# LIGO only measures of diagonal components of metric tensor?

In a paper from Paik et al. 2016, they state (Section 3) that, a terrestrial gravitational wave (GW) detector measures only one off-diagonal component of the metric tensor.

Can anyone further expand on why this is so? From my copy of Misner, Thorne, and Wheeler, I understand that we work with a linearized theory and adopt the transverse traceless (TT) gauge, which reduces the perturbed metric to be written just in terms of 2 degrees of freedom corresponding to the two GW polarizations.

In this case, is a terrestrial detector (e.g. LIGO) not detecting two components, and they are not necessarily off-diagonal?

Thanks

• As someone who's not too familiar with GR but very familiar with the detectors themselves, I can at least tell you that a single interferometer outputs exactly one number as a function of time. So from this you will only be able to derive one independent parameter of the metric tensor. But if you combine the time series data of more than one detector at different locations and orientations, you will learn something about the direction and polarization of the passing wave. – Emil May 5 '17 at 10:14
• Have you seen this dcc.ligo.org/public/0106/T1300666/003/Whelan_notes.pdf – anna v May 5 '17 at 10:21
• Why don't you ask the LIGO experts at question@ligo.org? They will give you the best possible answers to questions like these. – Wrichik Basu May 7 '17 at 19:34
• To expand the point Emil says, that seems exactly what they talk about. Terrestrial detector has only two arms and detects only one component so you have to combine data from several detectors. In contrast they talk about detecting everything with single detector. I don't think such a simple point by itself deserves bounty so won't post this as an answer. – OON May 8 '17 at 13:27

## 1 Answer

It does not matter if you linearize GR or not but in transverse-traceless gauge there are only 2 polarizations ($h_+$ and $h_\times$). In linearized gravity you have the freedom to write these two polarizations as, $$h_{\alpha \beta}(t,z) = \left( \begin{array}{c c c c} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) h_+(t-z) + \left( \begin{array}{c c c c} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) h_\times(t-z) \, ,$$ You may have more polarizations in a different gauge, but, eventually as you say there are only two independent DOF. Considering that the detectors response to an incoming wave is given as $$h = F_+h_+ + F_\times h_\times$$ So with one or two detector you cant say much. But, if you have 3 detectors you can tell the polarization of an incoming wave i.e. is it linear, circular, elliptic etc. polarized ? You can triangulate the source and tell about the skymap.