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Regauging an explicit harmonic-gauge plane-wave amplitude to transverse-traceless form

Example

Work in the inertial chart of the flat model with the wave covector ka=ω(1,0,0,1), ω=1, and let X be the space of symmetric amplitudes of Plane-wave perturbations of the flat linearization: planar profiles, wave covectors, amplitudes, TT form and the linearized field equations clause (c). Consider the amplitude

X=(−2−200−210200−100202),

whose components in the orthonormal basis (e0,e1,e2,e3) are displayed in the order (0,1,2,3). Then:

  1. X lies in V0={X:kaXˉab=0} and is not TT: X(e0,e0)=−2≠0 and its trace is tr⁡X=−X00+X11+X22+X33=2+1−1+2=4≠0.
  2. The unique Z of Harmonic-gauge plane-wave amplitudes split uniquely into a pure plane-wave gauge part and a transverse-traceless representative, with a two-dimensional space of polarizations clause (c) with Xab′:=Xab−kaZb−kbZa TT is Z=(1,2,0,1), and X′=(0000010000−100000), so the amplitudes of the TT representative are A+=1, A×=0.
  3. With ξb=Zbsin⁡(k⋅x) the perturbation hab=Xabcos⁡(k⋅x) satisfies hab′:=hab−∂aξb−∂bξa=Xab′cos⁡(k⋅x), which is a TT plane wave in harmonic gauge.

Verification

Given: The inertial chart, ka=(1,0,0,1), ka=(−1,0,0,1), the amplitude X above and the vector Z=(1,2,0,1).

[L1] A symmetric amplitude X is in V0 exactly when kaXˉab=0 for every b. For X∈V0 and ω≠0 in the adapted frame, its unique TT representative is Xab′=Xab−kaZb−kbZa, where Z0=−X00/(2ω), Z1=−X01/ω, Z2=−X02/ω, Z3=Z0−X03/ω (Harmonic-gauge plane-wave amplitudes split uniquely into a pure plane-wave gauge part and a transverse-traceless representative, with a two-dimensional space of polarizations clauses (b)-(c)).

1.1L1algebra

(Membership in V0 and failure of TT.) The trace is tr⁡X=ηabXab=−X00+X11+X22+X33=2+1−1+2=4, so Xˉab=Xab−2ηab has entries Xˉ00=−2+2=0, Xˉ01=−2, Xˉ03=0, Xˉ11=1−2=−1, Xˉ13=2, Xˉ22=−1−2=−3, Xˉ33=2−2=0 and Xˉ12=Xˉ23=0. With k0=1, k3=1, k1=k2=0, the four contractions are kaXˉa0=Xˉ00+Xˉ30=0, kaXˉa1=Xˉ01+Xˉ31=−2+2=0, kaXˉa2=0, kaXˉa3=Xˉ03+Xˉ33=0; hence X∈V0. It is not TT because X(e0,e0)=X00=−2≠0 and tr⁡X=4≠0. This is clause 1.

2.1L1step 1.1algebra

(The TT representative.) Since X∈V0 by step 1.1, [L1] with ω=1 gives the unique Z as Z0=−X00/2=1, Z1=−X01=2, Z2=−X02=0, Z3=Z0−X03=1, which is the displayed Z. Forming kaZb+kbZa with ka=(−1,0,0,1) gives the matrix with entries (kZ+Zk)00=−2, (kZ+Zk)01=−2, (kZ+Zk)03=0, (kZ+Zk)11=0, (kZ+Zk)13=2, (kZ+Zk)22=0, (kZ+Zk)23=0, (kZ+Zk)33=2, and subtracting from X gives the displayed X′, whose only nonzero entries are X11′=1 and X22′=−1; hence A+=X11′=1 and A×=X12′=0. This is clause 2.

3.1L2step 2.1algebra∎

(The perturbation.) By [L2] the gauge-transformed perturbation of the plane wave hab=Xabcos⁡(k⋅x) is h′=X′cos⁡(k⋅x), and X′∈T⊆V0 is TT and in harmonic gauge in the sense of Plane-wave perturbations of the flat linearization: planar profiles, wave covectors, amplitudes, TT form and the linearized field equations clause (c). This is clause 3.

Remarks

Depends on

Dependency tree · 2 levels

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