9 ms·
The most powerful cosmic ray since the oh-my-god particle puzzles scientists
- mensetmanusman 3y agoCould these be a source of cancer if one hits you?
- saagarjha 3y agoIt would be unlikely that they hit you because they’ll plow through the atmosphere first. You might get hit by a byproduct, though.
- cozzyd 3y agoCosmic rays are certainly a source of cancer (though, strangely, lack of cosmic rays is also potentially problematic; see https://www.frontiersin.org/articles/10.3389/fmicb.2017.00177/full https://www.frontiersin.org/articles/10.3389/fmicb.2017.0017... )
- Aardwolf 3y agoI see this all over the news, not only on HN. But the signal was detected in May 2021. What makes this newsworthy now instead of then?
- sandworm101 3y agoA slower news cycle. In recent days US politics and the various wars have hit relatively steady states. So the news outlets are casting out for interesting science content, stuff that doesn't require math to understand.
- exegete 3y agoThe findings were published now in a scientific journal and not then.
- moh_maya 3y agoPerhaps because the peer-reviewed paper was published only recently (23rd Nov) - while not a perfect system (see the recent issues with the retraction of the super-conductivity paper by Nature), I think it not a bad thing that journalists are covering such reports once they have undergone the peer-review process - however flawed it may be, and not just based on the university or research group published press-release announcing the results before the peer-review process is complete. [1] https://www.science.org/doi/10.1126/science.abo5095 https://www.science.org/doi/10.1126/science.abo5095
- jordanpg 3y agoAlso, while the final result can be simply stated in a headline, performing the analysis that rules out all known sources of error is very, very difficult and can take years to validate.
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- ivan_gammel 3y ago320 EeV = 3.2e20 eV ~ 50 J ~ energy required to lift 5 kg by 1 m or energy sufficient to warm a cup of water from 0°C to 40°C Edit: to launch 100t of payload to LEO with Starship you will need ~60 millions of such particles, a tiny fraction of the number of protons in human DNA.
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- Szpadel 3y agoI wonder what impact on eg human body such particle would have. would it just go through or boil some cells? could someone unlucky hit by such particle just drop dead?
- pletnes 3y agoProbably pass through while generating secondary radiation also powerful enough to leave the body. Nothing you’d notice. Might cause cancer but so could other cosmic rays or radioactive rocks you encounter day to day.
- nine_k 3y agoThe particle won't dissipate all its energy inside the body, so likely it won't even hurt you noticeably. It it hit you eye though you'd likely see a very bright flash. 50 J is about as much as an airsoft gun pellet carries; not going to hurt you even if it manages to dump all its energy on your body.
- closewith 3y agoAirsoft is typically ~1 J. Even the most powerful are below 5 J. 50 J is enough to hunt small pests in an air rifle. Easily enough to injure or blind a human.
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- melagonster 3y agothe link of research article is not work now.
- JKCalhoun 3y agoI wonder how complicated those detectors are and if amateurs could set up backyard detectors and contribute to cosmic ray detection.
- spacecadet 3y agoGreat Question. I dabble in some back yard Radio Astronomy, but now you have me wondering this too.
- getoffmycase 3y agoPut a DSLR in the freezer with a long exposure time.[1] Instant cosmic ray detector. [1] https://www.reddit.com/r/astrophotography/comments/m1w4wv/i_captured_some_cosmic_rays/ https://www.reddit.com/r/astrophotography/comments/m1w4wv/i_...
- IAmGraydon 3y agoFor those who might be considering doing this, don’t unless you’re ok with destroying the camera.
- CamperBob2 3y agoHow does it destroy the camera?
- IAmGraydon 3y agoMost DSLRs are designed to withstand down to 0C. Most freezers are around -20C. You risk damage to the battery, condensation on interior components, contraction and mechanical damage of components with tight tolerances, etc.
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- dist-epoch 3y agoCan someone explain why this particle couldn't have been gravitationally accelerated by taking a turn around a super-massive black hole just on the edge of the event horizon? Is there a limit to how much acceleration a black hole can give?
- WendyTheWillow 3y agoThe article says there are no known quasars in the area of space where the energy is seemingly from, or at least that’s my interpretation. So yeah it could be from a black hole, just an unknown one. Or it’s from a different region of space than we think, and our math is off.
- maxnoe 3y agoMass is of a single particle is extremely small (proton has a mass of 1.6e-27 kg), speed is very high and already very close to the speed of light at all but the lowest energies, so gravity "assists" are not really anything with which you can gain energy. These are charged particles (protons and fully ionized heavier nuclei), so electromagnetism is a much more efficient way to reach high energies. All currently proposed mechanisms for cosmic ray acceleration involve turbulent plasmas and shock fronts that reflect particles magnetically, giving them a small bump in energy each time [1] or rotating magnetic fields [2] in sufficiently extreme environments. [1] https://en.wikipedia.org/wiki/Fermi_acceleration https://en.wikipedia.org/wiki/Fermi_acceleration [2] https://en.wikipedia.org/wiki/Centrifugal_acceleration_(astrophysics) https://en.wikipedia.org/wiki/Centrifugal_acceleration_(astr... Gravity just doesn't play a direct role for accelerating charged particles, it's much too weak compared to the electromagnetic force for a charged particle.
- saiya-jin 3y agoSo, a pulsar would do the trick? How far back do we simulate movement of milky way and all galaxies around to say for sure they come from an 'empty' place? I think our local supercluster look quite a bit different say 1 billion years ago
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- WendyTheWillow 3y agoLots of scientists are religious firstly, and secondly naming things is hard!
- alex201 3y agoThe 'Oh-My-God particle' refers to an actual observation made in 1991, which is known by this name. Besides, not all scientists are atheist.
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- falserum 3y ago> here to stay I'm afraid. Nothing to be afraid. Just accept people as they are. (Even if you do not want to join them)
- notbeuller 3y agoThey don’t specify which god - perhaps one of the Eldritch horror type gods is being invoked and awoken!
- eesmith 3y agoYour conclusion suffers from several issues. From the link: "The scientists nicknamed the particle ‘Amaterasu’, after a Japanese Sun goddess." Shall we conclude the Japanese Sun goddess is here to stay too? Various devout Christians consider "Oh my God" to be contrary to the third commandment: Exodus 20:7 “You shall not take the name of the LORD your God in vain, for the LORD will not leave him unpunished who takes His name in vain.” Madalyn Murray O'Hair, "America's Most Hated Woman" due to her advocacy of atheism, would say "oh my God" because it triggers Christians. https://youtu.be/pu5cqoSbeJA?t=832 https://youtu.be/pu5cqoSbeJA?t=832 .
- lamontcg 3y agoIrony exists. And I thought you weren't supposed to take the "Lords" name in vain in an irreverent manner as well.
- idlewords 3y agoThe big deal about these events is that, if you reasonably assume that it's a proton, then it violates a bound called the Greisen–Zatsepin–Kuzmin limit. Anything going that fast should be energetic enough to interact with the cosmic microwave background, which is hugely blueshifted in its frame of reference. The GZK is a kind of cosmic speed limit over long distances. So anything this energetic would need have a nearby source (in astrophysical terms) so it doesn't have time to slow down. But when we trace these things back, we see bupkus in the direction it came from. This means there is either (almost certainly) interesting new astrophysics, or (with tiny probability) new particle physics involved. Whatever is giving individual protons the energy of a thrown baseball is probably something worth studying.
- willis936 3y agoI'm relatively lay in astrophysics. Could the origin not be galactic magnetic field lines between Earth and the galactic nucleus? Like the magnetic mirror between Earth and the Sun but much larger.
- smueller1234 3y agoTo our best understanding, particles of such high energy do not originate in our own galaxy. We don't have a good understanding of what exact process might be putting this much energy into a single particle, but the commonly accepted acceleration mechanisms of, eg., supernovae don't reach anywhere near these energies. GP also said "nearby source (in astrophysical terms)" which in this case is code for "could be over 100 million light years".
- unsupp0rted 3y agoOr a nearby emanation source we can’t readily detect?
- bee_rider 3y agoI think that might fall under > This means there is either (almost certainly) interesting new astrophysics, possibly. An emanation source they can just barely pick up is a mystery they can puzzle out, right?
- alister 3y ago> Cosmic rays with energies of more than 100 EeV are rarely spotted — fewer than one of these particles arrives on each square kilometre of Earth each century. How do they explain detecting such a particle at all? I would assume that the surface of the Earth has much less than 1 square kilometer worth of detectors, so on average they shouldn't have detected any 100 EeV particles since the invention of cosmic-ray detectors.
- maxnoe 3y agoWe have instrumented much, much more area than a square km. The largest instrument to observe cosmic rays is the Pierre Auger Observatory in Argentina, which has detectors placed on an area of over 3000 km².
- itishappy 3y agoThey detect a cascade of secondary particles created from the collision of cosmic rays with the upper atmosphere. There are a lot more of them. https://en.wikipedia.org/wiki/Air_shower_(physics) https://en.wikipedia.org/wiki/Air_shower_(physics) https://en.wikipedia.org/wiki/Cosmic-ray_observatory https://en.wikipedia.org/wiki/Cosmic-ray_observatory
- eesmith 3y agoIt uses photomultiplier tubes to record interactions of the cosmic rays with a good-sized chunk of atmosphere, plus detectors for the shower of particles when a cosmic ray interacts with the air. https://en.wikipedia.org/wiki/Telescope_Array_Project https://en.wikipedia.org/wiki/Telescope_Array_Project says it uses "a 762 km2 grid array with 1.2 km between each unit". > The Telescope Array project ... is designed to observe air showers induced by ultra-high-energy cosmic ray using a combination of ground array and air-fluorescence techniques. ... When a cosmic ray passes through the Earth's atmosphere and triggers an air shower, the fluorescence telescopes measure the scintillation light generated as the shower passes through the gas of the atmosphere, while the array of scintillator surface detectors samples the footprint of the shower when it reaches the Earth's surface. See also https://en.wikipedia.org/wiki/High_Resolution_Fly%27s_Eye_Cosmic_Ray_Detector https://en.wikipedia.org/wiki/High_Resolution_Fly%27s_Eye_Co... , an earlier version. We have other observaatories which are also pretty big, in the km-sized range. There's IceCube, a neutrino detector observing events in a cubic kilometer of ice, at https://en.wikipedia.org/wiki/IceCube_Neutrino_Observatory https://en.wikipedia.org/wiki/IceCube_Neutrino_Observatory . And KM3NeT, under construction will be a neutrino detector using several cubic km of ocean, https://en.wikipedia.org/wiki/KM3NeT https://en.wikipedia.org/wiki/KM3NeT. It is the next generation after ANTARES, https://en.wikipedia.org/wiki/ANTARES_(telescope) https://en.wikipedia.org/wiki/ANTARES_(telescope) .
- loufe 3y agoKurzgesagt recently uloaded a video on hypethetical interplantary weapons (https://www.youtube.com/watch?v=tybKnGZRwcU https://www.youtube.com/watch?v=tybKnGZRwcU) which is the only reason I even know what the OMG particle is. They described an "Ultra Relativistic Electron Beam" which theoretically could travel much closer to the speed of light than the OMG particle. I'm left wondering if the proximity to the pure speed of light has any bearing on total delivered energy when comparing different particles.
- qubex 3y agoYou’re often taught that E=mc² but the fuller equation is actually E²=m₀²c⁴+p²c² where p is the momentum, so at high speeds mass fades into irrelevance and the velocity component of momentum becomes dominant. EDIT in classical physics p=mv so you might wonder what I’m banging on about when mass appears linearly on both sides and on the first term is multiplied by c⁴ whereas in the second only by c². However relativistic momentum is classical momentum adjusted by the Lorenz transformation γ=1/√(1-[v/c]² so it actually dominates in the limit when v tends to c. EDIT 2: the latter works out to be sigh c³vm₀/√(c²-v²) and it grows without bound as v→c so that’s the pedantic answer. The first term is linear the second term is anything but.
- mitthrowaway2 3y agoE = mc² is the full equation if 'm' is the relativistic mass, which grows with velocity. E²=m₀²c⁴+p²c² is the full equation if 'm₀' is the rest mass.
- E2P2c2throwaway 3y agoIf the particle has zero rest mass, like a photon, it still has energy. E^2 = p^2 c^2 + m_0^2 c^4 captures that correctly, while E = mc^2 where 'm' is the relativistic mass, does not.
- mitthrowaway2 3y agoNo, it does correctly capture that. The relativistic mass of a photon is not zero even though its rest mass is. However for a zero-mass particle you cannot use the Lorentz transformation to calculate relativistic mass, because v = c and so you would be dividing zero by zero. You can correctly calculate the relativistic mass of a photon in other ways, such as from its momentum or its gravitational interaction, and doing so gives you the energy you'd expect from E = mc².
- cylinder714 3y agoJohn Walker (of Autodesk) wrote this excellent piece on the Oh-My-God Particle: https://fourmilab.ch/documents/OhMyGodParticle/ https://fourmilab.ch/documents/OhMyGodParticle/
- jacquesm 3y agoGreat writer. And great programmer too... Nice quote: "If God's radar gun is slightly out of calibration, this puppy's gonna be doin' hard time for speeding."
- qubex 3y ago> one per kilometre squared per century Okay so build a hundred km² (10 km per side) observatory and you’ve got yourself a ‘telescope’ to the same degree that super-kamiokande is sometimes described as being a “neutrino telescope”.
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- maxnoe 3y agoThat's exactly the kind of detector that observed this event. It's been done. Multiple observatories exist that have effective observation areas of hundreds to thousands of square kilometers. Pierre Auger Observatory in Argentina is more than 3000 km²
- johndunne 3y agoIf we're talking about a single proton, and looking back at the direction the proton came from, we see nothing; what are the chances the proton passed close enough to a black hole to deflect it a significant angle from its original path, eventually landing on earth?
- ace2358 3y agoI often think about this. But then I also often think about synchrotron and how they create EM radiation by accelerating electrons to high speed then bending them in magnetic fields. When the electron bends in the field, it releases energy in the form of light. It does this as for ‘ever action the is an opposite reaction’. I can’t figure out if it emits light because it’s in a magnetic field, or if it’s because it’s a charged particle. Either way, when being bent by a large gravitational field, would a proton emit light and lose energy and slow down?
- johndunne 3y agoI think the electrons emit photons as a result of decelerating which is caused by the magnetic field (which can influence the charged electron). And I’ve never actually pondered whether a proton changing its direction due to gravity (technically decelerating along its original path and accelerating along its new path) would emit photons. I don’t think it would because from its frame of reference it’s not accelerating… I need to do some homework, very interesting! Edit: It seems a charged particle accelerating due to gravity will not emit photons. See https://www.mathpages.com/home/kmath528/kmath528.htm https://www.mathpages.com/home/kmath528/kmath528.htm
- ace2358 3y agoExcellent! So there is some interaction between the charged particle and the magnetic field which creates the em radiation. So does a proton release light in magnetic field? Say in the LHC?