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On page 238 of his second edition of Schutz's he writes the following:
and ##\delta \bar{h}^{TT}_{xx}=2\pi \sigma m \Omega \ell_0 R \sin [\Omega (z-t)-\phi]##.
Here's what I tried:
$$A\tan \psi / ( 2\pi \sigma m \Omega \ell_0 R )= \cos \phi $$
$$A\cos(\Omega (z-t))+2\pi \sigma m \Omega \ell_0 R \sin ((\Omega (z-t)-\phi)=A\cos (\Omega (z-t))+2\pi \sigma m \Omega \ell_0 R [\sin (\Omega (z-t))\cos \phi -\cos (\Omega (z-t))\sin \phi ] =$$
$$ (A-2\pi \sigma m \Omega \ell_0 R \sin \phi )\cos (\Omega (z-t)) + 2\pi \sigma m \Omega \ell_0 R \sin (\Omega (z-t)) \cos \phi = \frac{(A-2\pi \sigma m \Omega \ell_0 R \sin \phi )\cos (\Omega (z-t))\cos \psi + A \sin (\Omega (z-t)) \sin \psi}{\cos \psi}$$
How to proceed to get the above identity in equation (9.117), I don't see it.
Can you help me?
Thanks in advance!
Where Eq. (9.107) is: $$\bar{h}^{TT}_{xx}=A\cos (\Omega (z-t)) , \bar{h}_{yy}^{TT}=-\bar{h}^{TT}_{xx}$$If we now add this to the incident wave, Eq. (9.107), we get the net result, to first order in ##R##,
$$(9.117)\bar{h}^{net}_{xx}=\bar{h}_{xx}^{TT}+\delta \bar{h}^{TT}_{xx}=(A-2\pi \sigma m \Omega \ell_0 R \sin \phi)\cos [\Omega (z-t)-\psi]$$
where $$(9.118) \tan \psi = \frac{2\pi \sigma m \Omega \ell_0 R}{A} \cos \phi$$
and ##\delta \bar{h}^{TT}_{xx}=2\pi \sigma m \Omega \ell_0 R \sin [\Omega (z-t)-\phi]##.
Here's what I tried:
$$A\tan \psi / ( 2\pi \sigma m \Omega \ell_0 R )= \cos \phi $$
$$A\cos(\Omega (z-t))+2\pi \sigma m \Omega \ell_0 R \sin ((\Omega (z-t)-\phi)=A\cos (\Omega (z-t))+2\pi \sigma m \Omega \ell_0 R [\sin (\Omega (z-t))\cos \phi -\cos (\Omega (z-t))\sin \phi ] =$$
$$ (A-2\pi \sigma m \Omega \ell_0 R \sin \phi )\cos (\Omega (z-t)) + 2\pi \sigma m \Omega \ell_0 R \sin (\Omega (z-t)) \cos \phi = \frac{(A-2\pi \sigma m \Omega \ell_0 R \sin \phi )\cos (\Omega (z-t))\cos \psi + A \sin (\Omega (z-t)) \sin \psi}{\cos \psi}$$
How to proceed to get the above identity in equation (9.117), I don't see it.
Can you help me?
Thanks in advance!
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