14.5 Additional Formulas Involving Projectors 269
bμνPμν,μ(±^2 )′ν′aμ′ν′=1
2
bμνaμν−hμbμνaνκhκ+1
4
(hμbμνhν)(hμ′aμ′ν′hν′)∓
i
2[bμνHντaτμ−hμbμνHντaτκhκ]. (14.58)Now let the tensorabe constructed from the components of the unit vectorse,viz.
aμν=eμeν.Then the real and imaginary parts of (14.55) and (14.56)are
P(μν,μ^0 ) ′ν′eμ′eν′ =3
2
hμhν[
(h·e)^2 −1
3
]
, (14.59)
(Pμν,μ(^1 ) ′ν′+Pμν,μ(−^1 )′ν′)eμ′eν′ =
1
2
(hμeν+hνeμ)(h·e)−hμhν(h·e)^2 ,(Pμν,μ(^2 ) ′ν′+Pμν,μ(−^2 )′ν′)eμ′eν′ =eμeν − 2 hμeν(h·e)+
1
2
hμhν[ 1 +(h·e)^2 ],i(
Pμν,μ(^1 ) ′ν′−Pμν,μ(−^1 )′ν′)
eμ′eν′ =[hμ(h×e)ν+hν(h×e)μ](h·e),i(
Pμν,μ(^2 ) ′ν′−Pμν,μ(−^2 )′ν′)
eμ′eν′ =1
2
[(h×e)μeν+(h×e)νeμ]−1
2
[hμ(h×e)ν+hν(h×e)μ](h·e). (14.60)The cross product(h×e)stems fromHμτeτ =(h×e)μ.Fore=h, all terms
on the right hand side of (14.60) and in the second and third equations of (14.59)
vanish. This is obvious since a rotation about an axis parallel toedoes not change the
direction ofe.Foreperpendicular toh, the equations involvingP..(±^1 )yield zero,
the remaining equations reduce to
Pμν,μ(^0 ) ′ν′eμ′eν′ =−1
2
hμhν, (14.61)
(
Pμν,μ(^2 ) ′ν′+Pμν,μ(−^2 )′ν′)
eμ′eν′ =eμeν+1
2
hμhν,i(
Pμν,μ(^2 ) ′ν′−Pμν,μ(−^2 )′ν′)
eμ′eν′ =1
2
[(h×e)μeν+(h×e)νeμ].Next, the special caseaμν= eμuν is considered, where the unit vectors are per-
pendicular to each other. Then (14.55) and (14.56) lead to
Pμν,μ(^0 ) ′ν′eμ′uν′ =3
2
hμhν (h·e)(h·u), (14.62)
(
Pμν,μ(^1 ) ′ν′+P(μν,μ−^1 )′ν′
)
eμ′uν′ =1
4
[(hμuν+hνuμ)(h·e)+(hμeν+hνeμ)(h·u)]
− 2 hμhν(h·e)(h·u),