TRANSFORMATION OF THE DERIVATIVE OF A VECTOR
Consider a vector v that is being observed simultaneously from two frames Fa and Fh that have relative rotation—say Fh rotates with angular velocity ы relative to Fa, which we may regard as fixed. From (A.4,3)
The derivatives of ya and h are of course
(A.4,13)
where vai = (d/dt)(vai), and so forth. It is important to note that v„ and yb are not simply two sets of components of the same vector, but are actually two different vectors.
Now because Fb rotates relative to Fa, the direction cosines are changing with time, and the derivative of (A.4,3) is
ifh = L.„v„ + L h/yn
or alternatively
V, = L ahyb + L abyh
the second terms representing the effect of the rotation.
Since L must be independent of v, the matrix Lab can readily be identified by considering the case when vb is constant (see Fig. A.5.). For then, from the fundamental definitions of derivative and cross product, the derivative of v as seen from Fa is readily shown to be
dy
— = Ы X V dt
The corresponding result from (A.4,14) is
Va = tabv„ (A.4,17)
It follows from equating (A.4,16) and (A.4,17) that
L аЬУЬ = "aVa
or
Kb^b = &аКьУь (A.4,18)
for all yb. Whence — <UaLafe
^ab^ba
Finally if the above argument is repeated with Fb considered fixed, and Fa having angular velocity – to, we clearly arrive at the reciprocal result
Lba = – d>bLba (A.4,19)
From (A.4,18) and (A.4,19), recalling that to is skew-symmetric so that to’ = —to, the reader can readily derive the result
From (A.4,14), (A.4,18), and (A.4,19) we have the alternative relations
Vft = Lbaxa – 6>hxh xa = L ahvh + ыаха
with two additional permutations made possible by (A.4,20). A particular form we shall finally want for application is that which uses the components of xa transformed into Fh, viz.
L bJa = + 6>bxh (A.4,22)
TRANSFORMATION OF A MATRIX
Equation (A.4,20) is an example of the transformation of a matrix, the elements of which are dependent on the frame of reference. Generally the matrix of interest A occurs in an equation of the form
v = Au (A.4,23)
where the elements of the (physical) vectors u and v and of the matrix A are all dependent on the reference frame. We write (A.4,23) for each of the two frames Fa and Fh, that is,
v„ = A„u(,
v,, = Ahub
and transform the second to
= A,,Lfc„uu
Premultiplying by Lab we get
Va = f a/)A/)L/,„U„
By comparison with (A.4,24a) we get the general result
A a = LahAbLba