# Gravitational wave detection time difference between LIGO Livingston and LIGO Hanford

Quote from LIGO’s news release:

By looking at the time of arrival of the signals—the detector in
Livingston recorded the event 7 milliseconds before the detector in
Hanford—scientists can say that the source was located in the Southern
Hemisphere.

GPS coordinates of LIGO Livingston: 30.5630018,-90.7763949
GPS coordinates of LIGO Hanford: 46.4554032,-119.4092701

Distance: 3030.13 km

Now if the gravitational waves propagate outwards at the speed of light a difference of 7ms implies the following maximum possible detector distance:

299792458 * 0.007 / 1000 = 2099 km


Where am I wrong ?

Update:

As pointed out by james-k in the comments below, the calculated 2099 km is the minimum distance and not the maximum … these waves seem to have caused some mental derangement 😉

Astronomy Asked by gollum on November 19, 2021

The signal does not travel from Livingston to Hanford. The signal comes in with an angle of about 45 degrees to the line joining the two, and has the same amplitude on planes perpendicular to the direction of travel (the peaks of water waves are not along the direction of travel of the water waves but perpendicular to the direction of travel). Thus if the signal came in at 90 degrees of the line joining Hanford and Livingston, the peaks of the signal would arrive at both at the same time. If it came in in the direction along the path, then it would take the signal 10 ms later at Hanford than at Livingston. At 45 degrees, the peak at Livingston would arrive $$cos(45°) times 10$$ms $$= 7$$ms later at Hanford than at Livingston. Thus the direction of the signal is on a cone with apex angle very close to 45 degrees. That is all one could tell from the timing.

Hanford and Livingston detectors were build to be as close to parallel to each other as possible. If we lived on a flat earth, then the signal at both would be identical. The earth is curved. Thus the detector at Hanford is tilted from the straight line joining them by about 27 degrees with respect to Livingston. This means that the sensitivity of the detector is not the same for incoming directions all along the cone. In some directions, Hanford is about 50% of the sensitivity of Livingston while for others they are nearly the same. By comparing the amplitudes at Hanford and Livingston, one can get a very rough idea of where along that 45 degree cone the signal came from, limiting the directions to something like half the cone.

When Virgo came online, even though its sensitivity is much less than either of the LIGO detectors, that extra timing information is enough to strongly limit the directions the signal could have come from.

Answered by W Unruh on November 19, 2021

On Feb 12 James wrote:"A difference of 7ms would imply a minimum separation of the sites of about 2000 km, not a maximum."

The calculated minimum separation of the sites is around 3001.65 km compared with the estimated maximum of 3030.12 km indicated by gollum that same day. The angular separation between Livingston and Hanford detectors is around 27.25 degrees from the center of our planet using an average Earth's radius of 6371.0 Km and gives this result. If the detected GW150914 gravitational waves propagated outwards at more speed that the speed of light a difference of 7ms is in accord with this MINIMUM possible detector distance.

The GW's velocity on 2015/09/14/09:50:45 UT based on this argument would be:

(3001.65 km/2099.0 km)*c ~ 1.43 times the speed of light

Answered by Juan Zapata-Arauco on November 19, 2021

Updated to use reported timing confidence intervals instead of trying to infer those from reported sky location resolution uncertainties. The latter approach was misleading because the sky location has been further resolved using more information than just the arrival time difference.

Beware that I am neither an astronomer nor involved with aLIGO. I just use classic euclidean geometry here, combined with occasional transformations between cartesian and spherical coordinates.

We have two observatories, $L$ (Livingston) and $H$ (Hanford) and a far-away point $E$ emitting a (gravitational wave) signal. Let us assume that the signal propagates at constant speed $c$ through low-density matter, including planet Earth. The straight-line distance between $L$ and $H$ (passing through Earth) is $|HL| approx ccdot 10.0cdot10^{-3},mathrm{s}$.

$L$ receives the signal from $E$ about $t_{ELH}approx 7cdot10^{-3},mathrm{s}$ before $H$. This implies that $E$ is a distance $d = c,t_{ELH}$ closer to $L$ than to $H$.

Now assuming large-scale euclidean geometry and looking at the plane containing $E,L,H$, the set of all candidate points $E$ where $|EH|-|EL| = d$ is one branch of a hyperbola, namely the branch closer to $L$:

In fact, setting begin{align} r &= frac{d}{2} & r_1 &= frac{|HL|}{2} & R &= |LE| \ L &= (r_1,0) & H &= (-r_1,0) & E &= (x,y) end{align} we get the cartesian hyperbola equation $$frac{x^2}{r^2} - frac{y^2}{r_1^2-r^2} = 1$$ and in polar coordinates, viewed from $L$: $$R = frac{r_1^2 - r^2}{r - r_1cosalpha}$$ For very large $R/r_1$, $|alpha|$ must be very close to its asympotic limit: begin{align} cosalpha &approx frac{r}{r_1} = frac{d}{|HL|} = frac{t_{ELH}}{t_{HL}} \ text{where}quad t_{HL} &= frac{|HL|}{c} approx 10.0cdot 10^{-3},mathrm{s} qquadtext{(time directly from H to L)} end{align} In 3D this means that $E$ lies on one of the two sheets of a hyperboloid of revolution whose axis of revolution is the line $HL$. In particular, in $L$'s view of the sky, partially hidden under the horizon, there should be a fairly thin circular band that covers all projected candidate points for $E$.

The solid angle of view enclosed by the circular band is $$A_{text{ste}} = 2pi(1 - cosalpha) approx 2pileft(1 - frac{t_{ELH}}{t_{HL}}right)$$ Uncertainties in $t_{ELH}$ yield uncertainties in $cosalpha$ which widen the circular band. Thus the circle becomes an annulus. The solid angle of view covered by that annulus measures begin{align} A_{text{ste.max}} - A_{text{ste.min}} = 2pi(cosalpha_{text{min}} &{}- cosalpha_{text{max}}) approx 2pifrac{t_{ELHtext{.max}} - t_{ELHtext{.min}}}{t_{HL}} \ text{briefly:}quad Delta A_{text{ste}} &approx 2pifrac{Delta t_{ELH}}{t_{HL}} end{align} Substituting reported data for $t_{ELH}$, with $text{min}$ and $text{max}$ referring to 90% probability intervals, we have: $$begin{array}{rcrrr} text{entity} & text{unit} & text{nom} & text{min} & text{max} & Delta \hline t_{ELH} & 10^{-3},mathrm{s} & 6.9 & 6.5 & 7.4 & 0.9 \ cosalpha & 1 & 0.69 & 0.65 & 0.74 & 0.09 \ 90^circ - alpha & {}^circ & 44 & 41 & 48 & 7 end{array}$$ And thus the annulus, when projected to a unit (celestial) sphere, covers $$Delta A_{text{ste}} approx 0.18pi approx (1.9cdot10^3)^{circcirc}$$ (The ${}^{circcirc}$ means square degrees.) In contrast, the reported value of the credible region varies between $140^{circcirc}$ (50% probability) and $590^{circcirc}$ (90% probablity). This is because large parts of the annulus could be ruled out based on additional information such as signal strength ratios. But those refinements are not the topic of this thread.

For easier visualization, let us move to a point $B$ on Earth's surface where the center of the annulus is in the zenith, and the annulus itself is at constant altitude $90^circ - alpha$, as given in the above table. To that end, let $O$ be the location of Earth's center and require that the 3D ray $OB$ points in the same direction as the ray $HL$. This determines the latitude and longitude of $B$. Note that the parallel shift from $HL$ to $OB$ incurs a parallax, but that is negligible because the distance to $E$ is so large.

Idealizing our planet to a sphere and doing the calculations (transforming the locations of $H$ and $L$ from spherical coordinates to cartesian coordinates, computing the difference vector, normalizing it, and transforming back to spherical coordinates), I find $B$ to be at $$Bcolon 27^circ 21'54.22''mathrm{S} 38^circ 36'33.45''mathrm{W} = 27.365061^circmathrm{S} 38.609293^circmathrm{W}$$ which is some 700km southeast of Rio de Janeiro in the Atlantic. If you had been there at 2015-09-14 09:50:45 UTC and had spun around yourself, drawing a band at constant altitude $90^circ - alpha approx 44^circ$ and width $7^circ$ in the sky, that would have marked all the directions where the signal source $E$ could have been.

To get the sky view simulated, I have written a tiny script gw150914.sts for Nightshade:

# Display the sky at the time of the aLIGO GW150914 observation
# from an Earth-based position chosen such that the set of possible sky
# positions of the source form a circular annulus with azimuth-independent
# altitude = asin(delta_t/10ms), about 44 deg.
clear
timerate rate 0
date utc 2015-09-14T09:50:45
flag show_tui_datetime on
set home_planet "Earth"
moveto lat -27.365061 lon -38.609293 alt 0 heading 0 duration 2
wait duration 2
flag azimuthal_grid on
zoom fov 120 duration 2


This gives the view below. I have added dashed green circles to delimit the annulus.

But do not expect to find likely candidate spots within that annulus. The annulus hides the radial depth of the projected space volume. A range for that radial depth is given by the estimated bounds for the distance $R=|LE|$. The corresponding flat space volume is $$Delta V = frac{1}{3}Delta A_{text{ste}} left(R_{text{max}}^3 - R_{text{min}}^3right)$$ Plugging in the reported values, again from 90% probability intervals, begin{align} Delta A_{text{ste}} &approx 590^{circcirc} approx 0.0572pi & R_{text{min}} &approx 0.23cdot10^{9},mathrm{pc} & R_{text{max}} &approx 0.57cdot10^{9},mathrm{pc} end{align} yields $$Delta V approx 10^{25},mathrm{pc}^3$$ To get a very rough estimate of the number of galaxies in that volume, let us multiply $Delta V$ with a rough galaxy density estimate of $10^{-20},mathrm{pc}^{-3}$, which results in about $10^5$ galaxies in that volume. My conclusion from this back-of-the-envelope calculation is that we cannot hope to localize $E$ further, unless given more information.

Things would have been a bit different if we had had a third aLIGO observatory distant from the line $HL$ operating and capable of detecting the signal from $E$. Then we could draw three annuli, take intersections and thus determine the direction of the signal within one reasonably small spot on the celestial sphere.

• As this LIGO post tells us, parts of the annulus could be ruled out by considering additional features of the data such as varying signal strength.
• A more detailed description of the probability distributions for the results is given in this LIGO paper.
• More information about the methods used to further narrow the credible sky regions of the source location is given in this paper.

An equivalent script gw150914.ssc for Stellarium:

// Display the sky at the time of the aLIGO GW150914 observation
// from an Earth-based position chosen such that the set of possible sky
// positions of the source form a circular annulus with azimuth-independent
// altitude = asin(delta_t/10ms), about 44 deg.

var obs_lat = -27.365061;
var obs_long = -38.609293;
var obs_alt = 0;
var obs_locname = "Atlantic, 700km SE from Rio de Janeiro";
var obs_planet = "Earth";
var obs_date = "2015-09-14T09:50:45";
var duration = 2;

core.clear("deepspace");
core.setTimeRate(0);
core.setDate(obs_date, "utc");
core.setObserverLocation(obs_long, obs_lat, obs_alt, duration,
obs_locname, obs_planet);
core.wait(duration);
core.moveToAltAzi("90", "0", duration);
core.wait(duration);
GridLinesMgr.setFlagAzimuthalGrid(true);
StelMovementMgr.zoomTo(120, duration);


And yes, core.setObserverLocation expects longitude before latitude.

Answered by ccorn on November 19, 2021

The addition of more detectors will immensly help event localization. To get an understanding of the calculation look at the old LORAN navigation system. This system used synchronized transmitters, and the receiver measured the difference in time.

The receiving device measured the time between signals from multiple sources, that chirp at the same time. If you hear the chirps at the same time, from two of the synchronized transmitters then you are on the plane equidistant between the sources. The plane is perpendicular to the line between the sources, and at the halfway point. If you plot the intersection of the plane, and the surface of the earth it creates a great circle. You are on the great circle.

Signals that have a difference in time of reception form a curved surface. As it cuts through the earth it defines a curved line on the surface (hyperbolic).

If the signals come at different times then you are closer to one than the other. The charts used for determining position had lines of constant time difference plotted. You listened to one pair of transmitters, got your time difference, found the line on the chart that represents that delay. Do this for another pair of transmitters, get that difference, find the line for that pair with that delay, and where the lines cross is about where you are.

This is repeated for multiple pairs, and finding all the intersections (the best two lines are those crossing at right angles to reduce the error possible in your timers).

This is the same thing in reverse.

Get more listener pairs and we can start to isolate the region of space of the signal source.

I hope that helps.

Answered by Michael F. on November 19, 2021

Here's how my non-scientist mind envisions it. I draw a straight line between the two LIGO sites on a map. Then I take another straight line (like a straight edge/ruler) that represents the GW coming in. If the GW line is coming from the south exactly parallel with the line I drew on the map, then both sites would detect the "chirp" at exactly the same time (0 milliseconds difference). If the GW line comes from the south exactly perpendicular to my drawn line, hitting the LA site first, then chirp would be detected at each site at approx 10 milliseconds (?) apart (assuming that's how long it takes for light to travel between the two sites in a straight line).

So I envision the GW "line" came in from the south somewhere between parallel and perpendicular (closer to perpendicular) in relation to my drawn line. Someone else smarter than I can figure out the maths and correct angles, numbers, etc.

Does that make any sense?

Answered by iMerchant on November 19, 2021

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