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Rumours swell over new kind of gravitational-wave sighting
- deleted 9y ago[deleted]
- whoopdedo 9y ago> swell ... wave Someone had fun writing that headline.
- ChuckMcM 9y agoNo doubt they will "crest" just before the announcement :-). And opinions will "undulate" over whether or not they are valid results. Perhaps we just need more drama in science classes to be more inclusive, "Will this acid change the Ph level of the solution? Or will its buffering protect it? Find out after the break ..."
- smhost 9y agoAs a non-science person who occasionally watches PBS documentaries, could this have any implications for Hawking radiation?
- kobeya 9y agoI don't think so, no.
- snissn 9y agoPotentially this neutron - neutron star merger or ones that are detected in the future will result in the formation of a black hole and in that event we may get more data to help refine models of black holes and those refined models may give us insight into the mechanics of black hole radiation - but it would be a bit indirect and we'd have to get lucky. Also Hawking radiation is "very slow" and may never be measured directly in interstellar black holes
- carbocation 9y agoMy understanding is that black hole mergers are not expected to have any optical-wavelength emissions, whereas neutron star mergers should have emissions across the electromagnetic spectrum. Is that distinction part of the excitement here?
- m1el 9y agoThat is correct.
- pducks32 9y agoMostly but also because we haven’t “heard” the waves caused by a Neutron star interaction before. So experimentally we still have to confirm that they cause GWs.
- raverbashing 9y agoWhich would be surprising if they didn't produce GW
- pducks32 9y agoYea that would be very surprising.
- qubex 9y agoOf course they produce gravitational waves! The main point of curiosity is likely that unlike black holes, that are "hairless" entities of pure curvature, neutron stars are complex objects with multiple, variable layers... and consequentially (I presume) the gravitational waves occasioned by their merger would both carry more information about the objects' composition and likely be more complicated (as well as less powerful at any given distance). I'm just guessing though.
- pducks32 9y ago“Of course” is not a popular phrase in physics.
- saganus 9y agoThis might be a naive question, but I'd like to clear it out. Do gravitational waves travel at the speed of light? I know the theory says nothing can travel faster than light. I also know that photons can be seen as quanta or as waves. So my guess is that gravitational waves travel at most, at the speed of light. But do they? or do they travel slower? faster? Is there a doppler effect for GWs? I ask because I would think ripples in the space-time fabric itself might be a bit different than light waves or other more studied phenomena. Can anyone point me in the right direction?
- sleavey 9y agoThe most popular theories say that yes, they do travel at the speed of light. Only a coincident detection of a gravitational wave and an electromagnetic counterpart would confirm this (such as from a binary neutron star coalescence; the binary black holes seen so far by LIGO didn't emit any visible EM as far as we know).
- tzs 9y agoWould even a binary neutron star coalescence give us the information we'd need to answer this? Interstellar space is not actually empty. It contains a low density plasma, and so the speed of light in it is slightly slower than the speed of light in a vacuum. I have no idea if that plasma would also slow down gravity waves, but even if it does it wouldn't necessarily be the same amount [1]. Maybe from the differences in arrival time of light at different frequencies we could figure out enough about the intervening plasma to figure out when the light would have arrived if the space had been empty, and check if that is when the gravity waves arrived? [1] Edit: I have done some Googling. Gravity waves would not be slowed down by the interstellar plasma. The slowdown of light through a medium depends on the existence of positive and negative charges in the medium that can form dipoles in response to the passing electromagnetic wave. For gravity all the "charges" are positive so there is no dipole formation.
- yorwba 9y ago> For gravity all the "charges" are positive so there is no dipole formation. I'm no physicist, but couldn't lack of matter in some region otherwise filled with particles play the part of "virtual" negative matter?
- legohead 9y agoThe idea of two neutron stars colliding gives me shivers. To see something like that would be mind boggling..
- andrewflnr 9y agoNormally, the scientific community is pretty careful about not revealing results before they're fully baked (press notwithstanding). Seeing how that control has broken down for this incident, is it correct to infer that astronomers have pretty much lost their minds over the possibility of capturing a neutron star merger? Edit: calm down, people, I'm excited too.
- evanb 9y agoLIGO didn't announce because they want to be sure. But one of the most valuable aspects of LIGO is as a trigger for optical astronomy of transient, time-sensitive events. So, when they see something, they have to tell a wider astronomical community---the other observers who look in non-gravitational channels---though they might not be ready to make the information public. Each observatory comprises hundreds of people. Suddenly your message isn't contained to just the hundreds of people loyal to LIGO, but to thousands of people. Moreover, these observatories tend to be publicly owned and funded, and transparency and modern open-science practices often mean live updating of the status of these observatories. Finally, there are the scientists who have nothing to do with this physics, but fought hard for some observation time and had their scheduled observations interrupted for something high-priority. These people can infer (or are often told directly) why their allocation was preempted.
- Koshkin 9y agoWhile I would have a trouble imagining a quantum of the space-time curvature (the graviton), it is not hard to see that the changes in the curvature could propagate in the form of waves. So, while yes, an experimental discovery of these waves is an important event in history of science, I am left curious as to whether it adds anything to our present understanding of Nature...
- raattgift 9y ago> trouble imagining ... the graviton Let's start with a graviton as a gauge boson, mediating the gravitational interaction similarly to how the photon is a gauge boson mediating the electromagnetic interaction. First, let's start with "gauge". We can use as an analogy an air pressure gauge, one that measures relative air pressures. Let's take it to a place with a standard pressure (say, in conditions which are effectively STP) and tune our pressure gauge so that it reads "0" at that air pressure. As we wander to and fro reading our calibrated gauge, we'll see pressures that are zero, positive or negative. If we climb a tall hill we'll see a negative reading. If the temperature drops we'll see a positive reading. If we contrive things so that we can take a reading of a generalized pressure with our pressure gauge everywhere in the universe at all times, we can construct a (classical) gauge field. In deep space, the gauge will read strongly negative. At the bottom of the ocean, or deep in Jupiter's atmosphere, or at the core of the sun it will read strongly positive. In the early universe, it'll be strongly positive; in the far far distant future away from the black hole that will dominate our patch of de Sitter vacuum, it will read strongly negative. We'll get zero values in some places, like near the Earth's surface through a lot of Earth's history, or in the upper reaches of Jupiter's atmosphere through a lot of its history. Our choice of "0" is not ideal, because "0" is only rarely the value at any point in our gauge field that permeates all of spacetime. Instead we should set "0" as the value in extragalactic space, because then "0"s will dominate the field (indeed it is possible that all readings will then be non-negative). In effect, when we set our "0" at STP we normalized the gauge field; when we decided instead to set our "0" in extragalactic space, we renormalized it. We could obtain an ideal renormalization if we could sample the whole of spacetime and find the lowest reading of our pressure gauge, but we can certainly get rid of practically all negative values by taking far fewer samples in regions where we think the lowest readings might be. Once we have settled on a decent normalization, we could look at the propagation of nonzeros and study their statistics. If they follow the Bose-Einstein statistics, we'd call them "bosons". If they follow the Fermi-Dirac statistics, we'd call them "fermions". If they follow some other statistics, we'd assign them yet another name. (Our choice of generalized "air pressure" probably follows some odd statistics.) Perturbative quantum gravity works something like this. [2] We have a background spacetime with a metric; we have a gauge that measures the deviation from this metric. It'll be "0" at every point where the arrangement of stress-energy exactly matches the metric, and nonzero elsewhere. We are interested in modelling the gravitational interaction as the arrangement of nonzero values in our field. Patterns of nonzeros around gravitationally interacting matter themselves evolve (under a suitable decomposition of spacetime into 3+1 space and time) and interact like (classical) waves (made up of many molecules), and upon some study we can determine that these waves in our gauge field form patterns that strongly suggest they have a rotational symmetry of two, which we expect on theoretical grounds too because the metric is a rank-2 tensor field so particles representing the (change in the) metric field should be spin 2. Conveniently, in a quantum gauge group theory, a particle with spin 2 is attractive of a particle with the same charge and repulsive of a particle with the opposite charge. (Compare with spin 1, where same-charges repel and opposite-charges attract). [4] So we can identify the nonzero numbers in our metric gauge field with gravitons. This is amenable to study with perturbation theory. Unfortunately General Relativity is a non-linear theory and in our perturbatively quantized gravity, when you have a lot of high-energy gravitons they spawn more gravitons. We would want to apply Wilson's thinking on renormalization and reset our gauge to "0" in a cluster of these high-energy gravitons by finding some suitable ground value in the cluster. This is extremely successful up to a point [1], but as the energies of the gravitons increases we have to take more measurements to find a suitable ground value, and eventually we have to take an infinite number of measurements to find one. This is what is meant when you read "gravity is perturbatively non-renormalizable". There are, as you suggest with ("... space-time curvature ..."), other values related to the gravitational interaction that we can turn into a quantum field [3], but most suffer a highly similar fate: in some conditions we have to do an infinite amount of work to make our field values sensible and match observables. Finally the gauge field that we built on the metric is fully relativistic and generally covariant, so it works with any system of coordinates, choice of units, slicing of spacetime into 3+1, etc. that we want, up to diffeomorphisms (we have to remember that we chose a static background spacetime). So even though it gives us useless readings in some regions of a spacetime containing strong gravity, perturbative quantum gravity is a useful and standard tool. However, it is not considered a candidate for a fundamental theory (barring some unforessen advancement in renormalization theory) rather than an effective theory and moreover by implication it undercuts General Relativity's claim to be a fundamental theory too. - -- [1] Relativists tend to define "strong gravity" at this point, since we get correct results from renormalization at any energy lower than it. Strong gravity only appears very close to gravitational singularities (and in the case of black holes, that means well inside the horizon). If we are using the path-integral formalism then we'd find that we have "strong gravity" in this sense in every Feynman diagram containing at least one loop of gravitons. [2] https://arxiv.org/abs/gr-qc/0206071 https://arxiv.org/abs/gr-qc/0206071 [3] https://www.wikiwand.com/en/Canonical_quantum_gravity https://www.wikiwand.com/en/Canonical_quantum_gravity [4] One might ask, "is there oppositely-gravitationally-charged matter anywhere"? It's an OK question, and people have discussed it seriously. Sabine Hossenfelder has touched on this a few times on her blog, including http://backreaction.blogspot.com/2017/04/why-doesnt-anti-matter-anti-gravitate.html http://backreaction.blogspot.com/2017/04/why-doesnt-anti-mat... although while the photon has no electromagnetic charge, the graviton itself (in perturbative quantum gravity) has gravitational charge (this reflects the non-linearity of General Relativity).
- MilnerRoute 9y ago"We are working hard to assure that the candidates are valid gravitational-wave events, and it will require time to establish the level of confidence needed to bring any results to the scientific community and the greater public. We will let you know as soon we have information ready to share." -- Statement from Ligo http://www.ligo.org/news/index.php#O2end http://www.ligo.org/news/index.php#O2end "This did not, in fact, blow my sox off." -- Astronomer Peter Yoachim https://twitter.com/PeterYoachim/status/901175225819176961 https://twitter.com/PeterYoachim/status/901175225819176961
- perseusprime11 9y agoSomeday could we ride these waves to Interstellar travel?
- Panoramix 9y agoNo. The effect of the waves is so small that it required a gargantuan project to detect the strongest events. The LIGO detector is possibly the most sensitive instrument of any kind humans have made.
- colordrops 9y agoI wonder if the gravitational wave would be physically noticeable if one were near the event.