Measurement of the neutrino velocity with the OPERA detector in the CNGS beam
(Submitted on 22 Sep 2011)
The OPERA neutrino experiment at the underground Gran Sasso Laboratory has measured the velocity of neutrinos from the CERN CNGS beam over a baseline of about 730 km with much higher accuracy than previous studies conducted with accelerator neutrinos. The measurement is based on high-statistics data taken by OPERA in the years 2009, 2010 and 2011. Dedicated upgrades of the CNGS timing system and of the OPERA detector, as well as a high precision geodesy campaign for the measurement of the neutrino baseline, allowed reaching comparable systematic and statistical accuracies. An early arrival time of CNGS muon neutrinos with respect to the one computed assuming the speed of light in vacuum of (60.7 \pm 6.9 (stat.) \pm 7.4 (sys.)) ns was measured. This anomaly corresponds to a relative difference of the muon neutrino velocity with respect to the speed of light (v-c)/c = (2.48 \pm 0.28 (stat.) \pm 0.30 (sys.)) \times 10-5.
So, according to this CERN find http://arxiv.org/abs/1109.4897, the neutrinos go faster than light? How can that be explained? I have 3 options:
A. An error in measurement
B. The neutrinos indeed go faster than light
Above 2 are the current favourites of people. Option A is obvious. I don’t like option B so much. Sure it is an option but I don’t find it ‘elegant’ that those neutrinos travel only a very little bit faster than the speed of light. I would think that if the speed of light can be broken you break it by a large margin. Just as the speed of sound, once you break it you build Concordes and go mach 2 or 3, you don’t stick around mach 1,000001. So boring. Because why bother breaking the speed of light and not profiting of your freedom? Nope, I don’t find option B elegant.
Sure, most people think it must be either option A or B, but, like in the DSK affaire, they forget option C:
C. The neutrinos take a shorter route
Option C is so elegant. Why is everybody neglecting this? Beats me. Anyway I decided to think it over a bit. The basic idea is that the neutrinos are not bound by the curvature of Space-Time. Hence the neutrinos take the shorter (direct) way. Instead of following the curve they cut it short, a straighter line. I tried to get a grip on the anomaly by toying with some numbers I found on the internet.
I’ve found that the curvature caused by the mass of the sun at the distance of the earth is about 0.99999999 or in other words, it causes a deviation of 1 – 0,99999999 = 1E-08 (I got this numer from here, possibly there is a better number available)
The gravitation at the earth surface is about 1875 times as strong as the gravitation we (on earth) feel of the sun. Thus causing a deviation of 1875 x 1E-08 = 1,875E-05
A deviation of 1,875E-05 over 730 km = 13,7 meters
13,7 meters at the speed of light c is about 45 ns
And 45 ns is remarkably close to the 60 ns anomaly the guys at CERN are measuring:)
Ok, we have to explain how on earth the neutrinos can take that short route, and beat curved Space-Time to it. I don’t know, perhaps they use another dimension for travel, and perhaps that is why those neutrinos are so hard to detect. They are not in our ‘plane’ they just cross it.
Edit: I consulted somebody who I highly respect and regard as a great authority on the matter but who shall remain anonymous. His prompt reply made clear that my basic calculation is off. The effect would be much smaller. Fun thing is that the idea of ‘a shorter route’ had crossed his mind too, so in that way option C is not that bad. I still regard option C a valid option until disproved by option A or B.
Date: Sun, 25 Sep 2011 00:16:46 -0400
Subject: Re: neutrinos
Leuk je bericht te lezen, en ook om de foto’s van toen en nu te zien.
Ik had vanochtend (half slapend, net voor ik wakker werd) even hetzelfde idee, maar kwam er op uit dat het effect te klein is.
De kromming van de ruimte zorgt er inderdaad voor dat de kortste weg niet een rechte lijn is, maar ruwweg een deel van een cirkel met straal R = c^2/g = lichtsnelheid in het kwadraat gedeeld door g. De hoek die de neutrinos langs deze cirkel afleggen is dus ongeveer
phi = d/R = d g / c^2
met d = afstand tussen CERN en Gran Sasso. Dit is een heel kleine hoek.
De resulterende verkorting van de weg is ruwweg gelijk aan het verschil tussen de bijbehorende cirkelboog en een rechte lijn. Een schatting hiervan is
d x (2 sinus(phi/2) – phi) ~ d x phi^3 = d maal (phi tot de derde macht)
Dit is een veel te kleine afstand. Dus de verklaring moet toch ergens anders liggen…