GPS Relativistic Corrections

Every GPS satellite left the factory with its clock deliberately set wrong. Not calibrated — wrong, on purpose, by $-4.4647\times10^{-10}$ , so the caesium standard reading $10.229\,999\,995\,43$ MHz on the bench will read exactly $10.23$ MHz once it is $20\,184$ km up and moving at $3.874$ km/s. Everybody knows the two halves of that number. Motion costs the satellite $-7.2$ microseconds a day; altitude buys it $+45.8$ ; the net is $+38.6$ , and ignoring it drifts your position by $11.6$ kilometres per day. What almost nobody says out loud is that those three numbers do not come from the formula everyone writes down. The standard textbook expression — subtract the ground station's speed from the satellite's, subtract its potential from the satellite's — gives the right net. It gives $-7.1$ and $+45.7$ for the two halves. Not $-7.2$ and $+45.8$ . The difference is $0.104$ µs/day, and it is exactly the Earth's rotation: $465$ metres per second at the equator. It has not disappeared. It has been booked into the gravity column, because the ground clock's rate is quoted through the effective geopotential on the geoid — gravity plus centrifugal, as one number. That is not a bookkeeping nicety. It is the only formulation in which the phrase "the ground clock" means anything at all.

Key takeaways

Signature $(+,-,-,-)$ : $\mathrm{d}s^2 = c^2\mathrm{d}t^2 - \mathrm{d}x^2$ , $p\!\cdot\!p = +m^2c^2$ , timelike positive. The base is written in $(-,+,+,+)$ — its metric and its proper-time card are both re-derived here, not transcribed.
The budget, and this article is the corpus's canonical owner of it. $\;$ SR (velocity) $= \mathbf{-7.213}$ µs/day $\;\cdot\;$ GR (potential) $= \mathbf{+45.788}$ µs/day $\;\cdot\;$ net $= \mathbf{+38.575}$ µs/day. Factory detune $\mathbf{-4.4647\times10^{-10}}$ (Ashby, Living Reviews in Relativity 6 (2003) 1) — reproduced here from WGS-84 constants to six figures.
The owned result — the geoid theorem. That triple does not follow from $\Delta f/f = \Delta\Phi/c^2 - (v_s^2 - v_g^2)/2c^2$ . It follows only when the ground rate is written as $\Phi_0/c^2$ , the effective geopotential — gravity plus centrifugal — on the geoid. The Earth's rotation ( $465$ m/s, worth $\mathbf{0.1040}$ µs/day) then sits in the gravity column.
Why that is the right convention, not merely a convention. The geoid is defined as a surface of constant $\Phi_0$ , so every clock on it ticks at the same rate, at every latitude — machine-checked: pole and equator agree to $\mathbf{0.00009}$ µs/day, though $GM/r$ differs by $0.34\%$ and the ground speed by $465$ m/s. Split gravity from rotation and "the ground clock rate" becomes latitude-dependent, and there is no longer any such thing as the ground clock.
The metric, re-derived in our lock. $\;\mathrm{d}s^2 = \left(1 + \tfrac{2\Phi}{c^2}\right)c^2\mathrm{d}t^2 - \left(1 - \tfrac{2\Phi}{c^2}\right)\mathrm{d}\mathbf{x}^2$ , giving $\;\mathrm{d}\tau/\mathrm{d}t \approx 1 + \Phi/c^2 - v^2/2c^2$ .
The satellite's rate is one clean term, $-\tfrac32 GM/rc^2$ . For any circular orbit $v^2 = GM/r$ , so potential and velocity merge. All of GPS is $\Delta f/f = |\Phi_0|/c^2 - \tfrac32 GM/rc^2$ — and the null radius falls out: $r_0 = 3GM/2|\Phi_0| = \mathbf{9545.5}$ km ( $3167$ km altitude), below which an orbiting clock runs slow. A clock at infinity gains $\mathbf{60.21}$ µs/day; GPS has climbed $64\%$ of the way there.
Where the base is wrong: metric and proper-time card mostly-plus · its $\Delta f/f$ is the naive split, contradicting the $-7.2$ / $+45.8$ its own limits card quotes · its uncorrected-drift figure ( $\sim\!10$ km/day) should be $\mathbf{11.6}$ · its null-radius equation is printed with a literal " $\ldots$ " in it.

The core insight

A GPS clock is compared against "a clock on the ground" — but the Earth spins, so no two ground clocks are at the same radius, in the same potential, or moving at the same speed. The way out is not to pick one; it is to notice that the surface they sit on, the geoid, is by construction a surface of constant effective potential $\Phi_0$ — gravity and centrifugal summed. Every clock on it therefore ticks at one rate, $1 + \Phi_0/c^2$ , and that is what the satellite is measured against. The consequence is a bookkeeping fact the whole literature uses and almost nobody states: the Earth's rotational time dilation is already inside the "gravitational" term. Move it to the "special-relativistic" column and you get $-7.1$ and $+45.7$ — a correct net, two wrong halves.

Where this lives today

The correction is not optional, and the residuals are now the interesting part. Every operational GPS satellite carries an oscillator detuned by $-4.4647\times10^{-10}$ ; without it, ranging error accumulates at $11.6$ km per day. But that offset is a constant, and handles only the mean rate. A real orbit has $e \lesssim 0.02$ , so radius and speed oscillate once per orbit and so does the clock rate; the receiver applies $\Delta t_{\text{ecc}} = -2\sqrt{GM\,a}\,e\sin E/c^2$ , reaching $\pm\mathbf{45.8}$ ns — worth $13.7$ m of position. Underneath sits the Sagnac term, $\Delta t = 2\Omega A_z/c^2$ : what the Earth's rotation does to a signal rather than to a clock, and the subject of `aq-sr-sagnac-effect`.

And the geoid convention has outgrown GPS. Because $\Phi_0/c^2$ is the rate at which the Earth's surface runs slow, the IAU adopted it — $L_G = 6.969290134\times10^{-10}$ , now a defining constant — as the conversion between Terrestrial Time and Geocentric Coordinate Time. The number in this article's ground column is, by international agreement, a definition. Optical clocks now resolve $10^{-18}$ , a centimetre of geoid height: the equation that detunes a satellite is becoming a way to survey the ground.

The physical picture

Two clocks. One in a $20\,184$ km orbit, one bolted to a laboratory floor. Ask each what rate it runs at against a common coordinate time $t$ — the Earth-centred inertial frame, which does not rotate.

The satellite is easy. A clock at potential $\Phi$ moving at speed $v$ runs at $\mathrm{d}\tau/\mathrm{d}t \approx 1 + \Phi/c^2 - v^2/2c^2$ , and for a circular orbit Newton hands you $v^2 = GM/r$ for free, so the two terms collapse:

\left(\frac{\mathrm{d}\tau}{\mathrm{d}t}\right)_{\!\text{sat}} = 1 - \frac{GM}{rc^2} - \frac{GM}{2rc^2} = 1 - \frac{3GM}{2rc^2}.

One term. No velocity anywhere. The relativistic behaviour of every circular orbit ever flown is the number $\tfrac32 GM/rc^2$ .

The ground clock is the hard one, and this is the part the textbooks skate over. A clock at the pole is $21$ km closer to the Earth's centre than one at the equator, and is not moving at all. So which ground clock? It does not matter, and the reason is the geoid — defined, and this is its definition rather than a property of it, as a surface of constant effective potential

\Phi_{\text{eff}}(\mathbf r) = \underbrace{\Phi_{\text{grav}}(\mathbf r)}_{\text{gravity}} + \underbrace{\left(-\tfrac12\Omega^2 r^2\sin^2\theta\right)}_{\text{centrifugal}} = \Phi_0 = \text{const}.

And $-\tfrac12\Omega^2 r^2\sin^2\theta$ is precisely $-\tfrac12 v_g^2$ . The centrifugal potential and the kinematic time dilation of a rotating clock are the same object. So the geoid is the locus on which $\Phi_{\text{grav}}/c^2 - v_g^2/2c^2$ is constant — the locus on which clocks tick at one rate. The comparison is therefore

\boxed{\;\frac{\Delta f}{f} \;=\; \left(\frac{\mathrm{d}\tau}{\mathrm{d}t}\right)_{\!\text{sat}} - \left(\frac{\mathrm{d}\tau}{\mathrm{d}t}\right)_{\!\text{geoid}} \;=\; \frac{|\Phi_0|}{c^2} \;-\; \frac{3GM}{2rc^2}\;}

with $\Phi_0/c^2 = -6.969283\times10^{-10}$ , computed below from WGS-84 and nothing else.

Now the bookkeeping — the whole point of the article. People report this as "SR $-7.2$ , GR $+45.8$ ". Look at what those pieces must be:

\underbrace{-\frac{v_s^2}{2c^2}}_{\text{SR}\;=\;-7.213\;\mu\text{s/day}} \quad\text{and}\quad \underbrace{-\frac{GM}{rc^2} - \frac{\Phi_0}{c^2}}_{\text{GR}\;=\;+45.788\;\mu\text{s/day}}

The "SR" term holds the satellite's speed and nothing else — no $+v_g^2/2c^2$ . The ground station's $465$ m/s has been absorbed into $\Phi_0$ , and appears in the gravitational column. Write $-(v_s^2 - v_g^2)/2c^2$ instead and you get $-7.109$ ; the gravitational term becomes $+45.684$ ; and the net is unchanged at $+38.575$ , because $0.104$ µs/day simply moved pockets.

Both bookkeepings are right. Only one produces the numbers everybody quotes.

Figure

THE GEOID LEDGER — the owned result. Two ways to split the same GPS clock budget, and they are not the same split. LEFT, the NAIVE ledger: subtract the ground station's potential from the satellite's, and subtract the ground station's speed from the satellite's, exactly as the textbook formula Δf/f = ΔΦ/c² − (v_s²−v_g²)/2c² tells you to. You get GR = +45.684 and SR = −7.109 µs/day. RIGHT, the GEOID ledger, which is what Ashby and the whole GPS literature actually use: the ground clock's rate is written as one number, Φ₀/c² = −6.969283×10⁻¹⁰, the EFFECTIVE geopotential on the geoid — gravity AND centrifugal, summed. You get GR = +45.788 and SR = −7.213. THOSE ARE THE CANONICAL NUMBERS. The NET is identical in both columns, +38.575 µs/day, because all that happened is that 0.1040 µs/day — the Earth's rotation, 465.10 m/s at the equator — moved out of the velocity pocket and into the gravity pocket. It did not vanish; it was reclassified. And the reason the right-hand ledger is the correct one is written across the bottom: the geoid is BY DEFINITION a surface of constant Φ₀, so every clock on it ticks at the same rate. Machine-checked here — the pole and the equator agree to 0.00009 µs/day, one part in 1100 of the quantity being argued over, even though GM/r differs between them by 0.34% and the ground speed by the full 465 m/s. Take the rotation out of Φ₀ and 'the ground clock rate' becomes a function of latitude, and there is no longer any such thing as THE ground clock

The core equations

#	relation	what it says
(1)	$\mathrm{d}s^2 = \left(1 + \dfrac{2\Phi}{c^2}\right)c^2\mathrm{d}t^2 - \left(1 - \dfrac{2\Phi}{c^2}\right)\mathrm{d}\mathbf{x}^2$	weak-field metric, our lock. The base prints this with both signs flipped
(2)	$\mathrm{d}\tau = \dfrac{1}{c}\sqrt{g_{\mu\nu}\,\mathrm{d}x^\mu\mathrm{d}x^\nu}$	proper time. No minus sign — timelike intervals are already positive
(3)	$\dfrac{\mathrm{d}\tau}{\mathrm{d}t} \approx 1 + \dfrac{\Phi}{c^2} - \dfrac{v^2}{2c^2}$	any clock, first order. Both terms negative for a clock in a well
(4)	$\Phi_0 = -\dfrac{GM}{a_1}\left(1 + \dfrac{J_2}{2}\right) - \dfrac{\Omega^2a_1^2}{2}$	the effective geopotential on the geoid — gravity, oblateness, centrifugal
(5)	$\boxed{\;\dfrac{\Delta f}{f} = \dfrac{\|\Phi_0\|}{c^2} - \dfrac{3GM}{2rc^2}\;}$	the whole of GPS relativity. One ground term, one orbit term
(6)	$r_0 = \dfrac{3GM}{2\|\Phi_0\|} = 9545.5\ \mathrm{km}$	the null radius — an orbiting clock ticks at the ground rate
(7)	$\Delta t_{\text{ecc}} = -\dfrac{2\,\mathbf{r}\cdot\mathbf{v}}{c^2} = -\dfrac{2\sqrt{GM\,a}}{c^2}\,e\sin E$	the residual periodic term for a non-circular orbit
(8)	$\Delta t_{\text{Sagnac}} = \dfrac{2\Omega}{c^2}A_z$	the rotating-Earth correction, applied in the receiver

Equation (5) does more work than it looks. Everything about the satellite is in $3GM/2rc^2$ — potential and velocity merged, because a circular orbit ties them together. Everything about the Earth is in $|\Phi_0|/c^2$ — mass, oblateness and spin. Two numbers, and GPS is solved.

Deriving the budget

Derivation AThe weak-field metric in our lock, and the clock-rate formula that follows

The base writes its metric with the timelike component negative and the spatial components positive — the opposite convention to this corpus's, and a transcribed metric is a bug waiting to happen. So we build it from scratch. The goal: the first-order clock rate $\mathrm{d}\tau/\mathrm{d}t$ for a slow clock in a weak static field, in $(+,-,-,-)$ , checked against the two limits we already trust.

Step 1. Fix the convention and the proper time. In $(+,-,-,-)$ a timelike interval is positive: $\mathrm{d}s^2 = c^2\mathrm{d}t^2 - \mathrm{d}\mathbf{x}^2 > 0$ along a massive worldline. So

c^2\,\mathrm{d}\tau^2 = \mathrm{d}s^2 = g_{\mu\nu}\,\mathrm{d}x^\mu\mathrm{d}x^\nu, \qquad \mathrm{d}\tau = \dfrac{1}{c}\sqrt{g_{\mu\nu}\,\mathrm{d}x^\mu\mathrm{d}x^\nu}.

Why: the base's eq[0] carries a minus sign under the root. That minus is not physics, it is a signature declaration — it says timelike intervals are negative. Copy it into our lock and the root turns imaginary. Never transcribe a metric.

Step 2. Write the weak static field, in an Earth-centred inertial (non-rotating) frame, with $|\Phi|/c^2 \sim 7\times10^{-10} \lll 1$ :

\mathrm{d}s^2 = \left(1 + \frac{2\Phi}{c^2}\right)c^2\mathrm{d}t^2 - \left(1 - \frac{2\Phi}{c^2}\right)\left(\mathrm{d}x^2+\mathrm{d}y^2+\mathrm{d}z^2\right), \qquad \Phi < 0 .

Why: $g_{00}$ is fixed by demanding the geodesic equation reproduce Newton; $g_{ij}$ follows to the same order. In our lock the SIGNS of both corrections flip relative to the base, while the PHYSICS is unchanged — the physics lives in $\mathrm{d}\tau$ , not in $\mathrm{d}s^2$ .

Step 3. Put a moving clock on it. Divide by $c^2\mathrm{d}t^2$ , with $v^2 = |\mathrm{d}\mathbf{x}/\mathrm{d}t|^2$ :

\left(\frac{\mathrm{d}\tau}{\mathrm{d}t}\right)^{\!2} = \left(1 + \frac{2\Phi}{c^2}\right) - \left(1 - \frac{2\Phi}{c^2}\right)\frac{v^2}{c^2}.

Why: this is positive, as it must be. In the mostly-plus convention the same expression comes out with an overall minus — exactly why the base needs one under its root. Two conventions, one physics, and the sign of $\mathrm{d}\tau^2$ keeps score.

Step 4. Expand. The cross term $2\Phi v^2/c^4$ is of order $10^{-19}$ and dies. With $\sqrt{1+\epsilon}\approx 1+\epsilon/2$ ,

\boxed{\;\frac{\mathrm{d}\tau}{\mathrm{d}t} \approx 1 + \frac{\Phi}{c^2} - \frac{v^2}{2c^2}\;}

Why: both corrections are negative for a clock in a well and in motion, since $\Phi < 0$ . Deeper is slower; faster is slower. GPS is about which of those two losses the satellite escapes.

Step 5. Check the two limits we already own. $\Phi = 0$ gives $1 - v^2/2c^2$ , which is $1/\gamma$ to first order — the kinematic dilation of `aq-sr-time-dilation`, whose GPS number ( $\gamma-1 = 8.35\times10^{-11}$ at $v_s = 3.874$ km/s) is live canon and is quoted, not re-derived, here. $v = 0$ gives $1 + \Phi/c^2$ , the static gravitational rate. Why: trap 27. A weak-field metric you have glanced at is not one you have checked. It must reduce, in the two limits where the answer is independently known, to the two things it claims to unify. *Verified in verify_aq-sr-gps-relativistic-corrections.py: the metric is built symbolically in sympy in $(+,-,-,-)$ ; $(\mathrm{d}\tau/\mathrm{d}t)^2$ is formed from it, series-expanded, and asserted equal to $1 + \Phi/c^2 - v^2/2c^2$ ; both limits are asserted symbolically; and the base's mostly-plus form is built alongside and shown to give the negative of the same quantity — which is what forces its extra minus sign.*

Go deeper · Derivation B · 3 minThe geoid theorem, and the ledger that closes only because of it

Derivation BThe geoid theorem, and the ledger that closes only because of it

Now build the budget, and watch the Earth's rotation change pockets. The goal: the canonical GPS triple $(-7.2, +45.8, +38.6)$ — derived, every constant named, and with an explicit demonstration that the naive formulation gives the right net and two wrong halves.

Step 1. The satellite. Circular orbit of radius $r$ , so $v_s^2 = GM/r$ exactly (Newton suffices; corrections are $O(\Phi^2/c^4)$ ):

\left(\frac{\mathrm{d}\tau}{\mathrm{d}t}\right)_{\!s} = 1 - \frac{GM}{rc^2} - \frac{1}{2c^2}\cdot\frac{GM}{r} = 1 - \frac{3GM}{2rc^2}.

Why: the factor $\tfrac32$ is the whole orbital story. Two-thirds gravity, one-third speed — and for a circular orbit they are not independent, because the same $GM/r$ appears in both.

Step 2. The ground, and the step everything turns on. A clock on the rotating Earth at co-latitude $\theta$ , radius $r$ , has speed $v_g = \Omega r\sin\theta$ , so

\left(\frac{\mathrm{d}\tau}{\mathrm{d}t}\right)_{\!g} = 1 + \frac{\Phi_{\text{grav}}}{c^2} - \frac{\Omega^2r^2\sin^2\theta}{2c^2} = 1 + \frac{1}{c^2}\underbrace{\left[\Phi_{\text{grav}} - \tfrac12\Omega^2r^2\sin^2\theta\right]}_{\displaystyle \equiv\;\Phi_{\text{eff}}} .

Why: look at what just happened. The kinematic term $-v_g^2/2c^2$ and the centrifugal potential are literally the same expression. The velocity of a rotating clock is not an extra effect to add to the potential — it is a potential, and it belongs inside $\Phi_{\text{eff}}$ .

Step 3. Invoke the geoid. The geoid is defined as the level surface of $\Phi_{\text{eff}}$ , on which $\Phi_{\text{eff}} = \Phi_0 =$ const. Therefore every clock on the geoid has the same rate, $1 + \Phi_0/c^2$ , regardless of latitude. Evaluating at the equator ( $\theta = 90°$ , $r = a_1$ , $P_2(\cos 90°) = -\tfrac12$ ),

\Phi_0 = -\frac{GM}{a_1}\left(1 + \frac{J_2}{2}\right) - \frac{\Omega^2 a_1^2}{2}.

Why: the constancy is the theorem. A pole clock sits $21$ km deeper in the well and does not move; an equatorial clock sits higher and runs at $465$ m/s. They cancel because the Earth's surface is an equilibrium shape. Machine-checked below: the two rates agree to $\mathbf{0.00009}$ µs/day.

Step 4. Put in WGS-84 and nothing else. $GM = 3.986004418\times10^{14}$ m³/s², $a_1 = 6\,378\,137$ m, $J_2 = 1.0826300\times10^{-3}$ , $\Omega = 7.2921151467\times10^{-5}$ rad/s:

\Phi_0 = -6.2636796\times10^{7}\ \mathrm{m^2/s^2}, \qquad \frac{\Phi_0}{c^2} = \mathbf{-6.969283\times10^{-10}} .

Why: the IAU's defining constant for Terrestrial Time is $L_G = 6.969290134\times10^{-10}$ exactly — and four WGS-84 numbers reproduce it to one part in $10^6$ , the residual being the higher harmonics we truncated. The agreement is the check.

Step 5. The orbit radius, from the mission requirement and nothing else. GPS orbits twice per sidereal day, so the ground track repeats: $T = 86\,164.0905/2 = 43\,082.05$ s. Kepler:

a = \left(\frac{GM\,T^2}{4\pi^2}\right)^{1/3} = \mathbf{26\,561.76\ km}, \qquad h = a - a_1 = \mathbf{20\,183.6\ km}, \qquad v_s = \sqrt{GM/a} = \mathbf{3873.8\ m/s} .

Why: the repeating ground track is why GPS is at that altitude, and the altitude is why the budget is what it is.

Step 6. Close the ledger, in the geoid convention.

\underbrace{-\frac{v_s^2}{2c^2} = -8.34852\times10^{-11}}_{\textbf{SR} \;=\; \mathbf{-7.213}\ \mu\text{s/day}} \qquad \underbrace{-\frac{GM}{rc^2} - \frac{\Phi_0}{c^2} = +5.299579\times10^{-10}}_{\textbf{GR} \;=\; \mathbf{+45.788}\ \mu\text{s/day}}

\textbf{net} = +4.4647275\times10^{-10} \;=\; \mathbf{+38.575\ \mu s/day}

Why: that is Ashby's factory detune, $-4.4647\times10^{-10}$ , reproduced to six figures from four WGS-84 constants and a sidereal day — and why the oscillator is set to $f_0(1 - \Delta f/f) = \mathbf{10.229\,999\,995\,43}$ MHz on the bench.

Step 7. Now do it the naive way, and watch $0.104$ move.

\textbf{SR}_{\text{naive}} = -\frac{v_s^2 - v_g^2}{2c^2} = \mathbf{-7.109}\ \mu\text{s/day}, \qquad \textbf{GR}_{\text{naive}} = \frac{\Phi_s - \Phi_{\text{grav,geoid}}}{c^2} = \mathbf{+45.684}\ \mu\text{s/day}

\textbf{net} = \mathbf{+38.575}\ \mu\text{s/day} \quad \text{— identical.}

Why: the net is invariant and the split is not. The quantity that moved is $v_g^2/2c^2 = \mathbf{0.1040}$ µs/day — the Earth's $465.10$ m/s at the equator. Both ledgers are arithmetically correct. Only the geoid ledger reproduces the numbers the world quotes, and it is the one to use, because it is the only one in which "the ground clock" is a single latitude-independent object. *Verified in verify_aq-sr-gps-relativistic-corrections.py: $\Phi_0$ , $a$ , $v_s$ , both ledgers, the net, $f_{\text{set}}$ , $r_0$ and the drift are computed at 50 dps from WGS-84 alone; the two ledgers are asserted to give identical nets and to differ by exactly $v_g^2/2c^2$ ; the geoid's latitude-independence is checked by evaluating $\Phi_{\text{eff}}$ at the pole and at the equator and asserting they agree to $10^{-5}$ . The artefact guard perturbs $GM$ , $a_1$ , $J_2$ , $\Omega$ and $T$ and asserts every headline moves — and, inversely, that the net does not move when the ledger convention is switched.*

Figure

THE ALTITUDE CURVE. For a circular orbit of radius r, the entire relativistic rate offset against a ground clock is a single expression — Δf/f = |Φ₀|/c² − (3/2)GM/rc² — because for a circular orbit v² = GM/r ties the velocity term to the potential term. The curve is that expression, in microseconds per day, plotted against orbital radius. LOW ORBITS LOSE. A clock in a grazing circular orbit at the Earth's surface runs 29.9 µs/day SLOW: it is no deeper in the well than the ground, but it is doing 7.9 km/s to stay up, and speed is pure loss. As you climb, the potential term relaxes faster than the velocity term, and the curve crosses zero at the NULL RADIUS r₀ = 3GM/2|Φ₀| = 9545.5 km — 3167 km altitude — where an orbiting clock ticks at exactly the ground rate and no correction at all would be needed. Above it, gravity wins. GPS sits at 26 561.8 km with +38.575 µs/day; geostationary at 42 164 km with +46.58. And the curve has a CEILING: as r → ∞ the orbital terms vanish entirely and all that is left is the ground clock's own deficit, |Φ₀|/c² = +60.21 µs/day. That asymptote is the deepest fact on this plot — it is the rate at which a clock at rest at infinity gains on the Earth's surface, it depends on nothing but the Earth, and GPS has climbed 64% of the way to it

Limiting cases

Case	Result	Meaning
$v \to 0$	$\mathrm{d}\tau/\mathrm{d}t \to 1 + \Phi/c^2$	pure gravitational rate. The tower-clock experiment
$\Phi \to 0$	$\mathrm{d}\tau/\mathrm{d}t \to 1 - v^2/2c^2$	pure kinematic dilation — the $1/\gamma$ of `aq-sr-time-dilation`, to first order
$c \to \infty$	$\mathrm{d}\tau/\mathrm{d}t \to 1$	no offset at all. GPS needs no clock correction in a Newtonian universe — and does not work in one
$r \to r_0 = 9545.5$ km	$\Delta f/f = 0$	the null orbit. Velocity loss exactly cancels potential gain
$r \to \infty$	$\Delta f/f \to \\|\Phi_0\\|/c^2 = +60.21$ µs/day	the ceiling. Depends only on the Earth, not on the orbit
$\Omega \to 0$ (Earth stops)	the two ledgers merge	The whole ambiguity is made of the Earth's spin
$e \to 0$	$\Delta t_{\text{ecc}} \to 0$	circular orbits need no periodic term. Real GPS: $e \lesssim 0.02$ , up to $\pm45.8$ ns

Common trapThe base's metric is mostly-plus — and so is its proper time

The base prints

\mathrm{d}\tau = \sqrt{-\tfrac{1}{c^2}\,g_{\mu\nu}\,\mathrm{d}x^\mu\mathrm{d}x^\nu}, \qquad \mathrm{d}s^2 = -\left(1+\tfrac{2\Phi}{c^2}\right)c^2\mathrm{d}t^2 + \left(1-\tfrac{2\Phi}{c^2}\right)\mathrm{d}\mathbf{x}^2 .

Both are $(-,+,+,+)$ . In this corpus's lock they are wrong on their face: timelike intervals are positive here, so there is no minus under the root and the metric's signs are the other way round. A metric copied rather than derived is a corrupted metric — that minus sign will follow it into an article where it does not belong and turn a real proper time imaginary.

The tell that it is a convention and not an error: the base's own $\mathrm{d}\tau/\mathrm{d}t \approx 1 + \Phi/c^2 - v^2/2c^2$ is correct, and identical to ours. Two conventions, one physics. But you cannot mix them, and there is exactly one way not to: re-derive.

Common trapThe base's Δf/f contradicts the base's own limits card — by the Earth's rotation

The base's eq[3] is

\frac{\Delta f}{f} \;\overset{?}{=}\; \frac{\Phi_s-\Phi_g}{c^2} - \frac{v_s^2 - v_g^2}{2c^2},

the naive split. But its own limits card says the velocity term alone is " $\approx -8.3\times10^{-11}$ , i.e. roughly $-7.2$ µs/day" — and $-8.3\times10^{-11}$ is $-v_s^2/2c^2$ with no $v_g$ in it at all. Put $v_g$ back, as its eq[3] instructs, and the velocity term becomes $-8.228\times10^{-11} = -7.109$ µs/day. The equation and the number disagree by $0.104$ µs/day — the Earth's rotation to four figures. The same discrepancy, opposite sign, sits in the gravitational column: the card quotes $+45.8$ , which requires $\Phi_0$ (centrifugal included); the equation gives $+45.684$ .

The net is right either way — that is why nobody notices. The fix is not to change a number; it is to say which potential you meant. Write $\Phi_0$ , say "effective", and the ledger closes on the canonical triple. Two smaller defects. The base's null-radius entry is printed with a literal `\ldots` in the formula — unfinished; the correct statement is $r_0 = 3GM/2|\Phi_0| = 9545.5$ km. And its tldr puts the uncorrected drift at " $\sim\!10$ km per day"; the figure is $38.575\ \mu\mathrm{s} \times c = \mathbf{11.6}$ km/day, which is what live canon `aq-sr-time-dilation` already prints.

Worked examples

1 · Build the entire GPS budget from four constants and a sidereal day

Take only $GM = 3.986004418\times10^{14}$ m³/s², $a_1 = 6\,378\,137$ m, $J_2 = 1.0826300\times10^{-3}$ , $\Omega = 7.2921151467\times10^{-5}$ rad/s. Nothing else.

The ground. $\Phi_0 = -(GM/a_1)(1 + J_2/2) - \Omega^2a_1^2/2 = -6.2636796\times10^{7}$ m²/s², so $\Phi_0/c^2 = \mathbf{-6.969283\times10^{-10}}$ . Sanity: the IAU's defining value for Terrestrial Time is $6.969290134\times10^{-10}$ . Agreement to one part in $10^6$ .

The orbit. Two revolutions per sidereal day $\Rightarrow T = 43\,082.05$ s $\Rightarrow a = (GM T^2/4\pi^2)^{1/3} = \mathbf{26\,561.76}$ km, altitude $\mathbf{20\,183.6}$ km, $v_s = \sqrt{GM/a} = \mathbf{3873.8}$ m/s.

The budget. SR $= -v_s^2/2c^2 = -8.34852\times10^{-11} \Rightarrow \mathbf{-7.213}$ µs/day. GR $= -GM/ac^2 - \Phi_0/c^2 = +5.299579\times10^{-10} \Rightarrow \mathbf{+45.788}$ µs/day. Net $= +4.4647275\times10^{-10} \Rightarrow \mathbf{+38.575}$ µs/day.

The hardware. $f_{\text{set}} = 10.23\ \mathrm{MHz} \times (1 - 4.4647\times10^{-10}) = \mathbf{10.229\,999\,995\,43\ MHz}$ — the number physically dialled into every satellite before launch. The consequence of not doing it: $38.575\ \mu\mathrm{s} \times c = \mathbf{11\,565\ m}$ , or $\mathbf{11.6}$ km of ranging error per day.

Nothing was looked up. The factory detune of a $\$12$ -billion constellation is four numbers about the Earth and one about how long a day is.

2 · Where does the Earth's rotation actually go?

The ground station at the equator moves at $v_g = \Omega a_1 = \mathbf{465.10}$ m/s. Its kinematic time dilation is

\frac{v_g^2}{2c^2} = 1.20344\times10^{-12} \;\Longrightarrow\; \mathbf{0.10398\ \mu s/day} .

Now find that number in the budget. It is not in the SR column, which is $-v_s^2/2c^2$ , the satellite's speed alone. It is inside $\Phi_0$ , as the centrifugal term $-\Omega^2a_1^2/2$ — and $-\Omega^2a_1^2/2 = -v_g^2/2$ identically, so its contribution to $\Phi_0/c^2$ is $-1.20344\times10^{-12}$ , the same number to every digit. When the literature writes GR $= +45.8$ , that $+45.8$ contains $+0.104$ µs/day of the Earth spinning.

And here is why that is not an abuse. Evaluate the effective potential at the pole, using the WGS-84 semi-minor axis $b = 6\,356\,752.3$ m, where $P_2(\cos 0°) = 1$ and there is no rotation at all:

\Phi_{\text{pole}} = -\frac{GM}{b}\left[1 - J_2\left(\frac{a_1}{b}\right)^{\!2}\right] = -6.2636701\times10^7\ \mathrm{m^2/s^2} .

Against the equatorial $\Phi_0 = -6.2636796\times10^7$ , they agree to $1.5\times10^{-6}$ — in clock terms $\mathbf{0.00009}$ µs/day, a thousand times smaller than the $0.104$ under dispute. A polar clock and an equatorial clock tick at the same rate, though one is $21$ km deeper in the well and the other is doing $465$ m/s. That is the geoid theorem, and it is why "the ground clock" is a well-defined object.

Physics of the portrait

Origins · 2 minAtomic clock and how this was discovered

Atomic clock

1955–1977 · Essen and Parry build a clock good enough to see relativity · Hafele and Keating fly one round the world · NTS-2 puts one in orbit and the prediction is waiting for it

The clock had to come first. In 1955 Louis Essen and Jack Parry, at the National Physical Laboratory, ran the first practical caesium beam standard; within a few years the second itself was redefined in terms of it. What that bought was not merely precision but a portable precision: one could now carry a clock somewhere, bring it back, and ask what the journey had cost it. In 1971 they carried four. Hafele and Keating flew caesium clocks east and west on commercial airliners and compared them against the USNO ensemble. The predicted differences were tens of nanoseconds, with gravity and velocity pulling opposite ways and the Earth's rotation breaking the symmetry between the two flight directions — and the measurement matched. The same three-term budget as GPS, at a hundredth of the altitude. Then the prediction went up ahead of the system. On 23 June 1977 the US Navy launched NTS-2, carrying the first caesium clock ever put in orbit. Its synthesiser — the device that would apply the relativistic offset — was deliberately left switched off for the first twenty days, so the raw rate could be measured. General relativity predicted the orbiting clock would run fast by $+446.5$ parts in $10^{12}$ . It ran fast by $+442.5$ : a discrepancy of $4$ parts in $10^{12}$ , comfortably inside the clock's own stability. That was the moment GPS became possible. There had been a real argument inside the programme about whether the correction was needed at all; NTS-2 ended it. The theory is not a correction applied to GPS. It is a component of it, installed before the first operational satellite ever flew.

History

1955

— Essen and Parry operate the first caesium beam frequency standard at the NPL, and clocks become good enough to measure relativity directly.

1959

— Pound and Rebka measure the gravitational frequency shift over 22.5 m of a Harvard tower, confirming the term that will later dominate the GPS budget.

1971

— Hafele and Keating fly caesium clocks east and west around the world; the gravitational and kinematic terms are separated experimentally, and the rotation of the Earth breaks the symmetry between the two directions.

1977

— NTS-2 is launched on 23 June carrying the first caesium clock in orbit; with its synthesiser switched off, the clock runs fast by +442.5 parts in 10^12 against a general-relativistic prediction of +446.5.

1978

— The first Block I GPS satellite is launched, with the relativistic frequency offset built into the oscillator before flight.

1980

— The Sagnac correction for the rotating Earth is formalised in the GPS interface specification; receivers apply it in software.

2003

— Neil Ashby publishes "Relativity in the Global Positioning System" (Living Reviews in Relativity 6, 1), which fixes the effective-geopotential formulation used here and remains the standard reference.

Going further

Texts and papers. N. Ashby, Relativity in the Global Positioning System, Living Reviews in Relativity 6 (2003) 1 — the standard reference, and the source of the effective-geopotential formulation · Hafele & Keating, Science 177, 166 and 168 (1972) · Pound & Rebka, Phys. Rev. Lett. 4, 337 (1960) · Misner, Thorne & Wheeler, Gravitation, §§16, 25 · NIMA TR8350.2, Department of Defense World Geodetic System 1984.

Provenance. Signature $(+,-,-,-)$ . Every number here is recomputed at 50 decimal digits in verify_aq-sr-gps-relativistic-corrections.py, which reads this file, extracts the numbers from the prose, and asserts them. Inputs: the four WGS-84 constants ( $GM$ , $a_1$ , $J_2$ , $\Omega$ ), the sidereal day, and $c$ — nothing is looked up. The orbit radius, $v_s$ , $\Phi_0$ , both ledgers, the net, $f_{\text{set}}$ , $r_0$ and the drift all follow. Cross-checks: $\Phi_0/c^2$ against the IAU defining constant $L_G$ (agrees to $10^{-6}$ ), and the net against Ashby's factory detune $-4.4647\times10^{-10}$ (six figures).

Base defects repaired. (i) eq[0] and eq[1] are mostly-plus — metric and proper time re-derived here in $(+,-,-,-)$ , never transcribed. (ii) eq[3] prints the naive split, contradicting the base's own limits card by exactly $v_g^2/2c^2 = 0.104$ µs/day; the geoid formulation is derived and the discrepancy machine-demonstrated. (iii) The null-radius entry is printed with a literal " $\ldots$ " — the correct form is $r_0 = 3GM/2|\Phi_0| = 9545.5$ km. (iv) Its uncorrected-drift figure, " $\sim\!10$ km/day", should be $\mathbf{11.6}$ . Its clock-rate formula (eq[2]), its eccentricity term (eq[7], verified here symbolically from Kepler) and its Sagnac term (eq[8]) are correct and used as written.

Scope. The kinematic $-7.2$ µs/day — $\gamma - 1 = 8.35\times10^{-11}$ at $v_s = 3.874$ km/s — is owned by live canon `aq-sr-time-dilation`, which states explicitly that the full budget "belongs elsewhere". This is elsewhere. This article owns the metric, the effective-geopotential ground rate, the complete budget, and the geoid bookkeeping theorem. The rotating-frame signal correction is the subject of `aq-sr-sagnac-effect`. The gravitational potential term is used here and is the topic of `aq-sr-gravitational-time-dilation-heuristic`; this article makes no claim about what that article contains.

Builds on. Time Dilation · Proper Time · Gravitational Time Dilation. Leads to. The Sagnac Effect · Michelson–Morley Isotropy Tests.

Is GPS a special-relativity effect or a general-relativity effect?

Both, and the answer is unbalanced: the gravitational term is about six times larger than the kinematic one and has the opposite sign. Keep only the special-relativistic piece — as a startling number of popular accounts do — and you predict the satellite clock runs slow by $7.2$ µs/day. It runs fast by $38.6$ . GPS is a general-relativistic system with a special-relativistic correction, not the other way round.

Why does the ground station's motion end up in the "gravitational" column?

Because the centrifugal potential is the kinematic term. For a clock rotating with the Earth, $-v_g^2/2c^2 = -\Omega^2r^2\sin^2\theta/2c^2$ — exactly the centrifugal contribution to the effective potential. They are not two effects that happen to be equal; they are one expression written twice. And you must put it somewhere: either inside $\Phi_{\text{eff}}$ (the geoid convention, which the literature uses) or outside it. Both give $+38.575$ µs/day. Only the first gives $-7.2$ and $+45.8$ .

Does the answer depend on which ground station you compare against?

No — and that is the point of the geoid, not an accident. It is defined as the level surface of the effective potential, so $\Phi_{\text{eff}} = \Phi_0$ everywhere on it and every clock on it runs at $1 + \Phi_0/c^2$ . An equatorial station is $21$ km further from the centre than a polar one and is moving at $465$ m/s; those effects cancel, and they cancel because the Earth's surface is an equilibrium shape — the oceans found the surface on which they do. Machine-checked: pole and equator agree to $0.00009$ µs/day. Split the potential and the rotation apart and that cancellation dies, and "the ground clock rate" becomes a function of latitude.

What is the null radius, and could you fly a satellite there?

$r_0 = 3GM/2|\Phi_0| = 9545.5$ km — altitude $3167$ km — is where the two effects exactly cancel and an orbiting clock needs no correction at all. You could fly there; nobody does. It sits deep in the inner Van Allen belt, whose proton flux is hostile to spacecraft electronics, and the period is under three hours, so a constellation would need many more satellites for the same coverage. The relativity is free there and everything else is expensive. Note what happens below it: the speed needed to stay in orbit costs more than the altitude buys, so a low-orbit clock runs slow — $-29.9$ µs/day for a grazing circular orbit.

If the factory offset handles the rate, why does the receiver do any relativity at all?

Because the factory offset is a constant and two of the effects are not. Eccentricity: a real orbit has $e \lesssim 0.02$ , so $r$ and $v$ oscillate once per revolution and so does the clock rate; the receiver adds $\Delta t_{\text{ecc}} = -2\sqrt{GM a}\,e\sin E/c^2$ , reaching $\pm45.8$ ns — about $13.7$ m of position. Sagnac: the Earth turns while the signal is in flight, so the receiver is not where it was at transmission; the correction is $\Delta t = 2\Omega A_z/c^2$ . Neither is optional at the metre level. The constant part is solved in the factory; the time-varying part is solved in your phone.

How do we know any of this is right, and not just self-consistent arithmetic?

NTS-2 settled it before the system existed: launched 23 June 1977 with the first caesium clock ever flown, its relativistic synthesiser deliberately left off for twenty days so the raw rate could be measured. General relativity predicted the orbiting clock would gain $+446.5$ parts in $10^{12}$ ; it gained $+442.5$ — agreement to $4$ parts in $10^{12}$ , within the clock's own stability, and it ended a genuine internal argument about whether the correction was needed at all. Since then the evidence has become industrial rather than experimental: thirty satellites, five decades, and a navigation error that would grow at $11.6$ km/day if the effect were not there.