730 PROFESSOR CLERK MAXWELL ON THE GEOMETRICAL 



for the mean distance by taking the point at a great distance from the system, 

 in which case the mean distance must approximate to the distance of the 

 centre of gravity. 



Thus it is well known that the harmonic mean distance of two spheres, each 

 of which is external to the other, is the distance between their centres, and that 

 the harmonic mean distance of any figure from a thin shell which completely 

 encloses it is equal to the radius of the shell. 



I shall not discuss the harmonic mean distance, because the calculations 

 which lead to it are well known, and because we can do very well without it. 

 I shall, however, give a few examples of the geometric mean distance, in order 

 to show its use in electro-magnetic calculations, some of which seem to me to 

 be rendered both easier to follow and more secure against error by a free use 

 of this imaginary line. 



If the co-ordinates of a point in the first of two plane figures be x and y, 

 and those of a point in the second £ and -q, and if r denote the distance between 

 these points, then R, the geometrical mean distance of the two figures, is 

 defined by the equation 



log E .ffffdx dy d£ dr) = ffff log r dx dy dt; drj . 



The following are some examples of the results of this calculation : — 



(1.) Let AB be a uniform line, and O a point 

 not in the line, and let OP be the perpendicular 

 from O on the line AB, produced if necessary, then 

 if R is the geometric mean distance of O from the 

 line AB, 



AB. (log R + 1) = PB. log OB - PA log OA + OP. AOB . 



(2.) The geometrical mean distance of P, a point in the line itself, from AB 

 is found from the equation 



AB (log R + 1) = PB log PB - PA log PA . 



When P lies between A and B, PA must be taken negative, but in taking the 

 logarithm of PA we regard PA as a positive numerical quantity. 



(3.) If R is the geometric mean distance between two finite lines AB and 

 CD, lying in the same straight line, 



AB . CD (2 log R + 3) = AD 2 log AD + BC 2 log BC - AC 2 log AC 



- BD 2 log BD . 



