Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/69

At Fig. 3 we have an illustration of curl combined with convergence.

Fig. 3

Let us now consider the meaning of the equation

$V\nabla \sigma =0$

This implies that $\nabla \sigma$ is a scalar, or that the vector $\sigma$ is the slope of some scalar function $\Psi$. These applications of the operator $\nabla$ are due to Professor Tait^[1]. A more complete development of the theory is given in his paper 'On Green's and other allied Theorems'^[2] to which I refer the reader for the purely Quaternion investigation of the properties of the operator $\nabla$.

26.] One of the most remarkable properties of the operator $\nabla$ is that when repeated it becomes

$\nabla ^{2}=-({\frac {d^{2}}{dx^{2}}}+{\frac {d^{2}}{dy^{2}}}+{\frac {d^{2}}{dz^{2}}})$

an operator occurring in all parts of Physics, which we may refer to as Laplace's Operator.

This operator is itself essentially scalar. When it acts on a scalar function the result is scalar, when it acts on a vector function the result is a vector.

If, with any point $P$ as centre, we draw a small sphere whose radius is $r$, then if $q_{0}$ is the value of $q$ at the centre, and ${\bar {q}}$ the mean value of $q$ for all points within the sphere,

$q_{0}-{\overline {q}}={\tfrac {1}{10}}r^{2}\nabla ^{2}q$;

so that the value at the centre exceeds or falls short of the mean value according as $\nabla ^{2}q$ is positive or negative.

I propose therefore to call $\nabla ^{2}q$ the concentration of $q$ at the point $P$, because it indicates the excess of the value of $q$ at that point over its mean value in the neighbourhood of the point.

If $q$ is a scalar function, the method of finding its mean value is well known. If it is a vector function, we must find its mean value by the rules for integrating vector functions. The result of course is a vector.

1. Proceedings R. S. Edin., 1862
2. Trans. R. S. Edin., 1869-70.