the axes of a quadratic vector. 339 



6. Properties of a Quadratic Vector Possessing a Central 



Axis. 



The differential equations determined by this type of vector have 

 been studied by Darboux^; the connection between his viewpoint 

 and that of vectorial algebra is established by certain theorems now 

 to be proved. 



Theorem 1. If a quadratic vector has a central axis, it may be 

 thrown into binominal form in three distinct ways, a proper choice 

 of the vector 8 being made in each case. 



One way has already been shown in (32), namely for the value of 5 

 given by ai = —A32, (Vi = a^ = 0. 



A second way, by inspection of (31), is to take as = 0, ai = —A32, 

 and a2 = —^31. Since /1 33 is zero by hypothesis, the component 

 along jSs vanishes. 



Similarly, if 8 is determined by making no = 0, ai = —A23, as = 

 — A21, the component along 182 vanishes. The theorem is thus proved. 



Theorem 2. If a quadratic vector has a central axis, the three 

 values of 5 by virtue of which we may pass from one binomial form to 

 another are all perpendicular to the central axis. 



Proof. These values of 8 are the respective differences of the values 

 occurring in the proof of theorem 1. They must therefore be coplanar. 

 To determine their values explicitly, let the binomial forms be Fi, F2, 

 and F3 in the order above given. Let the values of 8 which changes Fi 

 into F2, F2 into F3, and F3 into Fi be, in order, 5i, 82, and 63. Subtract- 

 ing values of the a's as above found we have, for 61, ai = 0, 02 = —^31, 

 0,3 = 0; for 52, ai = 0, a^ = +^31, as = —^21; for 8s, ai = 0, 02 = 0, 

 as = -{-A21. Since ai is zero in all three cases, the theorem is proved. 



Theorem 3. If a quadratic vector having a central axis be thrown 

 into either of the three binomial forms, a set of rectangular components 

 X, Y, Z, can be found in terms of rectangular coordinates x, y, z, such 

 that both X and Y are independent of z. 



Proof. With the notation used in the proof of theorem 2, let the 

 quadratic vector be in the binomial form F2. Let j3i = k, and let 

 i and j be any two unit vectors such that i, j, k forms a rectangular 

 unit system. Since 82 is perpendicular to i we may write 62 = ci + c'j. 

 Also p = ix + jy + kz. Hence pS82p = —i(cx^ + c'xy) — j{cxy + 

 cV) + terms in k. It was shown that i^s = ■f'2 + pS82p. Now F2 

 contained no component along /Ss- By adding pS82P we removed 

 the component along ^2- But pS82P does not contain z. Therefore 



6 See note to C. Q. V. page 375. 



