Schwere, Elektricität und Magnetismus:408

VECTOR ANALYSIS.

<section begin=t1 /> Vorlage:IdtIn the case of a vector function which is discontinuous at a surface, the expressions $\mathrm{\nabla }\cdot \omega dv$ and $\mathrm{\nabla }\times \omega dv$, relating to the element of the shell which we substitute for the surface of discontinuity, are easily transformed by the priciple that these expressions are the direct and skew surface-integrals of $\omega$ for the element of the shell. (See Nos. 55, 56.) The part of the surface-integrals relating to the edge of the element may evidently be neglected, and we shall have

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Vorlage:IdtWhenever, therefore, $\omega$ is discontinuous at surfaces, the expressions $\mathrm{Pot}\mathrm{\nabla }\cdot \omega$ and $\mathrm{New}\mathrm{\nabla }\cdot \omega$ must be regarded as implicitly including the surface-integrals

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respectively, relating to such surfaces, and the expressions $\mathrm{Pot}\mathrm{\nabla }\times \omega$ and $\mathrm{Lap}\mathrm{\nabla }\times \omega$ as including the surface-integrals

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respectively, relating to such surfaces.

Vorlage:Idt101. We have already seen that if $\omega$ is the curl of any vector function of position, $\mathrm{\nabla }\cdot \omega =0$. (No. 68.) The converse is evidently true, whenever the equation $\mathrm{\nabla }\cdot \omega =0$ holds throughout all space, and $\omega$ has in general a definite potential; for then

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Vorlage:IdtAgain, if $\mathrm{\nabla }\cdot \omega =0$ within any aperiphractic space $A$, contained within finite boundaries, we may suppose that space to be enclosed by a shell $B$ having its inner surface coincident with the surface of $A$. We may imagine a function of position ${\omega }^{\prime }$, such that ${\omega }^{\prime }=\omega$ in $A,{\omega }^{\prime }=0$ outside of the shell $B$, and the integral $\iiint {\omega }^{\prime }\cdot {\omega }^{\prime }dv$ for $B$ has the least value consistent with the conditions that the normal component of ${\omega }^{\prime }$ at the outer surface is zero, and at the inner surface is equal to that of $\omega$. Then $\mathrm{\nabla }\cdot \omega =0$ throughout all space, (No. 90,) and the potential of ${\omega }^{\prime }$ will have in general a definite value. Hence,

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and $\omega$ will have the same value within the space $A$.

__________________ New Haven: Printed by Tuttle, Morehouse & Taylor, 1881.