10 ms·

Okay. This should be an easy one but somehow I'm getting stumped. I understand the difference in frames between the GPS satellites and the ground, but the sats

Okay. This should be an easy one but somehow I'm getting stumped. I understand the difference in frames between the GPS satellites and the ground, but the sats themselves are fixed to each other, right? And the ground stations are also fixed to each other. Each pair is in a separate frame. But the measurement was on the ground, and the ground stations are not accelerating relative to each other, not from the satellites. So is this saying that the ground stations set their clocks initially wrong because of their relative movement to the satellites? If so, wouldn't this be proven out by comparing the neutrinos time to the time of a photon?

InclinedPlane 15y ago
I haven't read the article yet. However, the ground stations are deeper in the Earth's gravity well than the satellites. That is equivalent to acceleration in relativity.
- route66 15y ago
  And the ground stations have a different potential with respect to each other, so their reading of elapsed time could differ. That's what I understood from another article on that subject.
- lutorm 15y ago
  That's negligible.
  - hebejebelus 15y ago
    When it comes to 60 nanoseconds, I'm not sure anything is negligible.
    - VladRussian 15y ago
      The GR effects here is on the scale of 1e-15s, i.e. several orders of magnitude less than observed 6e-8s. 7 orders of magnitude is "negligible" as GP correctly stated.
      - mrb 15y ago
        Actually gravitational time dilation on Earth's surface is on the order of 1e-9. But still negligible.
        VladRussian 15y ago
        >Actually gravitational time dilation on Earth's surface is on the order of 1e-9. But still negligible. it is compare to zero gravitation far from Earth. The effect i'm talking about is integral of gravitational dilation change along the path of neutrinos. This path is a chord under Earth surface - the 350 km down to 10km depth and the next 350km bringing back to surface. The dilation change is about 1e-16 per meter of depth.
- pmjordan 15y ago
  Gravity is not accounted for in special relativity, only general relativity, and this argument is purely based on special relativity. I can't speculate quantitatively on the magnitude of gen. rel. effects, but given the short times, high neutrino speeds and low gravity I suspect they're dwarfed by the already minor effect discussed here.
themgt 15y ago
The neutrino is going through the earth - there's no tunnel - so there's no way to compare to a photon. Basically Gran Sasso is moving towards the neutrino as it travels, from the perspective of the GPS
cube13 15y ago
>But the measurement was on the ground, and the ground stations are not accelerating relative to each other, not from the satellites. So is this saying that the ground stations set their clocks initially wrong because of their relative movement to the satellites? If so, wouldn't this be proven out by comparing the neutrinos time to the time of a photon? If so, wouldn't this be proven out by comparing the neutrinos time to the time of a photon? Actually, the satellites and ground stations are accelerating relative to each other, since they're moving in(roughly) circular paths. Acceleration only means that the velocity vector is changing relative to the frame of reference. It doesn't mean that the absolute velocity is changing. The problem is that the GPS satellites are the ones keeping time, not the ground. The ground stations are effectively "reading" the clocks on the satellites. So all of the relativistic time calculations should be done from the satellites' frame of reference, rather than from the ground stations' frame. The satellites are moving from west to east(with the Earth's rotation) with an orbital period of approximately 12 hours(http://en.wikipedia.org/wiki/Satellite_navigation#Comparison_of_systems http://en.wikipedia.org/wiki/Satellite_navigation#Comparison...). Their rotational velocity, relative to the center of the Earth, is twice that of the ground stations. In terms of absolute velocity, they're moving quite a bit faster. Since they are doing the timekeeping(and assuming that neutrinos move at c regardless of the frame of reference), the receiving ground station is moving towards the initial origin point of the neutrino stream. At that point, it's a doppler effect calculation taking into account the relative velocity of the ground stations to the satellite's frame of reference to find the offset. Also, the photon time measurement isn't possible due to the curvature of the Earth and the fact that the receiving end is underground.
dmm 15y ago
> but the sats themselves are fixed to each other, right? Whoa, why would you think that? GPS satellites are in 12-hr, medium-earth, circular orbits in six-orbital planes. How would satellites in different orbital planes maintain fixed distances from each other?
lutorm 15y ago
I think this is the issue: The time is given by the GPS satellites sending out a signal. This signal is received by the two ground stations, who reset their time accordingly. This means that the two ground station clocks will define two events as simultaneous in the frame of the satellite. But the measurement is done in the ground frame. Two events that are simultaneous in the satellite frame are not simultaneous in the ground frame, due to the relative motion of the two frames. In the ground frame, there will be an offset between the two clocks, and that would, unless corrected for, result in a skewed measurement of the time of flight. In reality it's more complicated, since there is no one "satellite" frame. The GPS system determines position and receiver time from a global solution to all the satellites. Since GPS time is claimed to be accurate to something like 10ns, it seems the only way this would be possible is if the GPS solution has been constructed to specifically refer the frame of the observer, which would be a ground stationary frame. Is there someone with more intimate knowledge of the GPS algorithm that might be able to comment on this?
- bermanoid 15y ago
  I'm super-skeptical about this paper - I agree that more info on the GPS stuff is needed (this author seems to be making assumptions about how the correction procedures work that I couldn't find documentation for), but on its face, this just seems way like a way too obvious error to possibly be right. First off, I encourage everyone to read the actual paper at http://arxiv.org/pdf/1110.2685v1 http://arxiv.org/pdf/1110.2685v1 [PDF], I think many here are seriously overestimating the complexity of the author's claim. I've mostly said this already on this thread, but it's buried near the bottom of the page, so I'll summarize the paper here, it's extremely simple: 0) If you're measuring time in the reference frame of the moving clock, you've got to transform the length between the stations in accord with the Lorentz transformation. According to this author CERN already did this, so it's entirely irrelevant to the discussion - I couldn't verify this, because it's not mentioned in the CERN paper he referenced. 1) What they did not do (according to this paper) is account for the fact that in the satellite frame, during the neutrino's flight the receiver station was moving towards the source. 2) time = distance / rate 3) CERN used rate = c, but they should have used rate = c + v because the receiver station was moving towards the neutrino. That is quite literally the entire thing, check the math in the paper if you don't believe me. This is not a claim that CERN somehow misapplied special relativity, it's a claim that they simply forgot to include the motion of the receiver. IMO this is far too elementary a mistake to have been missed for this long, so I'm suspicious. The CERN paper that the author references (http://arxiv.org/pdf/1109.4897v1 http://arxiv.org/pdf/1109.4897v1) doesn't discuss relativistic corrections at all, so I don't know why the author assumes this mistake has been made, perhaps he's reading another paper that he didn't reference? From the OPERA one that he links (though I haven't read it in too much detail) I get the sense that if it was this simple, one of the many calibration tests that they've done would have easily caught the error (they talk about a lot of bidirectional synchronization stuff), so for now, at least, I would take this with a grain of salt.
pmjordan 15y ago
Disclaimer: It's been a few years since uni, and I haven't read the actual paper, just the linked article. From what I can tell, the quoted passage, "From the perspective of the clock, the detector is moving towards the source and consequently the distance travelled by the particles as observed from the clock is shorter," is the core of the argument. I think this is what you need to consider/read into it: - There is relative rotation involved, so the frames are accelerating with respect to each other. - Since the times involved are very short with respect to the period of rotation, we can probably ignore the acceleration itself (?), BUT: - LHC and GS have differing velocity vectors at any given time due to their difference in longitude. This means we have to treat them as separate frames of reference. ("From the perspective of the clock, the detector is moving towards the source") - I think this is just about a single satellite (though since the satellites are in orbit, they too are accelerating wrt earth) - I'm not sure if the effect will vary slightly depending on the satellite's position, but as it has to be "visible" from both locations, the range of possible positions is fairly constrained. I'd have to work it through/read the paper. :-P And yeah, comparing to the time of a photon is going to be difficult through the earth. Even gamma radiation won't make it.
- VladRussian 15y ago
  >And yeah, comparing to the time of a photon is going to be difficult through the earth. Even gamma radiation won't make it. so instead of comparing to photon going through the fiber on the surface we're comparing against photon(s) going to the satellite and back. Even just using HAM radio bouncing from ionosphere would be less complicated and more straightforward than through GPS satellites here.
  - nitrogen 15y ago
    1. The distance traveled by a photon going through a fiber optic cable will be different from the distance through the ground. 2. The distance traveled by radio waves bouncing off the ionosphere would have much more uncertainty than the ~60 feet in question. 3. Radio waves in the air and photons in a fiber will move noticeably slower than c due to the fact that they are not traveling through a vacuum. Neutrinos, however, minimally interact with matter, so they would travel faster than the test photon(s).

InclinedPlane 15y ago

I haven't read the article yet. However, the ground stations are deeper in the Earth's gravity well than the satellites. That is equivalent to acceleration in relativity.

route66 15y ago
And the ground stations have a different potential with respect to each other, so their reading of elapsed time could differ. That's what I understood from another article on that subject.
lutorm 15y ago
That's negligible.
- hebejebelus 15y ago
  When it comes to 60 nanoseconds, I'm not sure anything is negligible.
  - VladRussian 15y ago
    The GR effects here is on the scale of 1e-15s, i.e. several orders of magnitude less than observed 6e-8s. 7 orders of magnitude is "negligible" as GP correctly stated.
    - mrb 15y ago
      Actually gravitational time dilation on Earth's surface is on the order of 1e-9. But still negligible.
      - VladRussian 15y ago
        >Actually gravitational time dilation on Earth's surface is on the order of 1e-9. But still negligible. it is compare to zero gravitation far from Earth. The effect i'm talking about is integral of gravitational dilation change along the path of neutrinos. This path is a chord under Earth surface - the 350 km down to 10km depth and the next 350km bringing back to surface. The dilation change is about 1e-16 per meter of depth.
pmjordan 15y ago
Gravity is not accounted for in special relativity, only general relativity, and this argument is purely based on special relativity. I can't speculate quantitatively on the magnitude of gen. rel. effects, but given the short times, high neutrino speeds and low gravity I suspect they're dwarfed by the already minor effect discussed here.

themgt 15y ago

The neutrino is going through the earth - there's no tunnel - so there's no way to compare to a photon. Basically Gran Sasso is moving towards the neutrino as it travels, from the perspective of the GPS

cube13 15y ago

>But the measurement was on the ground, and the ground stations are not accelerating relative to each other, not from the satellites. So is this saying that the ground stations set their clocks initially wrong because of their relative movement to the satellites? If so, wouldn't this be proven out by comparing the neutrinos time to the time of a photon? If so, wouldn't this be proven out by comparing the neutrinos time to the time of a photon? Actually, the satellites and ground stations are accelerating relative to each other, since they're moving in(roughly) circular paths. Acceleration only means that the velocity vector is changing relative to the frame of reference. It doesn't mean that the absolute velocity is changing. The problem is that the GPS satellites are the ones keeping time, not the ground. The ground stations are effectively "reading" the clocks on the satellites. So all of the relativistic time calculations should be done from the satellites' frame of reference, rather than from the ground stations' frame. The satellites are moving from west to east(with the Earth's rotation) with an orbital period of approximately 12 hours(http://en.wikipedia.org/wiki/Satellite_navigation#Comparison_of_systems http://en.wikipedia.org/wiki/Satellite_navigation#Comparison...). Their rotational velocity, relative to the center of the Earth, is twice that of the ground stations. In terms of absolute velocity, they're moving quite a bit faster. Since they are doing the timekeeping(and assuming that neutrinos move at c regardless of the frame of reference), the receiving ground station is moving towards the initial origin point of the neutrino stream. At that point, it's a doppler effect calculation taking into account the relative velocity of the ground stations to the satellite's frame of reference to find the offset. Also, the photon time measurement isn't possible due to the curvature of the Earth and the fact that the receiving end is underground.

dmm 15y ago

> but the sats themselves are fixed to each other, right? Whoa, why would you think that? GPS satellites are in 12-hr, medium-earth, circular orbits in six-orbital planes. How would satellites in different orbital planes maintain fixed distances from each other?

lutorm 15y ago

I think this is the issue: The time is given by the GPS satellites sending out a signal. This signal is received by the two ground stations, who reset their time accordingly. This means that the two ground station clocks will define two events as simultaneous in the frame of the satellite. But the measurement is done in the ground frame. Two events that are simultaneous in the satellite frame are not simultaneous in the ground frame, due to the relative motion of the two frames. In the ground frame, there will be an offset between the two clocks, and that would, unless corrected for, result in a skewed measurement of the time of flight. In reality it's more complicated, since there is no one "satellite" frame. The GPS system determines position and receiver time from a global solution to all the satellites. Since GPS time is claimed to be accurate to something like 10ns, it seems the only way this would be possible is if the GPS solution has been constructed to specifically refer the frame of the observer, which would be a ground stationary frame. Is there someone with more intimate knowledge of the GPS algorithm that might be able to comment on this?

bermanoid 15y ago
I'm super-skeptical about this paper - I agree that more info on the GPS stuff is needed (this author seems to be making assumptions about how the correction procedures work that I couldn't find documentation for), but on its face, this just seems way like a way too obvious error to possibly be right. First off, I encourage everyone to read the actual paper at http://arxiv.org/pdf/1110.2685v1 http://arxiv.org/pdf/1110.2685v1 [PDF], I think many here are seriously overestimating the complexity of the author's claim. I've mostly said this already on this thread, but it's buried near the bottom of the page, so I'll summarize the paper here, it's extremely simple: 0) If you're measuring time in the reference frame of the moving clock, you've got to transform the length between the stations in accord with the Lorentz transformation. According to this author CERN already did this, so it's entirely irrelevant to the discussion - I couldn't verify this, because it's not mentioned in the CERN paper he referenced. 1) What they did not do (according to this paper) is account for the fact that in the satellite frame, during the neutrino's flight the receiver station was moving towards the source. 2) time = distance / rate 3) CERN used rate = c, but they should have used rate = c + v because the receiver station was moving towards the neutrino. That is quite literally the entire thing, check the math in the paper if you don't believe me. This is not a claim that CERN somehow misapplied special relativity, it's a claim that they simply forgot to include the motion of the receiver. IMO this is far too elementary a mistake to have been missed for this long, so I'm suspicious. The CERN paper that the author references (http://arxiv.org/pdf/1109.4897v1 http://arxiv.org/pdf/1109.4897v1) doesn't discuss relativistic corrections at all, so I don't know why the author assumes this mistake has been made, perhaps he's reading another paper that he didn't reference? From the OPERA one that he links (though I haven't read it in too much detail) I get the sense that if it was this simple, one of the many calibration tests that they've done would have easily caught the error (they talk about a lot of bidirectional synchronization stuff), so for now, at least, I would take this with a grain of salt.

pmjordan 15y ago

Disclaimer: It's been a few years since uni, and I haven't read the actual paper, just the linked article. From what I can tell, the quoted passage, "From the perspective of the clock, the detector is moving towards the source and consequently the distance travelled by the particles as observed from the clock is shorter," is the core of the argument. I think this is what you need to consider/read into it: - There is relative rotation involved, so the frames are accelerating with respect to each other. - Since the times involved are very short with respect to the period of rotation, we can probably ignore the acceleration itself (?), BUT: - LHC and GS have differing velocity vectors at any given time due to their difference in longitude. This means we have to treat them as separate frames of reference. ("From the perspective of the clock, the detector is moving towards the source") - I think this is just about a single satellite (though since the satellites are in orbit, they too are accelerating wrt earth) - I'm not sure if the effect will vary slightly depending on the satellite's position, but as it has to be "visible" from both locations, the range of possible positions is fairly constrained. I'd have to work it through/read the paper. :-P And yeah, comparing to the time of a photon is going to be difficult through the earth. Even gamma radiation won't make it.

VladRussian 15y ago
>And yeah, comparing to the time of a photon is going to be difficult through the earth. Even gamma radiation won't make it. so instead of comparing to photon going through the fiber on the surface we're comparing against photon(s) going to the satellite and back. Even just using HAM radio bouncing from ionosphere would be less complicated and more straightforward than through GPS satellites here.
- nitrogen 15y ago
  1. The distance traveled by a photon going through a fiber optic cable will be different from the distance through the ground. 2. The distance traveled by radio waves bouncing off the ionosphere would have much more uncertainty than the ~60 feet in question. 3. Radio waves in the air and photons in a fiber will move noticeably slower than c due to the fact that they are not traveling through a vacuum. Neutrinos, however, minimally interact with matter, so they would travel faster than the test photon(s).