Krdytkyu

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.

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.