Saturday, February 13, 2016 ... Deutsch/Español/Related posts from blogosphere

The Gates Octahedral Gravitational-wave Observatory

According to Wikipedia, LIGO only cost $0.62 billion up to September 2015. It was a lot of money for a project that wasn't guaranteed to produce results but I believe that the cost-and-benefits counting is different now. The benefits have become more obvious.

Two more gravitational waves: TRF is the only place in the world where you can learn about another gravitational wave that was detected on October 12th, one called GW151012 (later named LVT151012, a weaker black hole merger with masses 13 and 23 Suns and a false positive rate 2.3 years), and a Christmas one from December 26th, GW151226. Web cache was later erase; a screenshot. Stay tuned.

González at an AAAS meeting: When LIGO has 4 detections, they will make alerts of detections public! And according to Dr Weiss, there were at least 4 detections by the end of January. What should the previous sentences imply? ;-) Incidentally, readers-Pythonists should be able to get all the code used by LIGO to manipulate with the GW150914 signal.
There are only two LIGO detectors in the U.S. Consequently, one of the coordinates of the source of the gravitational waves cannot be measured accurately. Also, LIGO is largely insensitive to waves coming from certain directions, with certain polarizations, and it cannot accurately measure the time profile of both independent polarizations independently.

It's therefore common sense for Bill Gates to reduce his spending at random charity projects – something that could be easily done by the Western central banks etc. – and build a $1 billion GOGO, the Gates Octahedral Gravitational-wave Observatory. I believe that because the gadgets may be clones of each other, the budget could be much smaller than a simple multiple of one gadget. And if a GOGO clone is built somewhere in Africa, it could be better for the local populace than just a pile of cash.

Platonic polyhedra are cool. I chose the vertices of an octahedron to define the optimum locations of the GOGO detectors. Where are they?

Note that an octahedron has 8 faces, 6 vertices, and 12 edges; it is the dual polyhedron (via a face-vertex exchange) to the cube which has 6 faces, 8 vertices, and 12 edges. The 6 vertices of the polyhedron may be imagined to be e.g. the 2 poles and 4 points equally spaced around the equator.

Alternatively, for another simple configuration, you may place three vertices at the circle of latitude \(\theta\) (in spherical coordinates conventions, \(\theta=0\) is the North pole) and longitudes \(\phi\) and \(\phi\pm 2\pi /3\); and the three antipodal points at the latitude \(\pi-\theta\) and longitudes \(\phi+\pi\) and \(\phi\mp \pi/3\). A trivial Mathematica code
a = {Cos[theta], Sin[theta], 0}
b = {Cos[theta], Sin[theta]*Cos[2*Pi/3], Sin[theta]*Sin[2*Pi/3]}
c = {-Cos[theta], Sin[theta]*Cos[Pi/3], Sin[theta]*Sin[Pi/3]}
Total[(a - b)^2]
Total[(a - c)^2]
Total[(b - c)^2]
is enough to see that the three squared distances at the bottom are \(3\sin^2\theta\) and \(4\cos^2\theta+\sin^2\theta\) (twice) and they're equal, as expected for a regular polyhedron, if \(4\cos^2\theta=2\sin^2\theta\). It follows that \(\tan^2\theta=2\), \(\tan\theta=\sqrt{2}\), \(\theta\approx 0.955\) or 54.7°. The normal latitudes (with zero at the equator) are 35.3°. Can you verify this calculation of mine?

Obviously, GOGO would work pretty well even if the locations were a bit different.

So you have these six LIGO-like devices at the places of the six vertices. What about the polarizations? Note that the planes of the pair of antipodal points are exactly parallel. You don't want these pairs to measure exactly the same signal. Instead, you ideally want them to measure exactly the opposite – the "cross" vs "plus" polarizations relatively rotated by 45° i.e. \(\pi/4\) radians from one another.

I think that there is a clever way to achieve this goal in a totally symmetric way.

Sit at one of the GOGO centers, at the vertex of the octahedron. There are 5 other vertices. One of them is the antipodal point. The other 4 are "neighbors". If you stand at the GOGO location and point your finger to the 4 neighbors, you get a cross (well, you want to keep your arms horizontal so raise them). You could build the GOGO arms along the 2 directions of the cross. But in this way, you would get the same polarizations on the antipodal stations.

Instead, if you rotate this cross by 22.5° i.e. \(\pi/8\) in the clockwise way (or counter-clockwise, but the direction is expressed by the same local word at each location), you will make the antipodal stations's crosses rotated by 45° relatively to one another: they can perfectly measure the signal in both possible "linear" polarizations independently.

The same comment applies to each of the 3 antipodal pairs of GOGO detectors.

I think that it's a pretty good idea to have 6 (or at least 5) detectors because all the six spatial components of the metric tensor \[


\] may be nonzero when a gravitational wave arrives. Well, the trace \(h_{xx}+h_{yy}+h_{zz}\) is unphysical and is usually set to zero by a (diffeomorphism) gauge choice which is why I mentioned the number 5 as well. But the octahedron is more symmetric than any arrangement with 5 locations. To say the least, there is no Platonic polyhedron with 5 vertices.

Note that one detector basically measures something like \(h_{xx}-h_{yy}\) as a function of time, in some generally \(SO(3)\) rotated local \(xyz\) coordinate system. So six detectors are enough to basically measure the six spatial components \(h_{ij}\) as functions of time and I believe that the mankind should be doing that soon.

You may calculate the minimum signal in the GOGO system one gets for a given direction of the gravitational wave.

If the gravitational wave goes in the direction of the line connecting a vertex and its antipodal friend, these two detectors are the most sensitive ones. Without a loss of generality, it is enough to consider the directions of the gravitational wave in one of the 8 faces of the octahedron. Well, 1/6 of the face is enough.

The other extreme points appear when the gravitational wave's direction goes along the center of the edge; or a center of a face (which is where the vertex of the dual cube would be located). In those situations, larger numbers of the detectors (4 or all 6) will share the strongest signal more democratically.

You may like the octahedral location but you're worried about the sensitivity. Will GOGO be more sensitive than LIGO? Won't it be too expensive? Good questions. GOGO will be better: that's where Warren Buffett enters. ;-)

I've tried to find the best 6 locations of the vertices of the octahedron but haven't decided about the locations yet LOL. 2/3 of the Earth's surface is ocean. So chances are that for a generic orientation of the octahedron, 4 of the 6 vertices would be in water. You can do better – and if you don't do better, you may move the locations a little bit relatively to the exact octahedron, not a big deal.

Add to Digg this Add to reddit

snail feedback (0) :

(function(i,s,o,g,r,a,m){i['GoogleAnalyticsObject']=r;i[r]=i[r]||function(){ (i[r].q=i[r].q||[]).push(arguments)},i[r].l=1*new Date();a=s.createElement(o), m=s.getElementsByTagName(o)[0];a.async=1;a.src=g;m.parentNode.insertBefore(a,m) })(window,document,'script','//','ga'); ga('create', 'UA-1828728-1', 'auto'); ga('send', 'pageview');