186 G. PLARR ON THE ESTABLISHMENT OF THE 
be calculated by the general expression (defined in § 1) for the tensor of a 
quaternion. Thus— 
TH) = C7 + (O. 
The first member is equal to. unit by the “ law of tensors,” and the conditions 
Ti =1,Tj=¥. Therefore we have 
(een SV (gy) 
As the second member is also the value of (f’)””, (f”)?”, &c., we conclude to 
the equality of all the quantities F, f’, &c., to one value = f. 
The equation (II.) becomes therefore 
0 Tp Ta cos eS (Up)? — (Pax + 7?by + Kez) 
+ [Dp War Siniper — (bz + cy + ca + az + ay + ba) | f. 
Remembering that 
Ts "Neots = ax + by + cz, 
the equation may receive the form 
0 =[(Up)? — 2] az + [(Up)? —7?] by + [(Up)? — #] cz + Df, 
where we put 
nx» es 
D = Tp Ta (cos pa + sin pa) — (@+ 646) (2@+y+4+2). 
By the expression of p = za + jb + ke, we see that Up cannot be made to depend — 
on a, y,z. The equation just written must, therefore, not establish any relation 
between (Up)? and a, y, z. This cannot be avoided unless we have the values 
zero, namely 
f =0, 
and 
(Up) —# = 0, (Up)? — 7? = 0, (Up)? —# =0 
for the coefficients in the equation. 
This conclusion, avoiding any multiplication, may be arrived at thus: calling 
D,, D,, D,, what D becomes respectively for y = 0, z = 0, and for z = 0, 2 = 90, 
and for « = 0, y = 0, we have, as the equation must hold good for any system 
of values of a, y, z, ‘7 
[(Up” —? lax + D,f=0, 
[(Up)’ —j*] 6y + Df =0, 
[(Up)? —#] cz + D,f = 0. 

