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How come: 14.2 + 7.5 == 21.7 != 20.8 ?

Does the gravitational wave contain 0.9 solar masses of energy?

Yes (and good question!). From the article: "The initial binary was composed of two stellar-mass black holes with a source-frame primary mass m1 14.2 M⊙, secondary mass m2 7.5 M⊙, and a total mass of 21.8 M⊙. The binary merged into a black hole of mass 20.8 M⊙, radiating 1.0M⊙ in gravitational waves."

For reference, one solar mass of radiation is about the same as the output from a gamma ray burst, or about 2000 times the output from a supernova.

Or in less sensical units, the energy from detonating 1% of our galaxy's mass worth of TNT.

Gamma-ray bursts emission is strongly beamed into a small solid angle. So, although their equivalent isotropic energies (that is the energy that would be needed to power gamma-ray bursts if the emission was isotropic) are huge, the actual energies released in gamma-ray bursts are more or less compatible with those of supernovae (~0.1% of a solar mass).

As usual, XKCD provides some help to understand how much energy a supernova is:

"Which of the following would be brighter, in terms of the amount of energy delivered to your retina:

1. A supernova, seen from as far away as the Sun is from the Earth, or

2. The detonation of a hydrogen bomb pressed against your eyeball?

Answer: the supernova, by nine orders of magnitude."


Because the event is extremely violent.

Just as a nuclear bomb converts a small amount of mass to radiant energy (grams?), a merger can convert mass into other forms of energy, too.

Gravitational waves, of which there are probably many passing through us all the time from many sources, carry a lot of power. The only reason we don't see them every day is that spacetime itself is extremely stiff.

Isn't it also that power decreases over the cube of the distance to the event? A truly gargantuan amount of power (1 solar mass equivalent) distributed across a 3-dimensional shell dozens or hundreds of light years across should be minuscule at any Earth-sized point along that shell.

They put the distance at 440−190+180  Mpc. 440 megaparsecs is is around 1.2 billion light years away(!).

Wouldn't it be inverse square, not inverse cubed? Or do gravity waves behave different than other things we're used to, like light?

Actually the nice thing about gravitational waves is that we are sensitive to their amplitude, which scales as 1/r, and not to their energy, which scales as 1/r^2.

When you say that spacetime itself is extremely stiff, I can't help myself but to visualize a lattice of discrete points through which the wave propagates through. I understand this would be the wrong way about thinking about it, but is there another way?

This isn't really a wrong picture to imagine. In a way, this is what the experiments are trying to measure. You have a mirror and a laser at two points of your lattice (the space itself), and the "region" between them compresses and expands, which makes the laser take a slightly different time to bounce back.

Far away in linear approximation (which is what we saw here), the lattice picture is quite appropriate as a first order analogy.

That's an open problem right now. Lots of proposals, insufficient evidence to resolve the question.

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.

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.

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.


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.

> 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.and...

[2] https://what-if.xkcd.com/73/

Yea, I mentioned this because I just read it a couple of days ago :)

Hey, let's do some math!

The formula for gravitational potential energy is just mgh (mass times gravitational acceleration times height), and g is just GM/r^2, so the potential energy of one black hole in the other's gravitational field would be GMm/r, which would be the same for the other, so the total gravitational potential energy would be twice that.

Also, the schwarzschild radius of a black hole is about 3km per solar mass (2e30 kg).

Which means that before their event horizons touch the two black holes should be separated by a distance of at least 65.1 km.

So, 2 * G * 14.2 * 7.5 * (2e30kg)^2 / 65.1 km is...

8.73e47 joules

divide by c^2:

9.7e30 kg or 4.86 solar mass

So the system actually had nearly 5 solar masses of gravitational potential energy in it, some of which was radiated away as gravitational waves.

You're just calculating how much kinetic energy the black holes had when they collided. This energy goes into the spin of the final black hole. The energy for the gravitational waves comes from loss of mass of the final black hole.

Yes. Gravitational waves transmit both energy (i.e., mass) and angular momentum.

So... build more detectors and pinpoint the direction the gravity wave is coming from?

Yes! There are 2 now, so they can give a very coarse estimate of the direction with a precision of 'somewhere in this swath of the sky' (using the relative location, the wave propagation speed [light speed], and the arrival times relative to each detector).

As more detectors come online they will be able to triangulate more accurately. The next one is supposed to come online in 2018 iirc.

That's the idea! I seem to remember that there are plans to build another detector in India. I'm not sure what its status is.


Note that the VIRGO detector, located near Pisa (Italy), has already taken a set of measurements a few years ago, and it would have been sensitive enough to detect waves like the one from last September. Unfortunately, Virgo was down for upgrades when GW150914 came. I recall that they plan to restart Virgo by early 2017, so it should be operational well before Ligo India. (And I like to underline that Ligo and Virgo joined together in the Ligo/Virgo collaboration and thus work together: as a matter of fact, the first guy who saw GW150914 in the raw data was from Virgo, and the papers presenting the detection were signed by people in both collaborations.)

Funding for its construction was approved by India. We are hoping that it might coming online in ~ 5 years. It takes a long time to build everything from the ground up.

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