4 ms·

My knowledge of physics is only very basic but it's still fun to ponder these things. Do they need to factor in that the endpoints are rotating with the earth

My knowledge of physics is only very basic but it's still fun to ponder these things. Do they need to factor in that the endpoints are rotating with the earth and the neutrinos are not?

breadbox 15y ago
I'm pretty sure I remember that being on the list of things that were double-checked. The sense I've gotten is that the most likely culprit is some sort of systemic error in the software somewhere, as that stuff is the most labor-intensive part of the experimental setup to analyze for errors.
- Avshalom 15y ago
  While it may have been checked it doesn't seem to be mentioned in the paper http://arxiv.org/abs/1109.4897v2 http://arxiv.org/abs/1109.4897v2
westbywest 15y ago
I was explaining to a friend my understanding that the emitter and receiver for these neutrino experiments basically needed to be synchronized to a single, fictional (superluminal, etc.) frame of reference. I'm positive there are many factors going into how this is done.
nikcub 15y ago
At their latitude (~45 deg), the distance the earth travels in the 2ms it takes light to travel the distance is around 76cm (someone check my calc if you will). They know the distance to within +-20cm, so the 76cm of Earth's rotation would be very significant (but not significant enough, again with my rough calculations, to bring these results back in line with c). The distance is significant enough that to get an accurate measure of c over that distance you would have to take it into account. ie. it would be obvious
- Sapient 15y ago
  Since the rotation of the earth has to be factored in, does the earth's orbit around the sun need to be as well? How about the suns orbit around the Galactic center? I suppose my question is - all things being relative, why aren't the neutrinos bound to the same frame as the emitter and detector?
  - lisper 15y ago
    You're thinking of the wrong kind of relativity. The neutrinos aren't "bound" to any frame. They just move from A to B. The (apparent) distance between A and B changes depending on the observer's frame of reference, as does the (apparent) time it takes them to get from A to B. The faster the observer goes, the shorter the apparent distance (Lorentz contraction) and the smaller the apparent time (time dilation) become. But the RATIO of the apparent distance from A to B to the apparent time it takes light to travel from A to B remains constant. So if the neutrinos are moving faster than light, they will appear to do so as measured from any frame of reference as long as you're consistent about using the same frame to measure time and distance. THAT is where it gets tricky because the GPS satellites that they use to establish both time and distance are all moving in different frames. You also have to take gravity into account because the emitter and detector are non-intertial frames.
    - Sapient 15y ago
      Thanks for taking the time to answer! You have helped me understand a few things better. However I also realise now that I phrased my question wrongly and should never have used the word "bound". I also see now that I misread the grand-parent post. Of course these motions would not affect the measurements, and if they did, the rotation of the earth alone would be obvious.
    - ced 15y ago
      By "frame of reference", I know you meant "inertial frame of reference", but of course the Earth moving around the sun and the solar system moving in the galaxy are not inertial. My weight is a bit stronger during the day than during the night because of the centrifugal force, isn't it? I'm sure that the effects are tiny for the LHC, but are they accounted for at all?

My knowledge of physics is only very basic but it's still fun to ponder these things. Do they need to factor in that the endpoints are rotating with the earth and the neutrinos are not?

breadbox 15y ago

I'm pretty sure I remember that being on the list of things that were double-checked. The sense I've gotten is that the most likely culprit is some sort of systemic error in the software somewhere, as that stuff is the most labor-intensive part of the experimental setup to analyze for errors.

Avshalom 15y ago
While it may have been checked it doesn't seem to be mentioned in the paper http://arxiv.org/abs/1109.4897v2 http://arxiv.org/abs/1109.4897v2

westbywest 15y ago

I was explaining to a friend my understanding that the emitter and receiver for these neutrino experiments basically needed to be synchronized to a single, fictional (superluminal, etc.) frame of reference. I'm positive there are many factors going into how this is done.

nikcub 15y ago

At their latitude (~45 deg), the distance the earth travels in the 2ms it takes light to travel the distance is around 76cm (someone check my calc if you will). They know the distance to within +-20cm, so the 76cm of Earth's rotation would be very significant (but not significant enough, again with my rough calculations, to bring these results back in line with c). The distance is significant enough that to get an accurate measure of c over that distance you would have to take it into account. ie. it would be obvious

Sapient 15y ago
Since the rotation of the earth has to be factored in, does the earth's orbit around the sun need to be as well? How about the suns orbit around the Galactic center? I suppose my question is - all things being relative, why aren't the neutrinos bound to the same frame as the emitter and detector?
- lisper 15y ago
  You're thinking of the wrong kind of relativity. The neutrinos aren't "bound" to any frame. They just move from A to B. The (apparent) distance between A and B changes depending on the observer's frame of reference, as does the (apparent) time it takes them to get from A to B. The faster the observer goes, the shorter the apparent distance (Lorentz contraction) and the smaller the apparent time (time dilation) become. But the RATIO of the apparent distance from A to B to the apparent time it takes light to travel from A to B remains constant. So if the neutrinos are moving faster than light, they will appear to do so as measured from any frame of reference as long as you're consistent about using the same frame to measure time and distance. THAT is where it gets tricky because the GPS satellites that they use to establish both time and distance are all moving in different frames. You also have to take gravity into account because the emitter and detector are non-intertial frames.
  - Sapient 15y ago
    Thanks for taking the time to answer! You have helped me understand a few things better. However I also realise now that I phrased my question wrongly and should never have used the word "bound". I also see now that I misread the grand-parent post. Of course these motions would not affect the measurements, and if they did, the rotation of the earth alone would be obvious.
  - ced 15y ago
    By "frame of reference", I know you meant "inertial frame of reference", but of course the Earth moving around the sun and the solar system moving in the galaxy are not inertial. My weight is a bit stronger during the day than during the night because of the centrifugal force, isn't it? I'm sure that the effects are tiny for the LHC, but are they accounted for at all?