4 Proceedings of tlie Royal Irish Academy. 



From equation (8) we get SUv{r\ -?)') = 0, so that 

 „-„' = - VUvVUv{ri -■,{), 

 and VUv VUv (., - .,') Sa Uv = - F. Uv Vc („ - „'). 



Hence r.Uv[t - t' 4 F<t(i, - .,')] = 0. 



If the negative side of the surface is a perfect conductor, 



i + Van = ; 

 and we thus get the surface condition to be, that the vector 



must be normal. 



Operating on (7) with SUv, we find 4n-c = - c'-SUv{e - t), which gives 

 us the surface-density e in terms of c and *', whilst the relative current lo is 

 given by « - ca. 



In the case of a conductor moving in a general manner as a rigid body 

 the electromotive intensity at each point inside is zero, so that 



I + Via + Vwp) ((' = 0, 



where a is the velocity of the origin, and <u is the angular rotation, p being 

 the distance of the point from the origin. Operating on this with FT ( ), 

 we find, by the aid of (.3) and (4), 



'!±-S{<x+Vo>p)V.„ = 0, 



which shows that each part of the conductor preserves in magnihide and 

 carries with it the magnetic force which was there originally. The problem 

 thus loses nothing in generality if we suppose that the initial magnetic field 

 is initially and always afterwards null. The internal electric force will then 

 also be zero, so tliat the boundary condition is that the vector t + Van is rurnnal 

 to the boundary. The general problem consists of calculating vectors t and ij 

 which satisfy this condition, and then the equations 



C-'tSaUv + -iiri = Wv„. (9) 



iTTC = - c^SiUv. (10) 



I = to + e<T (11) 



give us the electrical condition of the surface. 



