Schwere, Elektricität und Magnetismus:405

VECTOR ANALYSIS.

<section begin=t1 /> 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 ${\rho }^{\prime }$, and we may set

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

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We thus obtain

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

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

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Vorlage:Idt98. If $\mathrm{Pot}\omega$ has in general a definite value,

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Hence, by No. 71,

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That is,

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Vorlage:IdtIf we set

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we have

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where ${\omega }_1$, and ${\omega }_2$ are such functions of position that $\mathrm{\nabla }\cdot {\omega }_1=0$, and $\mathrm{\nabla }\times {\omega }_2=0$. This is expressed by saying that ${\omega }_1$ is solenoidal, and ${\omega }_2$ irrotational. $\mathrm{Pot}{\omega }_1$, and $\mathrm{Pot}{\omega }_2$, like $\mathrm{Pot}\omega$, will have in general definite values.

Vorlage:IdtIt 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 ${\omega }_1+\epsilon$, ${\omega }_2-\epsilon$ are two other such parts,

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