
50 Space-time splitting
Note that P (U, u) = P (u, U)
−1
shown as, from their representations,
P (U, u)=P (u)+γνU ⊗ ˆν(U, u),
P (U, u)
−1
= P (U )+νU ⊗ ˆν(u, U),
P (u, U)=P (U)+γνu ⊗ ˆν(u, U ),
P (u, U)
−1
= P (u)+νu ⊗ ˆν(U, u). (3.121)
Let us deduce each of the above relations.
(a) Relation (3.121)
1
is easily verified. Consider a vector X ∈ LRS
u
; then, by
definition,
P (U, u)X = P (U)P (u)X = P (U)X = X + U(U · X) ; (3.122)
but
U · X = γ(u · X + νˆν(U, u) · X)=γνˆν(U, u) ·X.
Hence
P (U, u)X = X + γνUˆν(U, u) · X =[P (u)+γνU ⊗ ˆν(U, u)]
X, (3.123)
which completes the proof.
(b) Let us now verify (3.121)
2
. Consider a vector X = P (U, u)Y ∈ LRS
U
, with
Y ∈ LRS
u
; then, by using the relation u = γ(U + νˆν(u, U)) and the property
Y · u =0,wehave
Y · U = −νY · ˆν(u, U).
Hence
Y = P (U)Y − U(U · Y )=P (U, u)Y + νU[Y · ˆν(u, U)]
= P (U, u)Y + νU[P (U, u)Y · ˆν(u, U)]. (3.124)
Substituting Y = P (U, u)
−1
X gives
P (U, u)
−1
X = P (U)X + νU[X · ˆν(u, U )]
=[P (U)+νU ⊗ ˆν(u, U)]
X, (3.125)
which completes the proof.
(c) Relations (3.121)
3
and (3.121)
4
are straightforwardly verified by exchanging
U with u in (3.121)
1
and (3.121)
2
respectively.
One can then show that
P (U, u)ˆν(U, u)=−γ ˆν(u, U ),
P (u, U)
−1
ˆν(U, u)=−
1
γ
ˆν(u, U ). (3.126)
Note that, from the property (3.3) of the projection operator P (U), we have
P (U, u)=P (U)
P (u)=P (U) P (U, u)=P (U, u) P (u). (3.127)