54 G, PLARR ON THE SOLUTION OF THE EQUATION Vpop=0. 
The vector e which we introduce here is the same as that which enters into the 
definition of ¢—9’. 
(28) [poy Aiea: 
because by (12) we have 
o—t= 2aSB(ay—w’y) , 
and asa=9—m,, wo =~ —m, and SBy=0, &c., we have by (28) : 
o—T= 2aSh2Vey = —2ZaSac= + 2c, 
which is in accordance with (27). 
We may at once also establish that «=—zae, because, by (27) and (12) we 
have 2¢=0,—7,= 2aS(Ba"y—ya' B) = ZaS (w Boy—a'ya8) , and replacing a’B ,. 
@ y respectively by aB—2VeB , ay—2Vey, we get: 
2e, = ZaS| (@B—2VeB)ay—(wy—2Vey)aB |= —22aSeV(Bay—ya8) . 
We invoke here a general theorem, whose demonstration is easy, namely, 
that for any integer value of the number x, and for any linear vector function 
© we have 
(29) V(0¢"0 —0'¢"0) =—[ 2Sag’a+ 9" |VOE'. 
In the case of a, and for »=1, as by (8) we have Saw’a=>Sawa=0, we get 
simply ai! 
V(Bay—yof)=—o'a, 
with two other similar results by permutation of a, B, y, in circular order. 
Thus we have, suppressing the factor 2: 
(30) €,= DaSew a= DaSawe= — we. 
We introduce now the vectors ¢, «, G, «, into the expressions (15), (23), 
(25), and because (27) gives us: 
Sor=C—é 
SoyG=*—e 
S(oy7 + 710) = 28566,—2See, 
we have: 
2P=7? +27 —2e 
2 ; 
(31) | pl = 20 oe 
3Q =S(n« + 204 —2eq) . 
§ 5. We might now substitute these values into the expression (20) of IT’, 
and develop. But, as the terms depending explicitly on «, «,, in the develop- 
ment, would form a part the true sign of which could not be decided upon as a 
