4 ms·

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 ce

by 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?
  - lisper 15y ago
    > you meant "inertial frame of reference" Actually no. The speed of light is constant in ALL frames of reference, whether inertial or not. The math just gets more complicated in non-inertial frames. (That's the difference between special and general relativity.) UPDATE: it's actually a little more complicated than that. See http://en.wikipedia.org/wiki/Propagation_of_light_in_non-inertial_reference_frames http://en.wikipedia.org/wiki/Propagation_of_light_in_non-ine... > I'm sure that the effects are tiny for the LHC Actually, gravity is a pretty significant factor. See e.g.: http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.html http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps....

by Sapient 15y ago

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?
- lisper 15y ago
  > you meant "inertial frame of reference" Actually no. The speed of light is constant in ALL frames of reference, whether inertial or not. The math just gets more complicated in non-inertial frames. (That's the difference between special and general relativity.) UPDATE: it's actually a little more complicated than that. See http://en.wikipedia.org/wiki/Propagation_of_light_in_non-inertial_reference_frames http://en.wikipedia.org/wiki/Propagation_of_light_in_non-ine... > I'm sure that the effects are tiny for the LHC Actually, gravity is a pretty significant factor. See e.g.: http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.html http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps....