In this paper:

http://wwwphy.princeton.edu/mumu/target/King/king_WEBR6_pac99.pdf "Potential Hazards from Neutrino Radiation at MUON Colliders"


High energy muon colliders, such as the TeV-scale conceptual designs now being considered, are found to produce enough high energy neutrinos to constitute a potentially serious off-site radiation hazard in the neighbourhood of the accelerator site. A general characterization of this radiation hazard is given, followed by an order-of-magnitude calculation for the off-site annual radiation dose and a discussion of accelerator design and site selection strategies to minimize the radiation hazard.

It suggests "muon colliders" could produce neutrino beams powerful enough to pose a potentially dangerous radiation hazard, but I don't recall it going into the details of why the danger existed. How is that possible? Neutrinos are not usually very interactive, and the only other time I heard of "deadly neutrinos" was in regard to a supernova explosion, where if you were within 1 AU, it would be enough to be deadly, but if you were within 1 AU, you would be inside the stellar envelope anyways, and thus the neutrinos would be the last thing you'd be concerned about. I can't possibly imagine that a man-made source could come anywhere close to the power of a supernova, much less in emitted neutrinos.

The only thing I can think of is that, apparently, as the neutrino energy increases, they become more interactive, and the muon collider would be generating neutrinos with per-particle energies far in excess of those produced by a supernova. Thus while the total neutrino areal energy density would be nothing compared to that in a supernova, the much higher per-particle energy would dramatically boost the interactivity. Is this the correct explanation?

  • $\begingroup$ Are you sure the paper did not mention what produced the neutrinos in the first place? $\endgroup$ – user140606 Jan 9 '17 at 2:28
  • $\begingroup$ @TáMéCeart It was supposed to be produced by the particle accelerator. I don't remember what the details were though. $\endgroup$ – The_Sympathizer Jan 9 '17 at 2:30
  • $\begingroup$ @TáMéCeart I looked up something similar and it says that muon decays produce neutrinos. So since the produced muons inevitably decay, then they would produce a shower of neutrinos. $\endgroup$ – The_Sympathizer Jan 9 '17 at 2:32
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    $\begingroup$ Obligatory XKCD Reference $\endgroup$ – Cort Ammon Jan 9 '17 at 3:30
  • $\begingroup$ I think @CortAmmon is, as usual ahead, of me at least, in his thinking. If the neutrinos don't interact with X km of lead, then why would they interact with whatever is around the facilities to produce, well anything really. It's 4.10 am here, that's my excuse. I think you should always put the citation and extract in your posts, if only to stop me rushing answers telling you things you already know. Sorry about that. $\endgroup$ – user140606 Jan 9 '17 at 4:11

Something not mentioned in the other answers, but very important to understanding this is that the neutrino-other stuff cross-section grows with neutrino energy. In the regime between solar energies (~1 MeV) and current accelerator energies (a few to a few tens of GeV) the growth is roughly linear a trend which continues some ways further up the energy scale.

The oft mentioned notion of a neutrino going through a light-year of lead with only a 50% chance of interacting refers to solar neutrinos. At the current accelerator scale this is down to around 1/1000 light-year, and at the TeV scale down to one millionth a light-year.

And of course, every muon in a storage ring results in both a $\nu_\mu$ and a $\bar{\nu}_e$ both of which will have an appreciable fraction of the muon's kinetic energy.

Then to get an appreciable interaction rate the muon current in the ring will have to be prodigious.

Then when those (anti-)neutrino interact with matter most of the products continue at high enough boost to be ionizing radiation on their own. Thus the concern about "dirt events" generating a measurable amount of conventional ionizing flux in the plane of the ring.

  • $\begingroup$ Nice -- a (semi-)quantitative description of the actual interaction increase w.r.t. neutrino energy. That's what I was after... But even at 1% of a light year that still seems like it would require an insane amount of neutrinos to deliver enough interaction to make real havoc possible. If that means it's about 100x more interactive, then you should still need something on the order of 1% of a supernova in particle density to get enough to do real damage, no? That's still a heck of a lot of intensity. $\endgroup$ – The_Sympathizer Jan 9 '17 at 10:05
  • $\begingroup$ Well, to start with I had a brainfart on those numbers. $\endgroup$ – dmckee Jan 9 '17 at 10:10

Edit This answer is at best, only partially correct, and I leave it here in case anybody else thinks along the same lines) End edit.

My point is that, reading the Wikipedia extract below, as no electric charge is involved with neutrinos, and as we already undergo a good healthy dose of them in normal circumstances, the neutrinos are taking the rap for damage that may have been caused by some other particles that we can produce and control, such as the positively charged proton proton LHC beam.

Neutrinos can be created in several ways, including in beta decay of atomic nuclei or hadrons, nuclear reactions such as those that take place in the core of a star, and supernova, and when accelerated particle beams or cosmic rays hit atoms. The majority of neutrinos in the vicinity of the Earth are from nuclear reactions in the Sun. About 65 billion $(6.5×10^{10})$ solar neutrinos per second pass through every square centimeter perpendicular to the direction of the Sun in the region of the Earth.

Here's a guy who stuck his head into a particle accelerator Things not to do #1

  • $\begingroup$ Just found the paper -- see other post here. $\endgroup$ – The_Sympathizer Jan 9 '17 at 2:51

I just found the paper: http://wwwphy.princeton.edu/mumu/target/King/king_WEBR6_pac99.pdf

It says:

Most of the ionization energy dose deposited in a person will come from interactions in the soil and other objects in the person’s vicinity rather than from the more direct process of neutrinos interacting inside a person. At TeV energy scales, much less than one percent of the energy flux from the daughters of such interactions will be absorbed in the relatively small amount of matter contained in a person, with the rest passing beyond the person.

This is apparently the mechanism. However, I'm nonetheless still rather surprised that there would be that much intense interaction overall, given that this is nothing in terms of energy compared to a supernova blast. Thus there must be still more that I'm missing in terms of understanding this completely. The paper doesn't say what the per-neutrino energy is, for example. But I'd suppose that if the muons are coming out with enough kinetic energy, then that energy will be transferred to the neutrinos produced in decay (more specifically: the muon explodes in its rest frame, sending particles in all directions, which when viewed from the ground frame will have the added relative velocity of the original muon. Furthermore, due to the relativistic beaming, these particles are concentrated along the direction of motion to make a jet, increasing the neutrino density further), which will make them much more energetic and thus much more interactive. However I'm not 100% sure.

  • $\begingroup$ I agree with you about the mechanism, I sincerely apologise for my incorrect answer, (which I will modify in a min) and I do appreciate the chance to remember to question my assumptions before answering. :) $\endgroup$ – user140606 Jan 9 '17 at 2:58
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    $\begingroup$ The problem is that the neutrinos exit the collider in a very narrow beam (which of course becomes conical). To make matters worse, this cone narrows with increasing energy. Neutrinos from a supernova on the other hand are spread out more or less uniformly over $4\pi$ steradians. This paper implies that for a 100 TeV collider, the cone would have a half-angle of $10^{-6}$ radians, making for a solid angle of $3\times10^{-12}$ steradians. $\endgroup$ – David Hammen Jan 9 '17 at 9:33

Please see (for example) R.B. Palmer, "Muon Colliders," Reviews of Accelerator Science and Technology 7 (2014) 137–159. The idea is that a muon collider would be a very interesting particle accelerator to build, would be much more compact, and probably cheaper, than the alternatives to follow up on the LHC, and the world's particle physicists may therefore want to build and operate one some time in the next twenty years or so. If and when that happens, the design will have to be safe for people who might live at locations where the neutrinos from decay of the muon beams intersect the earth's surface. There are ways to ensure that that's the case, and permission to build the accelerator would be granted only after experts review the design and certify it as safe.


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