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LIGO and Virgo announce the detection of a black hole binary merger

129 points by srikar 9y ago

spuz 9y ago
One question I've had about gravitational wave detections that I haven't yet been able to find an answer to is what is the mechanism by which the mass of an orbiting black hole pair converts its mass into a gravitational wave? Presumably the mass of the black holes is comprised of matter (in whatever form that may be) and kinetic energy. Is the gravitational wave energy while the two objects orbit purely a conversion of kinetic energy to gravitational wave energy or is some of the mass lost too? What about when they finally collide? If in this case 1 solar mass of matter was converted into gravitational energy then by what process?
- sulam 9y ago
  Mass is energy. As I understand it, the waves are created because of the rapid acceleration of large masses, either black holes or neutron stars. These waves do transmit energy and the final resulting body will be less massive at least partially because of that.
  - dwaltrip 9y ago
    I believe all accelerating masses emit gravitational waves, but that most do so on an absurdly small scale. As if the 10^-20 scale of the black hole merger gwaves isn't small enough...
    - lou1306 9y ago
      You are correct. In fact, Einstein predicted gravitational waves but considered them un-observable, and the LIGO interferometer is 1km long exactly to magnify the effect.
- evanb 9y ago
  When an apple falls to the earth, by what process is its potential energy converted to kinetic energy? Gravitational processes alone. Apple-earth collisions primarily radiate apple sauce, black hole mergers primarily radiate in gravitational energy. Minor nit: In general relativity black holes are not actually comprised of matter---they're entirely warping of spacetime. Whether that remains true in a quantum theory of gravity is unknown.
  - spuz 9y ago
    None of this actually answers the question. The question is by what process does energy in the binary black hole system get converted into gravitational wave energy?
    - T-A 9y ago
      If you accelerate an electric charge, it emits electromagnetic waves. If you accelerate a mass, it emits gravitational waves. Two masses orbiting their common center of gravity are undergoing centripetal acceleration, so they continuously emit gravitational waves, spiraling closer as they lose energy (that's how the emission of gravitational waves was originally confirmed [1]). [1] https://www.nobelprize.org/nobel_prizes/physics/laureates/1993/speedread.html https://www.nobelprize.org/nobel_prizes/physics/laureates/19...
    - evanb 9y ago
      The answer is "by the processes of general relativity". When things fall "down" their potential energy is lowered, so they get faster. When mass-energy accelerates it emits gravitational waves, in much the same way when an electron accelerates it emits electromagnetic waves. If you're satisfied that a radio works by sloshing electrons around, you should be satisfied that a black hole merger emits radiation by sloshing mass around. Where did the photons "come from"? Well, they weren't stored "inside" the electrons. By what process were the photons generated? Electrons accelerating radiate. That's essentially the answer. You can "math it up" if you want. It's not exactly an axiom, but it's pretty close to the bottom.
      - evanb 9y ago
        Maybe I should be more clear: at least in the generally relativistic conception of gravity, all of the mass of the black hole "comes from" the warping of spacetime already. That's why I keep dancing around the question of whether it's "really" the kinetic energy or the mass or a mix that gets converted---it's hard to distinguish and may not be meaningful to distinguish the pieces from one another, especially if I go to a co-rotating frame.
- solipsism 9y ago
  Anything spinning in an asymmetrical way (e.g. two bodies orbiting each other, but not a single body spinning) loses angular momentum as gravity waves. This is true whether it's two black holes or a tumbling dumbbell. In addition to that, the ultimate merger of the black holes releases the incredible amount of gravitational potential energy that existed between them when they were separate bodies. No mass escapes the event horizon. These are black holes after all :)
  - spuz 9y ago
    How much of the 1 solar mass of detectable gravitational energy is released before the moment of collision and how much is released at the moment of collision? Is the mechanism for release of this energy the same in both cases? If 1 solar mass of energy was released in total, and the mass of the combined black hole is less than the two before the merger, how is it possible that none of the mass of the black holes has escaped their respective event horizons?
    - deleted 9y ago
      [deleted]
    - improbable22 9y ago
      There isn't exactly one moment of collision like snooker balls, the two spiral and merge and then settle down to look like one bigger black hole. But the time during which most of the energy is radiated is quite short, like 0.1s. The missing mass is precisely the amount of energy that was radiated away. We should not think of the mass of the black hole as being the amount of matter stored inside, which may escape... regardless of how it was created, the black hole is just a ball of pure gravity. Its mass is defined by its effect on things far away. You can measure the mass of Jupiter by watching how fast a satellite orbits, and a black hole whose satellites had the same orbits would be said to have the same mass.
  - mirimir 9y ago
    > No mass escapes the event horizon. These are black holes after all :) OK, that makes sense (for whatever that's worth). But from the article: > ... the latest discovery was produced by the merger of two relatively light black holes, 7 and 12 times the mass of the sun ... The merger left behind a final black hole 18 times the mass of the sun, meaning that energy equivalent to about 1 solar mass was emitted as gravitational waves during the collision. And you say: > the ultimate merger of the black holes releases the incredible amount of gravitational potential energy that existed between them when they were separate bodies. I think that I get it. It's just that the stated masses of the merging black holes (7 and 12 solar masses) include gravitational potential energy. So the rest masses of the black holes didn't change, just their gravitational potential energy. Yes?
- raattgift 9y ago
  > what is the mechanism by which the mass of an orbiting black hole pair converts its mass into a gravitational wave? This is an excellent question, although the formal answer is essentially "mu", or alternatively in some useful approximations of General Relativity we have a research project along the lines of "what is the mechanism that generates the (final) metric of merged black holes?". I'll try to give you a more useful answer. The important things are that we go from observables relating to a periodically perturbed near-Schwarzschild spacetime sourced by the inspirallers (this requires us to remove other contributors to the "true" metric, so we're left only with the contributions from the inspirallers) to a much more stable near-Schwarzschild spacetime. The energy-momentum density implied at the origin before merger is higher than that afterwards. So we can ask your question: where did that energy-momentum go? Ultimately the search for an answer comes down to making a few choices about how to describe a collection of values that appear at various points of interest in a (nonvacuum) spacetime solution of the Einstein Field Equations of General Relativity that survive a degree of calculational simplification. A more technical response is that by choosing how to split spacetime into space and time, and by choosing to represent spacetime curvature at every point in space (for a slice of spacetime that all has the same time coordinate) as a perturbation of a fixed background metric, one can treat changes in the metric as propagating like massless waves. This is easier when the masses sourcing the metric move slowly compared to the speed of light, and when one is dealing with the increasingly-flat metric far from the source (at least tens of wavelengths from the inspiralling objects), since one can then use linearized gravity. So we're in fairly good shape far from some black holes (or neutron stars) embedded in a galaxy a few million light-years away. One can then decide that we are not obliged to treat the perturbations as being exclusively sourced by matter, that is, the propagating gravitational waves can induce a squash-strain on matter. With some further choices of gauge and a change to a formalism like linearized gravity, one can treat some components of the two relevant tensors (the Einstein tensor G and the stress-energy tensor T) as representing specific forms of energy, with some (e.g. kinetic energy or angular momentum) being dumped into others (e.g. gravitational potential energy). General Relativity has at its heart a relationship between matter and spacetime curvature. The former contributes to the stress-energy tensor (T) mentioned above. The components of the tensor relate to the flux of momentum-energy between a point p and its neighbouring points in the four spacetime directions. While the tensor value itself is the same for all observers, the values of the individual components (e.g. the numerical values if you write the tensor out in 4x4 matrix form) depend on choice of coordinates, different observers have almost total freedom when it comes to choosing coordinates. Spacetime curvature is represented by the Einstein tensor G, which is a non-linear function of the metric tensor g. Omitting constant factors and indices (which range from 0 to 3 for each of the four dimension of spacetime), we can write the core of General Relativity as G = T. That is, spacetime curvature is totally determined by matter. For example, G = T = 0 is the case where there is no matter, and thus flat spacetime; that's the spacetime of Special Relativity. When we add any matter at all, we deviate from flat spacetime, however because the contribution of small amounts of matter to the stress-energy tensor is small, it can be very hard to distinguish between true flat spacetime and very very slightly curved spacetime. While in general it is convenient to think in terms of G = T -- the Einstein Tensor and thus the metric is totally determined by the configuration of matter -- there is a history of vacuum solutions of the Einstein Field equations in which G != 0, T = 0, that is that there is spacetime curvature even without matter being present. An early example was the Schwarzschild vacuum solution ("Schwarzschild spacetime"), which is perfectly spherically symmetric about an eternal gravitational singularity. Some study revealed that for a perfectly spherical uncharged unrotating arrangement of mass gives you Schwarzschild spacetime at a bit of a distance from the mass. For slight deviations from this perfect arrangement of central matter, we still get something very similar to Schwarzschild at a sufficient distance. Indeed, if one goes to enormous distances, Schwarzschild-like spacetime also starts to look like flat spacetime -- it's "asymptotically flat". We can then ask: at a sufficient distance from a pair of inspiralling objects, is the spacetime very similar to Schwarzschild spacetime? The answer is almost "yes". If you consider these objects as if they were a barbell with an infinitesimally thin handle connecting the two weighted ends, and tumble the barbell around the middle, to a chosen distant observer stationary with respect to the centre of mass of our inspiralling objects, sometimes sometimes one or the other weight will be spatially closer. When the observer is trying to figure out if it is in Schwarzschild spacetime or something else, and using Schwarzschild coordinates, the changing proximity of the barbell ends compared to the centre of mass will become apparent. If we go back to my fourth paragraph, the observer can make a choice to consider the local measurements of the true metric compared to his or her wristwatch or atomic clock, and a further choice to compare that with the Schwarzschild metric. There will be a periodic change in difference from the Schwarzschild background, relating to the orbital period of the inspiralling objects sourcing the true approximately Schwarzschild metric. Near the end of the inspiral, the "up and down" of the measurements of the metric they source follows a predictable evolution as the orbital period increases until the objects inevitably collide. The relative deviation from the perfect Schwarzschild metric also increases prior to collision. But after the collision the waves cease (the collided entities settle into a configuration that is much more like the perfect uncharged non-rotating sphere that would source a true Schwarzschild metric, a process called "balding" if the end result is a black hole (with perhaps some debris around it, depending on what the inspiralled objects were). Inevitably, measurements will show a greater Schwarzschild distance from the mass, or equivalently, that the source of the new "truer" Schwarzschild-like metric has less mass-energy at the origin of Schwarzschild coordinates than was in the pre-collision configuration. We are pretty free to interpret these observations. Generally there is no reason to invoke any local magic rather than suggest that the reduction of energy-momentum is exactly balanced by the increase in the perturbations from Schwarzschild that all possible observers would record, and look for evidence of that. Finally, how does one measure the deviation from a background like a Schwarzschild metric? Well, one option is to do it in the style of the https://en.wikipedia.org/wiki/Cavendish_experiment https://en.wikipedia.org/wiki/Cavendish_experiment . Another is to use interferometry, LIGO and Virgo (and so on) style. Either way, the measurer is looking for a little change in the arrangement of matter "here-and-now" correlated with the big change in the arrangement in matter "there-and-then". With some deliberate choices one can think of the change "here-and-now" as being various types of energy from "there-and-then" carried to us by gravitational radiation from the source, and calculate robustly on that basis. That picture holds up pretty well under detailed analysis given current experimental results. Although this invites all sorts of analogies or statements like "spacetime is plastic" or "gravitational waves are real in the sense of being observer-independent", those are big stretches of the theoretical underpinnings (i.e., General Relativity). All we have is a metric that covers the whole of spacetime -- all points at all times -- that we can slice up so that we can think of the value of the metric changing over time. But generally different observers will prefer different slicings, and for a family of observers any inspiraller will shed no gravitational waves at all [consider a diagram BHA ---- x ---- BHB where x is an observer at rest at the centre-of-mass of a pair of black holes in a mutual circular orbit; x will not measure any gravitational radiation under any spacetime slicing]. So they are really a gauging effect. Conveniently, when one turns a gravitational wave into a bunch of particles, one finds that the particles almost certainly have to have the characteristics of a gauge boson. And now I'm out of space and time. :-) Hopefully this was helpful, or at least somewhat interesting.
krylon 9y ago
> Dubbed GW170608, the latest discovery was produced by the merger of two relatively light black holes, 7 and 12 times the mass of the sun, at a distance of about a billion light-years from Earth. The merger left behind a final black hole 18 times the mass of the sun, meaning that energy equivalent to about 1 solar mass was emitted as gravitational waves during the collision. I wonder if one would experience any macroscopic effects from the gravitational waves if one were close enough to the black holes during the merger. Or would one have to be so close that tidal effects from the black holes' gravity would mask any of those effects? I ask because one solar mass worth of energy sounds like ... a lot. At least to me as an astronomical layperson.
- ISL 9y ago
  You're right: A solar mass is a spectacular quantity of energy. One would need to be very close to the merger to experience any macroscopic effects from gravitational waves alone, if at all. Spacetime is very stiff.
- Sharlin 9y ago
  Apparently from 1 AU you might just be able to hear it! Even that close the stretching and squishing would be on the order of micrometers but the eardrum might be able to pick it up. The frequency would be comfortably in the audible range as well. [1] [1] https://www.reddit.com/r/askscience/comments/45n0sz/how_close_to_gw150914_the_black_hole_merger_would/ https://www.reddit.com/r/askscience/comments/45n0sz/how_clos...
  - krylon 9y ago
    Huh. IIRC, black hole mergers are also a likely source of short gamma ray bursts. So I suspect at 1 AU, one would need a lot of shielding to not get fried by the gamma rays. EDIT: What I meant to say was: It would probably be quite awesome to experience such an event up close, provided it is possible to do so safely. ;-)
    - dwaltrip 9y ago
      I believe that black hole mergers are dark in the entire electromagnetic (EM) spectrum. This is why the recent binary neutron star merger was so groundbreaking -- it emitted both gravitational waves and a broad spectrum of EM waves. Indeed, though, it would be awesome to witness the collision and resulting gwaves personally, if such a thing was possible :)
Jasugser 9y ago
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raverbashing 9y ago
Interesting This was a couple of days days before Virgo got online AND one of the Ligo detectors was undergoing a noise modelling test (its mirrors were being vibrated at the time)
- mstade 9y ago
  So no second opinions then, it could just as well be noise?
  - astrosi 9y ago
    No, because the noise modeling test in the second LIGO detector (LIGO Hanford) was being wobbled at a very different frequency to that of the gravitational wave they were able to filter out the test motion and were still able to detect the merger.
libeclipse 9y ago
> ...meaning that energy equivalent to about 1 solar mass was emitted as gravitational waves during the collision. Damn. That's something like 179 100 000 000 000 000 000 000 000 000 000 000 000 000 000 000 J That's an insane amount of energy. It's equivalent to what you would get if you converted the entire mass of the sun into pure energy.
- Simon_says 9y ago
  Well, yea - that's what they mean when they say "1 solar mass".
  - libeclipse 9y ago
    You and I might know that but not everyone has studied physics.
merraksh 9y ago
Six black hole merges observed in ~2 years! That's quite interesting. More observations will make for some useful statistical study. I wonder if this would give us any insights w.r.t. matter and its distribution across the Universe, and/or help us better understand/estimate dark matter/energy.
- T-A 9y ago
  Dark matter, quite possibly: https://astrobites.org/2017/08/31/could-dark-matter-be-black-holes/ https://astrobites.org/2017/08/31/could-dark-matter-be-black... http://aasnova.org/2017/09/27/can-ligo-find-the-missing-dark-matter/ http://aasnova.org/2017/09/27/can-ligo-find-the-missing-dark...
  - QAPereo 9y ago
    http://resonaances.blogspot.com/2016/06/black-hole-dark-matter.html http://resonaances.blogspot.com/2016/06/black-hole-dark-matt... http://4.bp.blogspot.com/-jnY74NBc7ic/V2UcuCswL-I/AAAAAAAAB78/vWC7nOxV53ALFy5YiXQRYrYxP631PrJDgCK4B/s1600/blackholedarkmatterconstraints.png http://4.bp.blogspot.com/-jnY74NBc7ic/V2UcuCswL-I/AAAAAAAAB7... The odds are still very low on MACHO's as a source of most dark matter.
- nerfhammer 9y ago
  Most interesting is that these have all been medium-sized black holes which weren't known to exist previously and we don't have a theory of how they form.
smortaz 9y ago
Posting this whenever there's a new detection! Dr Roy Williams who's part of the LIGO team has put the notebooks used for analyzing the data online for anyone to check out & run: https://notebooks.azure.com/roywilliams/libraries/LIGOOpenScienceCenter https://notebooks.azure.com/roywilliams/libraries/LIGOOpenSc... Click Clone to get your own copy, then edit/run/etc.
jcims 9y ago
A 1 kg block of plutonium 12,000km away has ~5 orders of magnitude stronger gravitational field than a solar mass at 1 billion light years. Presumably a sudden mass-energy conversion of said kilogram would generate a sharp gravitational wave. Assuming someone went back through LIGOs algorithms to fine tune them for such a detection, doesn't it seem plausible that it would be able to do so? And presumably even locate it?
- jcims 9y ago
  I’ll toss out a guess that the signal is there but it’s so short that there isn’t enough to correlate across locations. So you might not be able to detect a detonation with it, but given a window of seconds from seismograph data you might be able to pinpoint location and corroborate yield.
- scentoni 9y ago
  Only a small percentage of the plutonium's mass gets converted to energy during fission.
  - jcims 9y ago
    Indeed, this site says it's about 46g per megaton: http://www.jick.net/hr/skept/EMC2/node4.html http://www.jick.net/hr/skept/EMC2/node4.html Still, 1kg has a field strength that's 5 orders of magnitude greater than a solar mass a billion light years away...10mg would be in the neighborhood. If i didn't screw up the math, that's ridiculous.

129 points by srikar 9y ago

spuz 9y ago

One question I've had about gravitational wave detections that I haven't yet been able to find an answer to is what is the mechanism by which the mass of an orbiting black hole pair converts its mass into a gravitational wave? Presumably the mass of the black holes is comprised of matter (in whatever form that may be) and kinetic energy. Is the gravitational wave energy while the two objects orbit purely a conversion of kinetic energy to gravitational wave energy or is some of the mass lost too? What about when they finally collide? If in this case 1 solar mass of matter was converted into gravitational energy then by what process?

sulam 9y ago
Mass is energy. As I understand it, the waves are created because of the rapid acceleration of large masses, either black holes or neutron stars. These waves do transmit energy and the final resulting body will be less massive at least partially because of that.
- dwaltrip 9y ago
  I believe all accelerating masses emit gravitational waves, but that most do so on an absurdly small scale. As if the 10^-20 scale of the black hole merger gwaves isn't small enough...
  - lou1306 9y ago
    You are correct. In fact, Einstein predicted gravitational waves but considered them un-observable, and the LIGO interferometer is 1km long exactly to magnify the effect.
evanb 9y ago
When an apple falls to the earth, by what process is its potential energy converted to kinetic energy? Gravitational processes alone. Apple-earth collisions primarily radiate apple sauce, black hole mergers primarily radiate in gravitational energy. Minor nit: In general relativity black holes are not actually comprised of matter---they're entirely warping of spacetime. Whether that remains true in a quantum theory of gravity is unknown.
- spuz 9y ago
  None of this actually answers the question. The question is by what process does energy in the binary black hole system get converted into gravitational wave energy?
  - T-A 9y ago
    If you accelerate an electric charge, it emits electromagnetic waves. If you accelerate a mass, it emits gravitational waves. Two masses orbiting their common center of gravity are undergoing centripetal acceleration, so they continuously emit gravitational waves, spiraling closer as they lose energy (that's how the emission of gravitational waves was originally confirmed [1]). [1] https://www.nobelprize.org/nobel_prizes/physics/laureates/1993/speedread.html https://www.nobelprize.org/nobel_prizes/physics/laureates/19...
  - evanb 9y ago
    The answer is "by the processes of general relativity". When things fall "down" their potential energy is lowered, so they get faster. When mass-energy accelerates it emits gravitational waves, in much the same way when an electron accelerates it emits electromagnetic waves. If you're satisfied that a radio works by sloshing electrons around, you should be satisfied that a black hole merger emits radiation by sloshing mass around. Where did the photons "come from"? Well, they weren't stored "inside" the electrons. By what process were the photons generated? Electrons accelerating radiate. That's essentially the answer. You can "math it up" if you want. It's not exactly an axiom, but it's pretty close to the bottom.
    - evanb 9y ago
      Maybe I should be more clear: at least in the generally relativistic conception of gravity, all of the mass of the black hole "comes from" the warping of spacetime already. That's why I keep dancing around the question of whether it's "really" the kinetic energy or the mass or a mix that gets converted---it's hard to distinguish and may not be meaningful to distinguish the pieces from one another, especially if I go to a co-rotating frame.
solipsism 9y ago
Anything spinning in an asymmetrical way (e.g. two bodies orbiting each other, but not a single body spinning) loses angular momentum as gravity waves. This is true whether it's two black holes or a tumbling dumbbell. In addition to that, the ultimate merger of the black holes releases the incredible amount of gravitational potential energy that existed between them when they were separate bodies. No mass escapes the event horizon. These are black holes after all :)
- spuz 9y ago
  How much of the 1 solar mass of detectable gravitational energy is released before the moment of collision and how much is released at the moment of collision? Is the mechanism for release of this energy the same in both cases? If 1 solar mass of energy was released in total, and the mass of the combined black hole is less than the two before the merger, how is it possible that none of the mass of the black holes has escaped their respective event horizons?
  - deleted 9y ago
    [deleted]
  - improbable22 9y ago
    There isn't exactly one moment of collision like snooker balls, the two spiral and merge and then settle down to look like one bigger black hole. But the time during which most of the energy is radiated is quite short, like 0.1s. The missing mass is precisely the amount of energy that was radiated away. We should not think of the mass of the black hole as being the amount of matter stored inside, which may escape... regardless of how it was created, the black hole is just a ball of pure gravity. Its mass is defined by its effect on things far away. You can measure the mass of Jupiter by watching how fast a satellite orbits, and a black hole whose satellites had the same orbits would be said to have the same mass.
- mirimir 9y ago
  > No mass escapes the event horizon. These are black holes after all :) OK, that makes sense (for whatever that's worth). But from the article: > ... the latest discovery was produced by the merger of two relatively light black holes, 7 and 12 times the mass of the sun ... The merger left behind a final black hole 18 times the mass of the sun, meaning that energy equivalent to about 1 solar mass was emitted as gravitational waves during the collision. And you say: > the ultimate merger of the black holes releases the incredible amount of gravitational potential energy that existed between them when they were separate bodies. I think that I get it. It's just that the stated masses of the merging black holes (7 and 12 solar masses) include gravitational potential energy. So the rest masses of the black holes didn't change, just their gravitational potential energy. Yes?
raattgift 9y ago
> what is the mechanism by which the mass of an orbiting black hole pair converts its mass into a gravitational wave? This is an excellent question, although the formal answer is essentially "mu", or alternatively in some useful approximations of General Relativity we have a research project along the lines of "what is the mechanism that generates the (final) metric of merged black holes?". I'll try to give you a more useful answer. The important things are that we go from observables relating to a periodically perturbed near-Schwarzschild spacetime sourced by the inspirallers (this requires us to remove other contributors to the "true" metric, so we're left only with the contributions from the inspirallers) to a much more stable near-Schwarzschild spacetime. The energy-momentum density implied at the origin before merger is higher than that afterwards. So we can ask your question: where did that energy-momentum go? Ultimately the search for an answer comes down to making a few choices about how to describe a collection of values that appear at various points of interest in a (nonvacuum) spacetime solution of the Einstein Field Equations of General Relativity that survive a degree of calculational simplification. A more technical response is that by choosing how to split spacetime into space and time, and by choosing to represent spacetime curvature at every point in space (for a slice of spacetime that all has the same time coordinate) as a perturbation of a fixed background metric, one can treat changes in the metric as propagating like massless waves. This is easier when the masses sourcing the metric move slowly compared to the speed of light, and when one is dealing with the increasingly-flat metric far from the source (at least tens of wavelengths from the inspiralling objects), since one can then use linearized gravity. So we're in fairly good shape far from some black holes (or neutron stars) embedded in a galaxy a few million light-years away. One can then decide that we are not obliged to treat the perturbations as being exclusively sourced by matter, that is, the propagating gravitational waves can induce a squash-strain on matter. With some further choices of gauge and a change to a formalism like linearized gravity, one can treat some components of the two relevant tensors (the Einstein tensor G and the stress-energy tensor T) as representing specific forms of energy, with some (e.g. kinetic energy or angular momentum) being dumped into others (e.g. gravitational potential energy). General Relativity has at its heart a relationship between matter and spacetime curvature. The former contributes to the stress-energy tensor (T) mentioned above. The components of the tensor relate to the flux of momentum-energy between a point p and its neighbouring points in the four spacetime directions. While the tensor value itself is the same for all observers, the values of the individual components (e.g. the numerical values if you write the tensor out in 4x4 matrix form) depend on choice of coordinates, different observers have almost total freedom when it comes to choosing coordinates. Spacetime curvature is represented by the Einstein tensor G, which is a non-linear function of the metric tensor g. Omitting constant factors and indices (which range from 0 to 3 for each of the four dimension of spacetime), we can write the core of General Relativity as G = T. That is, spacetime curvature is totally determined by matter. For example, G = T = 0 is the case where there is no matter, and thus flat spacetime; that's the spacetime of Special Relativity. When we add any matter at all, we deviate from flat spacetime, however because the contribution of small amounts of matter to the stress-energy tensor is small, it can be very hard to distinguish between true flat spacetime and very very slightly curved spacetime. While in general it is convenient to think in terms of G = T -- the Einstein Tensor and thus the metric is totally determined by the configuration of matter -- there is a history of vacuum solutions of the Einstein Field equations in which G != 0, T = 0, that is that there is spacetime curvature even without matter being present. An early example was the Schwarzschild vacuum solution ("Schwarzschild spacetime"), which is perfectly spherically symmetric about an eternal gravitational singularity. Some study revealed that for a perfectly spherical uncharged unrotating arrangement of mass gives you Schwarzschild spacetime at a bit of a distance from the mass. For slight deviations from this perfect arrangement of central matter, we still get something very similar to Schwarzschild at a sufficient distance. Indeed, if one goes to enormous distances, Schwarzschild-like spacetime also starts to look like flat spacetime -- it's "asymptotically flat". We can then ask: at a sufficient distance from a pair of inspiralling objects, is the spacetime very similar to Schwarzschild spacetime? The answer is almost "yes". If you consider these objects as if they were a barbell with an infinitesimally thin handle connecting the two weighted ends, and tumble the barbell around the middle, to a chosen distant observer stationary with respect to the centre of mass of our inspiralling objects, sometimes sometimes one or the other weight will be spatially closer. When the observer is trying to figure out if it is in Schwarzschild spacetime or something else, and using Schwarzschild coordinates, the changing proximity of the barbell ends compared to the centre of mass will become apparent. If we go back to my fourth paragraph, the observer can make a choice to consider the local measurements of the true metric compared to his or her wristwatch or atomic clock, and a further choice to compare that with the Schwarzschild metric. There will be a periodic change in difference from the Schwarzschild background, relating to the orbital period of the inspiralling objects sourcing the true approximately Schwarzschild metric. Near the end of the inspiral, the "up and down" of the measurements of the metric they source follows a predictable evolution as the orbital period increases until the objects inevitably collide. The relative deviation from the perfect Schwarzschild metric also increases prior to collision. But after the collision the waves cease (the collided entities settle into a configuration that is much more like the perfect uncharged non-rotating sphere that would source a true Schwarzschild metric, a process called "balding" if the end result is a black hole (with perhaps some debris around it, depending on what the inspiralled objects were). Inevitably, measurements will show a greater Schwarzschild distance from the mass, or equivalently, that the source of the new "truer" Schwarzschild-like metric has less mass-energy at the origin of Schwarzschild coordinates than was in the pre-collision configuration. We are pretty free to interpret these observations. Generally there is no reason to invoke any local magic rather than suggest that the reduction of energy-momentum is exactly balanced by the increase in the perturbations from Schwarzschild that all possible observers would record, and look for evidence of that. Finally, how does one measure the deviation from a background like a Schwarzschild metric? Well, one option is to do it in the style of the https://en.wikipedia.org/wiki/Cavendish_experiment https://en.wikipedia.org/wiki/Cavendish_experiment . Another is to use interferometry, LIGO and Virgo (and so on) style. Either way, the measurer is looking for a little change in the arrangement of matter "here-and-now" correlated with the big change in the arrangement in matter "there-and-then". With some deliberate choices one can think of the change "here-and-now" as being various types of energy from "there-and-then" carried to us by gravitational radiation from the source, and calculate robustly on that basis. That picture holds up pretty well under detailed analysis given current experimental results. Although this invites all sorts of analogies or statements like "spacetime is plastic" or "gravitational waves are real in the sense of being observer-independent", those are big stretches of the theoretical underpinnings (i.e., General Relativity). All we have is a metric that covers the whole of spacetime -- all points at all times -- that we can slice up so that we can think of the value of the metric changing over time. But generally different observers will prefer different slicings, and for a family of observers any inspiraller will shed no gravitational waves at all [consider a diagram BHA ---- x ---- BHB where x is an observer at rest at the centre-of-mass of a pair of black holes in a mutual circular orbit; x will not measure any gravitational radiation under any spacetime slicing]. So they are really a gauging effect. Conveniently, when one turns a gravitational wave into a bunch of particles, one finds that the particles almost certainly have to have the characteristics of a gauge boson. And now I'm out of space and time. :-) Hopefully this was helpful, or at least somewhat interesting.

krylon 9y ago

> Dubbed GW170608, the latest discovery was produced by the merger of two relatively light black holes, 7 and 12 times the mass of the sun, at a distance of about a billion light-years from Earth. The merger left behind a final black hole 18 times the mass of the sun, meaning that energy equivalent to about 1 solar mass was emitted as gravitational waves during the collision. I wonder if one would experience any macroscopic effects from the gravitational waves if one were close enough to the black holes during the merger. Or would one have to be so close that tidal effects from the black holes' gravity would mask any of those effects? I ask because one solar mass worth of energy sounds like ... a lot. At least to me as an astronomical layperson.

ISL 9y ago
You're right: A solar mass is a spectacular quantity of energy. One would need to be very close to the merger to experience any macroscopic effects from gravitational waves alone, if at all. Spacetime is very stiff.
Sharlin 9y ago
Apparently from 1 AU you might just be able to hear it! Even that close the stretching and squishing would be on the order of micrometers but the eardrum might be able to pick it up. The frequency would be comfortably in the audible range as well. [1] [1] https://www.reddit.com/r/askscience/comments/45n0sz/how_close_to_gw150914_the_black_hole_merger_would/ https://www.reddit.com/r/askscience/comments/45n0sz/how_clos...
- krylon 9y ago
  Huh. IIRC, black hole mergers are also a likely source of short gamma ray bursts. So I suspect at 1 AU, one would need a lot of shielding to not get fried by the gamma rays. EDIT: What I meant to say was: It would probably be quite awesome to experience such an event up close, provided it is possible to do so safely. ;-)
  - dwaltrip 9y ago
    I believe that black hole mergers are dark in the entire electromagnetic (EM) spectrum. This is why the recent binary neutron star merger was so groundbreaking -- it emitted both gravitational waves and a broad spectrum of EM waves. Indeed, though, it would be awesome to witness the collision and resulting gwaves personally, if such a thing was possible :)

Jasugser 9y ago

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raverbashing 9y ago

Interesting This was a couple of days days before Virgo got online AND one of the Ligo detectors was undergoing a noise modelling test (its mirrors were being vibrated at the time)

mstade 9y ago
So no second opinions then, it could just as well be noise?
- astrosi 9y ago
  No, because the noise modeling test in the second LIGO detector (LIGO Hanford) was being wobbled at a very different frequency to that of the gravitational wave they were able to filter out the test motion and were still able to detect the merger.

libeclipse 9y ago

> ...meaning that energy equivalent to about 1 solar mass was emitted as gravitational waves during the collision. Damn. That's something like 179 100 000 000 000 000 000 000 000 000 000 000 000 000 000 000 J That's an insane amount of energy. It's equivalent to what you would get if you converted the entire mass of the sun into pure energy.

Simon_says 9y ago
Well, yea - that's what they mean when they say "1 solar mass".
- libeclipse 9y ago
  You and I might know that but not everyone has studied physics.

merraksh 9y ago

Six black hole merges observed in ~2 years! That's quite interesting. More observations will make for some useful statistical study. I wonder if this would give us any insights w.r.t. matter and its distribution across the Universe, and/or help us better understand/estimate dark matter/energy.

T-A 9y ago
Dark matter, quite possibly: https://astrobites.org/2017/08/31/could-dark-matter-be-black-holes/ https://astrobites.org/2017/08/31/could-dark-matter-be-black... http://aasnova.org/2017/09/27/can-ligo-find-the-missing-dark-matter/ http://aasnova.org/2017/09/27/can-ligo-find-the-missing-dark...
- QAPereo 9y ago
  http://resonaances.blogspot.com/2016/06/black-hole-dark-matter.html http://resonaances.blogspot.com/2016/06/black-hole-dark-matt... http://4.bp.blogspot.com/-jnY74NBc7ic/V2UcuCswL-I/AAAAAAAAB78/vWC7nOxV53ALFy5YiXQRYrYxP631PrJDgCK4B/s1600/blackholedarkmatterconstraints.png http://4.bp.blogspot.com/-jnY74NBc7ic/V2UcuCswL-I/AAAAAAAAB7... The odds are still very low on MACHO's as a source of most dark matter.
nerfhammer 9y ago
Most interesting is that these have all been medium-sized black holes which weren't known to exist previously and we don't have a theory of how they form.

smortaz 9y ago

Posting this whenever there's a new detection! Dr Roy Williams who's part of the LIGO team has put the notebooks used for analyzing the data online for anyone to check out & run: https://notebooks.azure.com/roywilliams/libraries/LIGOOpenScienceCenter https://notebooks.azure.com/roywilliams/libraries/LIGOOpenSc... Click Clone to get your own copy, then edit/run/etc.

jcims 9y ago

A 1 kg block of plutonium 12,000km away has ~5 orders of magnitude stronger gravitational field than a solar mass at 1 billion light years. Presumably a sudden mass-energy conversion of said kilogram would generate a sharp gravitational wave. Assuming someone went back through LIGOs algorithms to fine tune them for such a detection, doesn't it seem plausible that it would be able to do so? And presumably even locate it?

jcims 9y ago
I’ll toss out a guess that the signal is there but it’s so short that there isn’t enough to correlate across locations. So you might not be able to detect a detonation with it, but given a window of seconds from seismograph data you might be able to pinpoint location and corroborate yield.
scentoni 9y ago
Only a small percentage of the plutonium's mass gets converted to energy during fission.
- jcims 9y ago
  Indeed, this site says it's about 46g per megaton: http://www.jick.net/hr/skept/EMC2/node4.html http://www.jick.net/hr/skept/EMC2/node4.html Still, 1kg has a field strength that's 5 orders of magnitude greater than a solar mass a billion light years away...10mg would be in the neighborhood. If i didn't screw up the math, that's ridiculous.