# Neutrinos vs. Photons: Who wins the race across the galaxy?

Inspired by the wording of this answer, a thought occurred to me. If a photon and a neutrino were to race along a significant stretch of our actual galaxy, which would win the race?

Now, neutrinos had better not be going faster than the speed of light in vacuum. However, an energetic enough neutrino can have a velocity arbitrarily close to $c$. Say we took a neutrino from a typical core-collapse supernova. It would have a speed $$v_\nu = (1 - \epsilon_\nu) c$$ for some small $\epsilon_\nu > 0$. What is the order of magnitude for $\epsilon_\nu$?

At the same time, photons can also travel slower than $c$. The interstellar medium is not completely devoid of matter, and in fact much of this matter is ionized plasma. As such, it should have a plasma frequency $\omega_\mathrm{p}$, and so it should effectively have an index of refraction depending on the ratio $\omega/\omega_\mathrm{p}$. Then the speed of a photon will be $$v_\gamma = (1 - \epsilon_\gamma) c,$$ where $\epsilon_\gamma$ is in general frequency-dependent. What is the order of magnitude for this deviation? I know it comes into play at radio frequencies, where in fact even the variation of $v_\gamma$ with frequency is detected: Pulses from pulsars suffer dispersion as they travel over hundreds to thousands of parsecs to reach us.

For simplicity, let's assume there are no obstructions like giant molecular clouds or rogue planets to get in the way of the photon. Is it possible that some photons will be outpaced by typical neutrinos? How big is this effect, and how does it depend on photon frequency and neutrino energy?

• +1 This is a great question and also (I think) really complicated. Neutrinos also experience refraction-like behavior via the MSW effect: en.wikipedia.org/wiki/… however I suspect the weak interaction is way less significant than the interaction photons experience. – Brandon Enright May 2 '13 at 2:10
• Yep. Also, while it is known that at least two of the neutrino masses are nonzero, it is still possible that the lightest mass eigenstate has exactly zero mass, in which case that neutrino would presumably always beat a photon (except for a tie in perfect vacuum of course). – Michael Brown May 2 '13 at 2:26
• As a matter of coincidence I happen to be studying neutrino kinetics at the moment. A nice theory can be found in Sigl, G., & Raffelt, G. Nuclear Physics B, 406(1-2), 423–451 (paywalled) (hard to find pdf). I'm sure someone here is already up to speed on this stuff, but if not I hope to be able to answer this question soonish. – Michael Brown May 2 '13 at 2:35
• Note that with the recent Plank results we have a new, lower, limit for the sum of the masses of the 3 known flavors, which means that the lower limit on their speed is now higher. – dmckee May 2 '13 at 3:48
• Another note: You'll need to assume a neutrino energy range. You specify a "typical" neutrino. If you mean one like those generated in the sun, then you are looking at energies very near 1 MeV. – dmckee May 2 '13 at 3:49

Cute question!

For a neutrino with mass $m$ and energy $E\gg m$, we have $v=1-\epsilon$, where $\epsilon\approx (1/2)(m/E)^2$ (in units with $c=1$). IceCube has detected neutrinos with energies on the order of 1 PeV, but that's exceptional. For neutrinos with mass 0.1 eV and an energy of 1 PeV, we have $\epsilon\sim10^{-32}$.

The time of flight for high-energy photons has been proposed as a test of theories of quantum gravity. A decade ago, Lee Smolin was pushing the idea that loop quantum gravity predicted measurable vacuum dispersion for high-energy photons from supernovae. The actual results of measurements were negative: http://arxiv.org/abs/0908.1832 . Photons with energies as high as 30 GeV were found to be dispersed by no more than $\sim 10^{-17}$ relative to other photons. What this tells us is that interactions with the interstellar medium must cause $\epsilon \ll 10^{-17}$, or else those interactions would have prohibited such an experiment as a test of LQG.

According to WP, the density of the interstellar medium varies by many order of magnitude, but assuming that it's $\sim 10^{-22}$ times the density of ordinary matter, we could guess that it causes $\epsilon\sim 10^{-22}$. This would be consistent with the fact that it wasn't considered important in the tests of vacuum dispersion.

For a neutrino with a mass of 0.1 eV to have $\epsilon\sim 10^{-22}$, it would have to have an energy of 10 GeV. This seems to be within but on the high end of the energy scale for radiation emitted by supernovae. So I think the answer is that it really depends on the energy of the photon, the energy of the neutrino, and the density of the (highly nonuniform) interstellar medium that the particles pass through.

• Good discussion, and effective plausibility argument. As Brandon Enright mentioned in a comment there is also the MSW effect on neutrinos propagating through matter. I guess it's probably irrelevant here, but one should at least do an order of magnitude calculation of it before declaring a winner. :) – Michael Brown May 2 '13 at 3:22
• I'd be interested in seeing that; my particle physics chops aren't up to it. I guess we're talking about neutrinos that could be either above or below the electroweak interaction scale...? – Ben Crowell May 2 '13 at 3:31
• I suppose so. The effect is stronger for higher neutrino energies (up to a point, I suppose, where the energy is so high the medium may as well be vacuum), but $\epsilon$ decreases with energy as you note. So there could be an interesting play off of these effects with an energetic "sweet spot" somewhere. I couldn't tell you where. – Michael Brown May 2 '13 at 3:41
• Quick note on the kinematics: Assume that the ISM is mostly hydrogen at rest. Then to a good approximation the centre of mass energy of a neutrino-proton collision is $\sqrt{m_p (m_p + 2 E)}$, so the threshold for real W production is $E\sim3200\ \mathrm{GeV}$, which sets "the weak scale" for this question much higher than the naive $\sim80\ \mathrm{GeV}$. – Michael Brown May 2 '13 at 4:01

I think you also have to consider that in the real world, the photon would travel a less linear path than the neutrino. This is due to things like gravitational lensing and any particles the photon interacts with any particles. Thinking of the super nova, how long does it take a photon to get from the center of a star to the outer most layers vs. a neutrino? or is that beyond the scope of your question?

• I think that's beyond the scope. This question assumes they start at the same place so things like how long a photo takes to escape a star don't matter. – Brandon Enright Dec 10 '13 at 6:50
• Neutrinos make it out of a supernova first. Photons tend to be absorbed and re-emitted. I read that it takes years for a photon to make it out of a normal star. The neutrino pressure is what makes the supernova explode: by balancing the force of gravity for a short time, throughout the entire bulk of the star. This is something that EM radiation cannot do. But I think that for gravitational lensing, the neutrino path would be identical to the photon path. – user95006 Feb 11 '16 at 21:29