396 Mr. H. T. Flint on Integration Theorems 



It is to be noted that V (V 2 a£)(V 2 'a£) =0 on account of 

 the relation (2'5), so that the two planes may be said to be 

 normal in the same way that two vectors p 1 and p 2 are 

 normal if V /o 1 /o 2 = 0. 



But V P 2 P 2 '=£0 although the relation 



V P 2 P 2 ' = V P 2 'P 2 still holds. 



Sommerfeld has defined the components of a six-vector 

 with respect to a plane. We can very easihy obtain his 

 result in our notation. The component is — V P 2 (V 2 a/3). 



We must make Y 2 afi a unit plane and naturally shall 

 choose a and /3 as two unit perpendicular vectors. 



Write 



a = l x i x -t »ij/ 2 + »i*8 +pih, 



/3=l 2 h + m 2 i 2 + n 2 i z +p 2 i±. 



Since a and ft are unit perpendicular vectors, 



SZ! 2 = % 2 =1 



and X/i/ 2 = 0. 



Then -V P 2 V 2 a/3 



= — V (Xv 3 PyJ (S 8 t S (W*2)) 



= Yy^m^) + P^n^) + P 2y (Z 1 m 2 ) + Px«(/i/? 2 ) 



+ Py M (ra 1 p 2 )+P ZM (rc 1 p 2 ) • (5'1) 



This should be compared with Sommerfeld's result (Ann, 

 d. Phys. xxxii. p. 760). 



For a vector A of the directed volume type we have : 



= I ftA. + 1 2 A y + i 3 A z + t 4 A„) = IAj (say). 



The component of A along any direction n may be 

 written : 



-TV 4 A 3 ?i or -V (IA,)ti or -V Ain. 



The component of A 3 along Ox is by this definition A x , 

 and similarly A y A z A u are the components along the other 

 axes. 



A 3 is normal to any direction n if V (IA 3 )n = 0. 



If four vectors p l p 2 pzp^ bound a four-dimensional volume 



