520 PROCEEDINGS OF THE AMERICAN ACADEMY. 



+ 



tube of the vector 5E connecting the new position of dr to the old, 



+ 

 the flux of 5E being against the direction of 6r and of magnitude de. 



And to satisfy equations (3) and (5) we must also assume a certain 



+ 

 variation 6H which is uniquely' defined by these equations and the 

 variations assumed above. We may now assume no variations of the 



negative forces, and for the positive forces only the necessary varia- 



+ 

 tion of U and the variations specified above. 



With these assumptions (12) becomes 



6'H).5H — (E — 6'E)-5E — bU\drdt = 0. 



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(E — GE)'8EdT, 



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we need to integrate only over the tube of 5E defined above, so that, 

 if de is of infinitesimal size, we may take for the result, 



(35) (E — ok) 'Side. 

 To calculate 



(36) j j MH — GK)'8HdT, 



00 



+ + 



we may consider 5H as the H produced by a current of strength 



+ + 



pP'f/o- flowing around the edge of the parallelogram one side of which 



contains the old position of de and the other side the new position, 



while the remaining sides are the tubes of the vector 5E made neces- 

 sary by the motion of the parallelogram. We ma^' now evaluate the 



integral, splitting the space up into elements of each of which two 



+ 

 sides, dS, are surfaces whose normals are in the direction of 5H while 



the remaining dimension, ds, is in the same direction. The integral 



may now be written 



(37) j j j (K-GH)' 8HdS - ds, 



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