c}^{2}} + \lambda )} {({{c}^{2}} - {{a}^{2}})}}} \right. \kern-0em} {({{c}^{2}} - {{a}^{2}})}}} + 1}}} \right) \\ \end{gathered} $$

(B9)

or

$${{I}_{{xy}}} = \frac{1}{{{{{({{c}^{2}} - {{a}^{2}})}}^{{3/2}}}}}\left( {\frac{{\sqrt {({{c}^{2}} + \lambda )({{c}^{2}} - {{a}^{2}})} }}{{{{a}^{2}} + \lambda }}} \right.\left. { - \,\,\frac{1}{2}\log \frac{{1 + \sqrt {{{({{c}^{2}} - {{a}^{2}})} \mathord{\left/ {\vphantom {{({{c}^{2}} - {{a}^{2}})} {({{c}^{2}} + \lambda )}}} \right. \kern-0em} {({{c}^{2}} + \lambda )}}} }}{{1 - \sqrt {{{({{c}^{2}} - {{a}^{2}})} \mathord{\left/ {\vphantom {{({{c}^{2}} - {{a}^{2}})} {({{c}^{2}} + \lambda )}}} \right. \kern-0em} {({{c}^{2}} + \lambda )}}} }}} \right).$$

(B10)

Finally, using the intermediate result from the Appendix A, when calculated the integral I_z, we have:

$$\begin{gathered} {{I}_{z}} = \int\limits_\lambda ^\infty {\frac{{d\xi }}{{({{a}^{2}} + \xi ){{{({{c}^{2}} + \xi )}}^{{3/2}}}}}} = \frac{2}{{c({{a}^{2}} - {{c}^{2}})}}\left[ {\left. {\frac{1}{y}} \right|_{\infty }^{{\sqrt {1 + \lambda /{{c}^{2}}} }} - \frac{{{{c}^{2}}}}{{{{a}^{2}} - {{c}^{2}}}}\int\limits_{\sqrt {1 + \lambda /{{c}^{2}}} }^\infty {\frac{{dy}}{{1 + \frac{{{{c}^{2}}}}{{{{a}^{2}} - {{c}^{2}}}}{{y}^{2}}}}} } \right] = \frac{2}{{({{a}^{2}} - {{c}^{2}})}} \\ \times \,\left[ {\frac{1}{{\sqrt {{{c}^{2}} + \lambda } }} - \frac{c}{{{{c}^{2}} - {{a}^{2}}}}\int\limits_{\sqrt {1 + \lambda /{{c}^{2}}} }^\infty {\frac{{dy}}{{1 - \frac{{{{c}^{2}}}}{{{{c}^{2}} - {{a}^{2}}}}{{y}^{2}}}}} } \right] = \frac{2}{{({{a}^{2}} - {{c}^{2}})}}\left[ {\frac{1}{{\sqrt {{{c}^{2}} + \lambda } }} + \frac{1}{{\sqrt {{{c}^{2}} - {{a}^{2}}} }}\frac{{^{{^{{}}}}}}{{_{{_{{_{{_{{_{{}}}}}}}}}}}}} \right. \\ \times \,\,\left. {\int\limits_{\sqrt {1 + \lambda /{{c}^{2}}} }^\infty {\frac{{d{{(c} \mathord{\left/ {\vphantom {{(c} {\sqrt {{{c}^{2}} - {{a}^{2}}} }}} \right. \kern-0em} {\sqrt {{{c}^{2}} - {{a}^{2}}} }}y)}}{{{{{({c \mathord{\left/ {\vphantom {c {\sqrt {{{c}^{2}} - {{a}^{2}}} }}} \right. \kern-0em} {\sqrt {{{c}^{2}} - {{a}^{2}}} }}y)}}^{2}} - 1}}} } \right] = (\alpha = \frac{{{{c}^{2}}}}{{{{c}^{2}} - {{a}^{2}}}}) = - \frac{2}{{{{{({{c}^{2}} - {{a}^{2}})}}^{{3/2}}}}}\left[ {\sqrt {\frac{{{{c}^{2}} - {{a}^{2}}}}{{{{c}^{2}} + \lambda }}} + \int\limits_{\sqrt {1 + \lambda /{{c}^{2}}} }^\infty {\frac{{d(\sqrt \alpha y)}}{{{{{(\sqrt \alpha y)}}^{2}} - 1}}} } \right] \\ = - \frac{2}{{{{{({{c}^{2}} - {{a}^{2}})}}^{{3/2}}}}}\left[ {\sqrt {\frac{{{{c}^{2}} - {{a}^{2}}}}{{{{c}^{2}} + \lambda }}} \frac{{^{{^{{^{{^{{}}}}}}}}}}{{_{{_{{}}}}}}} \right.\left. { + \,\,\int\limits_{\sqrt {1 + \lambda /{{c}^{2}}} }^\infty {\frac{{d(\sqrt \alpha y)}}{2}\left( {\frac{1}{{\sqrt \alpha y - 1}} - \frac{1}{{\sqrt \alpha y + 1}}} \right)} } \right] = - \frac{2}{{{{{({{c}^{2}} - {{a}^{2}})}}^{{3/2}}}}} \\ \times \,\,\left[ {\sqrt {\frac{{{{c}^{2}} - {{a}^{2}}}}{{{{c}^{2}} + \lambda }}} + \frac{1}{2}\left. {\log \frac{{\sqrt \alpha y - 1}}{{\sqrt \alpha y + 1}}} \right|_{{\sqrt {1 + \lambda /{{c}^{2}}} }}^{\infty }} \right] = \frac{2}{{{{{({{c}^{2}} - {{a}^{2}})}}^{{3/2}}}}}\left[ { - \sqrt {\frac{{{{c}^{2}} - {{a}^{2}}}}{{{{c}^{2}} + \lambda }}} + \frac{1}{2}\log \frac{{\sqrt {c{{(1 + {\lambda \mathord{\left/ {\vphantom {\lambda {{{c}^{2}}}}} \right. \kern-0em} {{{c}^{2}}}})} \mathord{\left/ {\vphantom {{(1 + {\lambda \mathord{\left/ {\vphantom {\lambda {{{c}^{2}}}}} \right. \kern-0em} {{{c}^{2}}}})} {({{c}^{2}} - {{a}^{2}})}}} \right. \kern-0em} {({{c}^{2}} - {{a}^{2}})}}} + 1}}{{\sqrt {cc{{(1 + {\lambda \mathord{\left/ {\vphantom {\lambda {{{c}^{2}}}}} \right. \kern-0em} {{{c}^{2}}}})} \mathord{\left/ {\vphantom {{(1 + {\lambda \mathord{\left/ {\vphantom {\lambda {{{c}^{2}}}}} \right. \kern-0em} {{{c}^{2}}}})} {({{c}^{2}} - {{a}^{2}})}}} \right. \kern-0em} {({{c}^{2}} - {{a}^{2}})}}} - 1}}} \right] \\ \end{gathered} $$

(B11)

or

$${{I}_{z}} = \frac{2}{{{{{({{c}^{2}} - {{a}^{2}})}}^{{3/2}}}}}\left[ {\frac{1}{2}\log \frac{{1 + \sqrt {{{({{c}^{2}} - {{a}^{2}})} \mathord{\left/ {\vphantom {{({{c}^{2}} - {{a}^{2}})} {({{c}^{2}} + \lambda )}}} \right. \kern-0em} {({{c}^{2}} + \lambda )}}} }}{{1 - \sqrt {{{({{c}^{2}} - {{a}^{2}})} \mathord{\left/ {\vphantom {{({{c}^{2}} - {{a}^{2}})} {({{c}^{2}} + \lambda )}}} \right. \kern-0em} {({{c}^{2}} + \lambda )}}} }} - \sqrt {\frac{{{{c}^{2}} - {{a}^{2}}}}{{{{c}^{2}} + \lambda }}} } \right].$$

(B12)

APPENDIX C

1.1 TRANSITION FROM A COMPRESSED ELLIPSOID TO SPHERE

Let’s calculate the free term K₁ in the expression (48) for the potential outside the compressed ellipsoid in transition from ellipsoid to sphere and the coefficients K₂ and K₃ in the same equation. Assuming

$$c = a - \delta \,\,\,\,{\text{and}}\,\,\,\,\bar {\delta } = {\delta \mathord{\left/ {\vphantom {\delta {2a}}} \right. \kern-0em} {2a}} \ll 1$$

(C1)

we can calculate K₁ as

$$\begin{gathered} {{K}_{1}} = \frac{{2\arccos \sqrt {\frac{{{{{(a - \delta )}}^{2}} + \lambda }}{{{{a}^{2}} + \lambda }}} }}{{\sqrt {{{a}^{2}} - {{{(a - \delta )}}^{2}}} }} = \frac{{2\arccos \sqrt {\frac{{{{a}^{2}} + \lambda - 2a\delta (1 - \bar {\delta })}}{{{{a}^{2}} + \lambda }}} }}{{\sqrt {2a\delta (1 - \bar {\delta })} }} \\ = \frac{{2{{{(1 - \bar {\delta })}}^{{ - 1/2}}}}}{{\sqrt {2a\delta } }}\arccos {{\left( {1 - 2a\delta \frac{{1 - \bar {\delta }}}{{{{a}^{2}} + \lambda }}} \right)}^{{1/2}}} \approx ({\text{Zero's order over }}\bar {\delta }{\text{)}} \approx \frac{2}{{\sqrt {2a\delta } }}\arccos \left( {1 - \frac{{a\delta }}{{{{a}^{2}} + \lambda }}} \right). \\ \end{gathered} $$

(C2)

However, for small α $\ll$ 1, arccos(1 – α) = t, and 1 – α = cost ≈ 1 – t²/2, then

$$\begin{gathered} {{K}_{1}} \approx \frac{2}{{\sqrt {2a\delta } }}\arccos \left( {1 - \frac{{a\delta }}{{{{a}^{2}} + \lambda }}} \right) \\ \approx \frac{2}{{\sqrt {2a\delta } }}\sqrt {\frac{{2a\delta }}{{{{a}^{2}} + \lambda }}} = \frac{2}{{\sqrt {{{a}^{2}} + \lambda } }}. \\ \end{gathered} $$

(C3)

Then, by using the substitutions and the relations, obtained during the calculation of K₁, we calculate K₂:

$$\begin{gathered} {{K}_{2}} = \frac{{\arccos \sqrt {\frac{{{{{(a - \delta )}}^{2}} + \lambda }}{{{{a}^{2}} + \lambda }}} - \frac{{\sqrt {[{{a}^{2}} - {{{(a - \delta )}}^{2}}][{{{(a - \delta )}}^{2}} + \lambda ]} }}{{{{a}^{2}} + \lambda }}}}{{{{{[{{a}^{2}} - {{{(a - \delta )}}^{2}}]}}^{{3/2}}}}} = \frac{1}{{{{{(2a\delta )}}^{{3/2}}}{{{(1 - \bar {\delta })}}^{{3/2}}}}} \\ \times \,\,\left[ {\arccos {{{\left( {1 - 2a\delta \frac{{1 - \bar {\delta }}}{{{{a}^{2}} + \lambda }}} \right)}}^{{1/2}}} - \frac{{\sqrt {2a\delta (1 - \bar {\delta })[{{a}^{2}} + \lambda - 2a\delta (1 - \bar {\delta })]} }}{{{{a}^{2}} + \lambda }}} \right] \\ \approx \frac{{{{{(1 - \bar {\delta })}}^{{ - 3/2}}}}}{{{{{(2a\delta )}}^{{3/2}}}}}\left\{ {\arccos \left[ {1\underbrace { - a\delta \frac{{1 - \bar {\delta }}}{{{{a}^{2}} + \lambda }} - \frac{{{{a}^{2}}{{\delta }^{2}}{{{(1 - \bar {\delta })}}^{2}}}}{{2{{{({{a}^{2}} + \lambda )}}^{2}}}}}_{ = - \alpha }} \right]\sqrt {\frac{{2a\delta }}{{{{a}^{2}} + \lambda }}} {{{(1 - \bar {\delta })}}^{{1/2}}}{{{\left( {1 - 2a\delta \frac{{1 - \bar {\delta }}}{{{{a}^{2}} + \lambda }}} \right)}}^{{1/2}}}} \right\}. \\ \end{gathered} $$

(C4)

For small α $\ll$ 1, arccos(1 – α) = t, but now we will keep the quadratic terms, so that 1 – α = cost ≈ 1 – t²/2 + t⁴/24, and

$$\begin{gathered} {{K}_{2}} \approx \frac{{{{{(1 - \bar {\delta })}}^{{ - 3/2}}}}}{{{{{(2a\delta )}}^{{3/2}}}}}\left\{ {\sqrt {2a\delta \frac{{1 - \bar {\delta }}}{{{{a}^{2}} + \lambda }} + \frac{{{{a}^{2}}{{\delta }^{2}}{{{(1 - \bar {\delta })}}^{2}}}}{{{{{({{a}^{2}} + \lambda )}}^{2}}}}} } \right.\left[ {1 + \frac{1}{{12}}\left( {a\delta \frac{{1 - \bar {\delta }}}{{{{a}^{2}} + \lambda }} + \frac{{{{a}^{2}}{{\delta }^{2}}{{{(1 - \bar {\delta })}}^{2}}}}{{2{{{({{a}^{2}} + \lambda )}}^{2}}}}} \right)} \right] \\ - \,\,\left. {\sqrt {\frac{{2a\delta }}{{{{a}^{2}} + \lambda }}} {{{(1 - \bar {\delta })}}^{{1/2}}}{{{\left( {1 - 2a\delta \frac{{1 - \bar {\delta }}}{{{{a}^{2}} + \lambda }}} \right)}}^{{1/2}}}} \right\} \\ \approx \frac{{{{{(1 - \bar {\delta })}}^{{ - 3/2}}}}}{{{{{(2a\delta )}}^{{3/2}}}}}\sqrt {\frac{{2a\delta }}{{{{a}^{2}} + \lambda }}} {{(1 - \bar {\delta })}^{{1/2}}}\left\{ {{{{\left[ {1 + \frac{{a\delta (1 - \bar {\delta })}}{{2({{a}^{2}} + \lambda )}}} \right]}}^{{1/2}}}\left( {1 + \frac{{a\delta }}{{12}}\frac{{1 - \bar {\delta }}}{{{{a}^{2}} + \lambda }}} \right) - {{{\left( {1 - 2a\delta \frac{{1 - \bar {\delta }}}{{{{a}^{2}} + \lambda }}} \right)}}^{{1/2}}}} \right\} \\ \approx \frac{{1 - \bar {\delta }}}{{2a\delta \sqrt {{{a}^{2}} + \lambda } }}\left( {1 + \frac{{a\delta }}{4}\frac{{1 - \bar {\delta }}}{{{{a}^{2}} + \lambda }} + \frac{{a\delta }}{{12}}\frac{{1 - \bar {\delta }}}{{{{a}^{2}} + \lambda }} - 1} \right.\left. { + \,\,a\delta \frac{{1 - \bar {\delta }}}{{{{a}^{2}} + \lambda }}} \right) \approx \frac{{1 - \bar {\delta }}}{{2a\delta \sqrt {{{a}^{2}} + \lambda } }}\frac{4}{3}a\delta \frac{{1 - \bar {\delta }}}{{{{a}^{2}} + \lambda }}, \\ \end{gathered} $$

(C5)

or

$${{K}_{2}} \approx \frac{{{2 \mathord{\left/ {\vphantom {2 3}} \right. \kern-0em} 3}}}{{{{{({{a}^{2}} + \lambda )}}^{{3/2}}}}},$$

(C6)

finally, by using all these results and methods:

$$\begin{gathered} {{K}_{3}} = \frac{1}{{{{{[{{a}^{2}} - c{{{(a - \delta )}}^{2}}]}}^{{3/2}}}}}\left( {\arccos \sqrt {\frac{{{{{(a - \delta )}}^{2}} + \lambda }}{{{{a}^{2}} + \lambda }}} - \sqrt {\frac{{{{a}^{2}} - {{{(a - \delta )}}^{2}}}}{{{{{(a - \delta )}}^{2}} + \lambda }}} } \right) = \frac{{{{{(1 - \bar {\delta })}}^{{ - 3/2}}}}}{{{{{(2a\delta )}}^{{3/2}}}}} \\ \times \,\,\left[ {\arccos {{{\left( {1 - 2a\delta \frac{{1 - \bar {\delta }}}{{{{a}^{2}} + \lambda }}} \right)}}^{{1/2}}} - \frac{{\sqrt {2a\delta (1 - \bar {\delta })} }}{{{{a}^{2}} + \lambda - 2a\delta (1 - \bar {\delta })}}} \right] \approx \frac{{{{{(1 - \bar {\delta })}}^{{ - 3/2}}}}}{{{{{(2a\delta )}}^{{3/2}}}}} \\ \times \,\,\left\{ {\sqrt {2a\delta \frac{{1 - \bar {\delta }}}{{{{a}^{2}} + \lambda }} + \frac{{{{a}^{2}}{{\delta }^{2}}{{{(1 - \bar {\delta })}}^{2}}}}{{{{{({{a}^{2}} + \lambda )}}^{2}}}}} \left[ {1 + \frac{1}{{12}}\left( {a\delta \frac{{1 - \bar {\delta }}}{{{{a}^{2}} + \lambda }} + \frac{{{{a}^{2}}{{\delta }^{2}}{{{(1 - \bar {\delta })}}^{2}}}}{{2{{{({{a}^{2}} + \lambda )}}^{2}}}}} \right)} \right]} \right. - \sqrt {\frac{{2a\delta }}{{{{a}^{2}} + \lambda }}} \left. { - \,\,{{{(1 - \bar {\delta })}}^{{1/2}}}{{{\left( {1 - 2a\delta \frac{{1 - \bar {\delta }}}{{{{a}^{2}} + \lambda }}} \right)}}^{{ - 1/2}}}} \right\} \\ \approx \frac{{{{{(1 - \bar {\delta })}}^{{ - 3/2}}}}}{{{{{(2a\delta )}}^{{3/2}}}}}\sqrt {\frac{{2a\delta (1 - \bar {\delta })}}{{{{a}^{2}} + \lambda }}} \left[ {{{{\left( {1 + \frac{{a\delta (1 - \bar {\delta })}}{{2({{a}^{2}} + \lambda )}}} \right)}}^{{1/2}}}\left( {1 + \frac{{a\delta }}{{12}}\frac{{1 - \bar {\delta }}}{{{{a}^{2}} + \lambda }}} \right) - \left( {1 + a\delta \frac{{1 - \bar {\delta }}}{{{{a}^{2}} + \lambda }}} \right)} \right] \approx \frac{{1 - \bar {\delta }}}{{2a\delta \sqrt {{{a}^{2}} + \lambda } }} \\ \times \,\,\left[ {1 + \frac{{a\delta }}{4}\frac{{1 - \bar {\delta }}}{{{{a}^{2}} + \lambda }} + \frac{{a\delta }}{{12}}\frac{{1 - \bar {\delta }}}{{{{a}^{2}} + \lambda }} - 1 - a\delta \frac{{1 - \bar {\delta }}}{{{{a}^{2}} + \lambda }}} \right] = \frac{{1 - \bar {\delta }}}{{2a\delta \sqrt {{{a}^{2}} + \lambda } }}\left( { - \frac{2}{3}a\delta \frac{{1 - \bar {\delta }}}{{{{a}^{2}} + \lambda }}} \right) \approx \frac{{{{ - 1} \mathord{\left/ {\vphantom {{ - 1} 3}} \right. \kern-0em} 3}}}{{{{{({{a}^{2}} + \lambda )}}^{{3/2}}}}}. \\ \end{gathered} $$

(C7)

Or

$${{K}_{3}} = \frac{{{{ - 1} \mathord{\left/ {\vphantom {{ - 1} 3}} \right. \kern-0em} 3}}}{{{{{({{a}^{2}} + \lambda )}}^{{3/2}}}}}.$$

(C8)