Schwere, Elektricität und Magnetismus:138

Bernhard Riemann: Schwere, Elektricität und Magnetismus
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Fertig! Dieser Text wurde zweimal anhand der Quelle Korrektur gelesen. Die Schreibweise folgt dem Originaltext.

Zweiter Abschnitt. §. 28.

Ferner hat man

{\frac {\partial m}{\partial x}}=-2{\frac {x}{s}},\quad {\frac {\partial ^{2}m}{\partial x^{2}}}=-{\frac {2}{s}},

${\frac {\partial m}{\partial y}}=-2{\frac {y}{s+\beta ^{2}}},\quad {\frac {\partial ^{2}m}{\partial y^{2}}}=-{\frac {2}{s+\beta ^{2}}},$

${\frac {\partial m}{\partial z}}=-2{\frac {z}{s+\gamma ^{2}}},\quad {\frac {\partial ^{2}m}{\partial ^{2}z}}=-{\frac {2}{s+\gamma ^{2}}}.$ ^[1]

Daraus berechnet sich

(5)	${\frac {\partial ^{2}\left(m^{\frac {1}{2}}\right)}{\partial x^{2}}}+{\frac {\partial ^{2}\left(m^{\frac {1}{2}}\right)}{\partial y^{2}}}+{\frac {\partial ^{2}\left(m^{\frac {1}{2}}\right)}{\partial z^{2}}}$

{\begin{aligned}=&-m^{-{\frac {3}{2}}}\left\lbrace \left({\frac {x}{s}}\right)^{2}+\left({\frac {y}{s+\beta ^{2}}}\right)^{2}+\left({\frac {z}{s+\gamma ^{2}}}\right)^{2}\right\rbrace \\&-m^{-{\frac {1}{2}}}\left\lbrace {\frac {1}{s}}+{\frac {1}{s+\beta ^{2}}}+{\frac {1}{s+\gamma ^{2}}}\right\rbrace .\\\end{aligned}}

Man findet aber leicht

{\frac {\partial m}{\partial s}}=\left({\frac {x}{s}}\right)^{2}+\left({\frac {y}{s+\beta ^{2}}}\right)^{2}+\left({\frac {z}{s+\gamma ^{2}}}\right)^{2},

${\frac {d\,\lg D}{ds}}={\frac {1}{s}}+{\frac {1}{s+\beta ^{2}}}+{\frac {1}{s+\gamma ^{2}}}.$

Folglich vereinfacht sich die Gleichung (5). Man erhält nemlich

(6)	${\frac {\partial ^{2}\left(m^{\frac {1}{2}}\right)}{\partial x^{2}}}+{\frac {\partial ^{2}\left(m^{\frac {1}{2}}\right)}{\partial y^{2}}}+{\frac {\partial ^{2}\left(m^{\frac {1}{2}}\right)}{\partial z^{2}}}$

{\begin{aligned}&=-m^{-{\frac {1}{2}}}\left({\frac {1}{m}}{\frac {\partial m}{\partial s}}+{\frac {d\,\lg D}{ds}}\right)\\&=-m^{-{\frac {1}{2}}}{\frac {d\,\lg(m\,D)}{ds}}.\\\end{aligned}}

Mit Hülfe dieser Gleichung ergibt sich

(7)	${\frac {\partial ^{2}X}{\partial x^{2}}}+{\frac {\partial ^{2}X}{\partial y^{2}}}+{\frac {\partial ^{2}X}{\partial z^{2}}}$

{\begin{aligned}&=-\rho \int {\frac {1}{\sqrt {m\,D}}}{\frac {d\lg(m\,D)}{ds}}ds\\&=-\rho \int (m\,D)^{-{\frac {3}{2}}}d(m\,D).\\\end{aligned}}

Das Integral ist zu erstrecken durch die Linie $L\,$ (Fig. 18) von $\infty$