Schwere, Elektricität und Magnetismus:400

VECTOR ANALYSIS.

<section begin=t1 /> Vorlage:IdtSetting $\mathrm{\nabla }u=\tau -\omega$, we have $\mathrm{\nabla }\cdot \mathrm{\nabla }u=0$ throughout the space, and the normal component of $\mathrm{\nabla }u$ at the boundary equal to zero. Hence throughout the whole space $\mathrm{\nabla }u=\tau -\omega =0$.

Vorlage:Idt89. If throughout a certain space (which need not be continuous, and which may extend to infinity)

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and in all the bounding surfaces

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and at infinite distances within the space (if such there are)

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then throughout the whole space

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This will be apparent if we consider separately each of the scalar components of $\tau$ and $\omega$.

Minimum Values of the Volume-integral $\iiint u\omega \cdot \omega dv$. (Thomson’s Theorems.)

<section begin=t1 />Vorlage:Idt90. Let it he required to determine for a certain space a vector function of position $\omega$ subject to certain conditions (to be specified hereafter), so that the volume-integral

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for that space shall have a minimum value, $u$ denoting a given positive scalar function of position.

Vorlage:Idta. In the first place, let the vector $\omega$ be subject to the conditions that $\mathrm{\nabla }\cdot \omega$ is given within the space, and that the normal component of $\omega$ is given for the bounding surface. (This component must of course be such that the surface-integral of $\omega$ shall be equal to the volume-integral $\int \mathrm{\nabla }\cdot \omega dv$. If the space is not continuous, this must be true of each continuous portion of it. See No. 57.) The solution is that $\mathrm{\nabla }\times (u\omega )=0$, or more generally, that the line-integral of $u\omega$ for any closed curve in the space shall vanish.

Vorlage:IdtThe existence of the minimum requires that

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while $\delta \omega$ is subject to the limitation that

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and that the normal component of $\delta \omega$ at the bounding surface vanishes. To prove that the line-integral of $u\omega$ vanishes for <section end=t1 />