If you were orbiting a star undergoing a core-collapse supernova, what would be the delay between the neutrino pulse from the collapse and the first visible effects on the surface? Basically, I'm trying to get a scientifically-plausible gauge of the warning time given by said neutrino pulse.

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    $\begingroup$ How close are you to the supernova? Although there is indeed a pulse associated with core collapse, the neutrino flux ramps up significantly in the hours before that. $\endgroup$ – ProfRob Nov 7 '17 at 1:45
  • $\begingroup$ I'm thinking of the time it takes for the shock wave or light/heat pulse to reach the photosphere and escape. From there, I can calculate for myself. $\endgroup$ – Colin Paddock Nov 7 '17 at 2:20
  • $\begingroup$ No, I mean before the neutrino pulse, there would be an increase in neutrino flux that would be detectable if you were close enough. So my answer is a lower limit. $\endgroup$ – ProfRob Nov 7 '17 at 3:55
  • $\begingroup$ Planning to rewrite "Currents of Space" ? :-) $\endgroup$ – Carl Witthoft Nov 7 '17 at 16:16
  • $\begingroup$ See also astronomy.stackexchange.com/questions/18423/… $\endgroup$ – ProfRob Nov 9 '17 at 7:37

The neutrino pulse associated with the core-collapse supernova SN1987A in the Large Magellanic Cloud, were estimated to have arrived at the Earth about 3 hours before the rise of the supernova light curve (e.g. the review by Beall 2006).

If neutrinos are massless or have small masses (which they do), then this is your answer.

If they have masses (and they do), then the answer could depend on the neutrino mass and how far away you are from the supernova, since particles with mass will travel slower than the speed of light. Therefore the "delayed", but faster, optical emission will catch-up with the neutrinos.

The neutrino masses are probably of the order of 0.1 eV, whilst the energies of neutrinos from core-collapse supernovae are around 30 MeV in energy. i.e. The neutrinos travel with a Lorentz factor $\gamma = (1-v^2/c^2)^{-1/2} \sim 3\times 10^{7}$. Thus $$\frac{1}{1 - v^2/c^2} \sim 9\times 10^{14}$$ and so $v/c \sim 1 - 1/1.8\times 10^{15}$. That is, the neutrinos probably travel at about 1 part in $10^{15}$ slower than the speed of light.

At the distance of the Sun, this amounts to a delay of $5 \times 10^{-13}$ s, at the distance of the Andromeda galaxy about 0.006s, but for a galaxy at a distance of a Gpc, the delay would be about 100 s.

So actually, unless you are detecting the supernova on the other side of the observable universe, it makes no odds and certainly makes no difference at the distance of the Large Magellanic Cloud. So I say your answer is about 3 hours.

  • $\begingroup$ If neutrinos are massless, then this is your answer If I understand correctly this is because the neutrino flux starts increasing noticeably hours before the the core collapse pulse, and if they were massless the would be traveling at light speed. Whatever the reason, could I suggest you add that to the body of the answer for clarity. $\endgroup$ – StephenG Nov 7 '17 at 3:33
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    $\begingroup$ @stephenG The question asks for the time between the neutrino pulse and optical signature. It's 3 hours. SN1987a is not far enough away for the small neutrino mass to make any difference. $\endgroup$ – ProfRob Nov 7 '17 at 3:47

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