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Gravitational waves are not discovered directly (yet). But does this have any consequences? Should we suspect out theories are wrong or not yet?

I found some data about gravitational waves suspected spectrum, like here, and wonder, where is it taken from?

Where is the location of indirect measurements of neutron star energy loss on this plot?

If LISA won't find any waves, what will it mean?

Qmechanic
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Dims
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1 Answers1

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Eventually gravitational waves will have to be detected, or general relativity is wrong.

It is certainly not certain that we should be able to see gravitational waves yet, with the equipment we have -- devices like LIGO and bar detectors require a certain proximity to events that would create large GW sources, like black hole mergers and supernovae, and so, there really is no knowing whether or not one has happened, and therefore, whether we SHOULD be seeing a gravitational wave. Heuristic calculations indicate that Advanced LIGO should be able to detect something, but it does depend on the count of binary black hole systems that are out there.

Zo the Relativist
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    Can't we know some specific source, like binary systems, calculate their population, emission intensity, calculate fading and be sure, that some detector DEFINITELY should detect waves? – Dims Jan 16 '15 at 19:46
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    The problem is one of sensitivity. Gravity is an extremely, extremely weak force to try to detect as waves. Generation of detectable waves requires certain types of events, such as masses orbiting each other or large explosions. We can't feasibly build a gravity wave generator because the noise factors would drown out the gravity waves easily, and the only natural sources we know of are very far away. It would be like trying to count the flagella on a bacterium using only a magnifying glass from across the room – Asher Jan 16 '15 at 21:21
  • There must be a 2D graph with regions on it showing our capability to observe vs expected event strength? – Tom Andersen Jan 17 '15 at 02:01
  • @TomAndersen:

    http://www.nature.com/nphoton/journal/v7/n8/images/nphoton.2013.177-f3.jpg

    "Strain" roughly corresponds to the amplitude of the $h_{+x}\cos \theta$ terms in the TT-gauge wave solutions.

     – Zo the Relativist Jan 19 '15 at 22:51
  • @Dims: getting expected event counts is nontrivial, and researchers disagree over orders of magnitude, especially when you restrict it to the "visible" region. – Zo the Relativist Jan 19 '15 at 22:52
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    Jerry, thanks. It would be nice if that graph had some 50 parsec double black hole coalesce on it, though. – Tom Andersen Jan 21 '15 at 00:32