2 Mr Hargreaves, Examples illustrating 
To transform (1b), a and X must be interchanged and the 
sign of V altered; a second determination of X’...c’ is reached, 
valzee 
0 O28 , spre Gt eo , ot Ae 
d=a5,+VYx, V= =) VX, o=6, \ 
po weGs 0 Oe pL MOL atl OL mares Ob Oza On 22) 
Mk ge VV t pa 2-4 
which should be in agreement with the first. The agreement is 
realized, the first and second lines of (2a) become identical with 
the second and first lines respectively of (2 b), if 
Ge G3 Ge i Ge Gn ee a OF 
OH On Ge OG Gh Od T G7 ae 
These are the general conditions defining the relation between 
(t, z) and (t’, 2’) when the forms (1a) and (1b) are invariant. 
The first condition gives pete Z= & , where ¢ is a function of 
OZ’ ot 
(t’, 2’). The second condition gives Cbs Ae from ohh we 
oz V2 ot?’ 
infer that 
p=fe+Vt)+ Fh —Vet). 
The third condition is represented by 
14+ 4V2f" (2/4 Vt) FF! (¢ -Vt)= 
It is not admissible to suppose that any relation exists between 
¢ and 2’, and we are therefore compelled to make f” and F” each 
wornetienn’. Thus 
p= ap he + VEY (vey pee (4) 
with > constant is the only admissible solution. Seeing that for 
¢ and z we are limited to linear functions we may also write 
z=az+ Pt, t=yt +62’; 
and then a=, OE ci io = Il : 
or NOW 
zZ=y2 + Bt, mee, with 7? — 82/V2=1...(6 b): 
Vex 
Whether in this form or that of (4) we are left with the choice 
of one significant quantity, a constant, in terms of which the 
transformation is to be stated. If w= /y is taken for that 
