# How Does LIGO Point Back to a Source?

LIGO published the position of their gravitational wave on the sky when they announced that they had found something, so I was wondering how this could be done. While LIGO has more information that just timing, I decided to consider the case of reconstructing a position based just on the relative timing from two detectors.

I started with a coordinate system with the origin at the center of Earth (assumed to be a perfect sphere here). The z-axis points toward the north pole, the x-axis toward the equator at the Greenwich meridian and the y-axis 90 degrees east of the x-axis. I can then place the two detectors in this coordinate system (easiest using spherical coordinates). The time separation for a gravitational wave along some axis given by a unit vector u(θ,φ) is just

$\Delta t(\theta,\phi) = \frac{R_\oplus}{c}({\bf v_1}-{\bf v_2})\cdot {\bf u(\theta,\phi)}$

where v1 and v2 are the unit vectors pointing toward the two detectors, R is the radius of Earth, and c is the speed of light (the speed of the waves). If each detector can measure the time of the wave to some uncertainty σ and the relative timing uncertainty (typically from GPS) is negligible, I can define some likelihood that u is the correct vector as

$L(\theta,\phi) = e^{-\frac{1}{2}\chi^2(\theta,\phi)}$

where

$\chi^2(\theta,\phi) = \frac{(\Delta t_{\rm meas} - \Delta t(\theta,\phi))^2}{2\sigma^2}$

Using this function for some test values, we can try to understand how pointing works. If the time separation is 0, then we expect to see most likely values along the great circle between the two detectors.

I will also note that this is plotting position on the sky in Earth coordinates. Earth is not in an inertial reference frame: it rotates and moves in space, so actually mapping this against the stars in the sky gets very complicated and I’ll ignore that since it doesn’t affect what I want to show. The coordinates are also spherical coordinates, which are non-Euclidean, so shapes look strange if you’re not used to them. The curve above is actually the spherical version of a “straight” line. For a time separation of 8 ms, we see a thin ring that widens out as the resolution is degraded.

At 10 ms, we’re getting a large separation, and the ring has shrunk almost to just a blob.

Basically, what seems to be going on is:

1. The general shape of the likelihood distribution is a ring of positions corresponding to the measured time separation
2. As the time separation increases, the size of the ring on the sky decreases
3. When the separation is too large (requires a speed less than c), no position works well, so there is a sharp peak at the best (but still non-optimal) position.
4. The ring is characteristic of having two detectors. The center of the ring never changes (there are actually two centers depending on the sign of the separation) and its position is determined by the positions of the detectors
5. A third detector should break the degeneracy and allow for reconstruction of the position as a single point on the sky
6. Since the measured position looks like a blob, it looks like the time separation is large enough for the timing uncertainty to prevent us from seeing a ring-like shape.