
G. PLARR ON THE ELIMINATION OF a, 8, y, ETC. 271 
We differentiate a = pig, which gives 
a, = pig + pig, . 
But a = pig gives ig = ga, and pz = ap; and as pg; = — pig we get 
a, =pig.a— apg =—2Vapq, 
because p;q is its own vector, Sp,q being zero, owing to Ty = constant. 
Comparing this with (21), a, =— 5 Vaz, , we see that 
Fe 1 
[ DiT = 4: 
(61) 
ic PL =— [D- 
By forming with the second the expression of peg, and with the first that of 
<1p.q, we see, by expressions (18) and (19), that we have 
1 
(2) 4 1 iL 
The first relation (16) giving I in function II, cannot by simple means be 
expressed with the help of p, g. However, the equation, which shows that 
VI = 0, is comprised in (56), giving 
V(pb 9g) — V(4p.9) = 9. 
: hee il 
The expression of P, which is — 7 V*II, becomes now 
(63) P =—4V*(peq) =— 4V"(<1p.9). 
That of Q (30), may be put under the form 
Q=— 78 —42(@,) x (— a). 
For a, we take 4p/q , and for (— a,) we take 4pq,, according to (61). Their 
product is, owing to pg = 1: 
167.9 X py. = 16piqa. 
Thus 
Q =— 4S’ppqg— 427.0; 
or aiso, introducing the variables z, y, z, 
(64) Q=— 48'ppg —42p.g., 
We have in consequence — 
P—Q=4(Sippg — Vppg) + 42p.9,- 
But 
Sppg — Vipeg = (Kpp gq) x (pea) 
VOL. XXVII. PART ITI. 4B 
