EXAMPLES OF SOLENOIDAL FIELDS. 



35 



The line-integral of the normal component of the vector is easily found for any closed 

 curve having the form of a rectangle with sides parallel to the axes of coordinates. 

 If two of the sides are the coordinate axes, and the two others the lines x = x andy = y, 

 the line-integral taken over the closed curve will be A x y-\-AyX. Substituting the 

 values (a) of the components, we get the line-integral equal to axy+bxy, and the 

 solenoidal condition is seen to be fulfilled if b = a. Thus the formulae will be 



(b) A x = ax A y = ay 



Fig. 36 represents the components of this solenoidal vector. If the vector represents 

 velocity, the motion given by fig. 36 will be the simplest typical fluid motion pro- 

 ducing a deformation of the fluid masses without change of volume. 



A vector-line is determined by the condition that the projections dx and dy of 

 its line-element are proportional to the vector-components A x and A y . It is there- 

 fore given by the differential equation 



(c) dx ^dy 



A X Ay 



Substituting the values of A x and A y according to (b), and integrating, we find 



(d) xy = const. 



i. e., the vector-lines are equilateral hyperbolae, having the axes of coordinates as 

 asymptotes (fig. 37). These axes themselves belong to the system of lines of flow, 



Fig. 37. Hyperbolic lines of flow and circular curves 

 of equal intensity I, 2, 3, . . . of a plane 

 deformation-field. 



Fig. 38. Neutral point of higher order. 



and cut each other at the neutral point of the field where A x = A y = o. The intensity 

 A of the vector is 



A = v A 2 x +Ay = av'x 2 +y 2 = ar 

 Thus the curves for equal intensity are circles r = const, around the neutral point. 

 Through equal lengths of a line parallel to one of the axes there will go equal trans- 

 port. Drawing the hyperbolae through equidistant points on such a line as made in 



