140 Dr. G. Bakker on the 



If we neglect the quantities of the order ™ with respect 

 to unity, and put ^—E v = p ? we can conclude from (3) and (4) : 



R : R K =(p-^) : &-£) (5) 



In this consideration we might express the surface-tension 

 of Laplace by the integral: 



e.= C {pi-frydk. 



However, if we wish to be perfectly exact, we are to 

 determine what is to be understood by surface and unity of 

 surface of a curved capillary layer, and it is self-evident to 

 point out the sphere of radius R as " surface of the capillary 

 layer." The capillary energy H depends consequently at very 

 strong curvature on so many spherical capillary layers of 

 equal curvature that the mentioned surfaces have a total 

 surface of unity. Let us call the surface per unit of mass 

 of the capillary layer, S. 



Through each point of the spherical capillary layer we 

 imagine a sphere, concentric to the adequate little drops. 

 We call points, lor which the R of these spheres in the 

 different capillary layers has the same value, homologous 

 ones. If the totality of the surfaces of these spheres per 

 unit of mass is S', and R' the corresponding radius, we have: 



8' : Si = ( R, + hf :\ir, 



if S x represents the total surface of the spheres with radius 

 R t . If we new consider the capillary energy as the volume 

 integral of the departures from the law of Pascal, we have 

 consequently: 



HS=f 2 S'(y,. s - / , T )'W (6) 



If we put R'rrRi + Zi, then S' : S^^ + A) 2 : R x 2 or: 



„, a A , 2h A 2 \ 



and the capillary energy per unit of mass becomes: 



118 = 8: J' (py- P r)dl< + J- ( \pM-p*)d»+g[fc§'(p.-pJd*. 



- l • • • (7) 



