176 Proceedings of the Royal Society of Edinburgh. [Sess. 
These four classes of strains involve, respectively, nine, eight, seven, 
and six scalar constants for their complete determination. 
8. We may now examine the conditions that two strains (p and Q shall be 
commutative. Consider first this theorem — 
Theorem I. — If 0 and d are commutative, and if 0 has an axis which 
is not an axis of d, then 0 is reducible. 
Proof. — Let 0a = aa and assume a not an axis of d. We have 
0(f>a = 0{aa) 
= aOa, 
because scalars are commutative. But if 0 and 6 are commutative, d0a = 0da. 
Equating values of d0a, we have 
<p0a = a0a. 
By hypothesis, da is not parallel to a. Thus 0 has two axes, viz. a and da, 
with the same root a, and is therefore reducible. 
It follows at once from this theorem that if two strains 0 and d are 
commutative, they are of the same class, or else one of them must be of 
class IV. 
9. If 0 and d are of class I, they are completely determined by their 
axes and the corresponding roots of the symbolic cubics. That they may 
be commutative, it is necessary and sufficient that they have the same axes. 
If 0 is of class I and d of class IV, by Theorem I 0 can have no axis 
which is not an axis of d. Therefore d, if written in the normal form, 
has y along one axis of 0, and X at right angles to the plane of the other 
two axes of 0 (or else X vanishes). 
10. If 0 and d are both of class II, by theorem I neither can have an 
axis which is not an axis of the other, but they are no longer fully 
determined by their axes and their cubics. Let 0 be thrown into the 
normal form (14). To determine the most general d commutative with (p, 
we have first, because /5 and /S^ must be axes of d, 
0(^ = h(3, = (29) 
where h and h-^ are constant scalars. If we suppose da to be resolved 
along a, /3, and /5j, we may write 
0a = xa-\- yfS + zf3j, ..... (30) 
where x, y, and 2 ; are to be determined. We have 
<pOa — Opa, if the strains are to be commutative, 
= 0{ga + c ^), by (14), letting c = cSa/3j3-^, 
^g0a + ch[3, by (29), 
= gxa + gy^ + gz(5^ + by (30). 
