Self-gravity in curved mesh elements

Abstract

The local character of self-gravity along with the number of spatial dimensions are critical issues when computing the potential and forces inside massive systems like stars and disks. This appears from the discretisation scale where each cell of the numerical grid is a self-interacting body in itself. There is apparently no closed-form expression yet giving the potential of a three-dimensional homogeneous cylindrical or spherical cell, in contrast with the Cartesian case. By using Green’s theorem, we show that the potential integral for such polar-type 3D sectors—initially, a volume integral with singular kernel—can be converted into a regular line-integral running over the lateral contour, thereby generalising a formula already known under axial symmetry. It therefore is a step towards the obtention of another potential/density pair. The new kernel is a finite function of the cell’s shape (with the simplest form in cylindrical geometry), and mixes incomplete elliptic integrals, inverse trigonometric and hyperbolic functions. The contour integral is easy to compute; it is valid in the whole physical space, exterior and interior to the sector itself and works in fact for a wide variety of shapes of astrophysical interest (e.g. sectors of tori or flared discs). This result is suited to easily providing reference solutions, and to reconstructing potential and forces in inhomogeneous systems by superposition. The contour integrals for the 3 components of the acceleration vector are explicitely given.

Appendices

Appendix A: Values of $F(\beta ,k)$ and $E(\beta ,k)$ for any amplitude $\beta$

The computation of $F(\beta ,k)$ must be performed with caution as soon as the amplitude $\beta$ stands outside the range $[0,\frac{\pi }{2}]$. It is in particular necessary to use the following rules:

$$\begin{aligned} \left\{ \begin{array}{ll} F(\beta ,k)=-F(-\beta ,k), \quad \hbox {if } \beta < 0,\\ F(\beta ,k)=n {\mathbf K}(k)+F\left( |\beta |-n\frac{\pi }{2},k\right) , \quad \hbox {if } |\beta |-n\frac{\pi }{2} \in [0,\frac{\pi }{2}] \hbox { and even } n,\\ F(\beta ,k)=n {\mathbf K}(k)-F\left( n\frac{\pi }{2}-|\beta |,k\right) ,\quad \hbox {if } n\frac{\pi }{2}-|\beta |\in [0,\frac{\pi }{2}] \hbox { and odd } n, \end{array}\right. \end{aligned}$$

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and

$$\begin{aligned} \left\{ \begin{array}{ll} E(\beta ,k)=-E(-\beta ,k), \quad \hbox {if } \beta < 0,\\ E(\beta ,k)=n {\mathbf E}(k)+E\left( |\beta |-n\frac{\pi }{2},k\right) , \quad \hbox {if } |\beta |-n\frac{\pi }{2} \in [0,\frac{\pi }{2}] \hbox { and even } n,\\ E(\beta ,k)=n {\mathbf E}(k)-E\left( n\frac{\pi }{2}-|\beta |,k\right) , \quad \hbox {if } n\frac{\pi }{2}-|\beta |\in [0,\frac{\pi }{2}] \hbox { and odd}\, n. \end{array}\right. \end{aligned}$$

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Appendix B: Derivation of $M$ and $N$ in the axially symmetrical case

In order to determine $M$ and $N$ in the following equation

$$\begin{aligned} \partial _a N - \partial _z M = 2 \sqrt{\frac{a}{R}} k {\mathbf K}(k), \end{aligned}$$

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we set without loss of generality:

$$\begin{aligned} \left\{ \begin{array}{l} M = \zeta k {\mathbf E}(k) f(a) + \zeta k {\mathbf K}(k) g(a),\\ N = a k {\mathbf E}(k) h(a,\zeta ) + a k {\mathbf K}(k) l(a,\zeta ), \end{array}\right. \end{aligned}$$

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where $f$, $g$, $h$ and $l$ are four functions to be determined. From Eqs. (18) and (19), and given (Gradshteyn and Ryzhik 2007):

$$\begin{aligned} \partial _k k{\mathbf E}(k) = 2 {\mathbf E}(k) - {\mathbf K}(k), \end{aligned}$$

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and

$$\begin{aligned} \partial _k k{\mathbf K}(k) = \frac{{\mathbf E}(k)}{{k'}^2}, \end{aligned}$$

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we get:

$$\begin{aligned} -\partial _z M&= k{\mathbf E}(k) f(a) + k{\mathbf K}(k) g(a) + \zeta \partial _\zeta k \left[ \left( 2{\mathbf E}(k) - {\mathbf K}(k) \right) f + g \frac{{\mathbf E}(k)}{{k'}^2} \right] \nonumber \\&= k{\mathbf E}(k) \left[ \left( 2\frac{k^2}{m^2}-1\right) f - \left( 1 - \frac{k^2}{m^2} \right) \frac{g}{{k'}^2} \right] + k{\mathbf K}(k) \left[ \left( 1 - \frac{k^2}{m^2} \right) f + g \right] \end{aligned}$$

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and

$$\begin{aligned} \partial _a N&= k{\mathbf E}(k) (ah' + h) + k{\mathbf K}(k) (al'+l) + a \partial _a k \left[ \left( 2{\mathbf E}(k) - {\mathbf K}(k) \right) h + l \frac{{\mathbf E}(k)}{{k'}^2} \right] \nonumber \\&= k{\mathbf E}(k) \left[ ah'+h + \left( 2h + \frac{l}{{k'}^2} \right) \left( \frac{1}{2} - \frac{a}{a+R}\frac{k^2}{m^2} \right) \right] \nonumber \\&\qquad + k{\mathbf K}(k) \left[ al'+l - h \left( \frac{1}{2} - \frac{a}{a+R} \frac{k^2}{m^2} \right) \right] . \end{aligned}$$

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Forming $\partial _a N-\partial _z M$, and gathering terms, we get

$$\begin{aligned} 2h + ah' - f - \frac{g}{{k'}^2} + \frac{l}{{2k'}^2} + \frac{k^2}{m^2} \left( 2f + \frac{g}{{k'}^2} -\frac{2a}{a+R}h - \frac{l}{{k'}^2}\frac{a}{a+R} \right) \end{aligned}$$

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for the term multiplying $k{\mathbf E}(k)$, and

$$\begin{aligned} = f+g+l+al'-\frac{h}{2}+\frac{k^2}{m^2} \left( h \frac{a}{a+R}-f \right) \end{aligned}$$

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for the term multiplying $k{\mathbf K}(k)$. The solution by (Ansorg et al. 2003) which corresponds to

$$\begin{aligned} \left\{ \begin{array}{l} M = \zeta \sqrt{\frac{a}{R}} k {\mathbf K}(k)\\ N= \sqrt{\frac{a}{R}} k \left\{ (a+R) {\mathbf K}(k) - \frac{2R}{k^2} \left[ {\mathbf K}(k) - {\mathbf E}(k) \right] \right\} , \end{array}\right. \end{aligned}$$

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is obtained for the following settings:

$$\begin{aligned} \left\{ \begin{array}{l} f=0,\\ g=\sqrt{\frac{a}{R}},\\ ah=\sqrt{\frac{a}{R}} \frac{2R}{k^2},\\ al=\sqrt{\frac{a}{R}} \left( a+R-\frac{2R}{k^2}\right) = \frac{1}{2}\sqrt{\frac{a^3}{R}}- \frac{R^2+\zeta ^2}{2\sqrt{aR}}, \end{array}\right. \end{aligned}$$

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which eliminates the term $k {\mathbf E}(k)$ and produces the factor $2 \sqrt{\frac{a}{R}}$ for the term $k{\mathbf K}(k)$.

Appendix C: A basic Fortran 90 program

Fortran 90 routines and a driver program which computes the $M$ and $N$ functions for a cylindrical cell and the associated potential at one space point $(R,\theta ,Z)$ are available from the online version of the paper. The quadrature is performed from a Newton-Cotes, second-order quadrature scheme. External calls to functions IEF(BETA,K) and IEE(BETA,K) refer to the values of the incomplete elliptic integral $F(\beta ,k)$ and $E(\beta ,k)$ respectively which can be obtained from any mathematical library. For single (double) precision computations, change _AP into 4 (resp 8), and EPSMACH is the corresponding precision (about $2 \times 10^{-16}$ in double precision). These routines are not optimized. Running the code with the default parameters generates the following output:

Appendix D: Accelerations

Since $M$ and $N$ are build from the two elementary kernels $M_0$ and $N_0$ from Eq. (25), we will only expand $\partial _R M_0$, $\partial _R N_0$, $\partial _\theta M_0$, etc. After calculus, we get

$$\begin{aligned} \left\{ \begin{array}{l} \frac{\partial M_0 }{\partial R} = - \frac{1}{4}\sqrt{\frac{a}{R^3}} \zeta k^3 \left\{ \left[ E(\beta _0,k) - \frac{k^2 \sin ^2 \beta _0}{2 \sqrt{1-k^2\sin ^2 \beta _0}} \right] \right. \\ \left. \qquad \qquad \times \left[ 1-\frac{(a+R)k^2}{2a} \right] \frac{1}{{k'}^2}+F(\beta _0,k) \right\} + \frac{1}{2} \alpha \frac{\partial f }{\partial R}\\ \frac{\partial N_0 }{\partial R} = - \frac{1}{4}\sqrt{\frac{a}{R^3}} k \frac{a^2+R^2}{a}F(\beta _0,k) + \frac{1}{2}\frac{kE(\beta _0,k)}{{k'}^2} \left\{ \frac{a-R}{a+R}\sqrt{\frac{a}{R}}\left( \frac{a+R}{2R}-\frac{k^2}{m^2}\right) -\frac{{k'}^2}{m}\right\} \\ \qquad \qquad +\,\, \frac{1}{4}k \sqrt{\frac{a}{R}} \left( 1-\frac{k^2}{m^2}+\frac{a-R}{2R} \right) \left( {k'}^2-\frac{a-R}{a+R}\right) \frac{\sin (2\beta _0)}{{k'}^2\sqrt{1-k^2\sin ^2 \beta _0}}\\ \qquad \qquad +\,\, \frac{1}{2}(1-\alpha ) \frac{\partial g }{\partial R}\\ \frac{\partial M_0 }{\partial \theta } = \frac{1}{4}\sqrt{\frac{a}{R}} \zeta \frac{k}{\sqrt{1-k^2\sin ^2 \beta _0}} + \frac{1}{2} \alpha \frac{\partial f }{\partial \theta }\\ \frac{\partial N_0 }{\partial \theta } = \frac{1}{4}\sqrt{\frac{a}{R}} [a+R \cos (2 \beta _0)] \frac{k}{\sqrt{1-k^2\sin ^2 \beta _0}} + \frac{1}{2}(1-\alpha ) \frac{\partial g }{\partial \theta }\\ \frac{\partial M_0 }{\partial Z} = \frac{1}{2}\sqrt{\frac{a}{R}}k \left[ F(\beta _0,k)-\left( 1-\frac{k^2}{m^2}\right) \frac{E(\beta _0,k)}{{k'}^2}\right] \\ \qquad \qquad +\,\, \frac{1}{4}\sqrt{\frac{R}{a}}\left( 1-\frac{k^2}{m^2}\right) \frac{k^3}{{k'}^2\sqrt{1-k^2\sin ^2 \beta _0}} + \frac{1}{2}\alpha \frac{\partial f }{\partial Z}\\ \frac{\partial N_0 }{\partial Z} = - \frac{1}{2}\sqrt{\frac{a}{R}}\left\{ \left( a+R-\frac{2R}{k^2}\right) \left[ E(\beta _0,k)-\frac{k^2 \sin (\beta _0)}{2\sqrt{1-k^2\sin ^2 \beta _0}}\right] \right. \\ \qquad \qquad \left. +\,\,\frac{2R{k'}^2}{k^2}F(\beta _0,k)\right\} \frac{k}{\zeta {k'}^2}\left( 1-\frac{k^2}{m^2}\right) + \frac{1}{2}(1-\alpha )\frac{\partial g }{\partial Z} \end{array}\right. \end{aligned}$$

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where

$$\begin{aligned} \left\{ \begin{array}{l} \frac{\partial f }{\partial R} = \frac{f}{R}+ \frac{1}{2} \sin (2 \beta _0) \sqrt{\frac{R}{a}} \frac{1}{1-m^2 \sin ^2 \beta _0}\frac{k}{\sqrt{1-k^2\sin ^2 \beta _0}} \frac{\zeta [R+a \cos (2 \beta _0)]}{(a+R)^2}\\ \frac{\partial g }{\partial R} = \frac{g}{R}- \frac{1}{2} \sin (2 \beta _0) \sqrt{\frac{R}{a}} \frac{k}{\sqrt{1-k^2\sin ^2 \beta _0}} \frac{[\zeta ^2 \cos (2\beta _0)-aR \sin ^2(2\beta _0)]}{\zeta ^2+R^2 \sin ^2 (2\beta _0)} \\ \frac{\partial f }{\partial \theta } = -R \cos (2 \beta _0) \mathrm{asinh}\frac{\zeta }{(a+R)\sqrt{1-m^2 \sin ^2 \beta _0}} \\ \qquad \qquad - \frac{1}{2} R \sin (2 \beta _0) \frac{\zeta \sqrt{aR}}{(a+R)^2}\frac{ \sin (2\beta _0)}{1-m^2 \sin ^2 \beta _0}\frac{k }{\sqrt{1-k^2\sin ^2 \beta _0}}\\ \frac{\partial g }{\partial \theta } = -R \cos (2 \beta _0) \mathrm{asinh}\frac{a+R \cos (2\beta _0)}{\sqrt{\zeta ^2+R^2 \sin ^2 (2\beta _0)}} \\ \qquad \qquad + \frac{1}{2} R \sin (2 \beta _0) \sqrt{\frac{R}{a}}\frac{[\zeta ^2+R^2 +(a+R)R \cos (2\beta _0)] \sin (2\beta _0)}{\zeta ^2+R^2 \sin ^2 (2\beta _0)}\frac{k}{\sqrt{1-k^2\sin ^2 \beta _0}} \\ \frac{\partial f }{\partial Z} = -\frac{1}{2} \sin (2 \beta _0) \sqrt{\frac{R}{a}} \frac{k}{\sqrt{1-k^2\sin ^2 \beta _0}}\\ \frac{\partial g }{\partial Z} = + \frac{1}{2} \sin (2 \beta _0) \sqrt{\frac{R}{a}} \frac{k}{\sqrt{1-k^2\sin ^2 \beta _0}} \frac{\zeta [a+R\cos (2\beta _0)]}{\zeta ^2+R^2 \sin ^2 (2\beta _0)} \end{array}\right. \end{aligned}$$

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