52 THE MECHANICS OF THE EARTH's ATMOSPHERE. 



vauces parallel to the plane in the direction in which the liquid parti- 

 vhs, between it and the plane, themselves move, and with one-fourth 

 of the velocity possessed by the particles that are at the foot of the per- 

 pendicular drawn from the filament to the plane. 



The assumption of the infinitely small cross-section leads to no inad- 

 missible results, because each individual filament exerts no force upon 

 itself affecting its own progression, but is driven forwards only by the 

 influence of the other filaments that may be present for by the action 

 at the boundary]. Bat it is otherwise in the case of curved filaments. 



YI. CIRCULAR VORTEX FILAMENTS. 



In a liquid mass of indefinite extent let there be present only circu- 

 lar filaments whose planes are perpendicular to theaxisof 2;, and whose 

 centers lie in this axis, so that all are symmetrical about this axis. 

 Transform the coordinates by putting 



X = X cos £, ft^=9 cos e, 



y = X sin e, b = g sin e, 



z ■= z, c = c. 



Agreeably to the assumption just made, the velocity of rotation a is 

 only a function of j and z, or of g and c, and the axis of rotation is every- 

 where perpendicular to x (or 9) and to the axis of z. Therefore tbe rec- 

 tangular components or the rotation attbis point whose coordinates are 

 ^, e, and c become 



B, = — a sin f, 7/=(7 cos e, r=0. 



In the equation (5a) we now have, 



r'^={z—c)'^-\-X^-\-(f—2xg cos (f — e) 



'i¥ = -^J J J -—^— g dg de dc 



From the equations for L and M by multiplying by cos f and sin s 

 and adding we obtain 



^ . 1, ^ r r r(?cos(6-e) , , 



L sm e — M cos £ = — ,^ I I I g dg d{e—c) dc, 



^ ■ 1, • ^ r r r(> sin u—e) , , 



Xcosf+Msmf^ 1^ I j j ^ 9 (^(J (i{^—G)dc, 



In both these integrals the angles e and a occur only in the connec- 

 tion {e—e) and this quantity can therefore be considered as the variable 

 under the sign of integration. In the second integral the terms that 

 contain (f— e) = c balance those that contain (f— e)=2;T — c; therefore 

 this integral is equal to zero. 



