WILSON AND LEWIS. — RKLATIN ITV. 501 



of deriving the scalar O'^ 'i"'l t^^^' 2-vector <Q>>^t, which are functions 

 of it. 



By means of the same (lefming e(|iiation the operator <^ may be 

 applied to 2-vector functions of position. The result <0>P is then a 

 dyadic in which the first vectors of the dyads are 1-vectors and the 

 second vectors 2-vectors. If written out in terms of the coordinate 

 unit vectors, such a dyadic would consist of twenty -four terms, each 

 of the type kjkjj., j 9^ k. By inserting the dot or cross, the 1-vector 

 <3>*F and the 3-vector <0»'P are innnediately found. In case the 2- 

 vector F is given as a product i<g of two 1-vectors, the dyadic OF 

 may be obtained directly by means of the rules of difi'erentiation in 

 terms of the dyadics <C>f and <^g. For 



f/r-OF = (IF = f/(fxg) = Jfxg + f/fxr/g = fxg — f/gxf, 



(IT'C>F = (iT'OUg — (iT'Ogxt, 



OF = Ofxg-Offxf- 



It was such analysis which was used in § 44. It illustrates strikingly 

 the great advantage of the symbol <^ over such symbols as Div, Rot, 

 Grad, and 3)iD. 



If ^ is a dyadic function of position, the equation f/r-^^ = d"^ 

 may be used to define <0^j w^hich is a triadic, that is, a sum of formal 

 products of which each contains three vectors juxtaposed without 

 any sign of multiplication. By interposing a dot between the first 

 two of the three vectors in the triads, we find the 1-vector <C>*'^. The 

 expression <y' • SE^ corresponds to what Minkowski calls lor ^, where ^ 

 is for him a matrix. 



We may compute the expression <0>^ in the case where 



^ = I [(I. A). (I- A') + (I. A'*). (I. A*)]. (157) 



First we write 



rfr.O[(I-A)-(I-A')] = rf[(I-A).(I.A')l 



= M(I.A)].(I-A') + (l'A)'dil'A'). 



The second term may be transformed so that the differential comes to 

 the front. For by the equation found in the previous footnote, 



(I-A).(I.rfA') = — A.c/A'I+ (I.(/A'*)-(I-A*). 

 Hence 



rf[(I-A).(I.A')] = - (f/A-I).(I-A') - {dA'*'l)'{l'A*)-dA''Al. 



Now 



(f/A.I).(I.A') = f/A.(I.I.A') = f/A.(I-A'). 



