Mixing
Why isn't the GPS location calculated from the Schwarzschild metric?
The GPS uses the flat space light propagation formula to calculate the distance from the source (the satellite) to the receiver (observer on Earth):
$$ d=c cdot Delta t$$
where $c$ is the speed of light in Minkowski vacuum, $Delta t$ is the difference between the times of emission and absorption of the signal (corrected for relativistic time dilations) and $d$ is the Euclidean distance. This formula is fed with the data beamed from 4 satellites to solve for the location of the receiver.
My questions are: what is the rationale for using this formula? Shouldn't the distance be calculated in the curved geometry setting, e.g. using the Schwarzschild metric? What are the errors in using the Euclidean version $ d=c cdot Delta t$?
N.B.: The time difference $Delta t$ contains relativistic corrections to times. However, it is not clear to me why it is correct to use the flat space (Minkowski) formula for light propagation with just the value of $Delta t$ amended to account for gravity.
Please, try to be as clear as possible and support your statements with calculations/derivations.
ADDENDUM: I found really good papers discussing in depth all the relativistic details and effects to GPS(-like) navigation in spacetime.
They are Thomas B. Bahder's Navigation in Curved Space-Time, Clock Synchronization and Navigation in the Vicinity of the Earth and Relativity of GPS Measurement.
general-relativity special-relativity metric-tensor distance gps
asked Nov 21 '18 at 4:38
label
19828
add a comment |
The GPS uses the flat space light propagation formula to calculate the distance from the source (the satellite) to the receiver (observer on Earth):
$$ d=c cdot Delta t$$
where $c$ is the speed of light in Minkowski vacuum, $Delta t$ is the difference between the times of emission and absorption of the signal (corrected for relativistic time dilations) and $d$ is the Euclidean distance. This formula is fed with the data beamed from 4 satellites to solve for the location of the receiver.
My questions are: what is the rationale for using this formula? Shouldn't the distance be calculated in the curved geometry setting, e.g. using the Schwarzschild metric? What are the errors in using the Euclidean version $ d=c cdot Delta t$?
N.B.: The time difference $Delta t$ contains relativistic corrections to times. However, it is not clear to me why it is correct to use the flat space (Minkowski) formula for light propagation with just the value of $Delta t$ amended to account for gravity.
Please, try to be as clear as possible and support your statements with calculations/derivations.
ADDENDUM: I found really good papers discussing in depth all the relativistic details and effects to GPS(-like) navigation in spacetime.
They are Thomas B. Bahder's Navigation in Curved Space-Time, Clock Synchronization and Navigation in the Vicinity of the Earth and Relativity of GPS Measurement.
general-relativity special-relativity metric-tensor distance gps
asked Nov 21 '18 at 4:38
label
19828
9
Earth's local spacetime isn't even schwarzchild if you want to talk precision, due to Earth's oblate nature there are higher multipole moments to take into account for the gravitational field. You might be interested in this article ncbi.nlm.nih.gov/pmc/articles/PMC5253894
– Triatticus
Nov 21 '18 at 5:03
5
This is not only a physics question but an electronics one (and I do mean electronics, not programming). The early receivers were designed to work using the technology of the 1980s or earlier. They needed to be portable, which might mean vehicle mounted rather than handheld, which would limit the size and power supply that coule be used. That all impacts the recommended calculations. Modern recievers are welcome to calculate position however they please, and survey ones may include some of what you suggest to get their millimetre precisions.
– TafT
Nov 21 '18 at 11:42
add a comment |
The GPS uses the flat space light propagation formula to calculate the distance from the source (the satellite) to the receiver (observer on Earth):
$$ d=c cdot Delta t$$
where $c$ is the speed of light in Minkowski vacuum, $Delta t$ is the difference between the times of emission and absorption of the signal (corrected for relativistic time dilations) and $d$ is the Euclidean distance. This formula is fed with the data beamed from 4 satellites to solve for the location of the receiver.
My questions are: what is the rationale for using this formula? Shouldn't the distance be calculated in the curved geometry setting, e.g. using the Schwarzschild metric? What are the errors in using the Euclidean version $ d=c cdot Delta t$?
N.B.: The time difference $Delta t$ contains relativistic corrections to times. However, it is not clear to me why it is correct to use the flat space (Minkowski) formula for light propagation with just the value of $Delta t$ amended to account for gravity.
Please, try to be as clear as possible and support your statements with calculations/derivations.
ADDENDUM: I found really good papers discussing in depth all the relativistic details and effects to GPS(-like) navigation in spacetime.
They are Thomas B. Bahder's Navigation in Curved Space-Time, Clock Synchronization and Navigation in the Vicinity of the Earth and Relativity of GPS Measurement.
general-relativity special-relativity metric-tensor distance gps
asked Nov 21 '18 at 4:38
label
19828
The GPS uses the flat space light propagation formula to calculate the distance from the source (the satellite) to the receiver (observer on Earth):
$$ d=c cdot Delta t$$
where $c$ is the speed of light in Minkowski vacuum, $Delta t$ is the difference between the times of emission and absorption of the signal (corrected for relativistic time dilations) and $d$ is the Euclidean distance. This formula is fed with the data beamed from 4 satellites to solve for the location of the receiver.
My questions are: what is the rationale for using this formula? Shouldn't the distance be calculated in the curved geometry setting, e.g. using the Schwarzschild metric? What are the errors in using the Euclidean version $ d=c cdot Delta t$?
N.B.: The time difference $Delta t$ contains relativistic corrections to times. However, it is not clear to me why it is correct to use the flat space (Minkowski) formula for light propagation with just the value of $Delta t$ amended to account for gravity.
Please, try to be as clear as possible and support your statements with calculations/derivations.
ADDENDUM: I found really good papers discussing in depth all the relativistic details and effects to GPS(-like) navigation in spacetime.
They are Thomas B. Bahder's Navigation in Curved Space-Time, Clock Synchronization and Navigation in the Vicinity of the Earth and Relativity of GPS Measurement.
general-relativity special-relativity metric-tensor distance gps
general-relativity special-relativity metric-tensor distance gps
asked Nov 21 '18 at 4:38
label
19828
asked Nov 21 '18 at 4:38
label
19828
edited Nov 21 '18 at 20:32
asked Nov 21 '18 at 4:38
label
19828
asked Nov 21 '18 at 4:38
label
19828
asked Nov 21 '18 at 4:38
label
19828
19828
9
Earth's local spacetime isn't even schwarzchild if you want to talk precision, due to Earth's oblate nature there are higher multipole moments to take into account for the gravitational field. You might be interested in this article ncbi.nlm.nih.gov/pmc/articles/PMC5253894
– Triatticus
Nov 21 '18 at 5:03
5
This is not only a physics question but an electronics one (and I do mean electronics, not programming). The early receivers were designed to work using the technology of the 1980s or earlier. They needed to be portable, which might mean vehicle mounted rather than handheld, which would limit the size and power supply that coule be used. That all impacts the recommended calculations. Modern recievers are welcome to calculate position however they please, and survey ones may include some of what you suggest to get their millimetre precisions.
– TafT
Nov 21 '18 at 11:42
add a comment |
9
Earth's local spacetime isn't even schwarzchild if you want to talk precision, due to Earth's oblate nature there are higher multipole moments to take into account for the gravitational field. You might be interested in this article ncbi.nlm.nih.gov/pmc/articles/PMC5253894
– Triatticus
Nov 21 '18 at 5:03
5
This is not only a physics question but an electronics one (and I do mean electronics, not programming). The early receivers were designed to work using the technology of the 1980s or earlier. They needed to be portable, which might mean vehicle mounted rather than handheld, which would limit the size and power supply that coule be used. That all impacts the recommended calculations. Modern recievers are welcome to calculate position however they please, and survey ones may include some of what you suggest to get their millimetre precisions.
– TafT
Nov 21 '18 at 11:42
9
9
Earth's local spacetime isn't even schwarzchild if you want to talk precision, due to Earth's oblate nature there are higher multipole moments to take into account for the gravitational field. You might be interested in this article ncbi.nlm.nih.gov/pmc/articles/PMC5253894
– Triatticus
Nov 21 '18 at 5:03
Earth's local spacetime isn't even schwarzchild if you want to talk precision, due to Earth's oblate nature there are higher multipole moments to take into account for the gravitational field. You might be interested in this article ncbi.nlm.nih.gov/pmc/articles/PMC5253894
– Triatticus
Nov 21 '18 at 5:03
5
5
This is not only a physics question but an electronics one (and I do mean electronics, not programming). The early receivers were designed to work using the technology of the 1980s or earlier. They needed to be portable, which might mean vehicle mounted rather than handheld, which would limit the size and power supply that coule be used. That all impacts the recommended calculations. Modern recievers are welcome to calculate position however they please, and survey ones may include some of what you suggest to get their millimetre precisions.
– TafT
Nov 21 '18 at 11:42
This is not only a physics question but an electronics one (and I do mean electronics, not programming). The early receivers were designed to work using the technology of the 1980s or earlier. They needed to be portable, which might mean vehicle mounted rather than handheld, which would limit the size and power supply that coule be used. That all impacts the recommended calculations. Modern recievers are welcome to calculate position however they please, and survey ones may include some of what you suggest to get their millimetre precisions.
– TafT
Nov 21 '18 at 11:42
add a comment |
2 Answers
2
active
oldest
votes
The general-relativistic corrections are too small to matter.
The Schwarzchild metric has dimensionless corrections of order $GM/rc^2$. Here $G$ is the Newton's gravitational constant, $M$ the mass of the Earth, $r$ the distance from the center of the Earth, and $c$ the speed of light.
At the surface of the Earth, these metric corrections are about one part in a billion; higher up, near the satellites, they are even smaller. The Christoffel symbols determining the signal's geodesic path will have corrections of the same magnitude.
The signal takes about 0.1 s to travel to Earth from the satellite, so the GR correction in $Delta t$ would be of order $10^{-10} s$ and the GR correction in $d$ would be approximately 3 cm. This is below the accuracy of the GPS system.
The case where the GPS satellite is directly overhead is easy to solve analytically. Start with the Schwartzschild metric
$$ds^2=-(1-2M/r)dt^2+(1-2M/r)^{-1}dr^2 +r^2 dtheta^2+r^2 sin^2{theta};dphi^2$$
in geometrical units where $G$ and $c$ are 1.
The signal follows a null geodesic where $ds=0$. A radial null geodesic satisfies
$$(1-2M/r);dt^2=(1-2M/r)^{-1};dr^2$$
which is a differential equation from which we can obtain $t(r)$ as
$$t=r_0-r+2Mlogfrac{r_0-2M}{r-2M}.$$
The initial conditions are that at $t=0$ the signal starts at $r=r_0$, the orbital radius of the overhead GPS satellite. We have taken the incoming solution; as $t$ increases, $r$ decreases and at some time $t=t_E$ it hits the GPS receiver on the surface of the Earth at $r=R_E$.
For calculating $t_E$ in seconds, restore $G$ and $c$ to get
$$ct_E=r_0-r_E+R_slogfrac{r_0-R_s}{r_E-R_s}$$
where $R_s=2GM/c^2$ is the Schwarzschild radius of the Earth, which is 9.0 mm.
Putting in the radius at which the GPS satellites orbit, $r_0=20,000$ km, and the Earth's radius $r_E=6400$ km, we find $t_E=0.045333333368$ s. When we ignore the GR corrections by taking $R_s$ to be 0 rather than 9 mm, we get $t_E=0.045333333333$ s. Thus the GR corrections slow the signal by 34 picoseconds, and cause the calculation of the distance to the satellite to be off by 1.0 cm. A good analytic approximation is
$$Delta d=R_s logfrac{r_0}{r_E}.$$
Correction: The OP pointed out that 20,000 km is the altitude of the GPS satellites, not their orbital radius. Their orbital radius is thus about 26,400 km. Redoing the numbers, I get a $Delta t$ of 43 picoseconds and a $Delta d$ of 1.3 cm.
edited Nov 21 '18 at 23:49
user2357112
1053
answered Nov 21 '18 at 4:56
G. Smith
4,512919
4
For comparison, the errors on any individual pseudorange due to signal resolution, ionospheric effects, thermal noise, imperfect clocks on the satellites, imperfect ephemerides, etc. under good conditions are around 10 nanoseconds, ~30 times larger.
– hobbs
Nov 21 '18 at 15:55
10
What's surprising to me is not that the effect is insignificant, but how close it is to being significant. A decent commercial GPS receiver can be accurate to less than one meter under good conditions, so an error of 1 cm due to GR is only two orders of magnitude below that. I can easily imagine a future GPS-like system being accurate enough for those corrections to be needed.
– Ilmari Karonen
Nov 21 '18 at 18:46
3
To be clear, this specific general relativistic correction is too small to matter. The clock drift that would occur without general relativistic corrections is much more significant.
– Chris
Nov 21 '18 at 22:28
2
The general-relativistic correction to the satellites' clock rates is not a matter of "imperfect clocks" -- gravitational time dilation is straight from GR. But it's not something GPS receivers need to care about, because the signal is already corrected for it when it is generated on the satellite.
– Henning Makholm
Nov 22 '18 at 1:31
3
@IlmariKaronen cm-scale GPS receiver systems do exist, but fundamentally they operate based on measuring the signal at reference locations and then sending out a correction, either with a time error (pseudorange) or position error. This is known as dGPS and many systems exist (WAAS/SBAS, LAAS/GLS, Starfire, SAIF, etc). Therefore, there is no need to make a special correction for every single possible error source, instead measuring error on a per-satellite or per-location basis.
– user71659
Nov 22 '18 at 2:10
|
show 8 more comments
I have some general comments about principles; I haven't looked up specific numbers but these will be in the sources below.
Firstly, see Neil Ashby's (open access) Living Reviews in Relativity article "Relativity in the Global Positioning System". Ashby is the biggest name in this field.
Secondly, while the "exact solutions" like Schwarzschild are extremely important, such beautiful mathematical solutions are not always the most practical. For the Solar System generally, it is not described using Schwarzschild etc, but instead "post-Newtonian theory" which is a sort of midway approximation between Newton and Einstein. This is the official International Astronomical Union recommendation -- see Soffel et al (2003). For Earth specifically, the gravitational field is best modeled as an expansion in spherical harmonics. For example, for a future Chinese satellite for detecting gravitational waves, it will be done this way. See the textbook by Poisson & Will.
Finally, one might think that because the Earth is rotating, the Kerr metric might be a good approximation for say $r>6500km$. However the exterior of a rotating body does not match to the Kerr metric in this way. See also Hartle's textbook chapter 14 titled "A little rotation", for an intermediate approximation between Schwarzschild and Kerr.
answered Nov 28 '18 at 2:01
Colin MacLaurin
37711
add a comment |
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2 Answers
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2 Answers
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The general-relativistic corrections are too small to matter.
The Schwarzchild metric has dimensionless corrections of order $GM/rc^2$. Here $G$ is the Newton's gravitational constant, $M$ the mass of the Earth, $r$ the distance from the center of the Earth, and $c$ the speed of light.
At the surface of the Earth, these metric corrections are about one part in a billion; higher up, near the satellites, they are even smaller. The Christoffel symbols determining the signal's geodesic path will have corrections of the same magnitude.
The signal takes about 0.1 s to travel to Earth from the satellite, so the GR correction in $Delta t$ would be of order $10^{-10} s$ and the GR correction in $d$ would be approximately 3 cm. This is below the accuracy of the GPS system.
The case where the GPS satellite is directly overhead is easy to solve analytically. Start with the Schwartzschild metric
$$ds^2=-(1-2M/r)dt^2+(1-2M/r)^{-1}dr^2 +r^2 dtheta^2+r^2 sin^2{theta};dphi^2$$
in geometrical units where $G$ and $c$ are 1.
The signal follows a null geodesic where $ds=0$. A radial null geodesic satisfies
$$(1-2M/r);dt^2=(1-2M/r)^{-1};dr^2$$
which is a differential equation from which we can obtain $t(r)$ as
$$t=r_0-r+2Mlogfrac{r_0-2M}{r-2M}.$$
The initial conditions are that at $t=0$ the signal starts at $r=r_0$, the orbital radius of the overhead GPS satellite. We have taken the incoming solution; as $t$ increases, $r$ decreases and at some time $t=t_E$ it hits the GPS receiver on the surface of the Earth at $r=R_E$.
For calculating $t_E$ in seconds, restore $G$ and $c$ to get
$$ct_E=r_0-r_E+R_slogfrac{r_0-R_s}{r_E-R_s}$$
where $R_s=2GM/c^2$ is the Schwarzschild radius of the Earth, which is 9.0 mm.
Putting in the radius at which the GPS satellites orbit, $r_0=20,000$ km, and the Earth's radius $r_E=6400$ km, we find $t_E=0.045333333368$ s. When we ignore the GR corrections by taking $R_s$ to be 0 rather than 9 mm, we get $t_E=0.045333333333$ s. Thus the GR corrections slow the signal by 34 picoseconds, and cause the calculation of the distance to the satellite to be off by 1.0 cm. A good analytic approximation is
$$Delta d=R_s logfrac{r_0}{r_E}.$$
Correction: The OP pointed out that 20,000 km is the altitude of the GPS satellites, not their orbital radius. Their orbital radius is thus about 26,400 km. Redoing the numbers, I get a $Delta t$ of 43 picoseconds and a $Delta d$ of 1.3 cm.
edited Nov 21 '18 at 23:49
user2357112
1053
answered Nov 21 '18 at 4:56
G. Smith
4,512919
4
For comparison, the errors on any individual pseudorange due to signal resolution, ionospheric effects, thermal noise, imperfect clocks on the satellites, imperfect ephemerides, etc. under good conditions are around 10 nanoseconds, ~30 times larger.
– hobbs
Nov 21 '18 at 15:55
10
What's surprising to me is not that the effect is insignificant, but how close it is to being significant. A decent commercial GPS receiver can be accurate to less than one meter under good conditions, so an error of 1 cm due to GR is only two orders of magnitude below that. I can easily imagine a future GPS-like system being accurate enough for those corrections to be needed.
– Ilmari Karonen
Nov 21 '18 at 18:46
3
To be clear, this specific general relativistic correction is too small to matter. The clock drift that would occur without general relativistic corrections is much more significant.
– Chris
Nov 21 '18 at 22:28
2
The general-relativistic correction to the satellites' clock rates is not a matter of "imperfect clocks" -- gravitational time dilation is straight from GR. But it's not something GPS receivers need to care about, because the signal is already corrected for it when it is generated on the satellite.
– Henning Makholm
Nov 22 '18 at 1:31
3
@IlmariKaronen cm-scale GPS receiver systems do exist, but fundamentally they operate based on measuring the signal at reference locations and then sending out a correction, either with a time error (pseudorange) or position error. This is known as dGPS and many systems exist (WAAS/SBAS, LAAS/GLS, Starfire, SAIF, etc). Therefore, there is no need to make a special correction for every single possible error source, instead measuring error on a per-satellite or per-location basis.
– user71659
Nov 22 '18 at 2:10
|
show 8 more comments
The general-relativistic corrections are too small to matter.
The Schwarzchild metric has dimensionless corrections of order $GM/rc^2$. Here $G$ is the Newton's gravitational constant, $M$ the mass of the Earth, $r$ the distance from the center of the Earth, and $c$ the speed of light.
At the surface of the Earth, these metric corrections are about one part in a billion; higher up, near the satellites, they are even smaller. The Christoffel symbols determining the signal's geodesic path will have corrections of the same magnitude.
The signal takes about 0.1 s to travel to Earth from the satellite, so the GR correction in $Delta t$ would be of order $10^{-10} s$ and the GR correction in $d$ would be approximately 3 cm. This is below the accuracy of the GPS system.
The case where the GPS satellite is directly overhead is easy to solve analytically. Start with the Schwartzschild metric
$$ds^2=-(1-2M/r)dt^2+(1-2M/r)^{-1}dr^2 +r^2 dtheta^2+r^2 sin^2{theta};dphi^2$$
in geometrical units where $G$ and $c$ are 1.
The signal follows a null geodesic where $ds=0$. A radial null geodesic satisfies
$$(1-2M/r);dt^2=(1-2M/r)^{-1};dr^2$$
which is a differential equation from which we can obtain $t(r)$ as
$$t=r_0-r+2Mlogfrac{r_0-2M}{r-2M}.$$
The initial conditions are that at $t=0$ the signal starts at $r=r_0$, the orbital radius of the overhead GPS satellite. We have taken the incoming solution; as $t$ increases, $r$ decreases and at some time $t=t_E$ it hits the GPS receiver on the surface of the Earth at $r=R_E$.
For calculating $t_E$ in seconds, restore $G$ and $c$ to get
$$ct_E=r_0-r_E+R_slogfrac{r_0-R_s}{r_E-R_s}$$
where $R_s=2GM/c^2$ is the Schwarzschild radius of the Earth, which is 9.0 mm.
Putting in the radius at which the GPS satellites orbit, $r_0=20,000$ km, and the Earth's radius $r_E=6400$ km, we find $t_E=0.045333333368$ s. When we ignore the GR corrections by taking $R_s$ to be 0 rather than 9 mm, we get $t_E=0.045333333333$ s. Thus the GR corrections slow the signal by 34 picoseconds, and cause the calculation of the distance to the satellite to be off by 1.0 cm. A good analytic approximation is
$$Delta d=R_s logfrac{r_0}{r_E}.$$
Correction: The OP pointed out that 20,000 km is the altitude of the GPS satellites, not their orbital radius. Their orbital radius is thus about 26,400 km. Redoing the numbers, I get a $Delta t$ of 43 picoseconds and a $Delta d$ of 1.3 cm.
edited Nov 21 '18 at 23:49
user2357112
1053
answered Nov 21 '18 at 4:56
G. Smith
4,512919
4
For comparison, the errors on any individual pseudorange due to signal resolution, ionospheric effects, thermal noise, imperfect clocks on the satellites, imperfect ephemerides, etc. under good conditions are around 10 nanoseconds, ~30 times larger.
– hobbs
Nov 21 '18 at 15:55
10
What's surprising to me is not that the effect is insignificant, but how close it is to being significant. A decent commercial GPS receiver can be accurate to less than one meter under good conditions, so an error of 1 cm due to GR is only two orders of magnitude below that. I can easily imagine a future GPS-like system being accurate enough for those corrections to be needed.
– Ilmari Karonen
Nov 21 '18 at 18:46
3
To be clear, this specific general relativistic correction is too small to matter. The clock drift that would occur without general relativistic corrections is much more significant.
– Chris
Nov 21 '18 at 22:28
2
The general-relativistic correction to the satellites' clock rates is not a matter of "imperfect clocks" -- gravitational time dilation is straight from GR. But it's not something GPS receivers need to care about, because the signal is already corrected for it when it is generated on the satellite.
– Henning Makholm
Nov 22 '18 at 1:31
3
@IlmariKaronen cm-scale GPS receiver systems do exist, but fundamentally they operate based on measuring the signal at reference locations and then sending out a correction, either with a time error (pseudorange) or position error. This is known as dGPS and many systems exist (WAAS/SBAS, LAAS/GLS, Starfire, SAIF, etc). Therefore, there is no need to make a special correction for every single possible error source, instead measuring error on a per-satellite or per-location basis.
– user71659
Nov 22 '18 at 2:10
|
show 8 more comments
The general-relativistic corrections are too small to matter.
The Schwarzchild metric has dimensionless corrections of order $GM/rc^2$. Here $G$ is the Newton's gravitational constant, $M$ the mass of the Earth, $r$ the distance from the center of the Earth, and $c$ the speed of light.
At the surface of the Earth, these metric corrections are about one part in a billion; higher up, near the satellites, they are even smaller. The Christoffel symbols determining the signal's geodesic path will have corrections of the same magnitude.
The signal takes about 0.1 s to travel to Earth from the satellite, so the GR correction in $Delta t$ would be of order $10^{-10} s$ and the GR correction in $d$ would be approximately 3 cm. This is below the accuracy of the GPS system.
The case where the GPS satellite is directly overhead is easy to solve analytically. Start with the Schwartzschild metric
$$ds^2=-(1-2M/r)dt^2+(1-2M/r)^{-1}dr^2 +r^2 dtheta^2+r^2 sin^2{theta};dphi^2$$
in geometrical units where $G$ and $c$ are 1.
The signal follows a null geodesic where $ds=0$. A radial null geodesic satisfies
$$(1-2M/r);dt^2=(1-2M/r)^{-1};dr^2$$
which is a differential equation from which we can obtain $t(r)$ as
$$t=r_0-r+2Mlogfrac{r_0-2M}{r-2M}.$$
The initial conditions are that at $t=0$ the signal starts at $r=r_0$, the orbital radius of the overhead GPS satellite. We have taken the incoming solution; as $t$ increases, $r$ decreases and at some time $t=t_E$ it hits the GPS receiver on the surface of the Earth at $r=R_E$.
For calculating $t_E$ in seconds, restore $G$ and $c$ to get
$$ct_E=r_0-r_E+R_slogfrac{r_0-R_s}{r_E-R_s}$$
where $R_s=2GM/c^2$ is the Schwarzschild radius of the Earth, which is 9.0 mm.
Putting in the radius at which the GPS satellites orbit, $r_0=20,000$ km, and the Earth's radius $r_E=6400$ km, we find $t_E=0.045333333368$ s. When we ignore the GR corrections by taking $R_s$ to be 0 rather than 9 mm, we get $t_E=0.045333333333$ s. Thus the GR corrections slow the signal by 34 picoseconds, and cause the calculation of the distance to the satellite to be off by 1.0 cm. A good analytic approximation is
$$Delta d=R_s logfrac{r_0}{r_E}.$$
Correction: The OP pointed out that 20,000 km is the altitude of the GPS satellites, not their orbital radius. Their orbital radius is thus about 26,400 km. Redoing the numbers, I get a $Delta t$ of 43 picoseconds and a $Delta d$ of 1.3 cm.
edited Nov 21 '18 at 23:49
user2357112
1053
answered Nov 21 '18 at 4:56
G. Smith
4,512919
The general-relativistic corrections are too small to matter.
The Schwarzchild metric has dimensionless corrections of order $GM/rc^2$. Here $G$ is the Newton's gravitational constant, $M$ the mass of the Earth, $r$ the distance from the center of the Earth, and $c$ the speed of light.
At the surface of the Earth, these metric corrections are about one part in a billion; higher up, near the satellites, they are even smaller. The Christoffel symbols determining the signal's geodesic path will have corrections of the same magnitude.
The signal takes about 0.1 s to travel to Earth from the satellite, so the GR correction in $Delta t$ would be of order $10^{-10} s$ and the GR correction in $d$ would be approximately 3 cm. This is below the accuracy of the GPS system.
The case where the GPS satellite is directly overhead is easy to solve analytically. Start with the Schwartzschild metric
$$ds^2=-(1-2M/r)dt^2+(1-2M/r)^{-1}dr^2 +r^2 dtheta^2+r^2 sin^2{theta};dphi^2$$
in geometrical units where $G$ and $c$ are 1.
The signal follows a null geodesic where $ds=0$. A radial null geodesic satisfies
$$(1-2M/r);dt^2=(1-2M/r)^{-1};dr^2$$
which is a differential equation from which we can obtain $t(r)$ as
$$t=r_0-r+2Mlogfrac{r_0-2M}{r-2M}.$$
The initial conditions are that at $t=0$ the signal starts at $r=r_0$, the orbital radius of the overhead GPS satellite. We have taken the incoming solution; as $t$ increases, $r$ decreases and at some time $t=t_E$ it hits the GPS receiver on the surface of the Earth at $r=R_E$.
For calculating $t_E$ in seconds, restore $G$ and $c$ to get
$$ct_E=r_0-r_E+R_slogfrac{r_0-R_s}{r_E-R_s}$$
where $R_s=2GM/c^2$ is the Schwarzschild radius of the Earth, which is 9.0 mm.
Putting in the radius at which the GPS satellites orbit, $r_0=20,000$ km, and the Earth's radius $r_E=6400$ km, we find $t_E=0.045333333368$ s. When we ignore the GR corrections by taking $R_s$ to be 0 rather than 9 mm, we get $t_E=0.045333333333$ s. Thus the GR corrections slow the signal by 34 picoseconds, and cause the calculation of the distance to the satellite to be off by 1.0 cm. A good analytic approximation is
$$Delta d=R_s logfrac{r_0}{r_E}.$$
Correction: The OP pointed out that 20,000 km is the altitude of the GPS satellites, not their orbital radius. Their orbital radius is thus about 26,400 km. Redoing the numbers, I get a $Delta t$ of 43 picoseconds and a $Delta d$ of 1.3 cm.
edited Nov 21 '18 at 23:49
user2357112
1053
answered Nov 21 '18 at 4:56
G. Smith
4,512919
edited Nov 21 '18 at 23:49
user2357112
1053
edited Nov 21 '18 at 23:49
user2357112
1053
edited Nov 21 '18 at 23:49
user2357112
1053
1053
answered Nov 21 '18 at 4:56
G. Smith
4,512919
answered Nov 21 '18 at 4:56
G. Smith
4,512919
answered Nov 21 '18 at 4:56
G. Smith
4,512919
4,512919
4
For comparison, the errors on any individual pseudorange due to signal resolution, ionospheric effects, thermal noise, imperfect clocks on the satellites, imperfect ephemerides, etc. under good conditions are around 10 nanoseconds, ~30 times larger.
– hobbs
Nov 21 '18 at 15:55
10
What's surprising to me is not that the effect is insignificant, but how close it is to being significant. A decent commercial GPS receiver can be accurate to less than one meter under good conditions, so an error of 1 cm due to GR is only two orders of magnitude below that. I can easily imagine a future GPS-like system being accurate enough for those corrections to be needed.
– Ilmari Karonen
Nov 21 '18 at 18:46
3
To be clear, this specific general relativistic correction is too small to matter. The clock drift that would occur without general relativistic corrections is much more significant.
– Chris
Nov 21 '18 at 22:28
2
The general-relativistic correction to the satellites' clock rates is not a matter of "imperfect clocks" -- gravitational time dilation is straight from GR. But it's not something GPS receivers need to care about, because the signal is already corrected for it when it is generated on the satellite.
– Henning Makholm
Nov 22 '18 at 1:31
3
@IlmariKaronen cm-scale GPS receiver systems do exist, but fundamentally they operate based on measuring the signal at reference locations and then sending out a correction, either with a time error (pseudorange) or position error. This is known as dGPS and many systems exist (WAAS/SBAS, LAAS/GLS, Starfire, SAIF, etc). Therefore, there is no need to make a special correction for every single possible error source, instead measuring error on a per-satellite or per-location basis.
– user71659
Nov 22 '18 at 2:10
|
show 8 more comments
4
For comparison, the errors on any individual pseudorange due to signal resolution, ionospheric effects, thermal noise, imperfect clocks on the satellites, imperfect ephemerides, etc. under good conditions are around 10 nanoseconds, ~30 times larger.
– hobbs
Nov 21 '18 at 15:55
10
What's surprising to me is not that the effect is insignificant, but how close it is to being significant. A decent commercial GPS receiver can be accurate to less than one meter under good conditions, so an error of 1 cm due to GR is only two orders of magnitude below that. I can easily imagine a future GPS-like system being accurate enough for those corrections to be needed.
– Ilmari Karonen
Nov 21 '18 at 18:46
3
To be clear, this specific general relativistic correction is too small to matter. The clock drift that would occur without general relativistic corrections is much more significant.
– Chris
Nov 21 '18 at 22:28
2
The general-relativistic correction to the satellites' clock rates is not a matter of "imperfect clocks" -- gravitational time dilation is straight from GR. But it's not something GPS receivers need to care about, because the signal is already corrected for it when it is generated on the satellite.
– Henning Makholm
Nov 22 '18 at 1:31
3
@IlmariKaronen cm-scale GPS receiver systems do exist, but fundamentally they operate based on measuring the signal at reference locations and then sending out a correction, either with a time error (pseudorange) or position error. This is known as dGPS and many systems exist (WAAS/SBAS, LAAS/GLS, Starfire, SAIF, etc). Therefore, there is no need to make a special correction for every single possible error source, instead measuring error on a per-satellite or per-location basis.
– user71659
Nov 22 '18 at 2:10
4
4
For comparison, the errors on any individual pseudorange due to signal resolution, ionospheric effects, thermal noise, imperfect clocks on the satellites, imperfect ephemerides, etc. under good conditions are around 10 nanoseconds, ~30 times larger.
– hobbs
Nov 21 '18 at 15:55
For comparison, the errors on any individual pseudorange due to signal resolution, ionospheric effects, thermal noise, imperfect clocks on the satellites, imperfect ephemerides, etc. under good conditions are around 10 nanoseconds, ~30 times larger.
– hobbs
Nov 21 '18 at 15:55
10
10
What's surprising to me is not that the effect is insignificant, but how close it is to being significant. A decent commercial GPS receiver can be accurate to less than one meter under good conditions, so an error of 1 cm due to GR is only two orders of magnitude below that. I can easily imagine a future GPS-like system being accurate enough for those corrections to be needed.
– Ilmari Karonen
Nov 21 '18 at 18:46
What's surprising to me is not that the effect is insignificant, but how close it is to being significant. A decent commercial GPS receiver can be accurate to less than one meter under good conditions, so an error of 1 cm due to GR is only two orders of magnitude below that. I can easily imagine a future GPS-like system being accurate enough for those corrections to be needed.
– Ilmari Karonen
Nov 21 '18 at 18:46
3
3
To be clear, this specific general relativistic correction is too small to matter. The clock drift that would occur without general relativistic corrections is much more significant.
– Chris
Nov 21 '18 at 22:28
To be clear, this specific general relativistic correction is too small to matter. The clock drift that would occur without general relativistic corrections is much more significant.
– Chris
Nov 21 '18 at 22:28
2
2
The general-relativistic correction to the satellites' clock rates is not a matter of "imperfect clocks" -- gravitational time dilation is straight from GR. But it's not something GPS receivers need to care about, because the signal is already corrected for it when it is generated on the satellite.
– Henning Makholm
Nov 22 '18 at 1:31
The general-relativistic correction to the satellites' clock rates is not a matter of "imperfect clocks" -- gravitational time dilation is straight from GR. But it's not something GPS receivers need to care about, because the signal is already corrected for it when it is generated on the satellite.
– Henning Makholm
Nov 22 '18 at 1:31
3
3
@IlmariKaronen cm-scale GPS receiver systems do exist, but fundamentally they operate based on measuring the signal at reference locations and then sending out a correction, either with a time error (pseudorange) or position error. This is known as dGPS and many systems exist (WAAS/SBAS, LAAS/GLS, Starfire, SAIF, etc). Therefore, there is no need to make a special correction for every single possible error source, instead measuring error on a per-satellite or per-location basis.
– user71659
Nov 22 '18 at 2:10
@IlmariKaronen cm-scale GPS receiver systems do exist, but fundamentally they operate based on measuring the signal at reference locations and then sending out a correction, either with a time error (pseudorange) or position error. This is known as dGPS and many systems exist (WAAS/SBAS, LAAS/GLS, Starfire, SAIF, etc). Therefore, there is no need to make a special correction for every single possible error source, instead measuring error on a per-satellite or per-location basis.
– user71659
Nov 22 '18 at 2:10
|
show 8 more comments
I have some general comments about principles; I haven't looked up specific numbers but these will be in the sources below.
Firstly, see Neil Ashby's (open access) Living Reviews in Relativity article "Relativity in the Global Positioning System". Ashby is the biggest name in this field.
Secondly, while the "exact solutions" like Schwarzschild are extremely important, such beautiful mathematical solutions are not always the most practical. For the Solar System generally, it is not described using Schwarzschild etc, but instead "post-Newtonian theory" which is a sort of midway approximation between Newton and Einstein. This is the official International Astronomical Union recommendation -- see Soffel et al (2003). For Earth specifically, the gravitational field is best modeled as an expansion in spherical harmonics. For example, for a future Chinese satellite for detecting gravitational waves, it will be done this way. See the textbook by Poisson & Will.
Finally, one might think that because the Earth is rotating, the Kerr metric might be a good approximation for say $r>6500km$. However the exterior of a rotating body does not match to the Kerr metric in this way. See also Hartle's textbook chapter 14 titled "A little rotation", for an intermediate approximation between Schwarzschild and Kerr.
answered Nov 28 '18 at 2:01
Colin MacLaurin
37711
add a comment |
I have some general comments about principles; I haven't looked up specific numbers but these will be in the sources below.
Firstly, see Neil Ashby's (open access) Living Reviews in Relativity article "Relativity in the Global Positioning System". Ashby is the biggest name in this field.
Secondly, while the "exact solutions" like Schwarzschild are extremely important, such beautiful mathematical solutions are not always the most practical. For the Solar System generally, it is not described using Schwarzschild etc, but instead "post-Newtonian theory" which is a sort of midway approximation between Newton and Einstein. This is the official International Astronomical Union recommendation -- see Soffel et al (2003). For Earth specifically, the gravitational field is best modeled as an expansion in spherical harmonics. For example, for a future Chinese satellite for detecting gravitational waves, it will be done this way. See the textbook by Poisson & Will.
Finally, one might think that because the Earth is rotating, the Kerr metric might be a good approximation for say $r>6500km$. However the exterior of a rotating body does not match to the Kerr metric in this way. See also Hartle's textbook chapter 14 titled "A little rotation", for an intermediate approximation between Schwarzschild and Kerr.
answered Nov 28 '18 at 2:01
Colin MacLaurin
37711
add a comment |
I have some general comments about principles; I haven't looked up specific numbers but these will be in the sources below.
Firstly, see Neil Ashby's (open access) Living Reviews in Relativity article "Relativity in the Global Positioning System". Ashby is the biggest name in this field.
Secondly, while the "exact solutions" like Schwarzschild are extremely important, such beautiful mathematical solutions are not always the most practical. For the Solar System generally, it is not described using Schwarzschild etc, but instead "post-Newtonian theory" which is a sort of midway approximation between Newton and Einstein. This is the official International Astronomical Union recommendation -- see Soffel et al (2003). For Earth specifically, the gravitational field is best modeled as an expansion in spherical harmonics. For example, for a future Chinese satellite for detecting gravitational waves, it will be done this way. See the textbook by Poisson & Will.
Finally, one might think that because the Earth is rotating, the Kerr metric might be a good approximation for say $r>6500km$. However the exterior of a rotating body does not match to the Kerr metric in this way. See also Hartle's textbook chapter 14 titled "A little rotation", for an intermediate approximation between Schwarzschild and Kerr.
answered Nov 28 '18 at 2:01
Colin MacLaurin
37711
I have some general comments about principles; I haven't looked up specific numbers but these will be in the sources below.
Firstly, see Neil Ashby's (open access) Living Reviews in Relativity article "Relativity in the Global Positioning System". Ashby is the biggest name in this field.
Secondly, while the "exact solutions" like Schwarzschild are extremely important, such beautiful mathematical solutions are not always the most practical. For the Solar System generally, it is not described using Schwarzschild etc, but instead "post-Newtonian theory" which is a sort of midway approximation between Newton and Einstein. This is the official International Astronomical Union recommendation -- see Soffel et al (2003). For Earth specifically, the gravitational field is best modeled as an expansion in spherical harmonics. For example, for a future Chinese satellite for detecting gravitational waves, it will be done this way. See the textbook by Poisson & Will.
Finally, one might think that because the Earth is rotating, the Kerr metric might be a good approximation for say $r>6500km$. However the exterior of a rotating body does not match to the Kerr metric in this way. See also Hartle's textbook chapter 14 titled "A little rotation", for an intermediate approximation between Schwarzschild and Kerr.
answered Nov 28 '18 at 2:01
Colin MacLaurin
37711
answered Nov 28 '18 at 2:01
Colin MacLaurin
37711
answered Nov 28 '18 at 2:01
Colin MacLaurin
37711
answered Nov 28 '18 at 2:01
Colin MacLaurin
37711
37711
add a comment |
add a comment |
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9
Earth's local spacetime isn't even schwarzchild if you want to talk precision, due to Earth's oblate nature there are higher multipole moments to take into account for the gravitational field. You might be interested in this article ncbi.nlm.nih.gov/pmc/articles/PMC5253894
– Triatticus
Nov 21 '18 at 5:03
5
This is not only a physics question but an electronics one (and I do mean electronics, not programming). The early receivers were designed to work using the technology of the 1980s or earlier. They needed to be portable, which might mean vehicle mounted rather than handheld, which would limit the size and power supply that coule be used. That all impacts the recommended calculations. Modern recievers are welcome to calculate position however they please, and survey ones may include some of what you suggest to get their millimetre precisions.
– TafT
Nov 21 '18 at 11:42