8 Proceedings of the Royal Irish Academy. 



If we introduce a function g{^) such that /i(l)5'(p)/2(l) = /3(p), we have 

 /i(P) = /i(l)5' (^P). /2(p) = .9'(P)/2(1). aii"i the equation becomes 



^(P)5'(q) = 5'(pq)- 



This involves ,i7(p) = 1, and taking sealars S^g'(j{^ = (Spq so that 



g'9 = l. ' 



The general solution of this equation, subject to g{V) = 1, is* 



g{ ) = ± a( )a-'. 



Hence since /^(p) = a2ja-y2(l), we must have /.(I) = 1 : and we get 

 finally 



/,(p) = bpa-i, 



/^(p) = apa"', 

 /i(p) = bpa->. 



Let q be the quaternion p - hct; then the symbol of differentiation 



cl = - Sdq^Q-. 



If we suppose a linear transformation q' = /{q), we get 



d = - Sdq:a' = - sdqf'ial = - 'SWqr/, so that a = f'ia')- 



If the " force " a- is ti'ansformed by tr = F((t'), then the general expression 



(j(T becomes f'i(j')F((j'); 



and if we require this to be a linear function of (j'a', then, as above, we 

 must have /'( ) = b( )a"^ and F( ) = a( )a"' by the last section. 

 Hence (jo- = - 47re becomes (j'a = - 4/n-e where e = be'''a"'. 



"We can now deduce the usual equations of relativity from the following 

 principles: — 1°. A charge e moving with velocity X to transform into a charge 

 e' at rest. 2°. A charge c at rest to transform into a charge moving with 

 velocity - X. In 1° we have e = (1 - /ic"'A) e. In 2° we have e = c. 



We thus get c (1 - ^''C''X) = ba"'e', 



e' = b (1 + hc-'X) a,-'e, 

 and it easily follows that 



e/c' = 1/v/ 1 + c-'k- 

 andf 



b- = a = (1 - hc-'X)i{l + c-'X'yi, 



so that the relativity transformations are 



a = a-i(7-V', 

 (T = au a"', 

 e = a 'e'a"'. 



* Joly : Manual of Quateiiiions, p. 283. 



t We can easily see that b"' and a have the same axis, and tlie condition that Se must be re^l 

 inakes b"' = a. 



