Schwere, Elektricität und Magnetismus:405

Bernhard Riemann: Schwere, Elektricität und Magnetismus
Seite 33
<< Zurück	Vorwärts >>
fertig
Fertig! Dieser Text wurde zweimal anhand der Quelle Korrektur gelesen. Die Schreibweise folgt dem Originaltext.

 

VECTOR ANALYSIS.

To find the value of this integral, we may regard the point $\rho \,$, which is constant in the integration, as the center of polar coordinates. Then $r\,$ becomes the radius vector of the point $\rho '\,$, and we may set

$dv'=r^{2}dqdr,\,$

where $r^{2}dq\,$ is the element of a spherical surface having center at $\rho \,$ and radius $r\,$. We may also set

$\nabla u'\cdot {\frac {\rho '-\rho }{r}}={\frac {du'}{dr}}.$

We thus obtain

$\nabla \cdot \operatorname {New} u=\iiint {\frac {du'}{dr}}dqdr=4\pi \int {\frac {d{\overline {u'}}}{dr}}dr=4\pi {\overline {u'}}_{r=\infty }-4\pi {\overline {u'}}_{r=0},\,$

where ${\overline {u}}\,$ denotes the average value of $u\,$ hi a spherical surface of radius $r\,$ about the point $\rho \,$ as center.

     Now if $\operatorname {Pot} u\,$ has in general a definite value, we must have $u'=0\,$ for $r=\infty \,$. Also, $\nabla \cdot \operatorname {New} u\,$ will have in general a definite value. For $r=0\,$, the value of ${\overline {u'}}\,$ is evidently $u\,$. We have, therefore,

$\nabla \cdot \operatorname {New} u=-4\pi u,\,$ $\nabla \cdot \nabla \operatorname {Pot} u=-4\pi u.\,$

     98. If $\operatorname {Pot} \omega \,$ has in general a definite value,

$\nabla \cdot \nabla \operatorname {Pot} \omega =\nabla \cdot \nabla \lbrack ui+vj+wk\rbrack =\nabla \cdot \nabla ui+\nabla \cdot \nabla vj+\nabla \cdot \nabla wk,\,$ $\nabla \cdot \nabla \operatorname {Pot} \omega =-4\pi \omega .\,$

Hence, by No. 71,

$\nabla \times \nabla \times \operatorname {Pot} \omega -\nabla \nabla \cdot \operatorname {Pot} \omega =4\pi \omega .\,$

That is,

$\operatorname {Lap} \nabla \times \omega -\operatorname {New} \nabla \cdot \omega =4\pi \omega .\,$

     If we set

$\omega _{1}={\frac {1}{4\pi }}\operatorname {Lap} \nabla \times \omega ,\quad \omega _{2}={\frac {-1}{4\pi }}\operatorname {New} \nabla \cdot \omega ,\,$

we have

$\omega =\omega _{1}+\omega _{2},\,$

where $\omega _{1}\,$, and $\omega _{2}\,$ are such functions of position that $\nabla \cdot \omega _{1}=0\,$, and $\nabla \times \omega _{2}=0\,$. This is expressed by saying that $\omega _{1}\,$ is solenoidal, and $\omega _{2}\,$ irrotational. $\operatorname {Pot} \omega _{1}\,$, and $\operatorname {Pot} \omega _{2}\,$, like $\operatorname {Pot} \omega \,$, will have in general definite values.

     It is worth while to notice that there is only one way in which a vector function of position in space having a definite potential can be thus divided into solenoidal and irrotational parts having definite potentials. For if $\omega _{1}+\varepsilon \,$, $\omega _{2}-\varepsilon \,$ are two other such parts,

$\nabla \cdot \varepsilon =0\,$ and $\nabla \times \varepsilon =0.\,$