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I was curious if a proton collision is visible to the human eye.

(This might sound like a really basic question and forgive me if it is. I am very inexperienced in Physics and just wanted an answer to my curiosity)

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Someone has (inadvertently) tried this! en.wikipedia.org/wiki/Anatoli_Bugorski –  charliehorse55 Jun 11 '13 at 2:25
    
Interesting, but one should realize this was a beam, i.e. ten to the something protons, not "a proton". I knew people who would center a beam from the cerenkof light left in the eyeball, but those were very weak beams. –  anna v Jun 25 '13 at 3:22
    
@charliehorse55 the comment was on your link –  anna v Jun 25 '13 at 3:33
    
@annav It's relevant, but not exactly what the question was asking. That's why I put it in a comment instead of an answer. –  charliehorse55 Jun 25 '13 at 10:12

5 Answers 5

This is actually a really good question. (And I'm not one of these people who insists that there's no such thing as a dumb question; I just think we shouldn't be embarrassed to ask dumb questions. Anyway, this isn't a dumb question.)

As you may know, collisions between two protons (like those the LHC usually does) can produce many different types of particles, some of which are Higgs bosons, but most of which are more familiar. Photons are definitely among the familiar types of particles that get created, and since humans can see photons, it's at least possible in principle that we might be able to see the flash of light that results. The question is whether that light is bright enough and has the right "color" so that we could see it.

Now, the "color" has to do with the energy of the photon. The preferred unit of energy for particle physics is the electron-volt (eV), and in these units, we see photons with energies between about 1.65 and 3.27 eV. So only if a photon produced by the proton-proton collision has an energy in that range could we even possibly see it. The LHC is designed to collide protons with energies totaling 14 TeV (equivalently $1.4 \times 10^{13}$ eV, or 14 million million eV), and most (but not all) particles that result from a collision will have comparable energies. So we can expect that most photons produced by those collisions will have energies well over a million times too large for us to see. They can happen, but they'll be very rare. So that's color; what about brightness? Well, in those rare cases, we'll get one or maybe even two photons produced with the right energy for us to see. And this link suggests that we would need five to nine of them from a single event. So it would be very unlikely for a single collision to do this; they won't generally be bright enough.

On the other hand, these are just the direct results of the proton-proton collisions. High-energy collisions like this typically produce showers of other particles as soon as the products of the collision enter some matter (like an eye). It's kind of like a bowling ball hitting a bunch of pins. The energy of the ball gets distributed to the pins, each of which moves at lower energies, and can hit other things that will move at even lower energies, and so on. So such a thing would dump a lot of energy into your eye, much of which would be in the form of visible light. And this isn't just possible, it's downright likely. Of course, the many things produced in such a particle shower include various forms of radiation that would kill a human pretty quickly if that human were exposed to the entire power of the LHC. But in principle, yes the human eye could perceive the results of a single proton-proton collision.

A related topic is that of cosmic-ray visual phenomena, in which astronauts can see flashes of light, which are presumably due to individual cosmic-ray particles interacting with their eyes. A relatively recent study (behind a paywall) suggested that the protons produce a shower of particles, including visible photons, within the various structures of the eye. The authors suggest that heavier cosmic rays may directly stimulate the retina itself. There's also a nice little (free) overview of the study here.

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As I was reading the entire post, I was thinking "I hope he mentions the flashes of light astronauts see in space..." You made me wait for it though! –  tpg2114 Jun 10 '13 at 18:18
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Well, the reasonable reader would immediately click off and start reading about that because it's so interesting. I just wanted my post to be read. :) –  Mike Jun 10 '13 at 18:19
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Hi Mike, do you have a more quantitative estimate why it's "unlikely" that a generic pp collision would produce more than $O(5)$ relatively soft photons in the eV range? It seems like that would be a very inclusive observable, although of course it doesn't have a huge phase space available. And getting 10 of them in a small solid angle is of course a different story... –  Vibert Jun 10 '13 at 19:21
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I think 14 TeV is 14 trillion eV, not 14 billion –  Sahuagin Jun 11 '13 at 4:31
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@Sahuagin That depends on whether you're e.g. in the USA (where they have millions = $10^6$, billions = $10^9$, trillions = $10^{12}$) or e.g. Germany (where there are millions = $10^6$, milliards = $10^9$, billions = $10^{12}$) - one more reason why people should stop using ambiguous words when they could just mention the relevant power of ten (See en.wikipedia.org/wiki/Long_and_short_scales). Either way, the SI prefix tera- means $10^{12}$. –  Tobias Kienzler Jun 11 '13 at 12:53

These collisions don't produce significant amount of light in the visible range, so the easy answer is "no".

They also take place in a vacuum, inside a beampipe which is itself buried in a detector apparatus that is ten meters plus on a side and packed full of stuff with no room for a human.

That said, there are several ways in which a high energy ionizing particle could---in principle---make light in the visible range. In particular a electromagnetic shower impinging on the eye itself might produce enough Cerenkov light to pass the simple filters the brain imposes on retinal output and be consciously registered as a blue flash. (Each flash would represent a small fraction of the--considerable!--ionizing radiation dose which you would be exposed to in the course of this stunt.)

If we assume a interaction region without a detector and a physicist dumb enough to be in the hall while the beam is on (having carefully bypassed both hardware and administrative interlocks designed to prevent this) who then put his head near the interaction region beam pipe, you might be able to say "I saw one!" every once in a while.

But you won't catch me trying it: I plan to die of something other than radiation poisoning or cancer.

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It all depends on your definition of visible.

Elementary particle collisions have been made visible since the time of cloud chambers and bubble chambers. A good site for bubble chamber pictures exists . Unfortunately proton scattering is not as photogenic as scattering by other particles so I was unable to find a photo of a proton proton scatter in a bubble chamber though thousands must exist.

Here is one kaon from the beam hitting a proton in the bubble chamber and creating a number of new particles from its excess energy ( go to "would you like to see" for a list of particles).

proton in bubble chamber

Due to the kaon having interacted with a proton in the hydrogen, the rightmost beam track produces a spray of 4 tracks. The longer highlighted track is clearly dark – it has produced a higher number of bubbles per centimetre than, say, the beam tracks; this tells us that it is moving more slowly. (For details, click here.) Such tracks are a common feature in bubble chamber pictures and usually signify protons.

In the link there exists an explanation of how a bubble chamber works to make visible charged tracks coming from an interaction.

For LHC energies the detectors become more complicated and depend on electronics and computer outputs to get the same visibility. Vertex detectors and other track chambers give accurate images of the interactions, though the high multiplicity of vertices makes life more difficult.

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A spinthariscope is a simple device consisting of a minute amount of an alpha-emitter and a zinc-sulphide screen. It often includes a magnifier to view the screen. The amazing part is that a dark-adapted eye can clearly see the flashes of light produces as individual alpha-particles hit the screen! There's probably multiple visible-light photons from the collision, but the eye can detect the light produced by the energy of a single nuclear decay.

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The point, however, is that a scintillator/fluoror is used to make that happen. The energy is not at question. The way the energy is produced is. –  dmckee Jun 11 '13 at 16:10

Hmm, interesting question at least. Visible as in see two balls crash into each other; no, visible as in see the light emitted from the collision; probably not.

The emitted photons are of the order of $\rm{}MeV$, around $10^{-13}\rm{}$ joules. The human eye works at around $10^{-5}$. So individual collisions no.

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