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I read in wiki that the speed of light is 88km/s slower in air than it is in a vacuum.

Do neutrinos travel faster than light in air?

Jitter
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    Even if so, it is not that interesting: because neutrino cannot decay to photons, so perhaps there is no cherenkov radiation? – wonderich Jan 17 '14 at 19:42
  • It could be interesting if scientists forget to account for air when comparing speeds from various experiments. Do scientists account for air in all speed of light experiments and observations? – Jitter Jan 17 '14 at 19:55
  • Ps. Is the 88km/s just refering to the phase velocity? Does the group velocity still travel at c? – Jitter Jan 17 '14 at 20:05
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    Related: http://physics.stackexchange.com/questions/62974/neutrinos-vs-photons-who-wins-the-race-across-the-galaxy –  Jan 17 '14 at 20:19
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    If you're worrying about whether the scientists at CERN were accounting for air when measuring the neutrino speed, you needn't worry. They measured the speed of neutrinos travelling through the ground (see http://www.neontommy.com/sites/default/files/users/user718/beamtrajectory-en-71dd9-8be65.png). Anyway they weren't comparing the speed of neutrinos to the speed the light travels at in the same medium, they were comparing the speed of neutrinos to the speed of light in a vaccuum. – kd88 Jan 17 '14 at 21:18

2 Answers2

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The answer is yes. Neutrinos will travel faster than light in a medium with a refractive index ($n$) greater than one (which is the case of air). Indeed the speed of light in that medium will be $v_{\text{medium}}=c/n$ where $c=2.998\times10^8$ m/s and $n>1$.

Then, because neutrinos interacts only very weakly (only through the weak nuclear force) with the medium, neutrinos will barely be slowed compared to how much light is slowed and thus will go faster than light. Remember that neutrinos are almost massless and thus already travel to nearly the speed of light.

--- New Edit --- Indeed, the neutrino speed will depend on it's energy (as pointed out in comments). But I think that in most process in which neutrinos are produced (take for instance a beta-decay), the energy of a neutrino is enough to consider it as going to nearly the vacuum speed of light. So strictly speaking, the answer is that it depends on the neutrino energy and what type of medium you are in.

VanillaSpinIce
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    Note that the speed of a neutrino depends on it's mass and energy. The absolute masses are not currently known, though they are believed to be much less than 1 eV. – dmckee --- ex-moderator kitten Jan 17 '14 at 21:30
  • What about media with $n<1$? – ybeltukov Jan 19 '14 at 06:41
  • Well this depends on the specific type of media. In media with $n<1$ (which in general is frequency dependent), the phase velocity of light is indeed faster than $c=2.98\times10^8$ m/s. However the group velocity (which I guess is implicitly what the question is about) will always be equal or smaller than $c$. – VanillaSpinIce Jan 19 '14 at 16:07
  • given the refractive index of air is "very close" to 1 and the speed of a neutrino is typically "very close" to $c$ can you give some typical values. What you say sound intuitively right but some numbers would make me feel better. – By Symmetry Oct 11 '14 at 23:53
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    @VanillaSpinIce Just a quibble. It is the signal transmission velocity which stays below $c$. The group velocity is only an approximation to this (albeit an excellent one in most cases). What matters is that the medium's transfer function should be causal, i.e. no output before a time $t/c$ when the wavefront first reached the input of the medium and $t$ is the distance through it and such criteria can be reduced to things like the Paley-Wiener integral criterion: see discussion here – Selene Routley Oct 12 '14 at 03:02
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It looks so, as neutrino speed was measured to coincide with the light speed, and neutrino interacts very weakly with matter. However, as neutrino probably has mass, the answer to your question is positive only for neutrinos of sufficient energy.

Waffle's Crazy Peanut
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akhmeteli
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    "However, as neutrino probably has mass (http://en.wikipedia.org/wiki/Neutrino#Mass ), the answer to your question is positive only for neutrinos of sufficient energy" --> Neutrinos are known to have mass "the answer to your question is positive only for neutrinos of sufficient energy" --> Neutrino masses are however (currently) immeasurably small... which means even with the current best upper limits on neutrino masses you would only need a neutrino with energy > 10 eV to be categorised as ultra-relatavistic. – kd88 Jan 17 '14 at 20:59
  • @jk88: "Neutrinos are known to have mass"--> As far as I know, "some neutrinos are known to have mass" - only differences of masses are known to be non-zero. – akhmeteli Jan 17 '14 at 21:43
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    @jk88: I agree that neutrinos need to have very small energy to be ultrarelativistic, but they do need to have some energy for that. – akhmeteli Jan 17 '14 at 22:05
  • "only differences of masses are known to be non-zero" --> correct, however a massless neutrino cannot mix between species, so the fact that we observe non-zero mixing between all three species means that neutrino masses are non-zero. "they do need to have some energy for that" --> I would guess that a neutrino with energy < 10 eV has never been observed. – kd88 Jan 18 '14 at 10:32
  • @jk88: "a massless neutrino cannot mix between species" - this statement does not seem obvious - could you please give a reference? My understanding is that a massless neutrino can mix with a massive neutrino. – akhmeteli Jan 18 '14 at 16:05
  • @jk88: " I would guess that a neutrino with energy < 10 eV has never been observed" - maybe not, but that does not make my statement incorrect: massive neutrinos still need to have some energy to be ultrarelativistic. – akhmeteli Jan 18 '14 at 16:08
  • This wikipedia article is a good start, but more specifically you should look at the neutrino mixing matrix wikipedia entry. To summarise the latter entry: the neutrino mixing matrix is the theoretical framework used to describe the mixing of neutrino flavours ($\nu_{e}$, $\nu_{\mu}$, $\nu_{\tau}$) with the neutrino mass states ($\nu_{1}$, $\nu_{2}$, $\nu_{3}$). Even if only 2 of the mass states are massive, because there is a mixing between all... – kd88 Jan 18 '14 at 20:59
  • ...3 states (in the jargon, 'the mixing angles are non-zero') then all 3 observable mixed neutrino states are massive. – kd88 Jan 18 '14 at 21:01
  • @jk88: I respectfully disagree. If a neutrino flavor is a mix of a massless and a massive state, it does not mean the flavor is necessarily massive. According to quantum theory, that means there is a probability of finding it in a massless state. Furthermore, as you say yourself, this is the "theoretical framework", but there is no experimental evidence all neutrinos are massive, as only nonzero differences of squared masses have been measured. That's why I said "probably". – akhmeteli Jan 18 '14 at 21:46
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    "If a neutrino flavor is a mix of a massless and a massive state, it does not mean the flavor is necessarily massive" --> yes it does! – kd88 Jan 19 '14 at 00:31
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    @jk88: I gave my argument, you have not offered yours. – akhmeteli Jan 19 '14 at 00:42
  • @akhmeteli define mix . in my math it will be something like (am_1 +bm_2)/d where a and b and d are given by whatever functional relationship nature gives and theory describes . Evebn if m_1 is 0 ,m2 remains. – anna v Jan 19 '14 at 04:51
  • @anna v: in my math, mix is $a\psi_{m1}+b\psi_{m2}$, where, e.g., $\psi_{m1}$ is a state with mass $m1$. If you measure the mass, there is some probability that it is $m1$. – akhmeteli Jan 19 '14 at 05:08
  • @akhmeteli It is different if you are talking of wavefunctions and different if you are talking of particles that have wave functions. If you look at the table of elements particles have specific masses, not probability distributions.If you measure m1 then it is the neutrino with flavor and mass m1 you have measured, not the one with flavor and mass m2 – anna v Jan 19 '14 at 05:23
  • @akhmeteli our measurements always measure a flavor eigenstate, becasue we detect the corresponding flavor particle, electron, muon or tau. We experimentally found that the masses must be diferent because when we start wit a beam with only muon neutrinos, we end up in detecting also tau neutrinos. It is the ratio of detected flavors after traveling a distance that changes, and implies a difference in masses. – anna v Jan 19 '14 at 05:41
  • @akhmeteli the flavor neutrinos have to be orthogonal in the standard model, and if there were no masses a beam with a start of a one flavor cannot after a distance have a component of another flavor. If there are masses one can posit different eigenstates of the wavefunction that allow it consistent with the SM. – anna v Jan 19 '14 at 05:51
  • @anna v: "If you measure m1 then it is the neutrino with flavor and mass m1" - it looks like we agree here - so you can measure zero mass, if one of the mass is zero. If you believe not even one of the neutrino masses can be zero, could you please give a reference? My understanding is this possibility has not been ruled out so far. – akhmeteli Jan 19 '14 at 06:12
  • @akhmeteli I will try to refresh my memory on this. reading http://arxiv.org/pdf/physics/0303116v1.pdf . The neutrinos are produced as flavor eigenstates and travel as mass eigenstates. page 12, "one of the masses may be zero, not known if so" so it seems you can have oscillations where one of the three masses is zero. – anna v Jan 19 '14 at 08:28