The most powerful cosmic ray since the oh-my-god particle puzzles scientists

https://www.nature.com/articles/d41586-023-03677-0

by WithinReason • 3 years ago

172 120 3 years ago

Aardwolf• 3 years ago

I 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• 3 years ago

A 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• 3 years ago

The findings were published now in a scientific journal and not then.

moh_maya• 3 years ago

Perhaps 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

jordanpg• 3 years ago

Also, 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.

ivan_gammel4• 3 years ago

320 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.

Szpadel• 3 years ago

I 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• 3 years ago

Probably 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_k2• 3 years ago

The 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.

jfengel• 3 years ago

You wouldn't notice it. The energy is sufficient, but the momentum is too small. Energy goes up with square of velocity and momentum just linearly, so this particle gets more of its energy from velocity and less from mass compared to a macroscopic object. The result is lowish momentum.

It probably hits a water molecule and the heat is quickly dissipated. You don't recoil because the momentum is small.

It might kill a few cells, but far smaller than the ordinary cycle of life. So any effect is swamped.

simne3• 3 years ago

Most impressive about nuclear physics, is knowledge, that in classic rigid matter exists huge-huge emptiness - atom core is just about 1/100,000 of volume of atom.

And this emptiness is so huge, that to make safety blanket against cosmic particles for interstellar flights (on current level of radiology medicine), need few METERS of solid shield, or hundred KILOmeters of atmosphere.

Second important consideration, nuclear fission reactions are very slow if compare to speed of light, this is why possible nuclear bomb, as if cores explode immediately, all nuclear fuel will be just fly out and not react.

So, high probability, if SINGLE high energy particle will intersect human body, will nothing happen, this particle will just fly right through body without any visible effects.

But even if particle will be so fortunate to directly hit some core in human body, according to rule of impulse saving, this core will also got very high speed and will fly out of body before fission appear (when high energy particles hit core of air at high altitude, they have some time to fission, so on surface usually detected traces of fission).

bjelkeman-again• 3 years ago

I cup of water would be about 2dl (200g), with the specific heat capacity of water being (4.18 J/g°C, or 4180 J/kg°C) I think 50 J is not enough.

ivan_gammel4• 3 years ago

Yes, thanks, I did a napkin calculation but apparently missed it by orders of magnitude.

nine_k2• 3 years ago

A cup of water is 0.2 kg. Water specific heat is 4200 J/kg*°C, so it takes about 840 J to heat a cup of water by 1°C. 50 J is not going to make any noticeable difference.

OTOH a tennis ball is about 0.06 kg, so from e = m*v^2/2 would give about 41 m/s, a good serving that you will definitely feel if it hits you.

dustingetz• 3 years ago

removed

IAmGraydon3• 3 years ago

There’s a lot wrong here, starting with the energy of the newly discovered particle. The particle in question had an energy of 240 EeV, which comes to 38.45 J.

ivan_gammel4• 3 years ago

By accident I used the energy of OMG particle which is also mentioned in the article, but it is not completely wrong — it’s the same order of magnitude as the new one, so it still can be used to understand how much is that.

fallingknife• 3 years ago

That would only get you to 3km. 20K to get to 100t * 3000 to get you to 3km. Would need another 100x to get to orbital height. But that's trivial compared to the energy you will need to get up to orbital speed. And then there's gravity drag.

vkou• 3 years ago

you will need ~60 millions of such particles, a tiny fraction of the number of protons in human DNA.

But how many ant-sized Olympic swimming pools is it?

melagonster• 3 years ago

the link of research article is not work now.

JKCalhoun• 3 years ago

I wonder how complicated those detectors are and if amateurs could set up backyard detectors and contribute to cosmic ray detection.

spacecadet• 3 years ago

Great Question. I dabble in some back yard Radio Astronomy, but now you have me wondering this too.

getoffmycase• 3 years ago

Put 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_...

IAmGraydon3• 3 years ago

For those who might be considering doing this, don’t unless you’re ok with destroying the camera.

taylorportman• 3 years ago

I wonder if there is a common substance that could be scanned for historical interactions - like some salt flats.. that might preserve, like film, a history of energetic anomalies.

eesmith3• 3 years ago

One such is a Miyake event. https://en.wikipedia.org/wiki/Miyake_event

an observed sharp enhancement of the production of cosmogenic isotopes by cosmic rays. It can be marked by a spike in the concentration of radioactive carbon isotope 14C in tree rings, as well as 10Be and 36Cl in ice cores, which are all independently dated.

Wood and ice are pretty common. :)

I don't know if there are other examples.

maxnoe4• 3 years ago

It's relatively easy to detect cosmic rays if you are happy with secondary muons that are produced in the atmosphere when a high energy cosmic ray is absorbed.

Detectors are relatively cheap and come out of the box for citizen science projects, this project is relatively well known: http://www.cosmicwatch.lns.mit.edu/

For high energy cosmic rays, you need to observe many of these secondary particles that are produced during the absorption with high temporal resolution, few nanoseconds, to be able to tell anything about their properties.

There are essentially five or so detection principles for measuring cosmic rays at the ground, and many observatories combine multiple techniques.

First, you can observe the secondary particles that reach the ground, mainly muons and electrons. This is possible either using water tanks with photosensors inside detecting Cherenkov light or using scintillators like in the project I linked above.

Then you can also detect very short and faint light that is also emitted in the air shower. This also comes in two variants: Cherenkov light is emitted in a cone around the charged particles and results in a "light pool" of roughly 250m diameter on the ground. Fluorescence light is emitted in all directions. We build optical telescopes with extremely fast and sensitive cameras to detect Cherenkov or fluorescence light.

Last, there is also radio emission from air showers, you can detect with antennas.

Auger in Argentina combines water tanks, scintillators and fluorescence telescopes and is investigating adding in radio antennas.

Telescope Array, the experiment which measured this event here, is using scintillators and fluorescence telescopes.

eternauta3k• 3 years ago

Auger in Argentina combines water tanks, scintillators and fluorescence telescopes and is investigating adding in radio antennas.

Any thoughts about surviving the next 4 years?

fllsdf• 3 years ago

Not that hard, Marco Reps, an excellent maker YouTube channel has a video on how to create one https://www.youtube.com/watch?v=PCB8nv4fatc

antognini• 3 years ago

There are a couple of apps that let you detect cosmic rays on your smartphone. (One example: https://cosmicrayapp.com/)

They basically keep the camera shutter closed and look for streaks on the camera CCD.

somenameforme• 3 years ago

It's really easy to build a basic visible cosmic ray detector - they're called cloud chambers. Here [1] is a random video, but you can find countless sources just searching for 'how to build a cloud chamber.'

If you haven't seen one of these before, it might sound more cool in paper than in practice. Cosmic rays are absurdly abundant to the point that dozens to hundreds are passing through you per second. So a cosmic ray detector ends up turning more into something like a really neat art show than a search for a signal. Of course there is the search for 'the big one', but somehow it's not quite so romantic when you're getting plowed by these guys constantly.

[1] https://www.youtube.com/watch?v=gt3Ad5_Z5IA

cozzyd4• 3 years ago

The scintillator detectors are quite simple, but you need a whole ton of them working together to be useful.

The fluorescence detectors are more complicated.

dist-epoch• 3 years ago

Can 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• 3 years ago

The 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.

maxnoe4• 3 years ago

Mass 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

[2] 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-jin2• 3 years ago

So, 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

ars• 3 years ago

That's not how gravitational acceleration works.

If the particle accelerates towards black hole it will decelerate when the leaves the black hole.

To actually accelerate and keep it, you need a third particle. In a spaceship that third particle is the fuel.

In a solar system that third particle is the planets interacting with their sun.

My post is telling you what's needed without explaining it at all, please watch some YouTube videos to actually understand it.

NegativeK• 3 years ago

You can extract energy from a rotating black hole via slingshotting around it.

But it's not an amount relevant to this discussion.

pixl97• 3 years ago

So, I'd say it's a little more complicated than this even.

There are orbitals around black holes where particles could maintain stable orbits for long periods of time. Now when you get a complex environment around said black hole with lots matter attempting to infall but not having the correct momentum you will get a lot of interaction between particles of different velocities, hence why we actually see black holes at all. you get matter crashing into each other releasing gamma rays and such. From these interactions alone you can get gravitational particle acceleration.

It just gets more complicated from here as you add magnetic field interactions.

idlewords5• 3 years ago

The 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• 3 years ago

I'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.

smueller12345• 3 years ago

To 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• 3 years ago

Or a nearby emanation source we can’t readily detect?

bee_rider• 3 years ago

I 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?

newZWhoDis• 3 years ago

Probably a leaky warp reactor on a passing starship :)

araes• 3 years ago

Honestly, still mildly crazy. If:

- Whatever Star Trek says about warp reactors (matter / antimatter reaction) is valid. Reacts "some" amount of matter/antimatter every second.

- Klingon Bird of Prey can maintain an effectively infinite cloaking time from a human observation perspective, so it can be nearby.

- Warp reactor "occasionally" leaks (1% of 1% of 1% of reactions? I dunno...) so we might actually detect something. "Slightly" imperfect shielding.

- Problem: 5E1 J for OMG Particle (people say its a proton). 1.8E14 J for 1 gram of matter / antimatter annihilation. Except: 1 Proton = 1.6726231E-24 g. Proton rest mass energy is 1.503E−10 J. So the particle is more energetic than a Proton / Antiproton annihilation event (by a lot). It's been upshifted More than 100,000,000,000 from the rest mass energy.

- It "might" work if the one proton escaping represented a single proton gaining enough energy to overcome reactor core shielding confinement, which, in my opinion, seems somewhat plausible physics by Star Trek standards.

- PS: Personal guess is a Q-Clearance [3] experiment at a DOE lab we deny exists.

[1] https://en.wikipedia.org/wiki/Orders_of_magnitude_(energy)

[2] https://en.wikipedia.org/wiki/Proton

[3] https://en.wikipedia.org/wiki/Q_clearance

captainmuon• 3 years ago

I wonder if it could be a background source that we don't understand. Maybe some freak coincidence where particles come in in exactly the right angles to fake a hyperenergetic photon. Or somebody messing with laser pointers. Or birds or whatever.

I used to work in particle physics, and never shared the confidence of my colleagues in rare events. If you just have 3-4 signal events, considering the expected number of events, you might statistically have a discovery. But you can't be sure how your detector is going to behave for those extreme events, because you have no benchmark. You have to assume a proton is a proton, an electron is an electron, and no weird things happen at high/low energies and angles (or that you understand how things change).

It is even worse if you are trying to disprove some specific model. You see no events of a certain kind. Does that mean you disproved the model, or that your setup (detector, triggers, event selection ...) is just blind in this very narrow part of the parameter space?

idlewords5• 3 years ago

If you go out in space and look at what extrasolar stuff zaps you, you'll find a very standard power law distribution of charged particle energies, with no known upper limit. This is a better position to be in (from a detector confidence standpoint) than seeing a completely bananas event every few years. Here it's bananas (and progressively less bananas) all the way down.

jp57• 3 years ago

I assume that limit accounts for the time dilation, too. At the speed the particle was traveling it wouldn't have had much time in its reference frame to interact with the background radiation, even blue-shifted as it would be, right?

idlewords5• 3 years ago

Yeah, the model does account for it, and the time dilations are nuts. For the Oh-my-god particle (a similar early detection), I remember that it would have crossed a billion light years in what it perceived as a day.

whoopdedo• 3 years ago

either interesting new astrophysics or new particle physics

Or measurement error which historically has been the most frequent explanation of these rule-breaking observations. Someone else will look at the data and notice an anomaly which once accounted for will make everything fit within expected models.

lamontcg3• 3 years ago

In this case though they've been observed over the past 30 years at at least 4 different detectors, which provides independent confirmation across multiple different research groups. It is very unlikely to be a "oops we didn't plug the cable in firmly" kind of situation. It would need to be something fundamental about the design of these detectors, which has been replicated four independent times, which has completely escaped the notice of physicists for >3 decades (which includes 3 decades worth of grad students incentivized to take a look at the problem with fresh eyes and make a name for themselves by explaining it all away).

a_wild_dandan• 3 years ago

What initial speed must the proton have for consistency with our observed final “dragged down” speed (assuming some ballpark galactic distance)? Alternatively, could a “clump” of particles have smashed into our detector? Whatever the new physics winds up explaining this bonkers momentum, it’ll be fascinating.

cozzyd4• 3 years ago

It would certainly be convenient for me if it is a proton, given the experiments I work on (protons at the highest energies -> more ultrahigh energy neutrinos -> maybe my experiments will see something!), but, unfortunately for me, I don't think we know that. Nuclear species composition can typically only be done statistically, and it looks like this wasn't even measured in the fluorescence detector, which would give the best measurement of the penetration depth of the particle in the atmosphere (the best handle on particle id). In,the journal article, which admittedly I've just skimmed, they give direction reconstructions giving different species assumptions (and Milky Way models, which are important for heavier nuclei) and say that while they can exclude a photon (which... would be weird!), without data from the fluorescence detector, they can't tell the difference between a proton and a heavier nucleus.

Though given how TA mass-composition measurements turn light at the highest energies, they may perhaps privately argue it's likely a proton. But, alas, Pierre Auger Observatory would argue otherwise. I'll ask Toshihiro what he thinks next time I see him...

It is intriguing that the smaller TA has seen more high-energy events than the bigger PAO. Maybe there really are big differences between the northern and southern hemisphere at play here...

idlewords5• 3 years ago

If Santa handed you a hundred billion dollars or so to study this stuff, what would be your dream detector?

rthomas6• 3 years ago

Could it be a small black hole orbiting the sun?

simne3• 3 years ago

You missing important part - the larger rest mass of particle, the less it will interfere with cosmic vacuum.

So, high energy electrons must be from near source (mostly, inside our galaxy), because otherwise they will slow down by magnetic field of our galaxy.

For high energy protons, this eased, but I could not remember approximations for possible distance.

And for heavy nucleus, may be possible to travel from very distant source, even from borders of Universe.

simne3• 3 years ago

BTW, I remembered interest fact, that on some particle accelerators run experiments on accelerate macro-objects, like dust.

Latest accelerators are not very suited for such activity, but earlier linear accelerators works well on this, and give very interesting data.

darkerside• 3 years ago

Alien spacecraft?

mock-possum• 3 years ago

something akin to a standing wave, where an exceptional signal develops out of otherwise unremarkabley noisy surroundings?

opportune• 3 years ago

I think the simplest explanation is that we're just not accurate enough at determining the trajectory of the particle in question. They tried matching against sources "just outside its arrival direction" but without access to the paper (the DOI link just doesn't work for me, loads an empty page) or knowledge about the instruments being used it's hard to say how much that rules out.

At least looking into the Telescope Array's workings, it seems plausible that it'd struggle to produce accurate trajectory data. It works by analyzing the light produced by cascading collisions of particles with the incoming one: https://en.wikipedia.org/wiki/Air_shower_(physics). At first glance this seems like it would be very noisy for determining trajectory.

smueller12345• 3 years ago

My recollection from the Pierre Auger telescopes was 1-2 degrees uncertainty. TA is likely a similar order of magnitude. So matching a single event against candidate sources might not be the most reliable analysis. (Caveat: I didn't dig up papers to check my memory from 15 years ago.)

alister• 3 years 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.

itishappy• 3 years ago

They 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/Cosmic-ray_observatory

eesmith3• 3 years ago

It 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 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_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 .

And KM3NeT, under construction will be a neutrino detector using several cubic km of ocean, https://en.wikipedia.org/wiki/KM3NeT. It is the next generation after ANTARES, https://en.wikipedia.org/wiki/ANTARES_(telescope) .

smueller12345• 3 years ago

And the largest cosmic ray observatory, Pierre Auger Observatory in Argentina clocks in at around 3000 km^2. https://en.wikipedia.org/wiki/Pierre_Auger_Observatory

A very different and really neat concept that hasn't become real yet is JEM EUSO, a telescope that would be mounted on a space station, pointed at Earth, would detect air showers via fluorescence like Auger's fluorescence telescopes do on the ground. This could theoretically cover a much larger area than traditional CR observatories. https://en.wikipedia.org/wiki/JEM-EUSO

maxnoe4• 3 years ago

We 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.

prpl• 3 years ago

Fluorescence detectors are volumetric though only operate during moondown at night, and in the case of both Auger and TA, are paired surface detectors. Auger uses more water tanks with a PMT lined with Tyvek and detect electronic and muonic secondary particle components for the EAS. TA has scintillation detectors with PMTs, which are polyvinyl toluene sheets embedded in a steel casing. In both cases, they have individual triggers, usually around 1 MIP (minimum ionizing particle), which will trigger communication to a tower, which would then poll nearby detectors for events greater than 1/3 MIP.

Or, at least this is how it was setup 15 years ago. Both experiments have added new fluorescence and surface detectors since then.

nealabq• 3 years ago

A century is about 3 billion seconds, and Earth's surface area is about half-a-billion square kilometres, so one of these hits Earth about every 6+ seconds.

loufe• 3 years ago

Kurzgesagt recently uloaded a video on hypethetical interplantary weapons (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.

qubex4• 3 years ago

You’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.

mitthrowaway26• 3 years ago

E = 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.

cylinder714• 3 years ago

John Walker (of Autodesk) wrote this excellent piece on the Oh-My-God Particle:

https://fourmilab.ch/documents/OhMyGodParticle/

jacquesm3• 3 years ago

Great 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."

qubex4• 3 years 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”.

maxnoe4• 3 years ago

That'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

mensetmanusman• 3 years ago

Could these be a source of cancer if one hits you?

saagarjha• 3 years ago

It would be unlikely that they hit you because they’ll plow through the atmosphere first. You might get hit by a byproduct, though.

cozzyd4• 3 years ago

Cosmic 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.2017.0017... )

johndunne2• 3 years ago

If 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?

ace23582• 3 years ago

I 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?

johndunne2• 3 years ago

I 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

The most powerful cosmic ray since the oh-my-god particle puzzles scientists

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