3 ms·

Since fusion is only happening in the center of the Sun, and the outer layers are almost entirely transparent to neutrinos, this is actually a direct image of t

• yummybear 6y ago
  So they can use this image to directly measure the width of the core?
  • lopsidedBrain 6y ago
    To an extent. I'm actually curious about its angular resolution.
  • 0PingWithJesus 6y ago
    In principle yes....although the specifics of the physics involved kinda make the question itself not well posed. There is no hard boundary to the core of the sun. The "core" is by definition where nuclear fusion reactions occur. However, those reactions don't just stop at a certain radius...but instead just occur at a lower and lower rate. So even if you could determine with 100% precision where a neutrino came from within the sun, you would still measure some exponential-like decay as a function of radius. But to add even more complexity there's ~10 different nuclear processes within the sun that produce neutrinos. Those processes all have different radial profiles. So even if you measure with 100% accuracy the radial profile of neutrinos associated with one or two nuclear processes...you still haven't really measure the core of the sun...you've just measured it for a few specific reactions. And for the neutrinos produced by many of the reactions this method cannot work, those neutrinos are too low in energy to provide direction information. And beyond that there are a handful of nuclear reactions that occur within the sun that don't produce neutrinos. So there doesn't really exist any way to measure the radial profile of those nuclear processes. And this all assume you can perfectly tell where the neutrino came from within the sun, which is also impossible. There will always be some relatively poor "resolution associated with your ability to place a neutrinos origin. Here is the "hard" physics limit to your angular resolution for a relatively high solar energy neutrino...it only gets worse as the energy goes down https://i.imgur.com/h3n8c4V.png https://i.imgur.com/h3n8c4V.png. But getting to even that resolution is impossible b/c an interaction will only produce so many photons from Cherenkov radiation (think 100s of photons). Then it becomes a statistics problem...what's the best angular resolution you could possibly achieve given an average number of photons that's around (say) 500. It ends up the answer is "pretty good" but far from perfect. And all of that is assuming the electron scattered from the solar neutrino will travel in only one direction...that's extremely untrue, the electron will always bounce off of other electrons & atoms after scattering. This multiple-scattering leads to even worse angular resolution. Here's a paper on the subject if you'd like further detail https://arxiv.org/pdf/1606.02558.pdf https://arxiv.org/pdf/1606.02558.pdf
    • sudhirj 6y ago
      Most lines of this kind are drawn based on some value we agree on, not any intrinsic rule in physics. The boundary of the atmosphere, earth's crust, the extent of the solar system, etc.
    • cp_mlreef 6y ago
      Thank you for sharing! Extremly interesting and mind boggling - what a time to be alive!
  • frumiousirc 6y ago
    Absolutely not. The angular resolution is nowhere near good enough. Every time the Internet regurgitates this link it is misinterpreted by essentially everyone that comments on it.
• ahazred8ta 6y ago
  The core is about 20% of the width of the sun, so about 0.1 degree wide as seen from Earth.
• zyxzevn 6y ago
  The circle in the centre of the image is much wider than the sun. Sadly, we can not see any meaningful imagery from it.