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The flux of cosmic ray muons is about 1 muon per square centimeter per minute. I was trying to see if I can detect these using a glass of water, putting it in a totally dark room and take a picture of that with a camera mounted on a tripod. I used the highest ISO setting available and exposed for about half a minute, and I let the camera perform a dark frame subtraction. Muons moving through the water should produce photons, some of which will be detected by the camera sensor.

I then converted the raw files using the dcraw program to tiff files without doing any demosaicing (so each pixel corresponds to the actual gray value detected by a sensor photosite). But I'm now a bit bogged down in analyzing the raw files (a problem here is that the Sony ARW files are not real raw files, they have build in lossy compression which is not ideal for doing this sort of analysis).

So far I have not seen any interesting, if I look at gray values that are many standard deviations above the noise then I don't see any obvious (broken) straight line patterns. So, I'm wondering if one would actually expect any signal at all. It seems to me that you would because from close up you would detect some photons from the water, but perhaps I'm too optimistic.

Count Iblis
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  • Nice experiment! Unfortunately your camera is not nearly sensitive enough for this (you would need photomultiplier tubes) and the muon tracks will not show up as straight lines, either. You are trying to make a tiny water Cherenkov detector there. The pattern in such a detector (using a larger water tank of cubic meter size) would be that of Cherenkov rings. If you want to detect tracks directly with reasonably cheap means, you will have to make a cloud chamber. The PMTs plus electronics for a small Cherenkov experiment would cost tens, if not hundreds of thousands of dollar, easily. – CuriousOne May 19 '15 at 18:47
  • We use PMTs to detect the photons because sometimes you might only get 1 photon. I doubt your camera can detect just 1 photon, and that's even when the photon is emitted in the direction of the camera – Jim May 19 '15 at 18:47
  • Possibly interesting link: http://quarknet.fnal.gov/toolkits/new/crdetectors.html. I worry that the noise you get while integrating for such a long time dwarfs the Cerenkov light from the muons. – Floris May 19 '15 at 18:48
  • If you want to detect cosmic rays directly, a single large PMT (8 inch plus) will do. If you can get your hands on one cheaply (which I doubt), you could see a large pulse every few minutes or so. That would be a cosmic ray shower going trough you and the detector... See http://hamamatsu.sitesearch.jp/api/pbimg/cc8/b92d4f9933e7dafa3f65227882851/0006-l.jpg for the kind of equipment it takes to do this stuff. – CuriousOne May 19 '15 at 18:54
  • I have updated my answer to the early question detailing the progress of my students work. @Count: I'm not sure why you are expecting straight features in your data. You expect (1) point-like features where the muon actually excites the focal plane directly and (optically distorted, pixelated and sparse) annuli where Cerenkov light intersects the detector plane; if the detector comes all the way to the surface of the detector you expect a disk. So yes, a DICH detector. Or if you take note that the detector is Barely able to detect the signal a ... ::cough:: Uhm, yeah. – dmckee --- ex-moderator kitten May 19 '15 at 21:16
  • @dmckee (and others), I was expecting to see "broken" lines (points that are on a line). I was not considering Cerenkov radiation, I was considering the 2 MeV per g/cm^2 energy loss due to ionization, so I was thinking that the muons cause ionization in the water and photons will be emitted due to recombination all along the muon's path. Some of these photons will move in the direction of the camera. – Count Iblis May 19 '15 at 23:38
  • Recombination photons are emitted into all $4\pi$ steradians of solid angle. You can more or less ignore them because (a) most will miss the detector and (b) those that do hit will have the registered position very weakly correlated with the emitted position (i.e. probability goes by $1/r^2$ and no other dependence). Full coverage calorimeters like KamLAND or Super-K could in principle get some use out of them, but they are swamped with the intended signals (scintillation or Cerenkov). – dmckee --- ex-moderator kitten May 19 '15 at 23:41
  • @dmckee Can you explain why the correlation would be weak? In my set up the glass of water was in reasonably good focus when I checked with the lights on. – Count Iblis May 19 '15 at 23:47
  • Ah...I think I see. I had misunderstood your geometry, but it has a very low geometric acceptance: most of the light simple misses the camera. In the case of the RICH my students built we can collect a large fraction of the emitted Cerenkov light and we are using the 2-3 least significant bits of the ADC (4-5, maybe 6 channels with dark noise of 1--2 ADC channels for the 10 second exposures we used). Another question to ask about recombination is where it falls wavelength wise relative the bands of the lens and detector. With the RICH geometry and our radiator most of the light was collected. – dmckee --- ex-moderator kitten May 19 '15 at 23:57
  • I should say that I spent three afternoons reading data sheets, calculating, estimating and designing before I was 90% certain we could get a signal. You are working---at best---right at the edge of feasibility here. – dmckee --- ex-moderator kitten May 20 '15 at 00:00

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