Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/62

we might with equal right set the indefinite expression equal to other definite expressions, and then be misled into supposing these definite expressions to be equal to one another. It will be safe to say that the above equations will hold, provided that the potential of $u$ or $\omega$ has a definite value. It will be observed that whenever ${\text{Pot }}u$ or ${\text{Pot }}\omega$ has a definite value in general (i.e., with the possible exception of certain points, lines, and surfaces),^[1] the first members of all these equations will have definite values in general, and therefore the second members of the equation, being necessarily equal to the first members, when these have definite values, will also have definite values in general. 94. Again, whenever Potu has a definite value we may write

$\nabla {\text{Pot }}u=\nabla \iiint {\frac {u'}{r}}dv'=\iiint \nabla {\frac {1}{r}}u'\,dv',$

where $r$ stands for $[\rho '-\rho ]_{0}.$ But

$\nabla {\frac {1}{r}}={\frac {\rho '-\rho }{r^{3}}},$

Moreover, ${\text{New }}u$ will in general have a definite value, if ${\text{Pot }}u$ has.

95. In like manner, whenever ${\text{Pot }}\omega$ has a definite value,

$\nabla \times {\text{Pot }}\omega =\nabla \times \iiint {\frac {\omega '}{r}}dv'=\iiint \nabla \times {\frac {\omega '}{r}}dv'=\iiint \nabla {\frac {1}{r}}\times \omega '\,dv'.$

Substituting the value of $\nabla {\frac {1}{r}}$ given above we have

$\nabla \times {\text{Pot }}\omega ={\text{Lap }}\omega .$

${\text{Lap }}\omega$ will have a definite value in general whenever ${\text{Pot }}\omega$ has.

96. Hence, with the aid of No. 93, we obtain

${\begin{aligned}\nabla \times {\text{Lap }}\omega &={\text{Lap }}\nabla \times \omega ,\\\nabla .{\text{Lap }}\omega &=0,\end{aligned}}$

whenever ${\text{Pot }}\omega$ has a definite value.

97. By the method of No. 93 we obtain

$\nabla .{\text{New }}u=\nabla .\iiint {\frac {\rho '-\rho }{r^{3}}}u'\,dv'=\iiint \nabla u'.{\frac {\rho '-\rho }{r^{3}}}dv'.$

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

$dv'=r^{2}\,dq\,dr,$

1. Whenever it is said that a function of position in space has a definite value in general, this phrase is to be understood as explained ahove. The term definite is intended to exclude both indeterminate and infinite values.