in Eq. (4-100). Also, from Eqs. (4-93) and (4-95) we derive

 1
 HABCD =2 (ABC, D + BAC,D ), (B12)

 HABCDE ( ABC, DE + BAC,DE) (B13)
 6 oC 0

where FABC 7IAF Bc. Then, substituting Eqs. (B3), (B4), (Bl1), (B12) and

(B13) into Eqs. (B8)-(B10), and exploiting the facts that rABn(j 0 and that

hABB = 0, we obtain

 0 0 A,60 A C D 0D K rL
 g(FN) -R- ACDB o0 i (K)I (L)XFN FN
 1 AA0BC iD iE XK L M
 R ACDB, E U0 7o(K) 7o(L) o(M) FNXFNAFN





 9(FN) RACDB U 0 n o(K) o(L) FNXFN
 1 j Ai(I)B C 0D iE xK xL xM
 + RACDB, E o0 o(K) o(L) O(M)XFNXFNXFN

 +O(X4/NR4), (B15)


 I I 1 ACDB o (I)Ain(J)BC iD XK XL
 9(FN) 3 ACDB 0 o(K) o(L) FNX FN
 1 D ii o(I)A (J)B C 0D iE vK vL vM
 -6 ACDB, E o o o(K) o(L) o(M)XFNXFNXFN

 +O(XN/NR4), (B16)

where the identities used are


 RABCD 0 ABD,C ABC,D (B17)
 0 0 0
 RABCD,E { ABD,CE ABC, DE (B8)
 0 0 0

 The local tetrad vectors Ap A A( p A A(3) where P (0,1, 2, 3)

is the label for each vector of the tetrad, essentially yield the inverse-Lorentz boost

between the Fermi normal frame and the static inertial frame via Eqs. (B1) and