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Gravitational waves from a binary black hole merger observed by LIGO and Virgo

528 points by nickcw 9y ago

rcthompson 9y ago
> about 3 solar masses were converted into gravitational-wave energy during the coalescence This is such a staggering amount of energy to be released in such a short time.
- leeoniya 9y ago
  more or less than https://en.wikipedia.org/wiki/Gamma-ray_burst https://en.wikipedia.org/wiki/Gamma-ray_burst ?
  - throwaway613834 9y ago
    I think less? When a gamma-ray burst is pointed towards Earth, the focusing of its energy along a relatively narrow beam causes the burst to appear much brighter than it would have been were its energy emitted spherically. When this effect is taken into account, typical gamma-ray bursts are observed to have a true energy release of about 10^44 J, or about 1/2000 of a Solar mass (M) energy equivalent. [1] [1] https://en.wikipedia.org/wiki/Gamma-ray_burst#Energetics_and_beaming https://en.wikipedia.org/wiki/Gamma-ray_burst#Energetics_and...
    - rcthompson 9y ago
      I think another sentence in that article might need revision: > No known process in the universe can produce this much energy [1 solar mass equivalent] in such a short time.
  - danharaj 9y ago
    The wikipedia article says that if a GRB were a spherical release of energy, they would be on the same order. However, GRB's are highly anisotropic; most of the energy is expelled in a smaller cone. One of the estimates on the page is that a typical GRB expels 1/2000th a solar mass of energy. So these gravity events are thousands of times more powerful. These scales are incomprehensible, at least to me.
  - ntumlin 9y ago
    From the wiki page: > When this effect is taken into account, typical gamma-ray bursts are observed to have a true energy release of about 1044 J, or about 1/2000 of a Solar mass (M) energy equivalent So more, by about 6000 times. This is consistant with WolframAlpha's output for amount of energy: 5.361e47 J. I tried to convert this to "number of years we could power the Earth", but it just changed it from one mind boggling number into another.
    - dragonwriter 9y ago
      > > When this effect is taken into account, typical gamma-ray bursts are observed to have a true energy release of about 1044 J, Copy-paste error here from losing sup/sub formatting, that's 10^44 J, not 1044 J.
  - deleted 9y ago
    [deleted]
- jobu 9y ago
  Didn't realize that energy could be converted into gravitational waves. It does make sense I suppose, but I always thought gravity and gravitational waves were just a side-effect of the way mass interacted with space-time.
  - SomeStupidPoint 9y ago
    Can't you radiate potential gravitational energy as gravitational waves without any sort of particle changes, because potential gravitational energy is stored in the configuration of the system? If I drop my bowling ball, it has lower potential energy, but can't some of that be compensated for by the Earth-ball system radiating a gravitational wave (rather than all being converted into kinetic energy)? Similarly, don't orbiting bodies emit gravitational waves? I forget the exact dynamics, but the way that they drag just a little bit as they orbit creates waves that radiate power out of the system -- I think the Sun and Jupiter radiate something like 55 watts of gravitational waves. If you take two things that are each tens of times larger than the Sun and set them spinning an appreciable amount of the speed of light very nearby -- to the point they eventually collide -- it's not surprising they emit quite a bit more than our solar system. Think of it this way: Two large objects start orbiting slowly far apart. But because they're large and the system is big, they have a lot of orbital momentum. They pull each other close, but as they get closer, that orbital momentum has nowhere to dissipate, so they orbit faster and faster!... ...Except that because they're so massive, they slightly tug space around with them as they orbit. And because they're orbiting so fast, space doesn't have a good way to smooth itself out. So some of the momentum from the heavy things spinning far apart gets stored in wrinkles in space as they come together. Which then radiates away from there to us, ...and slightly stretches one direction of reality as it passes by. Which we detect by measuring how long two perpendicular lines are very accurately a few different places on a nearly spherical object. ...So you can model it as just the way that mass interacts with spacetime if you remember to include things like the distribution of your mass as a source of potential energy that can itself be used to generate gravitational waves. However, this does not make the whole situation any less weird to explain. Reports just tend to convert the amount of energy released through mechanisms like that in terms of mass, because it's the only thing vaguely comprehensible. (eg, 3 suns)
- wlesieutre 9y ago
  EDIT - Thanks to bonzini for catching a mistake, I'd typod 47 as 74 on the first line and drastically inflated everything after that. It's still huge though. Wolfram Alpha gives a solar mass as 1.988435×10^30 kg. Multiply that by (299,292,458 m/s)^2, you get ~1.78×10^47 Joules. The most powerful manmade explosion in history was Tsar Bomba, a Soviet hydrogen bomb with 210 Petajoule (210×10^15 J) yeild. Amount of energy released by the black holes merging is 8.48×10^29 times greater, so you'd need to set off that many Tsar Bombas to get equal total yeild. The universe is 4.3×10^17 seconds old, meaning if you set off a Tsar Bomba every second since the beginning of time, you'd still be off by a factor of 1.953×10^12. You need about 2,000,000,000,000,000,000 Tsar Bombas every second since the big bang to match it. The mind boggles.
  - wlesieutre 9y ago
    EDIT - originally compared the last factor to number of grains of sand on Earth squared, but after catching the error above that's much too large. So how big is the remaining 2x10^12 bombs per seconds? Wikipedia has some other suggestions: ~10^12 stars in the Andromeda galaxy 1.98x10^12 links on Wikipedia 3.04×10^12 trees on Earth in 2015 3.5×10^12 estimated fish in the ocean So take every link on Wikipedia (or two thirds of the trees on Earth), turn each one into the largest hydrogen bomb ever detonated, and blow them all up every second since the beginning of the universe, and that's how much energy we're talking about. Thanks again to bonzini for spotting the rather large error!
    - rcthompson 9y ago
      Funnily enough, the past two XKCD comics are both relevant here: https://xkcd.com/1894/ https://xkcd.com/1894/ https://xkcd.com/1895/ https://xkcd.com/1895/
    - bas 9y ago
      Nicely (back of) napkinned!
      - wlesieutre 9y ago
ChuckMcM 9y ago
Pretty awesome, nice to see they confirmation with Virgo. I realize that I've never really considered what it would be like to be 'near' (say a few 10's of thousands of light years away) from an event like this. With all that energy, at what level would it pulverize a planetary body?
- sandworm101 9y ago
  Nothing. If it happened that close you would still feel/see nothing. These detectors are seeing movements far smaller than the width of a proton. A planet would have to be orbiting one of these binaries to be close enough to potentially feel much of anything.
  - deleted 9y ago
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  - exhilaration 9y ago
    Well the x-rays and gamma rays would kill you if you were close. Aren't those released when these events happen?
    - sandworm101 9y ago
      Yes but there are plenty of more likely events that emit those. The merger of two BHs is special for the gravity waves it generates. Rays can also be blocked given enough mass, gravity waves no.
  - gajomi 9y ago
    I was about to reply the same based on a naive Fermi estimation argument concerning the energy density three solar masses distributed equally over a shell with thickness proportional to the effective width of the pulse. But I wonder, what would be the actually mechanism that would break up a planetary body? Would sufficiently strong grativitational waves at wavelengths smaller than the planet be able to stress/shear enough to cause a breakup? Or is there some kind of gravity/gravity interaction that would be needed?
netcraft 9y ago
I know there were a lot of people hoping this was the rumored neutron star merger (with possibly visual observations!) but I don't think thats not still on the table, this was just a different announcement. Still quite an achievement and always great to get further confirmation of the observations. Imagine what we could do with a much larger observatory.
ricardobeat 9y ago
What are the practical applications of technology that could be derived from these discoveries? Any chance of us harvesting this energy?
- throwaway613834 9y ago
  You mean from this distance? Or do you mean if a closer one occurred?
- stefco_ 9y ago
  Hi, I'm a grad student working on LIGO. There is no practical way to harvest this energy. Gravitational waves interact very weakly with matter. This is part of why they are useful for observations: GWs can travel from the core of violent events unmodified, unlike light, which gets blocked by surrounding matter. You can't see the center of a core collapse supernova with light, but you can see it with GWs and neutrinos. They also won't get bent by EM or gravitational fields, meaning that the direction they come from in the sky points straight back to the source. And unlike very high energy gamma rays, they don't pair produce into electrons. This all makes them great for astronomy and totally useless for extracting energy.
  - chebucto 9y ago
    What happens to that energy, though? Energy is conserved, even with gravitational waves, correct? So, what happens to the energy - does some of it get converted into 'normal' potential energy by changing the (amount of gravity? gravitational fields?) of the stuff that it passes through - the weak interaction that you refer to? And, if it interacts very weakly, does all of the energy eventually get used up this way - do GWs get 'used up' before they travel across the universe? (Probably a meaningless question): can GWs hit the edge of the universe, and if so, what happens then?
    - stefco_ 9y ago
      > What happens to that energy, though? Energy is conserved, even with gravitational waves, correct? Yup. It's like ocean waves carrying energy. The difference is that it's hard to interact with GWs, so instead of "riding" them and capturing their energy, you find them mostly passing through you. They have tons of energy, but very little of it is dissipated back into matter.
dpcx 9y ago
A coworker earlier mentioned a paper questioning the validity of the findings around gravitational waves. Can anyone point me to them and (as a non-scientist) explain why this paper would think those findings are inaccurate?
- batmansmk 9y ago
  Can you coworker provide some info instead?
  - dpcx 9y ago
    He could provide information about the paper, sure. The explanation, I'm fairly certain, he could not.
- ISL 9y ago
  Context (too busy getting physics done today to expound further, except to state that I presently agree with LIGO): https://www.wired.com/story/strange-noise-in-gravitational-wave-data-sparks-debate/ https://www.wired.com/story/strange-noise-in-gravitational-w...
- sleavey 9y ago
  LIGO made an unofficial response via the blog of Sean Carroll [1]. The problem with the analysis in the paper that questioned LIGO's results was that they misinterpreted the filtering that had been applied to the data shown in the original GW150914 paper, which led them to think there was noise produced by the detectors themselves. [1] http://www.preposterousuniverse.com/blog/2017/06/18/a-response-to-on-the-time-lags-of-the-ligo-signals-guest-post/ http://www.preposterousuniverse.com/blog/2017/06/18/a-respon...
- InclinedPlane 9y ago
  Your coworker might be confusing the LIGO observations (widely considered to be some of the most meticulously done research in science) with the BICEP2 early big bang gravitational wave research, which has since been called into question substantially.
Diederich 9y ago
Can anyone expand on the meaning of this? "we find that the data strongly favor pure tensor polarization of gravitational waves, over pure scalar or pure vector polarizations" ..taken from the paper. Thanks!
- thethirdone 9y ago
  I'm not an expert by any means, but from reading a lecture notes [1] on it, I think I have a decent idea. Any gravitational wave polarization can be considered as a tensor. There are several constraints that reduce the dimensional of that tensor though. For example, it must be symmetric and thus reducing its dimension from 16 to 10. Further constraints reduce it to 6. For physically meaningful waves (under the most accepted theory of gravitational waves), all but 2 of those are 0. This leaves us with 2 polarizations of waves. One of the polarization applies only on the diagonal (of the tensor) and the other on the anti diagonal. From this background, I interpret "we find that the data strongly favor pure tensor polarization of gravitational waves, over pure scalar or pure vector polarizations" to mean that the polarization was not purely of the diagonal polarization and was instead a combination of the two. Alternatively, it could be a property that the tensor representation is more robust to noise or that a non-standard theory is correct and the polarization doesn't fit cleanly into the accepted theory. [1]: http://eagle.phys.utk.edu/guidry/astro616/lectures/lecture_ch21.pdf http://eagle.phys.utk.edu/guidry/astro616/lectures/lecture_c...
- andrewla 9y ago
  I'll take a shot at this. Essentially they're saying that they've found evidence that alternative theories about gravitation (non-Einsteinian theories) that are strictly simpler (like classical Newtonian gravity) are not consistent with the observations. Here scalar means that the field that describes the phenomenon can be described as a single scalar value at each point of space, like density, as is the case of sound waves. Vector is more complex; although "charge" is a scalar, electromagnetic waves do not propagate by inducing charge in the surrounding space, but by propagating through the free-space Maxwell equation (that is, where the charge density is zero). So the polarization of an electromagnetic wave can be described by a single vector at a point, and the measurement of the magnitude of the wave will depend on how closely the measurement axis corresponds to the polarization axis for a polarized wave. For a general wave, it can be described as some linear combination of the orthogonal polarizations. Although "tensor" is a general linear object of which vectors are a subset, in this case of gravitational waves, we require a 2-tensor to describe the polarization of the wave. Effectively that means the polarization (the strength of the signal with respect to a given detector) can only be described by two independent vector that describe the evolution of the wave. Since the three detectors (Ligo Hanford, Ligo Livingston, and Virgo) are not exactly aligned with one another, it should be possible to get good evidence that the polarization cannot be adequately described by the superposition of two orthogonal vectors, but instead two independent pairs of orthogonal vectors.
Florin_Andrei 9y ago
If you turn a detector on - and, as soon as you do that, you detect some event, chances are those events are happening, like, all the freaking time. There's got to be some interesting implications here for cosmology.
- perlgeek 9y ago
  As an armchair scientist, I mostly wonder about mass estimations. If there's a whole class of objects (black holes around 20 to 50 solar masses) that were previously thought to be virtually non-existent, and now turn out to pretty common, does that mean we can explain away at least some of the proposed dark matter? What if we keep finding more non-radiant objects that still (presumably) made of ordinary matter? Can these classes of objects account for the "missing" matter in galaxy rotation curves? Or would their distribution be simply wrong? Exciting times to live in, for sure!
  - btouellette 9y ago
    The word you're looking for is MACHO (massive compact halo object) and using microlensing observations you can make all kinds of statistical arguments for how likely objects of different masses are to make up the missing matter to match galaxy rotation curves to observations. Basically it doesn't seem likely that there are nearly enough intermediate mass black holes to explain dark matter. Because microlensing is gravitational it doesn't matter what kind of object the MACHO is. https://en.wikipedia.org/wiki/Massive_compact_halo_object https://en.wikipedia.org/wiki/Massive_compact_halo_object http://iopscience.iop.org/article/10.1086/319636 http://iopscience.iop.org/article/10.1086/319636
    - spchampion2 9y ago
      It would be interesting to take the modeled number of MACHO black holes based on lensing studies and compare it to the modeled numbers of intermediate black holes based on LIGO data. I'd love to know if those two numbers substantially differ.
      - 9y ago
autocorr 9y ago
From an observational astronomy point of view, perhaps the most interesting thing about the report is that by including the Virgo detector with both LIGO detectors, the uncertainty in the direction the gravitational wave event was reduced by a factor of 10! The "error ellipse" on the sky is now ~60 square degrees as opposed to ~600. This makes it much more feasible to do rapid observational follow-up with other observatories at all wavelengths (radio, optical, X-ray, etc.), to search for the counterparts. Many observatories like the VLA, Chandra, Fermi, Gemini, have a rapid response trigger for the detection events, where within minutes they will stop their current operations. Observations in the EM spectrum are crucial for learning about a huge number of questions in astrophysics, like what kinds of galaxies do the merger events originate from, where are they located in those galaxies, how does the luminosity decay at different wavelengths, etc. It will be seriously exciting when the first counterpart is discovered!
- sleavey 9y ago
  Here's a skymap [1] that really drives home the point about how much having a third detector helps. [1] http://www.virgo-gw.eu/skymap.html http://www.virgo-gw.eu/skymap.html
  - stefco_ 9y ago
    Having extra detectors also helps hugely with uptime. We're currently only active and collecting clean data ~50% of the time for each detector. We need at least two detectors active for a detection, and need three detectors for this kind of beautiful directional isolation. So having extra detectors decreases our skymap areas significantly, which helps in EM followup, but it also significantly reduces GW detector network downtime. And it allows for graceful degradation; with 5 detectors, you can afford to have a couple detectors down and still detect an event with good direction reconstruction. We also have tons of downtime for upgrades. Now that our second observing run is over, we have a ~1.5yr upgrade cycle. There's no point in running VIRGO without LIGO (except for engineering reasons); having more detectors will increase flexibility in this regard, too. When KAGRA and LIGO India come online, we should see much better skymaps, but we will also see much higher uptimes (and hence higher detection rates). It's especially exciting to see improved direction reconstruction given that VIRGO has faced setbacks this year with their mirrors. I wasn't expecting to see such beautiful skymaps until our third observing run! Needless to say it was an exciting summer :) - grad student working on LIGO
komali2 9y ago
I'm googling but it's all above my head - what's a gravitational wave, and how do they detect them, if anybody is willing to take a bit of time to explain to a turbo-layman? :D
- aeleos 9y ago
  There are many good videos on youtube that do a good job explaining what gravitational waves are and how we detect them. Many of these were made in response to the first detection, and they do a pretty good job explaining it in laymans terms. I personally really like veritasiums video here. https://www.youtube.com/watch?v=iphcyNWFD10 https://www.youtube.com/watch?v=iphcyNWFD10
- deleted 9y ago
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ivoras 9y ago
So, what are the implication on General relativity if it's shown that gravity changes move in waves, with a finite speed?
- vorotato 9y ago
  The speed of gravity is the speed of light, because it's the speed of causality. That's a pretty well understood thing, and isn't really something being explored by these.
- kuschku 9y ago
  Well, basically "confirmed", nothing more. Einstein’s theories predicted all this.
gus_massa 9y ago
> Black holes produce gravitational waves but not light. Is this correct? I thought that it would produce a lot of light, but the light is more easily absorbed by interstellar gas and dust, so it's more difficult to see it. Perhaps not the black holes doesn't emit light directly, but the surrounding gas will be extremely hot. And when the two event horizons collide, the virtual particles should be very confused and scape in huge numbers, in a process like the Hawking radiation, but much bigger.????? I'm far from being an expert in this, so this is only a conjecture and unsupported handwaving. But I would be very surprised to confirm that the event has no emission of light.
- azag0 9y ago
  > Perhaps not the black holes doesn't emit light directly, but the surrounding gas will be extremely hot. Yes, but not every black hole is surrounded by gas. > the virtual particles should be very confused In general, physics does not work like this.
- cyphar 9y ago
  Black holes themselves cannot emit light (Hawking radiation is a separate topic). Gas that is in accretion disks around a black hole can become hot and produce light, but that process has nothing to do with the black hole itself. By definition, light cannot escape a black hole so processes inside the black hole do not emit light (or rather, if they do we cannot see it). However, massive bodies experiencing rapid acceleration emit gravitational waves. This is something that happens due to a black hole's mass and not it's event horizon properties (in principle you can also detect neutron star interactions). But this is emitted due to the motion of the objects themselves, not from accretion disks (which may not be present for some black holes). So the event itself does not produce light.
  - oldandtired 9y ago
    A problem with your explanation is that from our perspective (which is the only one that actually matters here), "black holes" will never undergo rapid acceleration due to the extreme time dilation effects according to the gravity model used. I find it interesting that much ado is made about "black holes" existing when by the theory by which they are proposed, they cannot exist in our finite universe. The time dilation effects, which is a part of the theory, will not allow any such theoretical entity to be formed in our finite time universe. With that said, the observations of LIGO do not confirm the existence of such entities as "black holes". What these observation do do is indicate that there is something happening. We just don't know yet what this is. Going down the rabbit hole of sticking to the "black hole" model only engenders ignorance. These are interesting experiments and results and it behooves an open mind to try to further our understanding of what is causing these observational event. Unfortunately, the "dogma" of worshipping Einstein and his "holy works" only locks us into a position of not furthering our understanding. Note: I have used "worshipping" and "holy works" specifically because of the attitudes of those who will not countenance any different views from Generally Relativity et al. Whether or not, GR and SR have long term merit is still a question being pursued. Some variation of these may be more suitable. But that doesn't mean that GR and SR are correct (as is the usual defence given). The last couple of decades have been an interesting case study in terms of an unwitting increase in dogmatic push for specific theories and models in quite a few areas. When it boils down to it, these theories and models are subject to change and even throwing them out for different and better models. To get emotionally caught up in defending a specific theory and model as the "Truth" does not put science in a good light for the general populous. I have said it elsewhere, science is about the discovery of the universe around us and finding applicable theories and models that are useful for our understanding. If you want "Truth" then go and study philosophy and religion. Every theory you want to follow needs to have a good defence, but not to the extent of getting so emotionally caught up in it that you ignore discrepancies and results that disagree with your preferred theory and model. How often has the "Truth" that the general science community has promulgated been overturned? We need to be careful not to get so entrenched in a specific viewpoint that we cannot see the merit in other theories or models. It is history that tells us to be careful here.
c-smile 9y ago
Why black hole merges happen so frequently? We just started registering them and got three events already... Mass of Universe is estimated to be of 6e51kg. Mass of Sun is 1e25kg. This event is about 50 Sun masses. So that one event is about morphing 1e-26 part of Universe. I suspect that matter collapsing into black holes happens significantly more frequently than that. Do we have enough non-black-hole matter available for our (humanity) life time? Yet worth to mention that black holes "return" mass by not only Hawking–Zel'dovich radiation but also by such events - this particular merge emitted 3 solar masses as energy.
MikeGale 9y ago
Don't you look forward to the time when these events need to named like this GW1712231509236732?
ssijak 9y ago
How do they know exact sizes of the 2 black holes, and the remaining size also? Seems like they say it as 100% truth, which maybe it is, but my uneducated mind cannot think how they could calculate that. Only those sizes fit in the mathematical model of the gravitational waves compared to the observations?
- MAGZine 9y ago
  It is trivial to obtain the size of objects, so long as you have some information about a satellite orbiting it. It's how we get the mass of many interstellar objects, and it's how we know that dark matter exists. http://astro.physics.uiowa.edu/ITU/glossary/keplers-third-law/ http://astro.physics.uiowa.edu/ITU/glossary/keplers-third-la...
- 21 9y ago
  I imagine is like seeing a light source from two vantage points. Using the timing difference you can triangulate a distance. Using the computed distance and received brightness you can compute source brightness. Using source brightness you can compute source power.

528 points by nickcw 9y ago

> about 3 solar masses were converted into gravitational-wave energy during the coalescence This is such a staggering amount of energy to be released in such a short time.

leeoniya 9y ago
more or less than https://en.wikipedia.org/wiki/Gamma-ray_burst https://en.wikipedia.org/wiki/Gamma-ray_burst ?
- throwaway613834 9y ago
  I think less? When a gamma-ray burst is pointed towards Earth, the focusing of its energy along a relatively narrow beam causes the burst to appear much brighter than it would have been were its energy emitted spherically. When this effect is taken into account, typical gamma-ray bursts are observed to have a true energy release of about 10^44 J, or about 1/2000 of a Solar mass (M) energy equivalent. [1] [1] https://en.wikipedia.org/wiki/Gamma-ray_burst#Energetics_and_beaming https://en.wikipedia.org/wiki/Gamma-ray_burst#Energetics_and...
  - rcthompson 9y ago
    I think another sentence in that article might need revision: > No known process in the universe can produce this much energy [1 solar mass equivalent] in such a short time.
- danharaj 9y ago
  The wikipedia article says that if a GRB were a spherical release of energy, they would be on the same order. However, GRB's are highly anisotropic; most of the energy is expelled in a smaller cone. One of the estimates on the page is that a typical GRB expels 1/2000th a solar mass of energy. So these gravity events are thousands of times more powerful. These scales are incomprehensible, at least to me.
- ntumlin 9y ago
  From the wiki page: > When this effect is taken into account, typical gamma-ray bursts are observed to have a true energy release of about 1044 J, or about 1/2000 of a Solar mass (M) energy equivalent So more, by about 6000 times. This is consistant with WolframAlpha's output for amount of energy: 5.361e47 J. I tried to convert this to "number of years we could power the Earth", but it just changed it from one mind boggling number into another.
  - dragonwriter 9y ago
    > > When this effect is taken into account, typical gamma-ray bursts are observed to have a true energy release of about 1044 J, Copy-paste error here from losing sup/sub formatting, that's 10^44 J, not 1044 J.
- deleted 9y ago
  [deleted]
jobu 9y ago
Didn't realize that energy could be converted into gravitational waves. It does make sense I suppose, but I always thought gravity and gravitational waves were just a side-effect of the way mass interacted with space-time.
- SomeStupidPoint 9y ago
  Can't you radiate potential gravitational energy as gravitational waves without any sort of particle changes, because potential gravitational energy is stored in the configuration of the system? If I drop my bowling ball, it has lower potential energy, but can't some of that be compensated for by the Earth-ball system radiating a gravitational wave (rather than all being converted into kinetic energy)? Similarly, don't orbiting bodies emit gravitational waves? I forget the exact dynamics, but the way that they drag just a little bit as they orbit creates waves that radiate power out of the system -- I think the Sun and Jupiter radiate something like 55 watts of gravitational waves. If you take two things that are each tens of times larger than the Sun and set them spinning an appreciable amount of the speed of light very nearby -- to the point they eventually collide -- it's not surprising they emit quite a bit more than our solar system. Think of it this way: Two large objects start orbiting slowly far apart. But because they're large and the system is big, they have a lot of orbital momentum. They pull each other close, but as they get closer, that orbital momentum has nowhere to dissipate, so they orbit faster and faster!... ...Except that because they're so massive, they slightly tug space around with them as they orbit. And because they're orbiting so fast, space doesn't have a good way to smooth itself out. So some of the momentum from the heavy things spinning far apart gets stored in wrinkles in space as they come together. Which then radiates away from there to us, ...and slightly stretches one direction of reality as it passes by. Which we detect by measuring how long two perpendicular lines are very accurately a few different places on a nearly spherical object. ...So you can model it as just the way that mass interacts with spacetime if you remember to include things like the distribution of your mass as a source of potential energy that can itself be used to generate gravitational waves. However, this does not make the whole situation any less weird to explain. Reports just tend to convert the amount of energy released through mechanisms like that in terms of mass, because it's the only thing vaguely comprehensible. (eg, 3 suns)
wlesieutre 9y ago
EDIT - Thanks to bonzini for catching a mistake, I'd typod 47 as 74 on the first line and drastically inflated everything after that. It's still huge though. Wolfram Alpha gives a solar mass as 1.988435×10^30 kg. Multiply that by (299,292,458 m/s)^2, you get ~1.78×10^47 Joules. The most powerful manmade explosion in history was Tsar Bomba, a Soviet hydrogen bomb with 210 Petajoule (210×10^15 J) yeild. Amount of energy released by the black holes merging is 8.48×10^29 times greater, so you'd need to set off that many Tsar Bombas to get equal total yeild. The universe is 4.3×10^17 seconds old, meaning if you set off a Tsar Bomba every second since the beginning of time, you'd still be off by a factor of 1.953×10^12. You need about 2,000,000,000,000,000,000 Tsar Bombas every second since the big bang to match it. The mind boggles.
- wlesieutre 9y ago
  EDIT - originally compared the last factor to number of grains of sand on Earth squared, but after catching the error above that's much too large. So how big is the remaining 2x10^12 bombs per seconds? Wikipedia has some other suggestions: ~10^12 stars in the Andromeda galaxy 1.98x10^12 links on Wikipedia 3.04×10^12 trees on Earth in 2015 3.5×10^12 estimated fish in the ocean So take every link on Wikipedia (or two thirds of the trees on Earth), turn each one into the largest hydrogen bomb ever detonated, and blow them all up every second since the beginning of the universe, and that's how much energy we're talking about. Thanks again to bonzini for spotting the rather large error!
  - rcthompson 9y ago
    Funnily enough, the past two XKCD comics are both relevant here: https://xkcd.com/1894/ https://xkcd.com/1894/ https://xkcd.com/1895/ https://xkcd.com/1895/
  - bas 9y ago
    Nicely (back of) napkinned!
    - wlesieutre 9y ago

ChuckMcM 9y ago

Pretty awesome, nice to see they confirmation with Virgo. I realize that I've never really considered what it would be like to be 'near' (say a few 10's of thousands of light years away) from an event like this. With all that energy, at what level would it pulverize a planetary body?

sandworm101 9y ago
Nothing. If it happened that close you would still feel/see nothing. These detectors are seeing movements far smaller than the width of a proton. A planet would have to be orbiting one of these binaries to be close enough to potentially feel much of anything.
- deleted 9y ago
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- exhilaration 9y ago
  Well the x-rays and gamma rays would kill you if you were close. Aren't those released when these events happen?
  - sandworm101 9y ago
    Yes but there are plenty of more likely events that emit those. The merger of two BHs is special for the gravity waves it generates. Rays can also be blocked given enough mass, gravity waves no.
- gajomi 9y ago
  I was about to reply the same based on a naive Fermi estimation argument concerning the energy density three solar masses distributed equally over a shell with thickness proportional to the effective width of the pulse. But I wonder, what would be the actually mechanism that would break up a planetary body? Would sufficiently strong grativitational waves at wavelengths smaller than the planet be able to stress/shear enough to cause a breakup? Or is there some kind of gravity/gravity interaction that would be needed?

netcraft 9y ago

I know there were a lot of people hoping this was the rumored neutron star merger (with possibly visual observations!) but I don't think thats not still on the table, this was just a different announcement. Still quite an achievement and always great to get further confirmation of the observations. Imagine what we could do with a much larger observatory.

ricardobeat 9y ago

What are the practical applications of technology that could be derived from these discoveries? Any chance of us harvesting this energy?

throwaway613834 9y ago
You mean from this distance? Or do you mean if a closer one occurred?
stefco_ 9y ago
Hi, I'm a grad student working on LIGO. There is no practical way to harvest this energy. Gravitational waves interact very weakly with matter. This is part of why they are useful for observations: GWs can travel from the core of violent events unmodified, unlike light, which gets blocked by surrounding matter. You can't see the center of a core collapse supernova with light, but you can see it with GWs and neutrinos. They also won't get bent by EM or gravitational fields, meaning that the direction they come from in the sky points straight back to the source. And unlike very high energy gamma rays, they don't pair produce into electrons. This all makes them great for astronomy and totally useless for extracting energy.
- chebucto 9y ago
  What happens to that energy, though? Energy is conserved, even with gravitational waves, correct? So, what happens to the energy - does some of it get converted into 'normal' potential energy by changing the (amount of gravity? gravitational fields?) of the stuff that it passes through - the weak interaction that you refer to? And, if it interacts very weakly, does all of the energy eventually get used up this way - do GWs get 'used up' before they travel across the universe? (Probably a meaningless question): can GWs hit the edge of the universe, and if so, what happens then?
  - stefco_ 9y ago
    > What happens to that energy, though? Energy is conserved, even with gravitational waves, correct? Yup. It's like ocean waves carrying energy. The difference is that it's hard to interact with GWs, so instead of "riding" them and capturing their energy, you find them mostly passing through you. They have tons of energy, but very little of it is dissipated back into matter.

dpcx 9y ago

A coworker earlier mentioned a paper questioning the validity of the findings around gravitational waves. Can anyone point me to them and (as a non-scientist) explain why this paper would think those findings are inaccurate?

batmansmk 9y ago
Can you coworker provide some info instead?
- dpcx 9y ago
  He could provide information about the paper, sure. The explanation, I'm fairly certain, he could not.
ISL 9y ago
Context (too busy getting physics done today to expound further, except to state that I presently agree with LIGO): https://www.wired.com/story/strange-noise-in-gravitational-wave-data-sparks-debate/ https://www.wired.com/story/strange-noise-in-gravitational-w...
sleavey 9y ago
LIGO made an unofficial response via the blog of Sean Carroll [1]. The problem with the analysis in the paper that questioned LIGO's results was that they misinterpreted the filtering that had been applied to the data shown in the original GW150914 paper, which led them to think there was noise produced by the detectors themselves. [1] http://www.preposterousuniverse.com/blog/2017/06/18/a-response-to-on-the-time-lags-of-the-ligo-signals-guest-post/ http://www.preposterousuniverse.com/blog/2017/06/18/a-respon...
InclinedPlane 9y ago
Your coworker might be confusing the LIGO observations (widely considered to be some of the most meticulously done research in science) with the BICEP2 early big bang gravitational wave research, which has since been called into question substantially.

Diederich 9y ago

Can anyone expand on the meaning of this? "we find that the data strongly favor pure tensor polarization of gravitational waves, over pure scalar or pure vector polarizations" ..taken from the paper. Thanks!

thethirdone 9y ago
I'm not an expert by any means, but from reading a lecture notes [1] on it, I think I have a decent idea. Any gravitational wave polarization can be considered as a tensor. There are several constraints that reduce the dimensional of that tensor though. For example, it must be symmetric and thus reducing its dimension from 16 to 10. Further constraints reduce it to 6. For physically meaningful waves (under the most accepted theory of gravitational waves), all but 2 of those are 0. This leaves us with 2 polarizations of waves. One of the polarization applies only on the diagonal (of the tensor) and the other on the anti diagonal. From this background, I interpret "we find that the data strongly favor pure tensor polarization of gravitational waves, over pure scalar or pure vector polarizations" to mean that the polarization was not purely of the diagonal polarization and was instead a combination of the two. Alternatively, it could be a property that the tensor representation is more robust to noise or that a non-standard theory is correct and the polarization doesn't fit cleanly into the accepted theory. [1]: http://eagle.phys.utk.edu/guidry/astro616/lectures/lecture_ch21.pdf http://eagle.phys.utk.edu/guidry/astro616/lectures/lecture_c...
andrewla 9y ago
I'll take a shot at this. Essentially they're saying that they've found evidence that alternative theories about gravitation (non-Einsteinian theories) that are strictly simpler (like classical Newtonian gravity) are not consistent with the observations. Here scalar means that the field that describes the phenomenon can be described as a single scalar value at each point of space, like density, as is the case of sound waves. Vector is more complex; although "charge" is a scalar, electromagnetic waves do not propagate by inducing charge in the surrounding space, but by propagating through the free-space Maxwell equation (that is, where the charge density is zero). So the polarization of an electromagnetic wave can be described by a single vector at a point, and the measurement of the magnitude of the wave will depend on how closely the measurement axis corresponds to the polarization axis for a polarized wave. For a general wave, it can be described as some linear combination of the orthogonal polarizations. Although "tensor" is a general linear object of which vectors are a subset, in this case of gravitational waves, we require a 2-tensor to describe the polarization of the wave. Effectively that means the polarization (the strength of the signal with respect to a given detector) can only be described by two independent vector that describe the evolution of the wave. Since the three detectors (Ligo Hanford, Ligo Livingston, and Virgo) are not exactly aligned with one another, it should be possible to get good evidence that the polarization cannot be adequately described by the superposition of two orthogonal vectors, but instead two independent pairs of orthogonal vectors.

Florin_Andrei 9y ago

If you turn a detector on - and, as soon as you do that, you detect some event, chances are those events are happening, like, all the freaking time. There's got to be some interesting implications here for cosmology.

perlgeek 9y ago
As an armchair scientist, I mostly wonder about mass estimations. If there's a whole class of objects (black holes around 20 to 50 solar masses) that were previously thought to be virtually non-existent, and now turn out to pretty common, does that mean we can explain away at least some of the proposed dark matter? What if we keep finding more non-radiant objects that still (presumably) made of ordinary matter? Can these classes of objects account for the "missing" matter in galaxy rotation curves? Or would their distribution be simply wrong? Exciting times to live in, for sure!
- btouellette 9y ago
  The word you're looking for is MACHO (massive compact halo object) and using microlensing observations you can make all kinds of statistical arguments for how likely objects of different masses are to make up the missing matter to match galaxy rotation curves to observations. Basically it doesn't seem likely that there are nearly enough intermediate mass black holes to explain dark matter. Because microlensing is gravitational it doesn't matter what kind of object the MACHO is. https://en.wikipedia.org/wiki/Massive_compact_halo_object https://en.wikipedia.org/wiki/Massive_compact_halo_object http://iopscience.iop.org/article/10.1086/319636 http://iopscience.iop.org/article/10.1086/319636
  - spchampion2 9y ago
    It would be interesting to take the modeled number of MACHO black holes based on lensing studies and compare it to the modeled numbers of intermediate black holes based on LIGO data. I'd love to know if those two numbers substantially differ.
    - 9y ago

autocorr 9y ago

From an observational astronomy point of view, perhaps the most interesting thing about the report is that by including the Virgo detector with both LIGO detectors, the uncertainty in the direction the gravitational wave event was reduced by a factor of 10! The "error ellipse" on the sky is now ~60 square degrees as opposed to ~600. This makes it much more feasible to do rapid observational follow-up with other observatories at all wavelengths (radio, optical, X-ray, etc.), to search for the counterparts. Many observatories like the VLA, Chandra, Fermi, Gemini, have a rapid response trigger for the detection events, where within minutes they will stop their current operations. Observations in the EM spectrum are crucial for learning about a huge number of questions in astrophysics, like what kinds of galaxies do the merger events originate from, where are they located in those galaxies, how does the luminosity decay at different wavelengths, etc. It will be seriously exciting when the first counterpart is discovered!

sleavey 9y ago
Here's a skymap [1] that really drives home the point about how much having a third detector helps. [1] http://www.virgo-gw.eu/skymap.html http://www.virgo-gw.eu/skymap.html
- stefco_ 9y ago
  Having extra detectors also helps hugely with uptime. We're currently only active and collecting clean data ~50% of the time for each detector. We need at least two detectors active for a detection, and need three detectors for this kind of beautiful directional isolation. So having extra detectors decreases our skymap areas significantly, which helps in EM followup, but it also significantly reduces GW detector network downtime. And it allows for graceful degradation; with 5 detectors, you can afford to have a couple detectors down and still detect an event with good direction reconstruction. We also have tons of downtime for upgrades. Now that our second observing run is over, we have a ~1.5yr upgrade cycle. There's no point in running VIRGO without LIGO (except for engineering reasons); having more detectors will increase flexibility in this regard, too. When KAGRA and LIGO India come online, we should see much better skymaps, but we will also see much higher uptimes (and hence higher detection rates). It's especially exciting to see improved direction reconstruction given that VIRGO has faced setbacks this year with their mirrors. I wasn't expecting to see such beautiful skymaps until our third observing run! Needless to say it was an exciting summer :) - grad student working on LIGO

komali2 9y ago

I'm googling but it's all above my head - what's a gravitational wave, and how do they detect them, if anybody is willing to take a bit of time to explain to a turbo-layman? :D

aeleos 9y ago
There are many good videos on youtube that do a good job explaining what gravitational waves are and how we detect them. Many of these were made in response to the first detection, and they do a pretty good job explaining it in laymans terms. I personally really like veritasiums video here. https://www.youtube.com/watch?v=iphcyNWFD10 https://www.youtube.com/watch?v=iphcyNWFD10
deleted 9y ago
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ivoras 9y ago

So, what are the implication on General relativity if it's shown that gravity changes move in waves, with a finite speed?

vorotato 9y ago
The speed of gravity is the speed of light, because it's the speed of causality. That's a pretty well understood thing, and isn't really something being explored by these.
kuschku 9y ago
Well, basically "confirmed", nothing more. Einstein’s theories predicted all this.

gus_massa 9y ago

> Black holes produce gravitational waves but not light. Is this correct? I thought that it would produce a lot of light, but the light is more easily absorbed by interstellar gas and dust, so it's more difficult to see it. Perhaps not the black holes doesn't emit light directly, but the surrounding gas will be extremely hot. And when the two event horizons collide, the virtual particles should be very confused and scape in huge numbers, in a process like the Hawking radiation, but much bigger.????? I'm far from being an expert in this, so this is only a conjecture and unsupported handwaving. But I would be very surprised to confirm that the event has no emission of light.

azag0 9y ago
> Perhaps not the black holes doesn't emit light directly, but the surrounding gas will be extremely hot. Yes, but not every black hole is surrounded by gas. > the virtual particles should be very confused In general, physics does not work like this.
cyphar 9y ago
Black holes themselves cannot emit light (Hawking radiation is a separate topic). Gas that is in accretion disks around a black hole can become hot and produce light, but that process has nothing to do with the black hole itself. By definition, light cannot escape a black hole so processes inside the black hole do not emit light (or rather, if they do we cannot see it). However, massive bodies experiencing rapid acceleration emit gravitational waves. This is something that happens due to a black hole's mass and not it's event horizon properties (in principle you can also detect neutron star interactions). But this is emitted due to the motion of the objects themselves, not from accretion disks (which may not be present for some black holes). So the event itself does not produce light.
- oldandtired 9y ago
  A problem with your explanation is that from our perspective (which is the only one that actually matters here), "black holes" will never undergo rapid acceleration due to the extreme time dilation effects according to the gravity model used. I find it interesting that much ado is made about "black holes" existing when by the theory by which they are proposed, they cannot exist in our finite universe. The time dilation effects, which is a part of the theory, will not allow any such theoretical entity to be formed in our finite time universe. With that said, the observations of LIGO do not confirm the existence of such entities as "black holes". What these observation do do is indicate that there is something happening. We just don't know yet what this is. Going down the rabbit hole of sticking to the "black hole" model only engenders ignorance. These are interesting experiments and results and it behooves an open mind to try to further our understanding of what is causing these observational event. Unfortunately, the "dogma" of worshipping Einstein and his "holy works" only locks us into a position of not furthering our understanding. Note: I have used "worshipping" and "holy works" specifically because of the attitudes of those who will not countenance any different views from Generally Relativity et al. Whether or not, GR and SR have long term merit is still a question being pursued. Some variation of these may be more suitable. But that doesn't mean that GR and SR are correct (as is the usual defence given). The last couple of decades have been an interesting case study in terms of an unwitting increase in dogmatic push for specific theories and models in quite a few areas. When it boils down to it, these theories and models are subject to change and even throwing them out for different and better models. To get emotionally caught up in defending a specific theory and model as the "Truth" does not put science in a good light for the general populous. I have said it elsewhere, science is about the discovery of the universe around us and finding applicable theories and models that are useful for our understanding. If you want "Truth" then go and study philosophy and religion. Every theory you want to follow needs to have a good defence, but not to the extent of getting so emotionally caught up in it that you ignore discrepancies and results that disagree with your preferred theory and model. How often has the "Truth" that the general science community has promulgated been overturned? We need to be careful not to get so entrenched in a specific viewpoint that we cannot see the merit in other theories or models. It is history that tells us to be careful here.

c-smile 9y ago

Why black hole merges happen so frequently? We just started registering them and got three events already... Mass of Universe is estimated to be of 6e51kg. Mass of Sun is 1e25kg. This event is about 50 Sun masses. So that one event is about morphing 1e-26 part of Universe. I suspect that matter collapsing into black holes happens significantly more frequently than that. Do we have enough non-black-hole matter available for our (humanity) life time? Yet worth to mention that black holes "return" mass by not only Hawking–Zel'dovich radiation but also by such events - this particular merge emitted 3 solar masses as energy.

MikeGale 9y ago

Don't you look forward to the time when these events need to named like this GW1712231509236732?

ssijak 9y ago

How do they know exact sizes of the 2 black holes, and the remaining size also? Seems like they say it as 100% truth, which maybe it is, but my uneducated mind cannot think how they could calculate that. Only those sizes fit in the mathematical model of the gravitational waves compared to the observations?

MAGZine 9y ago
It is trivial to obtain the size of objects, so long as you have some information about a satellite orbiting it. It's how we get the mass of many interstellar objects, and it's how we know that dark matter exists. http://astro.physics.uiowa.edu/ITU/glossary/keplers-third-law/ http://astro.physics.uiowa.edu/ITU/glossary/keplers-third-la...
21 9y ago
I imagine is like seeing a light source from two vantage points. Using the timing difference you can triangulate a distance. Using the computed distance and received brightness you can compute source brightness. Using source brightness you can compute source power.