Why? It captures information from a flux of particles (not photons, but neutrinos in this case) emitted by astrophysical objects. It allows us to study properties of those objects (and of the detected particles as well). It doesn't have a resolution high enough to give us an "image" of most of those objects, but Hubble can't image most single stars too. IceCube won't give you a pretty picture for APOD, but it will do everything else we can do with an optical telescope, or a charged particle telescope such as Auger.
It can infer the direction a neutrino came from, so (given enough time) it can make "images". In fact, they've seen the moon already, as a deficit of neutrinos coming from the moon's direction. It is a telescope, just one that doesn't "see" photons and that you don't have to point at a target to see it.
To be 100% fair, good American beer does exist. I remember I could buy a pretty good lager from Samuel Adams at most shops near Chicago, for example; but the really good ones are from microbreweries. Flying Dog has a fantastic selection, for example. Their imperial porter (called Gonzo) is one of the best beers I've ever tasted. I highly recommend it to anyone.
American beer isn't bad at all. Mass-marketed American beer is pure cooled piss.
I agree with you, but you misunderstand KISS. That's "keep it simple", as in "don't add unnecessary frills"; not as in "make it so easy my grandma could use". Slackware is the incarnation of KISS. You have very few automated systems, you can (and are encouraged to) configure everything using only ed/nano/vi/emacs. You can understand every failure easily, there are very few possible points of failure, and most failures can be easily contained and repaired. Contrast this with a distro like Ubuntu, which is easy to use, but not "simple" in the same sense as what KISS represents. In Ubuntu, everything is automated and every system is hidden behind a black box to the user. You get GUI tools to configure most stuff, but editing the files directly isn't encouraged (and could lead to some problems). If apt breaks due to a failed install/uninstall, you can't install software until you repair it. It isn't a simple system: it's a complex system whose parts lead to a generally easy to use interface, hiding a lot of unneeded complexity.
In other words, don't bash against KISS, it's not in any way related to your other points.
Betelgeuse won't go GRB. It's about 50% below the mass limit for GRBs. We'd see quite a lot more GRBs if 20 solar masses were enough to trigger one. Even more, its axis isn't pointing at us (Betelgeuse is one of the few stars we can actually see as a disc and study more deeply) If you want to worry, worry about Eta Carinae or WR 104, which are massive enough for a GRB and (in WR104's case) may have its axis pointing at us.
However, IceCube has an extension called Deep Core being deployed soon, which will increase the photomultiplier density in the centre of the detector and allow for a lower energy threshold. This was done exactly to detect supernova neutrinos.
Also, IceCube detects neutrinos starting at about 100 GeV (10^11 eV) with the default trigger. Still higher than the ~10 MeV (10^7 eV) from a SN, but it's more feasible to go down 4 orders of magnitude than 8.
So? Remember they're not offering it for unlimited accounts. How is 2GB of usage in a small screen with a small keyboard different (to the network) than 2GB of usage in a big screen with a big keyboard? Is 2GB of flash games somehow harder to transmit than 2GB of video?
But why? The iPhone has an OS based on OS X. Its network stack is able to route anything; that is enough for most of what most people would want to do online, and easy to configure for the other cases. Apple doesn't allow that, makes this carrier-dependent and opens the door for AT&T (and others) to screw their consumers. That's a business strategy (being closer to the carriers so that they get more benefits), but it's not in the best interest for consumers.
How is tethering any different from normal data usage? Or, why do you have to pay more for your device to route requests from a local network (created via WiFi, BT, Ethernet, Usb, whatever) to a wide area network?
Do you pay more to your ISP if you connect a router and two different computers? No? So why should you pay more to use the connection you pay for in any device you want?
The point here is that the iPhone is (obviously) capable of routing. It shouldn't matter what the network says. You connect it to your computer, your computer should be able to use the network through the phone. Every smartphone I know of does that. Apple is at fault here just because they don't let you do what your phone can do unless you get permission from the network operator. But hey, that's another revenue stream for everybody, right? Who cares if the consumer gets screwed? AFAIK, American consumers are already used to being screwed by the mobile companies, so it's all right!
That's the classical deduction using Schrödinger's equation. The only "relativistic" approximation is the dispersion relation, but they use Schrödinger's equation to get the kets. We use spinors, not bras/kets, when you solve the (relativistic) Dirac equation; in that case, you get the same probabilities.
See this for a reasonably simple deduction using Schrödinger's equation, that gives you the exact same formulae as your Wiki link. If you use a fully relativistic approach, you get the same final results but no dependency on the mass (only on mass splitings).
We say neutrinos have mass because a quantum superposition of states with slightly different masses oscillate in a very specific way; and we see the neutrino flavour changing in exactly that same way, with the same dependency on L/E (distance divided by energy). Since that is a direct effect of the different mass eigenstates moving at slightly different speeds for the same energy, it's very hard to get a theory that shows the same dependency without resorting to neutrino masses. Being not sure of what courses you've taken, I redirect you to the wiki for some more info and possibly more advanced links in your level of understanding.
The problem with massive neutrinos in the SM is in the anomaly cancellation. If you add right-handed neutrinos (in order to have neutrino mass), each fermion family ends up with a net anomaly (at least in 3+1 dimensions), while the "normal" SM is anomaly-free in each family.
Proofreading is a good idea, people should do that... Of course I meant CKM angles are so SMALL that you can treat it as an identity matrix plus a perturbation.
Mass eigenstates don't oscillate. n1 is always n1, unless you try to measure it, in which case its eigenfunction collapses into the interaction base (ne+nm+nt). That's quantum weirdness for you.
The interactions (production and detection) happen in the flavour base. The propagation happens in the mass base. This means you never oscillate "from massless to massive": you are created with a mixture of massive and massless states, which travel differently, changing the probability of each flavour.
The time-dependent Schrödinger's equation doesn't apply for massless particles. It was never intended to. It isn't relativistic. Try to apply a simple boost and you'll see it's not Poincaré invariant. The main point is that you get the same probabilities if you use a relativistic theory, but you need A LOT of work to get there.
Oscillations work and happen in QFT, which is Poincaré-invariant and assumes special relativity. I can't find any references in a quick search, but I've done all the (quite painful) calculations a long time ago to make sure it works. It's one of those cases where the added complexity of relativistic quantum field theory doesn't change the results from a simple Schrödinger solution.
No. All flavour eigenstates MUST be massive: they are superpositions of the three mass eigenstates, one of which can have zero mass. Calling the three mass eigenstates n1, n2 and n3; and the three flavour eigenstates ne, nm and nt, we'd have:
ne=Ue1*n1+Ue2*n2+Ue3*n3
nm=Um1*n1+Um2*n2+Um3*n3
nt=Ut1*n1+Ut2*n2+Ut3*n3
So, if any of n1, n2 or n3 has a non-zero mass (and at least two of them MUST have non-zero masses, since we know two different and non-zero mass differences), all three flavour eigenstates have non-zero masses.
Also, remember that the limit for the neutrino mass is at about 1eV, while it's hard to have neutrinos travelling with energies under 10^6 eV. In other words, the gamma factor is huge, and they're always ultrarelativistic, travelling practically at "c".
Another point is that the mass differences are really, really small; of the order of 0.01 eV. This is ridiculously small; so small that the uncertainty principle makes it possible for one state to "tunnel" to the other.
I really can't go any deeper than that without resorting to quantuim field theory. I can only say that standard QM is not compatible with relativity: Schrödinger's equation comes from the classical Hamiltonian, for example. To take special relativity into account, you need a different set of equations (Dirac's), which use the relativistic Hamiltonian. In this particular case, the result is the same using Dirac, Schrödinger or the full QFT, but the three-line Schrödinger solution becomes a full-page Dirac calculation, or ten pages of QFT. In this particular case, unfortunately, the best I can do is say "trust me, it works; you'll see it when you get more background".
~13 degrees is small, compared to the two main angles in the PMNS matrix (ok, \theta_13 is smaller, but the atmospheric and solar angles are really big). In fact, CKM angles are so big that you can treat the matrix as an identity matrix with a perturbation; in the neutrino sector, the mass and flavour eigenstates are so different that this type of treatment is meaningless.
Light doesn't oscillate in this way. A photon is a photon, and remains a photon. Electric and magnetic fields oscillate, but the particle "photon" doesn't. Neutrinos start as one particle (say, as muon-neutrinos) and are detected as a completely different particle (say, as a tau-neutrino).
The explanation for that is that what we call "electron-neutrino", "muon-neutrino" and "tau-neutrino" aren't states with a definite mass; they're a mixture of three neutrino states with definite, different mass (one of those masses can be zero, but at most one). Then, from pure quantum mechanics (and nothing more esoteric than that: pure Schrödinger equation) you see that, if those three defined-mass states have slightly different mass, you will have a probability of creating an electron neutrino and detecting it as a tau neutrino, and every other combination. Those probabilities follow a simple expansion, based on only five parameters (two mass differences and three angles), and depend on the energy of the neutrino and the distance in a very specific way. We can test that dependency, and use very different experiments to measure the five parameters; and everything fits very well. Right now (specially after MINOS saw the energy dependency of the oscillation probability), nobody questions neutrino oscillations. This OPERA result only confirms what we already knew.
No, they're not the same. Mass-induced oscillation is a known fact in particle physics (search for "neutron kaon oscillation" for background), and neutrinos behave in exactly the predicted way; only with big mixing angles, unlike the almost-zero angles in the quark sector's CKM matrix.
The point is that, if two different theories have the exact same predictions, they are for all intents and purposes the same theory, and describe the same universe. If that is the case, why would you spend more time teaching and learning the more complex one, when a simple explanation is enough and (by definition, since they have the same predictions) you can't tell which one is correct?
Of course, if the new theory offers a good explanation to current data, but has a different prediction than the standard model in other, still-non-tested scenarios, the theory is more interesting. You can test it at the new scenario, and you'll be able to tell them apart. This is why* we study, for example, supersymmetry and extra dimensions theories: they behave just like the standard model where we have tested it, but can be different in other cases such as the LHC.
* = of course there are other motivations to develop the theories, but they are taken seriously because they are compatible with the SM and are testable. A theory whose predictions were exactly the same as the SM for every case wouldn't be worth studying, simply because you'd never be able to see if it is right.
No, it may not. The neutrino's energy (which is the exact analog to a photon's "colour") is conserved. The rest energy changes, but this means the original neutrino and the resulting one are on different rest referentials (and the change in mass is very small, inside the uncertainty principle). There is nothing "yet to be discovered", except possibly what is the mass generation mechanism (Dirac, like all other particles, or Marjorana). "Chameleon-like behaviour" is just the "scientific" journalist's way of saying he understands nothing about what he is paid to write; but we all know journalists rarely know much of what they write anyway.
You'd need a pretty complex theory to get non-mass oscillations to match all the data we got over the past 12 years, which is very compatible with a three-state, mass-driven oscillation scenario. Besides, you'd have to explain more than what the current "new standard model" (the SM with added neutrino masses) does if you want your theory to be accepted. If two theories explain the same data equally well, the simplest is more likely.
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Why? It captures information from a flux of particles (not photons, but neutrinos in this case) emitted by astrophysical objects. It allows us to study properties of those objects (and of the detected particles as well). It doesn't have a resolution high enough to give us an "image" of most of those objects, but Hubble can't image most single stars too. IceCube won't give you a pretty picture for APOD, but it will do everything else we can do with an optical telescope, or a charged particle telescope such as Auger.
It can infer the direction a neutrino came from, so (given enough time) it can make "images". In fact, they've seen the moon already, as a deficit of neutrinos coming from the moon's direction. It is a telescope, just one that doesn't "see" photons and that you don't have to point at a target to see it.
To be 100% fair, good American beer does exist. I remember I could buy a pretty good lager from Samuel Adams at most shops near Chicago, for example; but the really good ones are from microbreweries. Flying Dog has a fantastic selection, for example. Their imperial porter (called Gonzo) is one of the best beers I've ever tasted. I highly recommend it to anyone.
American beer isn't bad at all. Mass-marketed American beer is pure cooled piss.
I agree with you, but you misunderstand KISS. That's "keep it simple", as in "don't add unnecessary frills"; not as in "make it so easy my grandma could use". Slackware is the incarnation of KISS. You have very few automated systems, you can (and are encouraged to) configure everything using only ed/nano/vi/emacs. You can understand every failure easily, there are very few possible points of failure, and most failures can be easily contained and repaired. Contrast this with a distro like Ubuntu, which is easy to use, but not "simple" in the same sense as what KISS represents. In Ubuntu, everything is automated and every system is hidden behind a black box to the user. You get GUI tools to configure most stuff, but editing the files directly isn't encouraged (and could lead to some problems). If apt breaks due to a failed install/uninstall, you can't install software until you repair it. It isn't a simple system: it's a complex system whose parts lead to a generally easy to use interface, hiding a lot of unneeded complexity.
In other words, don't bash against KISS, it's not in any way related to your other points.
There is some truth in what you say; but I'd say this is the beauty of the thread.
Betelgeuse won't go GRB. It's about 50% below the mass limit for GRBs. We'd see quite a lot more GRBs if 20 solar masses were enough to trigger one. Even more, its axis isn't pointing at us (Betelgeuse is one of the few stars we can actually see as a disc and study more deeply) If you want to worry, worry about Eta Carinae or WR 104, which are massive enough for a GRB and (in WR104's case) may have its axis pointing at us.
However, IceCube has an extension called Deep Core being deployed soon, which will increase the photomultiplier density in the centre of the detector and allow for a lower energy threshold. This was done exactly to detect supernova neutrinos.
Also, IceCube detects neutrinos starting at about 100 GeV (10^11 eV) with the default trigger. Still higher than the ~10 MeV (10^7 eV) from a SN, but it's more feasible to go down 4 orders of magnitude than 8.
So? Remember they're not offering it for unlimited accounts. How is 2GB of usage in a small screen with a small keyboard different (to the network) than 2GB of usage in a big screen with a big keyboard? Is 2GB of flash games somehow harder to transmit than 2GB of video?
But why? The iPhone has an OS based on OS X. Its network stack is able to route anything; that is enough for most of what most people would want to do online, and easy to configure for the other cases. Apple doesn't allow that, makes this carrier-dependent and opens the door for AT&T (and others) to screw their consumers. That's a business strategy (being closer to the carriers so that they get more benefits), but it's not in the best interest for consumers.
How is tethering any different from normal data usage? Or, why do you have to pay more for your device to route requests from a local network (created via WiFi, BT, Ethernet, Usb, whatever) to a wide area network?
Do you pay more to your ISP if you connect a router and two different computers? No? So why should you pay more to use the connection you pay for in any device you want?
The point here is that the iPhone is (obviously) capable of routing. It shouldn't matter what the network says. You connect it to your computer, your computer should be able to use the network through the phone. Every smartphone I know of does that. Apple is at fault here just because they don't let you do what your phone can do unless you get permission from the network operator. But hey, that's another revenue stream for everybody, right? Who cares if the consumer gets screwed? AFAIK, American consumers are already used to being screwed by the mobile companies, so it's all right!
That's the classical deduction using Schrödinger's equation. The only "relativistic" approximation is the dispersion relation, but they use Schrödinger's equation to get the kets. We use spinors, not bras/kets, when you solve the (relativistic) Dirac equation; in that case, you get the same probabilities.
See this for a reasonably simple deduction using Schrödinger's equation, that gives you the exact same formulae as your Wiki link. If you use a fully relativistic approach, you get the same final results but no dependency on the mass (only on mass splitings).
You can't define a referential moving at "c", so you can't really describe a "rest referential" for a photon or any massless particle.
We say neutrinos have mass because a quantum superposition of states with slightly different masses oscillate in a very specific way; and we see the neutrino flavour changing in exactly that same way, with the same dependency on L/E (distance divided by energy). Since that is a direct effect of the different mass eigenstates moving at slightly different speeds for the same energy, it's very hard to get a theory that shows the same dependency without resorting to neutrino masses. Being not sure of what courses you've taken, I redirect you to the wiki for some more info and possibly more advanced links in your level of understanding.
The problem with massive neutrinos in the SM is in the anomaly cancellation. If you add right-handed neutrinos (in order to have neutrino mass), each fermion family ends up with a net anomaly (at least in 3+1 dimensions), while the "normal" SM is anomaly-free in each family.
Proofreading is a good idea, people should do that... Of course I meant CKM angles are so SMALL that you can treat it as an identity matrix plus a perturbation.
Mass eigenstates don't oscillate. n1 is always n1, unless you try to measure it, in which case its eigenfunction collapses into the interaction base (ne+nm+nt). That's quantum weirdness for you.
The interactions (production and detection) happen in the flavour base. The propagation happens in the mass base. This means you never oscillate "from massless to massive": you are created with a mixture of massive and massless states, which travel differently, changing the probability of each flavour.
The time-dependent Schrödinger's equation doesn't apply for massless particles. It was never intended to. It isn't relativistic. Try to apply a simple boost and you'll see it's not Poincaré invariant. The main point is that you get the same probabilities if you use a relativistic theory, but you need A LOT of work to get there.
Oscillations work and happen in QFT, which is Poincaré-invariant and assumes special relativity. I can't find any references in a quick search, but I've done all the (quite painful) calculations a long time ago to make sure it works. It's one of those cases where the added complexity of relativistic quantum field theory doesn't change the results from a simple Schrödinger solution.
No. All flavour eigenstates MUST be massive: they are superpositions of the three mass eigenstates, one of which can have zero mass. Calling the three mass eigenstates n1, n2 and n3; and the three flavour eigenstates ne, nm and nt, we'd have:
ne=Ue1*n1+Ue2*n2+Ue3*n3
nm=Um1*n1+Um2*n2+Um3*n3
nt=Ut1*n1+Ut2*n2+Ut3*n3
So, if any of n1, n2 or n3 has a non-zero mass (and at least two of them MUST have non-zero masses, since we know two different and non-zero mass differences), all three flavour eigenstates have non-zero masses.
Also, remember that the limit for the neutrino mass is at about 1eV, while it's hard to have neutrinos travelling with energies under 10^6 eV. In other words, the gamma factor is huge, and they're always ultrarelativistic, travelling practically at "c".
Another point is that the mass differences are really, really small; of the order of 0.01 eV. This is ridiculously small; so small that the uncertainty principle makes it possible for one state to "tunnel" to the other.
I really can't go any deeper than that without resorting to quantuim field theory. I can only say that standard QM is not compatible with relativity: Schrödinger's equation comes from the classical Hamiltonian, for example. To take special relativity into account, you need a different set of equations (Dirac's), which use the relativistic Hamiltonian. In this particular case, the result is the same using Dirac, Schrödinger or the full QFT, but the three-line Schrödinger solution becomes a full-page Dirac calculation, or ten pages of QFT. In this particular case, unfortunately, the best I can do is say "trust me, it works; you'll see it when you get more background".
~13 degrees is small, compared to the two main angles in the PMNS matrix (ok, \theta_13 is smaller, but the atmospheric and solar angles are really big). In fact, CKM angles are so big that you can treat the matrix as an identity matrix with a perturbation; in the neutrino sector, the mass and flavour eigenstates are so different that this type of treatment is meaningless.
Light doesn't oscillate in this way. A photon is a photon, and remains a photon. Electric and magnetic fields oscillate, but the particle "photon" doesn't. Neutrinos start as one particle (say, as muon-neutrinos) and are detected as a completely different particle (say, as a tau-neutrino).
The explanation for that is that what we call "electron-neutrino", "muon-neutrino" and "tau-neutrino" aren't states with a definite mass; they're a mixture of three neutrino states with definite, different mass (one of those masses can be zero, but at most one). Then, from pure quantum mechanics (and nothing more esoteric than that: pure Schrödinger equation) you see that, if those three defined-mass states have slightly different mass, you will have a probability of creating an electron neutrino and detecting it as a tau neutrino, and every other combination. Those probabilities follow a simple expansion, based on only five parameters (two mass differences and three angles), and depend on the energy of the neutrino and the distance in a very specific way. We can test that dependency, and use very different experiments to measure the five parameters; and everything fits very well. Right now (specially after MINOS saw the energy dependency of the oscillation probability), nobody questions neutrino oscillations. This OPERA result only confirms what we already knew.
No, they're not the same. Mass-induced oscillation is a known fact in particle physics (search for "neutron kaon oscillation" for background), and neutrinos behave in exactly the predicted way; only with big mixing angles, unlike the almost-zero angles in the quark sector's CKM matrix.
The point is that, if two different theories have the exact same predictions, they are for all intents and purposes the same theory, and describe the same universe. If that is the case, why would you spend more time teaching and learning the more complex one, when a simple explanation is enough and (by definition, since they have the same predictions) you can't tell which one is correct?
Of course, if the new theory offers a good explanation to current data, but has a different prediction than the standard model in other, still-non-tested scenarios, the theory is more interesting. You can test it at the new scenario, and you'll be able to tell them apart. This is why* we study, for example, supersymmetry and extra dimensions theories: they behave just like the standard model where we have tested it, but can be different in other cases such as the LHC.
* = of course there are other motivations to develop the theories, but they are taken seriously because they are compatible with the SM and are testable. A theory whose predictions were exactly the same as the SM for every case wouldn't be worth studying, simply because you'd never be able to see if it is right.
Thanks, good catch.
No, it may not. The neutrino's energy (which is the exact analog to a photon's "colour") is conserved. The rest energy changes, but this means the original neutrino and the resulting one are on different rest referentials (and the change in mass is very small, inside the uncertainty principle). There is nothing "yet to be discovered", except possibly what is the mass generation mechanism (Dirac, like all other particles, or Marjorana).
"Chameleon-like behaviour" is just the "scientific" journalist's way of saying he understands nothing about what he is paid to write; but we all know journalists rarely know much of what they write anyway.
You'd need a pretty complex theory to get non-mass oscillations to match all the data we got over the past 12 years, which is very compatible with a three-state, mass-driven oscillation scenario. Besides, you'd have to explain more than what the current "new standard model" (the SM with added neutrino masses) does if you want your theory to be accepted. If two theories explain the same data equally well, the simplest is more likely.