\n
![\"\"](\"https://www.googleapis.com/download/storage/v1/b/kaggle-forum-message-attachments/o/inbox%2F1581556%2F335ef09b4d4ea6d7fbf388534d6ed843%2Fazimuth_distr.png?generation=1676299016003563&alt=media\")
I want to share a couple of interesting observations on the neutrino incident angles in the training data.
\nFirst, what I would expect to see. The simplest answer is symmetric, that is we might expect that the neutrinos come in equal numbers from all directions. Let's check if the actual distribution fits that model.
\nSpoiler: neither azimuth, nor zenith angles are distributed exactly uniformly.
\nHere is the histogram of the azimuth angle ϕ across all batches. It's indeed an almost uniform distribution in [0,2π]. We'll return to this \"almost\" in a moment.
\n
The histogram above is uniform, except for the small dips - and these dips correspond nicely with the 6 directions that connect the neighboring DOMs. I'd like to note that this is not effect of measurement (like, shading from the neighboring strings), but simulated distribution.
\n![\"\"](\"https://www.googleapis.com/download/storage/v1/b/kaggle-forum-message-attachments/o/inbox%2F1581556%2F0c310171712ed52bbaf7ab1bc27f9705%2Fazimuth_6n.png?generation=1676298794651769&alt=media\")
Now about the zenith angle. First of all, it is not the azimuth angle that we expect uniformly distributed. Why? Every unit of area of the sphere surrounding the detectors should contain the same number of neutrinos. We know that the area in spherical coordinates is given by dS = sin(θ)dθdϕ = -d(cos(θ))dϕ. That is, we need cos(θ) to be uniformly distributed, so that all directions are equal. Let's plot the histogram:
\n![\"\"](\"https://www.googleapis.com/download/storage/v1/b/kaggle-forum-message-attachments/o/inbox%2F1581556%2F8f2aac9d1dd91b9a84b73a526aebd571%2Fcos_zenith_distr.png?generation=1676299059341996&alt=media\")
This looks very interesting, with rather prominent maximums at both poles (for θ = 0 and π), and a less pronounced one at the equator (θ = π/2). It would be great, and probably help us come up with better solutions, if we could infer the reasons why the hosts selected such distribution to generate the simulated data.
", "messages": [ { "id": "2142461", "postDate": "02/13/2023 14:48:38", "content": "I want to share a couple of interesting observations on the neutrino incident angles in the training data.
\nFirst, what I would expect to see. The simplest answer is symmetric, that is we might expect that the neutrinos come in equal numbers from all directions. Let's check if the actual distribution fits that model.
\nSpoiler: neither azimuth, nor zenith angles are distributed exactly uniformly.
\nHere is the histogram of the azimuth angle ϕ across all batches. It's indeed an almost uniform distribution in [0,2π]. We'll return to this \"almost\" in a moment.
\n
The histogram above is uniform, except for the small dips - and these dips correspond nicely with the 6 directions that connect the neighboring DOMs. I'd like to note that this is not effect of measurement (like, shading from the neighboring strings), but simulated distribution.
\n
Now about the zenith angle. First of all, it is not the azimuth angle that we expect uniformly distributed. Why? Every unit of area of the sphere surrounding the detectors should contain the same number of neutrinos. We know that the area in spherical coordinates is given by dS = sin(θ)dθdϕ = -d(cos(θ))dϕ. That is, we need cos(θ) to be uniformly distributed, so that all directions are equal. Let's plot the histogram:
\n
This looks very interesting, with rather prominent maximums at both poles (for θ = 0 and π), and a less pronounced one at the equator (θ = π/2). It would be great, and probably help us come up with better solutions, if we could infer the reasons why the hosts selected such distribution to generate the simulated data.
", "rawMarkdown": "I want to share a couple of interesting observations on the neutrino incident angles in the training data.\n\nFirst, what I would expect to see. The simplest answer is symmetric, that is we might expect that the neutrinos come in equal numbers from all directions. Let's check if the actual distribution fits that model.\n\nSpoiler: **neither azimuth, nor zenith angles are distributed exactly uniformly.**\n\nHere is the histogram of the azimuth angle ϕ across all batches. It's indeed an almost uniform distribution in [0,2π]. We'll return to this \"almost\" in a moment.\n\n\nThe histogram above is uniform, except for the small dips - and these dips correspond nicely with the 6 directions that connect the neighboring DOMs. I'd like to note that this is not effect of measurement (like, shading from the neighboring strings), but simulated distribution.\n\n\nNow about the zenith angle. First of all, it is not the azimuth angle that we expect uniformly distributed. Why? Every unit of area of the sphere surrounding the detectors should contain the same number of neutrinos. We know that the area in spherical coordinates is given by dS = sin(θ)dθdϕ = -d(cos(θ))dϕ. That is, we need **cos(θ) to be uniformly distributed**, so that all directions are equal. Let's plot the histogram:\n\n\nThis looks very interesting, with rather prominent maximums at both poles (for θ = 0 and π), and a less pronounced one at the equator (θ = π/2). It would be great, and probably help us come up with better solutions, if we could infer the reasons why the hosts selected such distribution to generate the simulated data.", "votes": null }, { "id": "2142755", "postDate": "02/13/2023 18:52:52", "content": "Your last graph is quite interesting. I would have expected there to be some kind of dip around the zero mark. This would correspond to angle that comes thru most amount of earth, wouldn't it. While it wouldn't necessarily stop a neutrino, it would have at least caused some disruption in the path as it went thru core and what not. Just my thoughts, interesting graph though; I look forward to seeing what others think.
", "rawMarkdown": "Your last graph is quite interesting. I would have expected there to be some kind of dip around the zero mark. This would correspond to angle that comes thru most amount of earth, wouldn't it. While it wouldn't necessarily stop a neutrino, it would have at least caused some disruption in the path as it went thru core and what not. Just my thoughts, interesting graph though; I look forward to seeing what others think.", "votes": null }, { "id": "2143566", "postDate": "02/14/2023 10:57:00", "content": "Hi, you are right that an isotropic distribution is flat in azimuth and cos(zenith).
\nThe neutrinos were indeed simulated that way, but after simulation they undergo a long chain of processing, including:
\nThis means that even if we inject neutrinos isotopically, what we record (here simulated) in the detector may not be isotropic anymore. In other words: the acceptance of the detector is not uniform.
\nWhat you observe in the azimuth distribution is an imprint of the hexagonal detector geometry, which is also what we see in all of our data and a well-known effect. The zenith distribution is more complicated, but in short, we do have very good sensitivity at the horizon (cos(zenith) = 0), and for events below the horizon there is a certain absorption effect by the Earth for very high energy neutrinos. Also the DOMs themselves do not have isotropic sensitivity in zenith, and neither has the trigger logic.
So what you are seeing is totally expected, I hope my explanation helps.
", "rawMarkdown": "Hi, you are right that an isotropic distribution is flat in azimuth and cos(zenith).\n\nThe neutrinos were indeed _simulated_ that way, but after simulation they undergo a long chain of processing, including:\n- neutrino interaction process and particle showers, energy loss mechanisms and decays\n- photon propagation through the ice\n- PMT simulation and simulation of the detector electronics\n- Trigger logic, detector readout and event builder/splitter\n- online processing & filtering\n\nThis means that even if we inject neutrinos isotopically, what we record (here simulated) in the detector may not be isotropic anymore. In other words: the acceptance of the detector is not uniform.\nWhat you observe in the azimuth distribution is an imprint of the hexagonal detector geometry, which is also what we see in all of our data and a well-known effect. The zenith distribution is more complicated, but in short, we do have very good sensitivity at the horizon (cos(zenith) = 0), and for events below the horizon there is a certain absorption effect by the Earth for very high energy neutrinos. Also the DOMs themselves do not have isotropic sensitivity in zenith, and neither has the trigger logic.\n\nSo what you are seeing is totally expected, I hope my explanation helps.", "votes": null }, { "id": "2144255", "postDate": "02/14/2023 21:15:26", "content": "Thank you, I indeed suspected that the hex pattern may be the result of some events being dropped from simulation (if they produce zero pulses, for example).
\nI read about DOMs asymmetric sensitivity, but understood it so that this asymmetry depends on the direction of photons hitting the sensor, rather then the direction of neutrinos. As for the Earth absorption, one would expect it to lower the neutrinos flux density for zenith angles close to π. In fact it's grows there, so it looks the other effects make bigger impact.
\nThe more we dig into this competition, the more we see how incredibly complex is physics and engineering here.
", "rawMarkdown": "Thank you, I indeed suspected that the hex pattern may be the result of some events being dropped from simulation (if they produce zero pulses, for example). \n\nI read about DOMs asymmetric sensitivity, but understood it so that this asymmetry depends on the direction of photons hitting the sensor, rather then the direction of neutrinos. As for the Earth absorption, one would expect it to lower the neutrinos flux density for zenith angles close to π. In fact it's grows there, so it looks the other effects make bigger impact. \n\nThe more we dig into this competition, the more we see how incredibly complex is physics and engineering here.", "votes": null } ], "comments": [ { "id": 2142755, "author_name": "jaredvanderveen", "author_url": "", "post_date": "02/13/2023 18:52:52", "content": "Your last graph is quite interesting. I would have expected there to be some kind of dip around the zero mark. This would correspond to angle that comes thru most amount of earth, wouldn't it. While it wouldn't necessarily stop a neutrino, it would have at least caused some disruption in the path as it went thru core and what not. Just my thoughts, interesting graph though; I look forward to seeing what others think.
", "votes": null, "replies": [] }, { "id": 2143566, "author_name": "pellerphys", "author_url": "", "post_date": "02/14/2023 10:57:00", "content": "Hi, you are right that an isotropic distribution is flat in azimuth and cos(zenith).
\nThe neutrinos were indeed simulated that way, but after simulation they undergo a long chain of processing, including:
\nThis means that even if we inject neutrinos isotopically, what we record (here simulated) in the detector may not be isotropic anymore. In other words: the acceptance of the detector is not uniform.
\nWhat you observe in the azimuth distribution is an imprint of the hexagonal detector geometry, which is also what we see in all of our data and a well-known effect. The zenith distribution is more complicated, but in short, we do have very good sensitivity at the horizon (cos(zenith) = 0), and for events below the horizon there is a certain absorption effect by the Earth for very high energy neutrinos. Also the DOMs themselves do not have isotropic sensitivity in zenith, and neither has the trigger logic.
So what you are seeing is totally expected, I hope my explanation helps.
", "votes": null, "replies": [ { "id": 2144255, "author_name": "alexz0", "author_url": "", "post_date": "02/14/2023 21:15:26", "content": "Thank you, I indeed suspected that the hex pattern may be the result of some events being dropped from simulation (if they produce zero pulses, for example).
\nI read about DOMs asymmetric sensitivity, but understood it so that this asymmetry depends on the direction of photons hitting the sensor, rather then the direction of neutrinos. As for the Earth absorption, one would expect it to lower the neutrinos flux density for zenith angles close to π. In fact it's grows there, so it looks the other effects make bigger impact.
\nThe more we dig into this competition, the more we see how incredibly complex is physics and engineering here.
", "votes": null, "replies": [] } ] } ], "raw_markdown_by_id": { "2142461": "I want to share a couple of interesting observations on the neutrino incident angles in the training data.\n\nFirst, what I would expect to see. The simplest answer is symmetric, that is we might expect that the neutrinos come in equal numbers from all directions. Let's check if the actual distribution fits that model.\n\nSpoiler: **neither azimuth, nor zenith angles are distributed exactly uniformly.**\n\nHere is the histogram of the azimuth angle ϕ across all batches. It's indeed an almost uniform distribution in [0,2π]. We'll return to this \"almost\" in a moment.\n\n\nThe histogram above is uniform, except for the small dips - and these dips correspond nicely with the 6 directions that connect the neighboring DOMs. I'd like to note that this is not effect of measurement (like, shading from the neighboring strings), but simulated distribution.\n\n\nNow about the zenith angle. First of all, it is not the azimuth angle that we expect uniformly distributed. Why? Every unit of area of the sphere surrounding the detectors should contain the same number of neutrinos. We know that the area in spherical coordinates is given by dS = sin(θ)dθdϕ = -d(cos(θ))dϕ. That is, we need **cos(θ) to be uniformly distributed**, so that all directions are equal. Let's plot the histogram:\n\n\nThis looks very interesting, with rather prominent maximums at both poles (for θ = 0 and π), and a less pronounced one at the equator (θ = π/2). It would be great, and probably help us come up with better solutions, if we could infer the reasons why the hosts selected such distribution to generate the simulated data.", "2142755": "Your last graph is quite interesting. I would have expected there to be some kind of dip around the zero mark. This would correspond to angle that comes thru most amount of earth, wouldn't it. While it wouldn't necessarily stop a neutrino, it would have at least caused some disruption in the path as it went thru core and what not. Just my thoughts, interesting graph though; I look forward to seeing what others think.", "2143566": "Hi, you are right that an isotropic distribution is flat in azimuth and cos(zenith).\n\nThe neutrinos were indeed _simulated_ that way, but after simulation they undergo a long chain of processing, including:\n- neutrino interaction process and particle showers, energy loss mechanisms and decays\n- photon propagation through the ice\n- PMT simulation and simulation of the detector electronics\n- Trigger logic, detector readout and event builder/splitter\n- online processing & filtering\n\nThis means that even if we inject neutrinos isotopically, what we record (here simulated) in the detector may not be isotropic anymore. In other words: the acceptance of the detector is not uniform.\nWhat you observe in the azimuth distribution is an imprint of the hexagonal detector geometry, which is also what we see in all of our data and a well-known effect. The zenith distribution is more complicated, but in short, we do have very good sensitivity at the horizon (cos(zenith) = 0), and for events below the horizon there is a certain absorption effect by the Earth for very high energy neutrinos. Also the DOMs themselves do not have isotropic sensitivity in zenith, and neither has the trigger logic.\n\nSo what you are seeing is totally expected, I hope my explanation helps.", "2144255": "Thank you, I indeed suspected that the hex pattern may be the result of some events being dropped from simulation (if they produce zero pulses, for example). \n\nI read about DOMs asymmetric sensitivity, but understood it so that this asymmetry depends on the direction of photons hitting the sensor, rather then the direction of neutrinos. As for the Earth absorption, one would expect it to lower the neutrinos flux density for zenith angles close to π. In fact it's grows there, so it looks the other effects make bigger impact. \n\nThe more we dig into this competition, the more we see how incredibly complex is physics and engineering here." }, "source": "meta" }