of Four-Dimensional Vector Analysis. 391 



The vector V 2 p 1 /) 2 is the vector area of the parallelogram 

 of which the two adjacent sides are p x and p 2 . It can be 

 split up into six components, which we may denote by 

 A yz , A zx , etc., these being the coefficients of i 2 i 3 , iii 2 , etc., 

 and which are the projections of the area on the six planes 

 which may be chosen out of the four fundamental directions. 

 The vector area although it has six coefficients is defined 

 by five quantities, for a relation exists between the co- 

 efficients, viz. : 



A^A^ + A^A yu + A^A^=0, . . . (2-5) 



as can be seen by expansion of the determinants. 



The elementary area bounded by two short lines dp and Sp 

 may be written : 



hh (dy £~ — ty dz) + hh(dz 8x — Sz dx) -f (2*6) 



Minkowski uses the terms four- and six-vectors in a 

 particular sense, the vectors are subject to a particular 

 transformation, but the comparison of these vectors with 

 (2*3) and (2*4) is clear, e. g. (2*4) is a six-vector without the 

 limitations regarding the transformation. 



Thus the scalar and vector products of Minkowski's calculus 

 are component parts of the complete quaternion ic product 

 P1P2 We may not speak of a unique perpendicular to a plane 

 in four dimensions, for just as in three dimensions a per- 

 pendicular to a line is not determinate, so in three the 

 perpendicular to a plane is not definite. 



We have seen that a plane may be defined by two lines 

 a. and /?, and it is denoted by V 2 a/3. 



The magnitude of the area, or the tensor as it is called, is 

 defined to be the square root of the expression : 



A 2, A 2 i A 2. A 2, A 2i A 2 

 ■™-yz i -&-zx T^ -^Vry i ^ru IT -E^yu i^ ^-zu • 



If this is unity the plane is a unit plane. 



We may now show that the plane normal to Y 2 uj3 is 

 definite if we define it as the locus of lines perpendicular 

 both to a and to (3 and consequently to any line in their 

 plane. For let the unit plane normal to V 2 a/3 be Y07S. 



Then by definition : 



V a 7 = V «S = Vo/9 7 = V /3S = 0, 



and the tensor of V 2 yS = 1 



or TV 27 S=1. 



2 D 2 



