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 5 being made in each case. 



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

 given by ai = —A32, 0^2 — as = 0. 



A second way, by inspection of (31), is to take (13 = 0, ai = —A32, 

 and 02 = —An- Since .433 is zero by hypothesis, the component 

 along jSs vanishes. 



Similarly, if 8 is determined by making (to = 0, a\ = —^23, clz = 

 — Job "^be component along /80 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 5 are the respective differences of the values 

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

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

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

 into Fo, F2 into F3, and F3 into Fi be, in order, 61, 82, and 83. Subtract- 

 ing values of the a's as above found we have, for 81, oi = 0, ao = —A31, 

 03 = 0; for 82, fli = 0, as = +^31, 0.3 — —A21; for 83, ai = 0, 02 = 0, 

 03 = + J21. Since oi 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 Fo- Let /3i = 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 82 = ct -\- c j. 

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

 cV) + terms in k. It was shown that i^s = ■f'2 + P'S62p. Now F2 

 contained no component along ^3. By adding pS52P we removed 

 the component along ^2- But p88op does not contain z. Therefore 



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



