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I am reading this article and looking at eq. (4). I am not sure how the expressions in terms of the viewing angle $\iota$ and the phase $\Phi$ are achieved. It looks to me like it has nothing to do with general relativity but with geometrical definitions. Let the gravitational wave amplitude be $$ h_0 \equiv \frac{4c}{D} \left( \frac{G \cal{M}}{c^3} \right) \Omega^{2/3} .$$ Then the expressions are

$$h_+ = h_0 \cdot \frac{1}{2} (1+\cos^2\iota)\cos 2\Phi(t)$$

$$ h_x = h_0 \cdot \cos\iota \sin 2\Phi(t).$$

The angles are related to how we observe the waves, and I am not sure how they come about.

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  • $\begingroup$ Are you asking about what these angles represent or about how the dependence of the strain on these angles comes about? $\endgroup$
    – TimRias
    Commented Dec 28, 2021 at 21:49
  • $\begingroup$ I am asking about the exact dependence (cosine / sine expressions of the appropriate parameters). Thanks $\endgroup$
    – Halo
    Commented Dec 28, 2021 at 22:10
  • $\begingroup$ I found the following reference (eq. 3.22): dcc.ligo.org/public/0106/T1300666/003/Whelan_geometry.pdf a lot more involved than I thought... $\endgroup$
    – Halo
    Commented Dec 29, 2021 at 16:40

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