﻿792 Dr. L. Silber stein on the 



Finally, the time is transformed according to 



*'= 7 [)4(rv)]. 



To get the resultant r' take the sum of (a) and of (/3) x u. 

 Then write, for the sake of subsequent convenience, 



l = tct, i = v— 1, 

 and similarly l' = ict f . 



Thus, the relativistic formulae will become 

 r ' = r + (7 — 1) (ru)u -f ifiylu \ 



r = 7 [/-, /3 (ru)], J 



(2) 



quite independent of any system of coordinate-axes. 



Now, to obtain the required quaternionic representation 

 (1) of the whole transformation (2), let us introduce the 

 quaternion 



q = x + l = r-\-ict, ...... (3) 



and similarly q' = r' + V =r' + id! (3') 



Then the problem will consist in rinding a pair of quaternions 

 a, b such that 



r'-M' =«(> + % 



and will be solved by developing the right side of this 

 equation. 



Having done this, explicitly, and compared with (2), I 

 found immediately that the quaternions a, b can differ from 

 one another only by an ordinary scalar factor, and since this 

 may be distributed equally among a, b (their tensors entering 

 only by the product), we may as well take simply equal a, 6, 

 say, both = Q. In fact, then, the form (1) is much too general 

 for our purpose. Thus, to spare the reader any superfluous 

 complication, let us at once seek for 



<?' = Q?Q (la) 



as the quaternionic equivalent of (2). 



Denote the unknown vector and scalar parts of Q by w 

 and s respectively, i. e. write 



Q = w + s. ....... (4) 



