# Why do the neutrinos (with mass) from a supernova arrive before the light (no mass)?

I've already read the below questions (and their answers) regarding neutrinos vs. electromagnetic waves propagating through space, but I'm still not clear on something.

Given that

• Light from SN 1987A arrived 2 or 3 hours after its neutrinos, implying that it was "slowed down" relative to the neutrinos
• Light from SN Refsdal has been "lensed" multiple times to re-appear on time scales of several decades, implying that light interacts with matter (mass)
• Neutrinos interact extremely little with matter but are known to have mass and energy

Question

Why did neutrinos (with their mass and momentum) arrive before the light (considered to be massless) from SN 1987A? Considering SR and GR, this seems to be a contradiction. What am I missing?

Postscript

I've tried desperately to avoid using the word "photon" above (in reference to light) after learning of the Lamb Controversy™ (via related discussions here and here on Phys SE).

• I edited the title to avoid confusion between this and the "FTL neutrinos" incident from the OPERA experiment (which was entirely Earth-based). – David Z Sep 25 '17 at 17:58
• @DavidZ: Thank you for doing that... I was originally tempted to add some verbiage to my question to avoid all those OPERA Q&A's out there. – pr1268 Sep 27 '17 at 1:53

Both neutrinos and photons were produced in the core of the star but photons have a much stronger probability of interacting with the outer layer of the star than the neutrinos. Thus the photons were trapped whereas the neutrinos easily escaped. This has nothing to do with mass and all to do with the cross-section of interaction with protons/electrons for photons on one hand and for neutrinos on the other.

Reading @dmckee's answer made me realise that the phrasing of the previous paragraph makes it sound like the light flash we observe might be due to those photons eventually escaping. This is not what I meant: it would take millions of years for those photons to escape, as is well known for our own Sun. It is only because the outer layers of the star are eventually blown off that we see a light flash.

I should also have pointed out that electron neutrinos can escape only in the early stages of the collapse of type II supernovae. As the density increases beyond a few times $10^{11} \text{g}\ \text{cm}^{-3}$, the scattering of neutrinos with stellar matter is sufficient to make the timescale of the diffusion of neutrinos out of the star shorter than the collapse timescale. This is a combination of increasing density (and therefore increasing interactions) and accelerating collapse. So the neutrino flash measured on Earth came from the very beginning of the evolution into a supernova.

Let me add some orders of magnitude. The cross-section of photon-electron scattering is of the order of $10^{-24} \text{cm}^2$. Compare this with the neutrino-nucleon scattering. It varies as the square of the neutrino energy:

$$\sigma_\nu \approx 10^{-44} E_\nu^2\ \text{cm}^2$$

with the energy in MeV. So that's 20 orders of magnitude, give or take.

Where does this huge difference come from? Neutrinos interact solely through the weak interaction whereas photons interact through the electromagnetic interaction with charged nuclei and electrons in the star plasma. So this is just a reflection of the relative strength of both interactions. There is no reason it should be like that: it is just the way our universe is! We would not be here to discuss these matters if it were not, actually…

• Agreed, and dmckee's answer here and also here concur, but perhaps a deeper understanding of the "stronger probability to interact" part of your answer is the essence of what I'm seeking. – pr1268 Sep 24 '17 at 16:21
• I have a vague recollection of reading that the extreme densities reached during collapse are high enough to cause 10% of the neutrinos attempting to escape to be captured in collisions. Is the scattering delay you mentioned the same thing described differently? If not, how significant is it in reducing the total that are released? – Dan Is Fiddling By Firelight Sep 26 '17 at 14:29
• @DanNeely I am not sure. It is all mixed up in my head with the final stage: supernovae leaves a hot neutron star which cools down by emitting neutrinos. I'll try to search… – user154997 Sep 26 '17 at 15:35

The situation with supernova is not about speed of flight but about time of emergence. A type IIa supernova candidate is big, even with the vastly powerful explosion of the core it takes hours to blow the envelope off and expose the violence of the interior—and it is only after that happens that the star becomes brighter in the electromagnetic spectrum. But the neutrinos escape essentially immediately.

No exotic physics needed.

In principle—with a good enough model of how supernovas work—this could be a probe of the absolute neutrino mass.

• Per the last sentence - at some distance the photons are going to catch up with and pass the neutrinos emitted earlier, yes? And knowing the times involved would allow us to calculate neutrino speed and thus mass? – Michael Sep 25 '17 at 17:59
• That's the idea. If you know the envelope of emission times for photons and neutrinos from a model, know the distance distance and measure the arrival times for photons and neutrinos you can deduce the neutrino velocity and thus mass. – dmckee --- ex-moderator kitten Sep 25 '17 at 18:53
• Even better if you can measure gravitational waves, which should arrive at the speed of light and are produced at almost exactly the same time as the neutrino pulse. – Rob Jeffries Sep 25 '17 at 20:09
• @RobJeffries This is a recent article about the correlation of gravitational wave and neutrino arrival for SN1987A: link.springer.com/article/10.1140%2Fepjc%2Fs10052-016-4277-4 – DavePhD Sep 26 '17 at 12:08

General relativity does not have much to do with this.

The special theory of relativity does not say that light moves at speed $c$ unconditionally; that would contradict experiment: it says that light moves at speed $c$ in vacuum. In glass and water, light moves slower because it interacts with matter. While the light was coming from the supernova, of course, it passed through vacuum: but before that, it was formed at the center of a collapsing star and it had to make its way through the star, which is a place rich with a plasma of free electrons and a sea of nuclei that causes similar slowdown. Heck, in our Sun it takes 171,000 years for the produced light to scatter out, though admittedly that is inelastic scattering and above we probably need to consider elastic scattering, which would speed up the process.

The special theory of relativity also does not say that particles must move with some maximum speed $v_\text{max} < c$. In fact as you dump more and more energy into a particle it moves faster and faster, and $c$ itself is the speed limit. The relativistic formula is that if you give a kinetic energy $K$ to a particle then $v = c \sqrt{1 - 1/(K/mc^2 - 1)^2},$ which is a rather large mouthful to deal with, but if this ratio $\kappa = K/mc^2$ is very large this becomes just $v\approx c\sqrt{1 - 1/\kappa^2} \approx c~(1 - \frac12 \kappa^{-2}).$ For these neutrinos we can imagine kinetic energies in the keV or MeV scales whereas their masses are perhaps on the eV scale -- we don't know the exact details but these two numbers likely cause very large $\kappa$ of perhaps 100,000 or more, which would mean these things are going at least at speed $c$ minus about one part in 5 billion. Travelling over only 160,000 years, this only works out to about 15 minutes of time, not enough to cancel out the whole three hours between when the neutrinos got out of the star and when the light finally did.

• I would think that most of the neutrinos leave the star immediately, without interacting at all. IIUC only perhaps 1% of the neutrino energy is transferred to the surrounding matter. That's still enough to drive the explosion but seems to indicate that most neutrinos do not interact, even with the dense matter of a collapsing core shock front; as is their custom. That is certainly not the case with light. – Peter - Reinstate Monica Sep 25 '17 at 20:13
• Exactly, the neutrinos leave immediately as collapse begins, but the photons we see don't leave until a little the surface starts to explode, which takes a nontrivial amount of time as the energy from the interior starts to make its way out. – CR Drost Sep 25 '17 at 20:28