The Rindler Metric and Position-Dependent Clock Rates

Put two atomic clocks in a rocket — one on the floor, one in the nose — and light the engine. The clocks drift apart, and they never stop. The nose clock runs fast, permanently, at a fixed fractional rate that depends only on how far apart they are and how hard the engine is burning. This is not a malfunction and it is not the twin paradox: both clocks are at rest with respect to each other, forever. And there is no gravity. The rocket is in deep space, the spacetime is Minkowski's, and the Riemann curvature tensor is zero at every event. That is the Rindler metric — flat spacetime, written in the coordinates an accelerating family would naturally use, in which the coefficient multiplying $\mathrm{d}\tau^2$ depends on where you are. Einstein saw this in 1907, and it is why he went looking for gravity in geometry rather than in a force law. Half a century later Pound and Rebka sent a gamma ray up a tower at Harvard and measured the number he had predicted: $2.46\times10^{-15}$ . Everything below happens in flat space.

Key takeaways

Signature $(+,-,-,-)$ : $\mathrm{d}s^2 = c^2\mathrm{d}t^2 - \mathrm{d}x^2$ , timelike intervals positive. Every sign below follows from that, and flipping it turns a real clock rate into an imaginary one.
The Rindler metric is Minkowski spacetime in the coordinates of a uniformly accelerated family. From the map $cT = x\sinh(g\tau/c)$ , $X = x\cosh(g\tau/c)$ it is $\;\mathrm{d}s^2 = +\left(\frac{gx}{c^2}\right)^2 c^2\mathrm{d}\tau^2 - \mathrm{d}x^2$ , equivalently $\;\mathrm{d}s^2 = \rho^2\mathrm{d}\eta^2 - \mathrm{d}\rho^2\;$ with $\eta$ dimensionless. Its time-time coefficient depends on position: a clock at $x$ ticks at $\mathrm{d}\tau_{\text{proper}} = (gx/c^2)\,\mathrm{d}\tau$ . Higher clocks run faster; at $x = 0$ they stop, and that is the Rindler horizon.
Owned result: the clock-rate ratio is $\Delta\tau_2/\Delta\tau_1 = x_2/x_1$ — a ratio of distances to the horizon, and it is EXACT. The familiar $1 + gh/c^2$ is not an approximation to it; it is it, provided $g$ is the proper acceleration at the lower clock.
Owned result: this is gravitational time dilation with $R^{\lambda}{}_{\mu\nu\kappa} = 0$ — computed here, not asserted. The clock gradient belongs to the observers, not the geometry.
Pound–Rebka: $gh/c^2 = 2.46\times10^{-15}$ over $h = 22.5$ m. The uniform-field prediction and Earth's real $1/r$ field differ by $8.7\times10^{-21}$ — tens of thousands of times below their error bar. They tested the equivalence principle; they could not have seen curvature.
Key idea: a position-dependent clock rate does not mean spacetime is curved. It means you are accelerating. The whole of gravitational redshift is already present in flat space — which is exactly why Einstein believed gravity had to be geometry.

The core insight

A metric is not something spacetime has — it is what a particular set of rulers and clocks reports. Hand the same flat Minkowski spacetime to an inertial observer and to an accelerating one and you get two different-looking line elements, both exactly correct, both describing the identical geometry. The accelerating one has $g_{\tau\tau}$ varying with height. Nothing has been added to the world: a different family of worldlines is doing the measuring, and their "at the same time, over there" is a fan of lines through a point rather than a stack of parallel slices. Everything strange that follows is that fan.

Where this lives today

The near-horizon limit of every black hole is a Rindler wedge. Expand the Schwarzschild metric about a point just outside the event horizon and to leading order it becomes $\mathrm{d}s^2 = \rho^2\mathrm{d}\eta^2 - \mathrm{d}\rho^2$ , with the surface gravity playing the reference acceleration and the proper distance to the horizon playing $\rho$ . This is why the flat-space calculation above is not a toy. A static observer hovering outside a black hole is, locally, a Rindler observer: the infinite redshift they see from the horizon is the $x_{\text{emit}} \to 0$ limit of (5), and the divergent thrust they need to hover is $\alpha = c^2/x \to \infty$ . Horizon physics can be rehearsed in flat spacetime first, and that is exactly what the field does.

Optical clocks now measure this in a laboratory. Modern optical lattice clocks reach fractional uncertainties below $10^{-18}$ and therefore resolve a centimetre of height. Relativistic geodesy — clock rate as an altimeter — has stopped being a thought experiment, and the redshift has now been resolved across a height difference of about a millimetre.

Rindler coordinates are the arena of the Unruh effect. Quantise a field on the wedge, trace over the region behind the horizon, and the Minkowski vacuum appears to the accelerated observer as a thermal bath at $T = \hbar\alpha/2\pi ck_B$ . The classical scaffolding is exactly the chart derived above: a horizon, a timelike Killing vector, and a notion of "static" only an accelerated observer possesses. That is `aq-sr-unruh-effect`'s business — but the reason the effect is possible is on this page, and it is the horizon.

The physical picture

A uniformly accelerated observer traces a hyperbola. Take a whole family of them, one through each point of the positive $x$ -axis, all asymptotic to the same pair of light rays through the origin: that is the Rindler congruence, and the wedge it fills, $X > |cT|$ , is the right Rindler wedge. Label each hyperbola by its distance $x$ from the origin at $T = 0$ . Fixed $x$ gives $X^2 - c^2T^2 = x^2$ : a worldline of constant proper acceleration. Fixed $\tau$ gives $X/cT = \coth(g\tau/c)$ : a straight line through the origin. Those straight lines are the family's surfaces of simultaneity, and they are the whole article. In an inertial frame the "now" slices are parallel and evenly spaced, and every clock at rest ticks alike. Here they are a fan, hinged at the origin — splayed wide at large $x$ , pinched near $x = 0$ . A clock far out must live through a lot of proper time between one slice and the next; one near the origin barely ages. The tick rate is set by how widely the fan opens where you stand, and it opens in proportion to $x$ .

Figure

PUNCHLINE — flat spacetime, and yet the clock rate depends on height. The coloured hyperbolae are the accelerated observers, one per Rindler coordinate x; the grey straight lines through the origin are their shared surfaces of simultaneity. The fan of nows is pinched near the origin and splayed far away, so equal steps of Rindler time buy unequal proper time: dτ_proper = (gx/c²)·dτ. The bar at the bottom is the tick rate — 0 at the horizon, exactly 1 on the reference worldline x = c²/g, unbounded beyond. Everything you see is Minkowski spacetime. R = 0 at every event on this diagram

The reference worldline is the one at $x = c^2/g$ : it, and only it, has proper acceleration exactly $g$ , and on it proper time equals coordinate time. Everyone else is calibrated against it — so $\tau$ is a bookkeeping label, not anybody's watch. That is the first thing that trips people, and it is the same convention Schwarzschild coordinates use, for the same reason.

The core equations

The map from Minkowski to Rindler, on the right wedge $X > |cT|$ :

cT = x\,\sinh\!\left(\frac{g\tau}{c}\right), \qquad X = x\,\cosh\!\left(\frac{g\tau}{c}\right), \qquad x > 0 \qquad (1)

where: $x$ — the spatial Rindler coordinate, constant along each accelerated worldline; $\tau$ — the shared Rindler coordinate time; $g$ — the reference proper acceleration, fixing the scale of $\tau$ . Physical reading: a change of coordinates on flat spacetime, nothing more, one-to-one on the wedge. Substituting (1) into $\mathrm{d}s^2 = c^2\mathrm{d}T^2 - \mathrm{d}X^2$ :

\boxed{\;\mathrm{d}s^{2} \;=\; +\left(\frac{g x}{c^{2}}\right)^{2} c^{2}\,\mathrm{d}\tau^{2} \;-\; \mathrm{d}x^{2}\;} \qquad\Longleftrightarrow\qquad g_{\tau\tau} = +\left(\frac{gx}{c^{2}}\right)^{2}c^{2}, \quad g_{xx} = -1 \qquad (2)

where: the $+$ on time and $-$ on space are the signature $(+,-,-,-)$ . Physical reading: $g_{xx} = -1$ says $x$ is proper distance along a slice of constant $\tau$ — Rindler rulers are honest. $g_{\tau\tau}$ depends on $x$ : Rindler clocks are not. That single asymmetry is the article. With a dimensionless time $\eta = g\tau/c$ and $\rho \equiv x$ , every constant disappears:

cT = \rho\sinh\eta, \quad X = \rho\cosh\eta \qquad\Longrightarrow\qquad \boxed{\;\mathrm{d}s^{2} = \rho^{2}\,\mathrm{d}\eta^{2} - \mathrm{d}\rho^{2}\;} \qquad (3)

where: $\eta$ is a rapidity, not a time — a dimensionless hyperbolic angle. Physical reading: the Minkowski analogue of polar coordinates' $\mathrm{d}s^2 = \mathrm{d}\rho^2 + \rho^2\mathrm{d}\theta^2$ , hyperbolic angle in place of circular. Nobody thinks polar coordinates curve the plane; nobody should think Rindler coordinates curve spacetime. Setting $\mathrm{d}x = 0$ in (2) and reading off $c^2\mathrm{d}\tau_{\text{proper}}^2 = \mathrm{d}s^2$ :

\boxed{\;\frac{\mathrm{d}\tau_{\text{proper}}}{\mathrm{d}\tau} = \frac{g x}{c^{2}}\;} \qquad\Longrightarrow\qquad \boxed{\;\frac{\Delta\tau_{2}}{\Delta\tau_{1}} = \frac{x_{2}}{x_{1}}\;} \qquad (4)

where: $\Delta\tau_i$ is the proper time logged at fixed $x_i$ over a shared coordinate interval. Physical reading: the ratio of clock rates is the ratio of the clocks' distances to the horizon, and $g$ has cancelled out of it entirely. Exact for any $x_1, x_2 > 0$ — there is nothing to expand. Proper acceleration, and the redshift of light climbing between two heights:

\alpha(x) = \frac{c^{2}}{x}, \qquad\qquad \boxed{\;\frac{\nu_{\text{rec}}}{\nu_{\text{emit}}} = \frac{x_{\text{emit}}}{x_{\text{rec}}}\;} \qquad (5)

where: $\alpha(x)$ is what an accelerometer at $x$ reads. Physical reading: $\alpha$ diverges at the horizon while the clock rate vanishes there — two faces of the same $1/x$ . Light climbing outward arrives redshifted, because the receiver's clock runs fast and counts fewer crests per second of its own time. The profile $\alpha = c^2/x$ is owned by `aq-sr-born-rigidity`; used here, not re-derived.

Where the metric comes from

Derivation AFrom Minkowski to Rindler and the sign that every base gets wrong

The goal is to obtain the line element (2) by direct substitution with every sign carried explicitly, because it is the one equation in this chapter most often printed wrong. The AllQFT base for this article — and for three of its four neighbours — prints it with both signs reversed: the $(-,+,+,+)$ convention wearing our notation. In our signature the time part carries the plus. The derivation is three lines, and there is no excuse for transcribing it from anywhere.

Step 1. Differentiate the map. Writing $S \equiv \sinh(g\tau/c)$ , $C \equiv \cosh(g\tau/c)$ ,

\mathrm{d}(cT) = \frac{gx}{c}\,C\,\mathrm{d}\tau + S\,\mathrm{d}x, \qquad \mathrm{d}X = \frac{gx}{c}\,S\,\mathrm{d}\tau + C\,\mathrm{d}x .

Why: the factor $g/c$ comes out of $\partial_\tau\sinh(g\tau/c)$ and is the only place the reference acceleration enters.

Step 2. Form the interval in our signature, $\mathrm{d}s^2 = \mathrm{d}(cT)^2 - \mathrm{d}X^2$ — time positive, space negative:

\mathrm{d}s^{2} = \frac{g^{2}x^{2}}{c^{2}}\big(C^{2} - S^{2}\big)\mathrm{d}\tau^{2} \;+\; \big(S^{2} - C^{2}\big)\mathrm{d}x^{2} \;+\; 2\frac{gx}{c}\big(CS - SC\big)\mathrm{d}\tau\,\mathrm{d}x .

Why: the cross terms cancel identically, $CS - SC = 0$ , which is the entire reason this chart is any use — its time axis is exactly orthogonal to its space slices.

Step 3. Apply $C^2 - S^2 = 1$ :

\mathrm{d}s^{2} = \frac{g^{2}x^{2}}{c^{2}}\,\mathrm{d}\tau^{2} - \mathrm{d}x^{2} = +\left(\frac{gx}{c^{2}}\right)^{2}c^{2}\,\mathrm{d}\tau^{2} - \mathrm{d}x^{2} .

Why: $(gx/c^2)^2c^2 = g^2x^2/c^2$ — the same number written two ways, and the second makes it obvious that $g_{\tau\tau}$ is dimensionally a squared velocity, as it must be when $\tau$ is a time.

Step 4. Sanity-check the signs against physics. A clock at rest at $x$ has $\mathrm{d}x = 0$ and $\mathrm{d}s^2 = (gx/c^2)^2c^2\mathrm{d}\tau^2 > 0$ — positive, which in $(+,-,-,-)$ is what timelike means, giving a real $\mathrm{d}\tau_{\text{proper}} = \sqrt{\mathrm{d}s^2}/c$ . Flip the signs the way the base does and $\mathrm{d}s^2 < 0$ for a clock sitting still: its worldline becomes spacelike and its proper time imaginary. Why: a signature is a choice, but using one convention's formula inside another's is an error — and it announces itself in four seconds if you ask what a stationary clock does.

Step 5. Repeat with $\eta = g\tau/c$ dimensionless: same three lines give $\mathrm{d}s^2 = \rho^2\mathrm{d}\eta^2 - \mathrm{d}\rho^2$ . Why: this is the form worth memorising, and it makes the "singularity" at $\rho = 0$ plainly the same kind of thing as the one at the origin of a polar chart. *Verified symbolically in verify_aq-sr-rindler-metric.py: sympy substitutes the map into $\mathrm{d}(cT)^2 - \mathrm{d}X^2$ and returns all three charts exactly, including the Møller form $+(1+a\xi/c^2)^2c^2\mathrm{d}\tau^2 - \mathrm{d}\xi^2$ ; the cross term is confirmed identically zero; and the script greps this article to confirm the mostly-plus string appears nowhere except where it is named as the base's defect.*

Go deeper · Derivation B · 2 minThe clock-rate law is exact and it is a statement about the horizon

Derivation BThe clock-rate law is exact and it is a statement about the horizon

Every textbook writes gravitational time dilation as $1 + gh/c^2$ and calls it the weak-field limit. In a uniform field that is misleading: the law is exact, and calling it an approximation hides what it is really about. The goal here is the exact statement — from which $g$ will drop out entirely — and then an honest account of what is actually being approximated when someone writes a " $\dots$ " after the linear term. The answer is: not the field strength, but whose $g$ you used.

Step 1. Hold a standard clock at fixed $x$ . Set $\mathrm{d}x = 0$ in (2): $\mathrm{d}s^2 = (gx/c^2)^2c^2\mathrm{d}\tau^2$ , so $\mathrm{d}\tau_{\text{proper}} = (gx/c^2)\,\mathrm{d}\tau$ . Why: a metric coefficient is a conversion factor between a coordinate and a measurement, and here it is a function of position. That is the whole physical content.

Step 2. Compare two clocks over the same coordinate interval $\Delta\tau$ , legitimate because $\tau$ is shared:

\frac{\Delta\tau_{2}}{\Delta\tau_{1}} = \frac{(g x_{2}/c^{2})\,\Delta\tau}{(g x_{1}/c^{2})\,\Delta\tau} = \frac{x_{2}}{x_{1}} .

The reference acceleration has cancelled. Why: this is the owned result. The ratio does not know about $g$ ; it knows only the two clocks' distances from the horizon. Change $g$ and you merely re-label the time coordinate — and a re-labelling cannot change a ratio of clock rates, which is exactly what you can watch fail to happen here.

Step 3. Translate into the language of a "field". Lower clock at $x_1$ , upper at $x_2 = x_1 + h$ . The proper acceleration at the lower clock is $g_1 = c^2/x_1$ , by (5). Then

\frac{\Delta\tau_{2}}{\Delta\tau_{1}} = \frac{x_{1} + h}{x_{1}} = 1 + \frac{h}{x_{1}} = \boxed{\,1 + \frac{g_{1}h}{c^{2}}\,} \qquad\text{— exactly, with no higher-order terms.}

Why: this is the point. The famous $1 + gh/c^2$ is not the first term of a series; in a uniform field it is the whole answer, provided $g$ is the local proper acceleration at the emitter. A series appears only if you insist on the acceleration at the upper clock, $g_2 = c^2/x_2$ , giving $(1 - g_2h/c^2)^{-1} = 1 + g_2h/c^2 + (g_2h/c^2)^2 + \cdots$ . Same physics, different bookkeeping. The "weak field" being assumed is really "I do not care which end I measured $g$ at."

Step 4. Get the redshift by counting crests. A source at $x_{\text{emit}}$ emits $N$ crests in its proper time $\Delta\tau_{\text{emit}}$ ; the wave is static in the Rindler frame, so the same $N$ crests arrive at $x_{\text{rec}}$ over the same coordinate interval, hence over the receiver's proper time $\Delta\tau_{\text{rec}}$ . Frequencies are $N/\Delta\tau$ :

\frac{\nu_{\text{rec}}}{\nu_{\text{emit}}} = \frac{\Delta\tau_{\text{emit}}}{\Delta\tau_{\text{rec}}} = \frac{x_{\text{emit}}}{x_{\text{rec}}} .

Why: crest-counting shows nothing happens to the photon en route — it does not "lose energy climbing", having no potential energy to lose. Emitter and receiver simply keep different time. And as $x_{\text{emit}} \to 0$ the ratio goes to zero: infinite redshift from the horizon, where the clock rate vanishes and the thrust needed to hover diverges. The next card proves that spacetime itself does nothing at all at that surface. *Verified numerically in verify_aq-sr-rindler-metric.py: $\Delta\tau_2/\Delta\tau_1 = x_2/x_1$ checked with $g$ symbolic and confirmed $g$ -independent by differentiation; $1 + g_1h/c^2$ shown exact; $(1 - g_2h/c^2)^{-1}$ shown to be its geometric series; and the redshift $\nu_{\text{rec}}/\nu_{\text{emit}} = x_{\text{emit}}/x_{\text{rec}}$ checked against the clock ratio.*

Go deeper · Derivation C · 2 minThe Riemann tensor of the Rindler metric is zero and here is the computation

Derivation CThe Riemann tensor of the Rindler metric is zero and here is the computation

The thesis of this chapter is that special relativity handles acceleration perfectly well, and that the folklore "you need general relativity for accelerated frames" is false. Nothing settles it more cleanly than computing the curvature of the metric just derived and finding it zero. This card does the computation rather than asserting the answer — because a metric with position-dependent $g_{\tau\tau}$ looks like a gravitational field, and the only way to be sure it is not one is to differentiate.

Step 1. Use the clean chart, $\mathrm{d}s^2 = \rho^2\mathrm{d}\eta^2 - \mathrm{d}\rho^2$ : $g_{\eta\eta} = \rho^2$ , $g_{\rho\rho} = -1$ , $g_{\eta\rho} = 0$ ; inverse $g^{\eta\eta} = 1/\rho^2$ , $g^{\rho\rho} = -1$ . Why: curvature is a property of the geometry, not the chart, so compute in whichever chart hurts least.

Step 2. Christoffels, from $\Gamma^{\lambda}{}_{\mu\nu} = \tfrac12 g^{\lambda\sigma}(\partial_\mu g_{\sigma\nu} + \partial_\nu g_{\sigma\mu} - \partial_\sigma g_{\mu\nu})$ . The only non-zero derivative is $\partial_\rho g_{\eta\eta} = 2\rho$ , and it produces exactly two independent symbols:

\Gamma^{\eta}{}_{\eta\rho} = \Gamma^{\eta}{}_{\rho\eta} = \frac{1}{\rho}, \qquad \Gamma^{\rho}{}_{\eta\eta} = \rho .

Why: non-zero Christoffels are not curvature. They are non-zero in polar coordinates on a flat plane too. A connection coefficient measures how your coordinate basis twists; a curvature tensor measures whether parallel transport around a closed loop returns you rotated. Confusing the two generates the folklore.

Step 3. Assemble $R^{\lambda}{}_{\mu\nu\kappa} = \partial_\nu\Gamma^{\lambda}{}_{\mu\kappa} - \partial_\kappa\Gamma^{\lambda}{}_{\mu\nu} + \Gamma^{\lambda}{}_{\nu\sigma}\Gamma^{\sigma}{}_{\mu\kappa} - \Gamma^{\lambda}{}_{\kappa\sigma}\Gamma^{\sigma}{}_{\mu\nu}$ . In two dimensions there is one independent component, and it is this:

R^{\rho}{}_{\eta\rho\eta} = \partial_\rho\Gamma^{\rho}{}_{\eta\eta} - \partial_\eta\Gamma^{\rho}{}_{\eta\rho} + \Gamma^{\rho}{}_{\rho\sigma}\Gamma^{\sigma}{}_{\eta\eta} - \Gamma^{\rho}{}_{\eta\sigma}\Gamma^{\sigma}{}_{\eta\rho} = \underbrace{1}_{\partial_\rho(\rho)} - 0 + 0 - \underbrace{\rho\cdot\tfrac{1}{\rho}}_{\Gamma^{\rho}{}_{\eta\eta}\Gamma^{\eta}{}_{\eta\rho}} = 1 - 1 = 0 .

Why: the derivative and quadratic terms cancel exactly. Not luck — this metric came from a flat one by a change of coordinates, and curvature is a tensor: vanishing in one chart, it vanishes in all.

Step 4. Draw the conclusion, and draw it hard. $R^{\lambda}{}_{\mu\nu\kappa} = 0$ everywhere on the wedge. There is no gravitational field here and no tidal force: two test particles released side by side inside the rocket do not converge — they drift off along straight inertial lines. Everything the metric exhibits is present in flat spacetime. Why: general relativity is not needed for acceleration. It is needed for curvature, that is for tidal effects, that is for exactly the quantity we have just computed to be zero. That is the boundary, and `aq-sr-equivalence-principle` makes it precise. *Verified symbolically in verify_aq-sr-rindler-metric.py: sympy builds $g_{\mu\nu}$ for both $\rho^2\mathrm{d}\eta^2 - \mathrm{d}\rho^2$ and $(gx/c^2)^2c^2\mathrm{d}\tau^2 - \mathrm{d}x^2$ , computes every Christoffel symbol and all sixteen Riemann components from the definition, and confirms each is identically zero; the two non-vanishing Christoffels are checked against the values printed above.*

Figure

The tick-rate curve dτ_proper/dτ = gx/c² over the whole Rindler wedge — a straight line through the origin: zero at the horizon, exactly 1 on the reference worldline x = c²/g = 9.16×10¹⁵ m for g = 1 gee, climbing without bound beyond. The punchline is the red sliver. The ENTIRE Pound–Rebka experiment — a 22.5 m tower at Harvard — lives inside a slice of this curve 22.5 m wide out of 9.16×10¹⁵ m, and the fractional clock-rate difference they measured is gh/c² = 2.46×10⁻¹⁵. The curve is STRAIGHT, so the effect is exactly linear in height: there is no second-order term to find, and the tower could have been a thousand times taller without one appearing

Limiting cases

Case	Result	Meaning
$h \ll x_1$	$\Delta\tau_2/\Delta\tau_1 = 1 + g_1h/c^2$	the "weak-field" formula — exact in a uniform field when $g_1$ is the emitter's value
$x \to 0$	$g_{\tau\tau} \to 0$ , $\alpha \to \infty$	the Rindler horizon. Clocks stop; the thrust needed to hover diverges
$g \to 0$	$x = c^2/g \to \infty$	the horizon recedes to infinity and the chart becomes Minkowski's
Riemann	$R^{\lambda}{}_{\mu\nu\kappa} = 0$	everywhere, identically. No curvature, no tidal forces, no gravity — and a clock gradient regardless

Common trapSpecial relativity handles acceleration perfectly well — this article is the proof

The folklore that "you need general relativity to deal with accelerated frames" is false. Every result on this page — metric, clock gradient, redshift, horizon — was obtained by substituting a change of variables into the Minkowski line element. That is all. The spacetime is flat, the Riemann tensor is zero, and you may transform back to inertial coordinates whenever you like, whereupon the whole apparent field evaporates. What general relativity supplies is curvature, which shows up as tidal effects: the relative acceleration of neighbouring freely falling particles, which no choice of coordinates removes. In the Rindler wedge that relative acceleration is exactly zero. The line where SR stops and GR begins is not "acceleration" — it is the tidal tensor of `aq-sr-equivalence-principle`.

Common trapThe signature, and what happens if you take it from the base

In $(+,-,-,-)$ the Rindler metric is $\mathrm{d}s^2 = +(gx/c^2)^2c^2\mathrm{d}\tau^2 - \mathrm{d}x^2$ . The AllQFT base prints $\mathrm{d}s^2 = -(gx/c^2)^2c^2\mathrm{d}\tau^2 + \mathrm{d}x^2$ — in its definition string, its first summary bullet, its metric-components card and its derivation — and so do the bases for the Rindler horizon, Born rigidity and the equivalence principle. That is $(-,+,+,+)$ , and importing it into a $(+,-,-,-)$ corpus is not a stylistic difference: a clock at rest in the Rindler frame acquires $\mathrm{d}s^2 < 0$ , becomes spacelike, and reports an imaginary proper time. Note $g_{xx}$ flips too — it is $-1$ , not $+1$ . The check takes four seconds: ask what a stationary clock does. Never transcribe a metric. Re-derive it from the map.

Worked examples

1 · Pound and Rebka, done properly

Gamma rays of $14.4$ keV from $^{57}\mathrm{Fe}$ are sent up the $22.5$ m Jefferson tower at Harvard. What shift does the Rindler clock-rate law predict, what did the experiment measure, and what quantity did it measure?

Step 1 — identify and apply. Emitter and receiver sit at $x_1 = c^2/g$ and $x_2 = x_1 + h$ , so by (5), $\dfrac{\nu_{\text{rec}}}{\nu_{\text{emit}}} = \dfrac{x_1}{x_1 + h} \Longrightarrow \dfrac{\Delta\nu}{\nu} = -\dfrac{h}{x_1 + h} \approx -\dfrac{gh}{c^{2}}$ , the approximation good to one part in $10^{15}$ .

Step 3 — the number. $\dfrac{gh}{c^{2}} = \dfrac{(9.80665)(22.5)}{(2.99792458\times10^{8})^{2}} = \dfrac{220.65}{8.98755\times10^{16}} = \mathbf{2.46\times10^{-15}}$ — about $8.6$ kHz on a $3.5\times10^{18}$ Hz gamma ray, resolvable only because the Mössbauer effect gives the $^{57}\mathrm{Fe}$ line a fractional width of order $10^{-12}$ .

Step 4 — what was actually measured, and it is not that number. The experiment ran both ways — source at the top, then at the bottom — and the observable was the difference, which cancels every shift that does not reverse with the direction of travel. So the quantity Pound and Rebka print is twice the one-way shift: a predicted $\mathbf{-4.92\times10^{-15}}$ (their tower is 74 feet, $22.56$ m — that is where the last digit comes from) against a measured $\mathbf{-(5.13 \pm 0.51)\times10^{-15}}$ , a published ratio of observed to predicted of $\mathbf{1.05 \pm 0.10}$ — agreement to about $10\%$ . Pound and Snider reached about $1\%$ in 1964–65. The widely-repeated one-way figure " $2.57\times10^{-15}$ " appears nowhere in the paper: it is half of $5.13$ , a derived number rather than a measured one.

Step 5 — the boundary number. Now ask the question this chapter exists to ask. The Rindler prediction assumes a uniform field; Earth's goes as $1/r^2$ , so the true potential difference $\Delta\Phi = GMh/(r_1r_2)$ falls short of $g_1h$ by the fraction $h/(r_1+h) = 3.53\times10^{-6}$ . In absolute terms the uniform-field answer overshoots the real one by $\mathbf{8.7\times10^{-21}}$ — while their one-sigma error bar, $0.51\times10^{-15}$ two-way, is $\mathbf{2.9\times10^{4}}$ times larger. They could not possibly have distinguished Earth's real field from a uniformly accelerating rocket. That is not a criticism: it is the precise sense in which theirs is a test of the equivalence principle and not of curvature. Curvature is a different measurement, and it is `aq-sr-equivalence-principle`'s subject.

2 · How slowly can a clock in a rocket be made to run?

A rocket's nose feels $1\,g$ . A clock sits on the hull half-way between the nose and the Rindler horizon. How fast does it tick, and how hard is it accelerating?

Step 1 — identify. The nose is the reference worldline: $\alpha_{\text{nose}} = g$ , so $x_{\text{nose}} = c^2/g = 9.16\times10^{15}$ m, and half-way to the horizon is $x = 4.58\times10^{15}$ m.

Step 2 — apply the exact law. From (4), $\Delta\tau_{\text{clock}}/\Delta\tau_{\text{nose}} = x/x_{\text{nose}} = \tfrac12$ . The clock runs at exactly half the nose clock's rate, permanently — not approximately half. Over a nose-year it logs six months.

Step 3 — the acceleration there. $\alpha = c^2/x = 2g = \mathbf{19.6\ m/s^{2}}$ . Halving the distance to the horizon doubles the felt weight and halves the tick rate: the same $1/x$ , read two ways.

Step 4 — check against the "weak-field" formula. Naively, $1 + gh/c^2$ with $h = -\tfrac12 c^2/g$ gives $\tfrac12$ . It agrees exactly, at a field strength nobody would call weak — because it was never an approximation, so long as $g$ is the emitter's value, and here the emitter is the nose.

Step 5 — the boundary number. This rocket is $4.6\times10^{15}$ m long, about half a light-year, and nothing has broken. But push the clock to $x = 0$ and its tick rate hits zero while its required thrust diverges: a rocket cannot be longer than $c^2/\alpha_{\text{nose}}$ , or its tail would sit behind the horizon. That bound is owned by `aq-sr-born-rigidity` and sharpened for the body itself in `aq-sr-born-rigidity-s6`. Here it is simply where the metric runs out.

Physics of the portrait

Gravitational redshift is the oldest surviving prediction of general relativity, and it was made from an accelerating box with no field equation in sight.

Origins · 2 minGravitational redshift and how this was discovered

Gravitational redshift

1907–1965 · Einstein's accelerating box · Rindler's wedge · the Jefferson tower at Harvard · a gamma ray weighed against a Mössbauer line

The prediction. In 1907, invited to review relativity for the Jahrbuch der Radioaktivität und Elektronik (4, 411–462), Einstein gave its fifth and final part to gravitation and did something nobody had asked for: he assumed a uniformly accelerated frame to be completely equivalent to a frame at rest in a homogeneous gravitational field, and computed the consequences. Out came the position-dependent clock rate, and out came the redshift of light climbing a potential — eight years before the field equations, twelve before Eddington's eclipse. The argument was the accelerating rocket, and it is the one reproduced above. He returned to it in 1911, sharpening the redshift and adding the deflection of starlight, which the 1907 paper does not contain. The measurement. Robert Pound and his graduate student Glen Rebka Jr. mounted a $^{57}\mathrm{Fe}$ source at one end of the 74-foot Jefferson tower at Harvard and an absorber at the other, using the Mössbauer effect — recoil-free gamma emission, found by Rudolf Mössbauer in 1958 — for a line sharp enough to resolve a few parts in $10^{15}$ . They ran it both ways and took the difference. Announced in 1960, it was the first laboratory measurement of the gravitational redshift, and it agreed with Einstein's 1907 number to about ten per cent.

History

1907

— Einstein, in his Jahrbuch der Radioaktivität und Elektronik review (vol. 4, 411–462), postulates the complete physical equivalence of a uniformly accelerated frame and a homogeneous gravitational field, and derives from it both gravitational time dilation and the redshift of light climbing a potential.

1911

— Einstein returns to the equivalence argument in "Über den Einfluss der Schwerkraft auf die Ausbreitung des Lichtes", sharpening the redshift and adding the deflection of starlight, which the 1907 paper did not contain.

1958

— Mössbauer discovers recoil-free nuclear resonance absorption, making a gamma line sharp enough to weigh a photon against a 22 m tower.

1960

— Pound and Rebka measure the gravitational redshift over the 74-foot Jefferson tower at Harvard, running the beam both up and down and publishing the two-way difference: predicted −4.92 × 10⁻¹⁵, measured −(5.13 ± 0.51) × 10⁻¹⁵, a ratio of 1.05 ± 0.10.

1964–65

— Pound and Snider repeat the experiment with better systematics and reach roughly one per cent.

1966

— Rindler's Am. J. Phys. treatment of the uniformly accelerated frame and its causal boundary fixes the coordinates, and the horizon, that carry his name; the term "event horizon" is his too, coined a decade earlier in a cosmological paper.

Going further

Texts. Rindler, Introduction to Special Relativity, 2nd ed. · Misner, Thorne & Wheeler, Gravitation, §6.6 · Wald, General Relativity, §6.4 · Pound & Rebka, Phys. Rev. Lett. 4, 337 (1960) · Pound & Snider, Phys. Rev. 140, B788 (1965).

Provenance. Signature $(+,-,-,-)$ . Every equation and number above is recomputed in verify_aq-sr-rindler-metric.py, which also reads this file and asserts each printed value against what it computed. Base defects repaired: (i) the AllQFT base prints the Rindler metric as $\mathrm{d}s^2 = -(gx/c^2)^2c^2\mathrm{d}\tau^2 + \mathrm{d}x^2$ — a $(-,+,+,+)$ artefact — in its definition string, its first summary bullet, its metric-components card and its derivation; corrected throughout, and its $g_{xx} = 1$ likewise becomes $g_{xx} = -1$ . (ii) the base's Pound–Rebka figure of $2.5\times10^{-15}$ comes from $g = 9.8$ , $c = 3\times10^{8}$ ; with proper constants it is $2.46\times10^{-15}$ . (iii) the base calls that "the value Pound and Rebka measured" — it is the value they predicted, and what they published was the two-way difference.

Scope. The hyperbolic worldline and the profile $\alpha(x) = c^2/x$ are owned by `aq-sr-relativistic-uniform-acceleration` and `aq-sr-born-rigidity`; used, not re-derived. The wedge's causal structure is `aq-sr-rindler-horizon`; the step to a gravitational field, and the tidal tensor that limits it, is `aq-sr-equivalence-principle`.

Builds on. Relativistic Uniform Acceleration · Rindler Coordinates · Proper Acceleration. Leads to. The Rindler Horizon · The Equivalence Principle · Born Rigidity.

If the spacetime is flat, why do the clocks disagree at all?

Because they are not comparing themselves against spacetime — they are comparing themselves against each other, along a shared notion of "now" they had to invent. Their simultaneity slices are a fan of lines hinged at the origin, not a stack of parallel ones: pinched near the hinge, splayed far from it. A clock at large $x$ therefore lives through more proper time between two neighbouring slices than one at small $x$ . The disagreement is about the slicing, and the slicing is a choice their own acceleration forced on them. Flatness is about curvature; this is about coordinates.

Does the photon lose energy climbing out of the well?

That is the standard story and it is a poor one, because it invites you to ask what a photon's potential energy is, and a photon does not have one. Crest-counting is better: the source emits $N$ crests, the receiver receives exactly those $N$ — none are created or destroyed en route — and the two disagree about how long that took, because their clocks run at different rates. Frequency is crests per second, the count is agreed, so the frequencies differ by the ratio of clock rates and by nothing else. Nothing happens to the light. Something is different about the two ends of its journey.

Is τ anybody's time?

Only the reference observer's, at $x = c^2/g$ . For everyone else $\tau$ is a coordinate — a shared label the accelerated observers agree to use so they can say "now" to one another, at the price of its not being their own watch. This is exactly the role Schwarzschild's $t$ plays for a static observer far from a black hole, and it is why "coordinate time" and "proper time" must be separate words.

Why does x measure real distance if τ does not measure real time?

Because $g_{xx} = -1$ while $g_{\tau\tau}$ is not $\pm c^2$ . Along a slice of constant $\tau$ the metric gives $\mathrm{d}s^2 = -\mathrm{d}x^2$ : a spacelike interval of proper length exactly $|\mathrm{d}x|$ . The coordinate difference between two Rindler observers is the proper distance between them on their common slice, and it never changes — which is precisely what makes this the natural chart for a Born-rigid body. The clocks carry all the position dependence instead.

Could you tell you were in a rocket rather than a gravitational field by measuring the clock gradient?

No — that is the point, and it is why the equivalence principle works. But you could tell by measuring the tidal field, which is why the principle is only local. Release two test masses side by side. In the rocket they drift along straight inertial lines and nothing surprising happens: the tidal tensor is zero, because $R^{\lambda}{}_{\mu\nu\kappa} = 0$ . In a real field they converge, because the field lines converge, and no change of coordinates removes that. The boundary is `aq-sr-equivalence-principle`'s, and the residue is real: for a 2 m box on Earth's surface it is $6.2\times10^{-6}\ \mathrm{m/s^2}$ — small, and not zero.

The metric coefficient vanishes at x = 0. Is that a real singularity?

No. It is a coordinate singularity of exactly the same species as the origin of a polar chart, where $g_{\theta\theta} = \rho^2$ also vanishes and the angular coordinate goes degenerate. The Riemann tensor is zero at $x = 0$ and everywhere else, and an inertial observer sails straight through the surface $X = cT$ noticing nothing. What is not an artefact is the causal structure: the surface really is a one-way boundary for the accelerated family. A coordinate singularity marking a genuine causal boundary is exactly the situation at a black hole's event horizon — which is why the analogy is worth taking seriously, and `aq-sr-rindler-horizon` makes it exact.