Schwere, Elektricität und Magnetismus:399

VECTOR ANALYSIS.

<section begin=t1 /> and in all the bounding surfaces the normal components of $\mathrm{\nabla }t$ and $\mathrm{\nabla }u$, are equal, and at infinite distances within the space (if such there are) $r^2(\frac{dt}{dr}-\frac{du}{dr})=0$, where $r$ denotes the distance from some fixed origin,—then throughout the space

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and in each continuous part of which the space consists

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Vorlage:Idt86. If throughout any continuous space (or in all space)

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and in any finite part of that space, or in any finite surface in or bounding it,

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then throughout the whole space

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Vorlage:IdtFor, since $\mathrm{\nabla }\times (\tau -\omega )=0$, we may set $\mathrm{\nabla }u=\tau -\omega$, making the space acyclic (if necessary) by diaphragms. Then in the whole space $u$ is single-valued and $\mathrm{\nabla }\cdot \mathrm{\nabla }u=0$, and in a part of the space, or in a surface in or bounding it, $\mathrm{\nabla }u=0$. Hence throughout the space $\mathrm{\nabla }u=\tau -\omega =0$.

Vorlage:Idt87. If throughout an aperiphractic* space contained within finite boundaries but not necessarily continuous

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and in all the bounding surfaces the tangential components of $\tau$ and $\omega$ are equal, then throughout the space

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Vorlage:IdtIt is evidently sufficient to prove this proposition for a continuous space. Setting $\mathrm{\nabla }u=\tau -\omega$, we have $\mathrm{\nabla }\cdot \mathrm{\nabla }u=0$ for the whole space, and $u=\text{constant}$ for its boundary, which will be a single surface for a continuous aperiphractic space. Hence throughout the space $\mathrm{\nabla }u=\tau -\omega =0$.

Vorlage:Idt88. If throughout an acyclic space contained within finite boundaries but not necessarily continuous

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and in all the bounding surfaces the normal components of $\tau$ and $\omega$ are equal, then throughout the whole space

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Vorlage:References <section end=t1 />