3 ms·

Stoset is right. The reason why we do not notice them is due to the inverse square law. This event was 1.3 billion light years away. 1 Solar mass of energy bein

Stoset is right. The reason why we do not notice them is due to the inverse square law. This event was 1.3 billion light years away. 1 Solar mass of energy being turned into a gravitational wave nearly instantly creates a very strong gravitational wave. But by the time it comes here, it deforms things by distances that we normally reserve for subatomic particles. What this finding shows is that this happens somewhat regularly in our Universe, it was just that we were previously unable to achieve the precision to measure such small deformations.

jessriedel 10y ago
Not really. You could have been within a few thousand miles of this collision and survived. In fact, it would be well below your ability to notice if it were 1 AU away. http://www.scottaaronson.com/blog/?p=2651 http://www.scottaaronson.com/blog/?p=2651 Needless to say, you would not survive if 1 solar mass of energy were released in any other form (even neutrinos). The key feature is that that tremendous amount of energy results in only a minuscule deformation of spacetime, i.e., spacetime is extremely stiff. By the way, the distances are way smaller than merely subatomic. The strain sensitivity is 10^-21, which for the ~ 1 km arms of LIGO is a shift of just 10^-8 the width of an atom.
- sampo 10y ago
  > you would not survive if 1 solar mass of energy were released in any other form (even neutrinos) But it is close! A supernova releases "few times 10^45 J of neutrino energy" [1], so let's say 5. 5e45 J is about 6e28 kg, while solar mass is 2e30 kg. And neutron radiation from a supernova would get fatal when closer than about 2.3 AU [2]. So we have a factor of 30 from the masses and a factor of 5 (1^2 AU vs. 2.3^2 AU) from the distance. So about 1/150 of solar mass released in neutron radiation would be survivable at the distance of 1 AU. [1] http://wayback.archive.org/web/20120313045458/http://www.andrewkaram.com/andy/pdf/HPJ.pdf http://wayback.archive.org/web/20120313045458/http://www.and... [2] https://what-if.xkcd.com/73/ https://what-if.xkcd.com/73/
  - jessriedel 10y ago
    Yea, I mentioned this because I just read it a couple of days ago :)
Certhas 10y ago
Not inverse square law, but simply inverse law. As energy radiates outwards, it becomes diluted on a sphere that increases its surface area as r^2. Thus the local energy density goes with inverse square law in the distance to the observer. But we don't measure the local energy density of gravitational waves, but the local amplitude. And in wave mechanics energy is the square of the amplitude. So the amplitude goes down with 1/r.

jessriedel 10y ago

Not really. You could have been within a few thousand miles of this collision and survived. In fact, it would be well below your ability to notice if it were 1 AU away. http://www.scottaaronson.com/blog/?p=2651 http://www.scottaaronson.com/blog/?p=2651 Needless to say, you would not survive if 1 solar mass of energy were released in any other form (even neutrinos). The key feature is that that tremendous amount of energy results in only a minuscule deformation of spacetime, i.e., spacetime is extremely stiff. By the way, the distances are way smaller than merely subatomic. The strain sensitivity is 10^-21, which for the ~ 1 km arms of LIGO is a shift of just 10^-8 the width of an atom.

sampo 10y ago
> you would not survive if 1 solar mass of energy were released in any other form (even neutrinos) But it is close! A supernova releases "few times 10^45 J of neutrino energy" [1], so let's say 5. 5e45 J is about 6e28 kg, while solar mass is 2e30 kg. And neutron radiation from a supernova would get fatal when closer than about 2.3 AU [2]. So we have a factor of 30 from the masses and a factor of 5 (1^2 AU vs. 2.3^2 AU) from the distance. So about 1/150 of solar mass released in neutron radiation would be survivable at the distance of 1 AU. [1] http://wayback.archive.org/web/20120313045458/http://www.andrewkaram.com/andy/pdf/HPJ.pdf http://wayback.archive.org/web/20120313045458/http://www.and... [2] https://what-if.xkcd.com/73/ https://what-if.xkcd.com/73/
- jessriedel 10y ago
  Yea, I mentioned this because I just read it a couple of days ago :)

Certhas 10y ago

Not inverse square law, but simply inverse law. As energy radiates outwards, it becomes diluted on a sphere that increases its surface area as r^2. Thus the local energy density goes with inverse square law in the distance to the observer. But we don't measure the local energy density of gravitational waves, but the local amplitude. And in wave mechanics energy is the square of the amplitude. So the amplitude goes down with 1/r.