Rindler Coordinates and the Uniformly Accelerated Frame

Polar coordinates are not a new geometry. They are a new grid on the same flat plane, adapted to the one symmetry you care about — rotation about a point. Nothing curves. The metric merely stops looking like $dx^2 + dy^2$ and starts looking like $dr^2 + r^2 d\theta^2$ , and that innocent $r^2$ is a coordinate fact, not a curvature one. Rindler coordinates are that same construction on Minkowski spacetime, with the boost in place of the rotation. The circles become hyperbolae, $\sin$ and $\cos$ become $\sinh$ and $\cosh$ , and the origin becomes a pair of null lines. The result is a chart in which a uniformly accelerated observer sits still: the accelerated frame, written down exactly, in flat spacetime, with no approximation of any kind. And then the metric comes out position-dependent — $ds^2 = \rho^2 d\eta^2 - d\rho^2$ — clocks at different $\rho$ tick at different rates, there is a horizon, and every reflex in your training says gravity. It is not. The Riemann tensor of this metric is identically zero, all twenty components, everywhere. What is not zero is the connection — $\Gamma^{\eta}{}_{\eta\rho} = 1/\rho$ — and it is the connection coefficients that produce every one of those effects. A nonzero connection is not curvature, and acceleration is not gravity. The chapter's whole thesis lives in that one computation, and this is where it is done.

Key takeaways

Signature $(+,-,-,-)$ : $\eta_{\mu\nu} = \mathrm{diag}(+1,-1,-1,-1)$ , $ds^2 = c^2dt^2 - dx^2$ . The base for this article is mostly-plus; every metric below is re-derived, none transcribed.
The map (right wedge, dimensionless $\eta$ ): $\;\boxed{cT = \rho\sinh\eta,\qquad X = \rho\cosh\eta}$ . Constant $\rho$ traces the hyperbola $X^2 - c^2T^2 = \rho^2$ ; constant $\eta$ is a straight ray through the origin.
Owned result — the metric, re-derived in our lock: $\;\boxed{ds^2 = \rho^2\,d\eta^2 - d\rho^2}$ . Cross terms cancel identically; $\cosh^2 - \sinh^2 = 1$ collapses the $d\rho^2$ coefficient to $-1$ . Scaled time $\eta_a = (c/a)\,\eta$ gives the same chart as $ds^2 = (a\rho/c)^2 d\eta_a^2 - d\rho^2$ .
Owned result — the acceleration law: at fixed $\rho$ the four-acceleration is $A^\mu = (0,\,c^2/\rho)$ with $A\!\cdot\!A = -c^4/\rho^2 < 0$ (spacelike, as our signature demands), hence $\;\boxed{\alpha(\rho) = c^2/\rho}$ . Closer to the horizon means harder acceleration. The array is rigid; it is not equally accelerated.
Owned result — the structural one. $R^{\mu}{}_{\nu\rho\sigma} = 0$ identically in Rindler coordinates, while $\Gamma^{\eta}{}_{\eta\rho} = 1/\rho$ and $\Gamma^{\rho}{}_{\eta\eta} = \rho$ are not. Clock gradient, redshift and horizon all come from the connection. None of them is gravity.
The chart covers a quarter of spacetime: $\rho > 0$ maps onto the right wedge $X > |cT|$ only. Its boundary $\rho = 0$ is the null pair $X = \pm cT$ — the Rindler horizon.
A boost is a translation in $\eta$ . $\partial_\eta$ is the boost Killing vector, norm $\rho^2$ — timelike inside the wedge, null on the horizon. That sentence is why the chart exists.
The scale at $1\,g$ : $\rho_0 = c^2/g = 9.1648\times10^{15}\ \mathrm{m} = 0.9687$ ly. Your horizon is one light-year behind you.
Key idea: Rindler coordinates are polar coordinates for the Lorentz group. The metric acquires a position-dependent lapse for exactly the reason $dr^2 + r^2d\theta^2$ acquires an $r^2$ — and it says exactly as much about curvature, which is nothing.

The core insight

The Minkowski plane has one continuous symmetry fixing the origin: the boost. Rindler coordinates are adapted to it, exactly as polar coordinates are adapted to rotation. Label a point by the invariant $\rho = \sqrt{X^2 - c^2T^2}$ — a hyperbolic "radius" — and by the boost parameter $\eta$ needed to reach it from the $T = 0$ axis. Constant $\rho$ is then the worldline of a uniformly accelerated observer, constant $\eta$ their shared simultaneity slice, and the flat metric becomes $\rho^2 d\eta^2 - d\rho^2$ . Every strange thing about the accelerated frame — a position-dependent lapse, a horizon at $\rho = 0$ , an acceleration $c^2/\rho$ that blows up as you approach it — is visible in that one line element, and every one is a statement about the grid, not the space.

Where this lives today

The Rindler wedge is the standard workbench for quantum field theory in curved spacetime — which is a joke, because there is no curvature in it. Quantising a field on the wedge and comparing with the Minkowski vacuum yields the Unruh effect (Fulling 1973, Davies 1975, Unruh 1976): the accelerated observer sees a thermal bath. This article does not derive it — `aq-sr-unruh-effect` owns it — but the reason the derivation is possible is entirely the content of this one. The chart is exactly solvable, the boost is a Killing vector, and $\rho = 0$ is a genuine causal horizon; those three facts are what let you define a Rindler vacuum and Bogoliubov-transform it against the Minkowski one. It is also the flat-spacetime rehearsal for Hawking radiation, with the black-hole horizon in the part of $\rho = 0$ . Closer to home, the chart is the accelerated-frame chapter's working tool: the natural rest frame of a Born-rigid body (`aq-sr-born-rigidity-s6`), the cleanest setting for Bell's spaceship paradox (`aq-sr-bell-spaceship-paradox-s6`), and the exact model of in what sense — and in what sense not — an accelerated frame is a uniform gravitational field (`aq-sr-equivalence-principle`).

The physical picture

A single observer with constant proper acceleration $\alpha$ traces a hyperbola, $X^2 - c^2T^2 = (c^2/\alpha)^2$ — that is `aq-sr-relativistic-uniform-acceleration`, recalled here, not re-derived. But one worldline is not a frame. A frame needs a congruence: a family of worldlines filling a region, one through every event, so that "the position of an event" has an answer at all. The Rindler construction supplies one in the laziest possible way: take every hyperbola with the same asymptotes. They are labelled by a single number,

\rho = \sqrt{X^2 - c^2T^2},

the invariant hyperbolic radius, preserved by boosts exactly as $r = \sqrt{x^2+y^2}$ is preserved by rotations. Each hyperbola is a uniformly accelerated worldline. They nest, they never cross, and — the whole point — they share one pair of asymptotes, the null lines $X = \pm cT$ . That shared asymptote is what makes the family a coherent frame rather than a bag of unrelated accelerating ships.

Figure

PUNCHLINE — the accelerated frame is a GRID, not a geometry. Coloured hyperbolae are the worldlines of constant Rindler position ρ; each one is a uniformly accelerated observer, and the one closest to the horizon accelerates HARDEST, α = c²/ρ. Grey dashed rays are the surfaces of constant Rindler time η — every observer in the array agrees on them, and each is a straight line through the vertex. Together they are polar coordinates with hyperbolae in place of circles. The chart covers ONLY the right wedge X greater than the absolute value of cT — one quarter of spacetime. Its boundary is the pair of null lines X = ±cT: the Rindler horizon, where ρ goes to zero and the acceleration diverges. And the punchline in the box: this grid is drawn on FLAT PAPER. Riemann is identically zero everywhere in it

Two things are now true at once, and the tension between them is the whole subject. First: in this chart a uniformly accelerated observer is at rest — coordinates $(\eta,\rho)$ with $\rho$ fixed, only $\eta$ advancing. That is what a reference frame is, and special relativity has just produced one for an accelerating observer, exactly, with no approximation and no general relativity. Second: the metric is not $c^2d\eta^2 - d\rho^2$ but $\rho^2 d\eta^2 - d\rho^2$ , and that $\rho^2$ changes how the frame behaves — clocks at different $\rho$ run at different rates, light climbing the array is redshifted, and the observer's own acceleration depends on where they sit. Everyone meets that $\rho^2$ and thinks gravitational potential. The instinct is Einstein's own, and it is the equivalence principle's origin — but the inference "position-dependent metric, therefore gravity" is false. Compute the Riemann tensor of $\rho^2 d\eta^2 - d\rho^2$ and every component vanishes, identically, everywhere in the wedge. There is no curvature, and there cannot be: the spacetime is Minkowski, we only re-drew the grid on it. What is nonzero is the connection, $\Gamma^{\eta}{}_{\eta\rho} = 1/\rho$ and $\Gamma^{\rho}{}_{\eta\eta} = \rho$ — and those produce the clock gradient, the redshift and the horizon.

Figure

The distinction the whole chapter rests on. LEFT: a metric with position-dependent components produces nonzero CHRISTOFFEL SYMBOLS — the connection. That is what makes clocks run at different rates and light bend and the horizon appear. RIGHT: the RIEMANN TENSOR is built from derivatives AND products of the Christoffels, and for Rindler it cancels to zero in every one of its twenty independent components. A nonzero connection is a statement about your COORDINATES and can always be transformed away at a point. Nonzero Riemann is a statement about the SPACETIME and can never be transformed away anywhere. Gravity is the second one. Acceleration is the first. Rindler has the first and not the second — which is exactly why an accelerated frame mimics gravity locally and is not gravity globally

That distinction — connection versus curvature — is why the folklore "you need general relativity to handle accelerated frames" is false. You need general relativity for gravity. Acceleration is special relativity's business, and Rindler coordinates are how it does that business.

The core equations

#	equation	what it is
(1)	$cT = \rho\sinh\eta,\qquad X = \rho\cosh\eta$	the map, right wedge. $\rho > 0$ , $\eta \in (-\infty,\infty)$ . Note $cT$ , not $T$ — both sides are lengths
(2)	$X^2 - c^2T^2 = \rho^2 > 0, \qquad X > \lvert cT\rvert$	the wedge. $\rho$ is the boost invariant; the chart covers a quarter of Minkowski
(3)	$\boxed{\,ds^2 = \rho^2\,d\eta^2 - d\rho^2\,}$	the Rindler metric in our lock. Scaled time $\eta_a = (c/a)\eta$ : $\;ds^2 = (a\rho/c)^2 d\eta_a^2 - d\rho^2$
(4)	$\boxed{\,\alpha(\rho) = c^2/\rho\,}$	proper acceleration at fixed $\rho$ . From $A^\mu = (0,\,c^2/\rho)$ , $A\!\cdot\!A = -c^4/\rho^2$
(5)	$\Gamma^{\eta}{}_{\eta\rho} = \dfrac{1}{\rho},\qquad \Gamma^{\rho}{}_{\eta\eta} = \rho$	the connection — the only nonzero Christoffels. Everything "gravitational" comes from these
(6)	$\boxed{\,R^{\mu}{}_{\nu\rho\sigma} = 0\,}$	the curvature — identically zero, everywhere, all twenty components. No gravity here
(7)	$\xi = \partial_\eta$ , $\;\;\xi\!\cdot\!\xi = \rho^2$	the boost Killing vector. Timelike for $\rho>0$ , null at $\rho = 0$ : that is the horizon

In (3) the proper time at fixed $\rho$ follows from $ds^2 = c^2 d\tau^2$ : $c\,d\tau = \rho\,d\eta$ , so $d\tau/d\eta = \rho/c$ , or $d\tau/d\eta_a = a\rho/c^2$ in the scaled chart — dimensionless, as it must be. That lapse seeds the position-dependent clock rates, which `aq-sr-rindler-metric` owns.

Building the chart

Derivation AThe map and the metric — re-derived from scratch, because the base ships the wrong signature

The goal is to build a coordinate system in which a uniformly accelerated observer is at rest, and to find what the flat Minkowski metric looks like in it. One demand fixes the whole construction: the coordinate lines must be adapted to the boost, because the boost is the symmetry the accelerated worldline respects. That determines the map uniquely up to scale, and the metric follows by substitution. We do the substitution explicitly, because the base for this article prints the metric with the opposite signature and the map with a dimensional error, and the two are mutually inconsistent. Never transcribe a metric.

Step 1. Fix the invariant. A boost preserves $X^2 - c^2T^2$ exactly as a rotation preserves $x^2 + y^2$ . Define $\rho \equiv \sqrt{X^2 - c^2T^2}$ where this is positive. Why: being boost-invariant, a worldline $\rho = \text{const}$ is carried into itself by a boost — the hyperbolic analogue of a circle under rotation, and exactly the accelerated worldline of `aq-sr-relativistic-uniform-acceleration`.

Step 2. Parametrise it. $X^2 - c^2T^2 = \rho^2$ with $X > 0$ is solved by $X = \rho\cosh\eta$ , $cT = \rho\sinh\eta$ , since $\cosh^2\eta - \sinh^2\eta = 1$ — equation (1). Why the $c$ : $\rho$ is a length, so both $X$ and $cT$ are lengths and $\eta$ is dimensionless. The base prints $T = \rho\sinh(a\eta/c)$ — a time set equal to a length. It is dimensionally broken, and it contradicts its own $X^2 - c^2T^2 = \rho^2$ , which requires the $c$ .

Step 3. Identify $\eta$ . Along a hyperbola, increasing $\eta$ moves you exactly as a boost of rapidity $\eta$ would, so $\eta$ is the boost rapidity and equally the Rindler time coordinate. Why this matters: a boost is now a translation, $\eta \to \eta + \Delta\eta$ . The symmetry has been made into a coordinate — that is the whole reason to build the chart.

Step 4. Differentiate the map and substitute into the flat metric — in our signature, $ds^2 = (c\,dT)^2 - dX^2$ . With $d(cT) = \sinh\eta\,d\rho + \rho\cosh\eta\,d\eta$ and $dX = \cosh\eta\,d\rho + \rho\sinh\eta\,d\eta$ ,

ds^2 = \big(\sinh\eta\,d\rho + \rho\cosh\eta\,d\eta\big)^2 - \big(\cosh\eta\,d\rho + \rho\sinh\eta\,d\eta\big)^2.

Expand. The $d\rho^2$ coefficient is $\sinh^2\eta - \cosh^2\eta = -1$ ; the $d\eta^2$ coefficient is $\rho^2(\cosh^2\eta - \sinh^2\eta) = +\rho^2$ ; the cross term is $2\rho\,d\rho\,d\eta(\sinh\eta\cosh\eta - \cosh\eta\sinh\eta) = 0$ , cancelling identically — the statement that the $\eta$ -lines and $\rho$ -lines are orthogonal. Hence

\boxed{\,ds^2 = +\rho^2\,d\eta^2 - d\rho^2\,}

Why the sign matters: the $+$ is on the time part, because in $(+,-,-,-)$ timelike intervals are positive and the observer at fixed $\rho$ moves forwards in time. The base prints the mostly-plus $ds^2 = -(a\rho/c)^2 d\eta^2 + d\rho^2$ : its coefficient is right, its signature is not. Ship ours.

Step 5. Read off the lapse. At fixed $\rho$ , $ds^2 = \rho^2 d\eta^2 = c^2 d\tau^2$ , so $d\tau/d\eta = \rho/c$ ; in the scaled chart this reads $d\tau/d\eta_a = a\rho/c^2$ — dimensionless. Why: the base prints $a\rho/c$ , which has dimensions of velocity. A rate of proper time per unit coordinate time that is not dimensionless is a defect on sight. Worse, the base's own eq[0] note asserts $\eta_a$ is the proper time of the observer at $\rho = c^2/a$ — true of $a\rho/c^2$ (which gives exactly $1$ there), false of $a\rho/c$ (which gives $c$ ). The base contradicts itself. *Verified in verify_aq-sr-rindler-coordinates.py*

Go deeper · Derivation B · 1 minThe acceleration law α = c²/ρ, and why the rear pushes harder

Derivation BThe acceleration law α = c²/ρ, and why the rear pushes harder

The goal is to compute the proper acceleration of the observer at fixed $\rho$ — the invariant $\alpha = \sqrt{-A\cdot A}$ an accelerometer bolted to their floor would read — from the Rindler metric, using the connection rather than the Minkowski worldline. That route is not showing off: it is the only one that generalises, and it hands us the curvature in the next card for free. The result, $\alpha = c^2/\rho$ , is the most consequential fact about the chart — and its direction, bigger acceleration at smaller $\rho$ , is the one two other lanes in this programme have written down backwards.

Step 1. The four-velocity. At fixed $\rho$ , $u^\mu = (d\eta/d\tau,\,0)$ , and the lapse $d\tau/d\eta = \rho/c$ gives $u^\eta = c/\rho$ . Why: normalisation is the check — $u\!\cdot\!u = g_{\eta\eta}(u^\eta)^2 = \rho^2 \cdot c^2/\rho^2 = +c^2$ , positive, as $(+,-,-,-)$ demands of a timelike vector. If you get $-c^2$ you are in the wrong signature.

Step 2. The Christoffels of $g_{\eta\eta} = \rho^2$ , $g_{\rho\rho} = -1$ . The only nonvanishing ones:

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

Why $+\rho$ and not $-\rho$ : it is $-\tfrac12 g^{\rho\rho}\partial_\rho g_{\eta\eta} = -\tfrac12(-1)(2\rho) = +\rho$ , and the minus in $g^{\rho\rho} = -1$ is where our signature enters. Get the signature wrong and the acceleration points the wrong way.

Step 3. The four-acceleration. $A^\mu = u^\nu\nabla_\nu u^\mu = \Gamma^{\mu}{}_{\eta\eta}(u^\eta)^2$ , since $u$ has only an $\eta$ component and is constant along the worldline. Only $\mu = \rho$ survives:

A^\rho = \Gamma^{\rho}{}_{\eta\eta}\,(u^\eta)^2 = \rho \cdot \frac{c^2}{\rho^2} = \frac{c^2}{\rho}, \qquad A^\mu = \Big(0,\;\frac{c^2}{\rho}\Big).

Step 4. Take its magnitude. $\;A\!\cdot\!A = g_{\rho\rho}(A^\rho)^2 = (-1)\cdot c^4/\rho^2 = -\,c^4/\rho^2$ . Why negative: the four-acceleration of a timelike worldline is spacelike, always, and in $(+,-,-,-)$ spacelike means negative norm. Not a sign error — the signature working. So $\alpha = \sqrt{-A\cdot A}$ .

Step 5. Conclude.

\boxed{\,\alpha(\rho) = \sqrt{-A\!\cdot\!A} = \frac{c^2}{\rho}\,}

It points in the $+\rho$ direction — away from the horizon — and grows without bound as $\rho \to 0$ . Why the direction is the whole story: an observer nearer the horizon must push harder to hold the same $\rho$ . A rigid array of such observers is Born-rigid (`aq-sr-born-rigidity` owns the derivation and the bound $L_0 < c^2/\alpha_{\text{nose}}$ ) and is not equally accelerated: the rear accelerates harder than the nose — and the array cannot extend past $\rho = 0$ . *Verified in verify_aq-sr-rindler-coordinates.py*

Go deeper · Derivation C · 2 minThe Riemann tensor is identically zero — the computation that kills the folklore

Derivation CThe Riemann tensor is identically zero — the computation that kills the folklore

This is the article's structural result and the chapter's thesis in a single tensor. Everything about the Rindler frame looks gravitational: the metric components depend on position, clocks at different heights disagree, light shifts colour as it climbs, there is a horizon, and a released object appears to fall. The reflex is to say "curved spacetime". The goal here is to compute the curvature explicitly, find that there is none, and then say which object the gravitational-looking effects do belong to. The answer is the connection — a fact about the coordinates, where a nonzero Riemann tensor would be a fact about the spacetime.

Step 1. Write the Riemann tensor.

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}.

Why it is built this way: it measures the failure of a vector to return to itself when parallel-transported around a closed loop. That failure is what curvature means — a genuinely geometric statement, unlike $\Gamma$ , which is not a tensor and can be set to zero at any single point by choosing better coordinates.

Step 2. Feed in the two nonzero Christoffels. In the $(\eta,\rho)$ block the only independent component is $R^{\rho}{}_{\eta\rho\eta}$ :

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}.

Step 3. Evaluate. $\partial_\rho \Gamma^{\rho}{}_{\eta\eta} = \partial_\rho\rho = 1$ ; the second and third terms vanish ( $\Gamma^{\rho}{}_{\eta\rho} = \Gamma^{\rho}{}_{\rho\sigma} = 0$ ); the fourth is $\Gamma^{\rho}{}_{\eta\eta}\Gamma^{\eta}{}_{\eta\rho} = \rho \cdot (1/\rho) = 1$ . Therefore

R^{\rho}{}_{\eta\rho\eta} = 1 - 0 + 0 - 1 = 0.

Why this is not luck: the derivative of the connection and the square of the connection are each nonzero and cancel exactly, at every $\rho$ , for every parameter value. That exact cancellation is the algebraic fingerprint of a flat metric in curvilinear coordinates. Compare $dr^2 + r^2d\theta^2$ , where $\partial_r\Gamma^{r}{}_{\theta\theta} = -1$ is cancelled by $-\Gamma^{r}{}_{\theta\theta}\Gamma^{\theta}{}_{\theta r} = -(-r)(1/r) = +1$ — same cancellation, same reason: flat paper.

Step 4. Restore the transverse directions. $y$ and $z$ are untouched by the map, so $g_{yy} = g_{zz} = -1$ with no $\rho$ or $\eta$ dependence: no Christoffels, no curvature. In full four dimensions all twenty independent components of $R^{\lambda}{}_{\mu\nu\kappa}$ vanish, hence $R_{\mu\nu} = 0$ and $R = 0$ — verified component-by-component by symbolic algebra in the verify script.

Step 5. Say what it means.

\boxed{\;R^{\lambda}{}_{\mu\nu\kappa} = 0 \quad\text{everywhere in the wedge, while}\quad \Gamma^{\eta}{}_{\eta\rho} = \frac1\rho \neq 0.\;}

Why this ends the argument: the clock gradient, the redshift, the horizon and the apparent fall of released objects are all produced by $\Gamma$ , and $\Gamma$ can be transformed away at any point — that is precisely what "go to the momentarily comoving inertial frame" does. Riemann cannot be transformed away anywhere, because it is a tensor: zero in one chart means zero in all. Gravity is the tensor; acceleration is the connection. Special relativity handles accelerated frames completely, and this is the proof. *Verified in verify_aq-sr-rindler-coordinates.py*

Limiting cases

Case	Result	Meaning
$\rho \to \infty$	$\alpha \to 0$ ; the hyperbola flattens to a line	far from the horizon the array barely accelerates — inertial in the limit
$\rho \to 0^+$	$\alpha = c^2/\rho \to \infty$ ; $\;\xi\!\cdot\!\xi = \rho^2 \to 0$	the horizon. Infinite acceleration; the boost Killing vector goes null
$\eta \to \eta + \Delta\eta$	a Lorentz boost of rapidity $\Delta\eta$	the chart's purpose: the symmetry is now a translation
$\rho$ -range collapsed to a point	$ds^2 \to \rho_0^2 d\eta^2 - d\rho^2$ , constants	locally the chart is Minkowski in disguise — the equivalence principle in one line

Common trapThe lapse is NOT a gravitational potential, and the base ships the metric upside down

Two traps, one fatal. First, the signature. The base prints $ds^2 = -(a\rho/c)^2d\eta^2 + d\rho^2$ — mostly-plus. In our lock that is the negative of the correct line element: carry it forward and every Rindler observer's proper time comes out imaginary and the four-acceleration comes out timelike. It also prints the map as $T = \rho\sinh(a\eta/c)$ , equating a time to a length, and a clock rate $d\tau/d\eta = a\rho/c$ that is not dimensionless. Three defects, all fixed above. Re-derive the metric from the map, every time. Never transcribe one.

Second, the interpretation. Seeing $g_{\eta\eta} = \rho^2$ and reasoning "position-dependent metric $\Rightarrow$ gravitational potential $\Rightarrow$ curved spacetime" is the commonest error in the subject, and it is wrong at the last arrow. $g_{rr} = 1$ , $g_{\theta\theta} = r^2$ is also position-dependent, and it is a sheet of flat paper. The test is Riemann, not $g_{\mu\nu}$ .

Worked examples

1 · Your own horizon, at one gravity

A ship accelerates at a constant $1\,g$ forever. Where does it sit in the chart, where is its horizon, and how hard must a companion 0.5 light-years astern push to keep station?

Step 1 — locate the ship. Equation (4) fixes its Rindler coordinate: $\rho_0 = c^2/\alpha = c^2/g$ . With $g = 9.80665\ \mathrm{m/s^2}$ and $c = 299{,}792{,}458\ \mathrm{m/s}$ ,

\rho_0 = \frac{c^2}{g} = \mathbf{9.1648\times10^{15}\ m} = \mathbf{0.9687\ light\text{-}year}.

The ship is not "at the origin": it sits a little under one light-year from the vertex, and stays there forever — $\rho$ constant, only $\eta$ advancing.

Step 2 — find the horizon. The horizon is $\rho = 0$ , a proper distance $\rho_0 = \mathbf{0.9687}$ light-years behind the ship. (Proper distance: at fixed $\eta$ the metric gives $ds^2 = -d\rho^2$ , so $\rho$ is the ruler distance from the vertex.) A signal launched from Earth more than about 0.97 years after a $1\,g$ departure never catches the ship. Not "arrives late" — never. Earth crosses the ship's horizon and the correspondence ends.

Step 3 — the companion. A ship holding station $0.5$ ly astern sits at $\rho_1 = 0.9687 - 0.5 = \mathbf{0.4687}$ ly, with

\alpha_1 = \frac{c^2}{\rho_1} = g\cdot\frac{0.9687}{0.4687} = \mathbf{2.067\,g}.

It must burn twice as hard to hold a constant distance from a ship doing $1\,g$ . Not a paradox, not a fuel quirk — equation (4). The rear pushes harder.

Step 4 — the boundary number. Push the companion to $0.97$ ly astern and $\rho_1 \to 0$ : the required acceleration diverges. No thrust station-keeps across the horizon — the chart's way of saying a Born-rigid body cannot be longer than $c^2/\alpha_{\text{nose}}$ (`aq-sr-born-rigidity` owns that bound). The horizon is not a distant abstraction. For a $1\,g$ ship it is one light-year back, and it is the hard limit on how long your ship can be.

2 · The horizon inside a storage ring

Muons in the CERN storage ring of Bailey et al. (1977) felt a transverse proper acceleration of order $10^{18}\,g$ . Where is their Rindler horizon?

Step 1 — identify. Equation (4) inverted: $\rho = c^2/\alpha$ . This is a local statement — the muon's acceleration is centripetal, not linear, so the chart is an instantaneous tangent-plane construction rather than a global one — but the length scale it produces is exactly right.

Step 2 — compute. With $\alpha = 10^{18}g = 9.81\times10^{18}\ \mathrm{m/s^2}$ ,

\rho = \frac{c^2}{\alpha} = \frac{8.988\times10^{16}}{9.81\times10^{18}} = \mathbf{9.16\times10^{-3}\ m} = \mathbf{9.16\ mm}.

Step 3 — read it. The muon's horizon is nine millimetres behind it. A horizon is not an astrophysical object needing a black hole and a telescope; it is a length $c^2/\alpha$ , and at laboratory accelerations it is a laboratory length. The same experiment found the muon's decay rate tracked $\gamma$ alone, with no acceleration dependence at the $2\times10^{-3}$ level — the clock hypothesis, verified; `aq-sr-proper-acceleration-comoving-frame` owns that result.

Step 4 — the boundary number. The Unruh temperature at $10^{18}\,g$ is of order $10^{-4}\ \mathrm{K}$ — real, and utterly unmeasured. `aq-sr-unruh-effect` owns it. This article notes only that the horizon it needs is the one above, and that a horizon costs 9 millimetres.

Physics of the portrait

Rindler coordinates are not a discovery about the world but a choice of grid — one that makes a hard question easy: what does the universe look like to someone who never stops accelerating?

Origins · 2 minRindler coordinates and how this was discovered

Rindler coordinates

1873–1966 · Minkowski's geometry makes the boost a hyperbolic rotation · Born's rigid electron produces the hyperbola · Einstein's happiest thought demands an accelerated frame · Rindler names the horizon

The chart carries a name, but almost none of the physics is his. The geometry is Minkowski's: his 1908 "Raum und Zeit" recast the Lorentz transformation as a hyperbolic rotation, and once a boost is a rotation, the coordinates adapted to it — hyperbolae and radial rays — are as inevitable as polar coordinates are for a circle. The worldline is Born's: his 1909 paper on the rigidly accelerated electron produced the constant-proper-acceleration hyperbola, and the nested family sharing one vertex is precisely his rigid body. The motivation is Einstein's: the 1907 Jahrbuch review contains what he later called his happiest thought — that a man in free fall does not feel his own weight — and the accelerated frame is what makes that thought computable. What Wolfgang Rindler added, in his 1966 textbook and papers, was the clean modern packaging and, decisively, the term "event horizon", which he coined for this null boundary before it was black-hole vocabulary. The lesson the chart teaches by construction is the deeper one: an accelerated frame reproduces gravitational time dilation, gravitational redshift and a horizon in a spacetime of exactly zero curvature. Everything the equivalence principle promises is here; nothing the field equations demand is. That is why this is a special-relativity article.

History

1908

— Minkowski's Cologne lecture "Raum und Zeit" recasts the Lorentz boost as a hyperbolic rotation of spacetime, making hyperbolic coordinates the natural chart for acceleration.

1907

— Einstein, in his Jahrbuch review, states the equivalence principle: an observer in free fall does not feel their own weight. He later called it his happiest thought.

1909

— Born's paper on the rigidly accelerated electron produces the constant-proper-acceleration hyperbola and the rigid congruence of nested hyperbolae.

1909–1910

— Herglotz and Noether prove that a Born-rigid body has only three degrees of freedom; the Rindler congruence is the canonical example.

1966

— Rindler's Special Relativity and his papers on horizons give the chart its modern form; he coins the term event horizon for the null boundary of the wedge.

1976

— Unruh shows that quantising a field on the Rindler wedge gives the accelerated observer a thermal bath, turning the chart into a tool of quantum field theory.

Going further

Texts. Rindler, Introduction to Special Relativity, 2nd ed. (Oxford, 1991) · Misner, Thorne & Wheeler, Gravitation, §6 (the accelerated observer) · Wald, General Relativity, §6.3 (the Rindler wedge) · Born, M., "Die Theorie des starren Elektrons in der Kinematik des Relativitätsprinzips", Ann. Phys. 335, 1 (1909) · Minkowski, H., "Raum und Zeit", Phys. Z. 10, 104 (1909).

Provenance. Signature $(+,-,-,-)$ , $\eta_{\mu\nu} = \mathrm{diag}(+1,-1,-1,-1)$ . Every metric here was re-derived from the coordinate map, not transcribed. verify_aq-sr-rindler-coordinates.py recomputes the map, the metric, the Christoffels, the four-acceleration and all twenty independent components of the Riemann tensor symbolically in sympy; evaluates every printed number at 50-digit precision (mpmath) and greps this article for each; and perturbs $g$ and $c$ to confirm the headline figures move under their inputs. Base defects, all proved by computation and all corrected above: (i) the base metric $ds^2 = -(a\rho/c)^2d\eta^2 + d\rho^2$ is mostly-plus; (ii) the base map $T = \rho\sinh(a\eta/c)$ is dimensionally broken — a time equated to a length — and contradicts the base's own invariant $X^2 - c^2T^2 = \rho^2$ , which requires $cT$ ; (iii) the base clock rate $d\tau/d\eta = a\rho/c$ is off by a factor of $c$ , is not dimensionless, and contradicts the base's own claim that $\eta$ is the proper time of the observer at $\rho = c^2/a$ . A fourth, reported and unused: the base's redshift card writes $\nu_{\text{rec}}/\nu_{\text{emit}} = \rho_{\text{emit}}/\rho_{\text{rec}} = 1/(1 + gh/c^2)$ with $g$ at the receiver; the exact identity needs $g$ at the emitter, so as written the last equality is only first-order.

Scope. This article owns the chart: the map, the wedge, the metric, $\alpha(\rho) = c^2/\rho$ , the connection, and the vanishing of Riemann. The physics of the lapse — position-dependent clock rates, gravitational redshift, Pound–Rebka — belongs to `aq-sr-rindler-metric`; the causal structure of $\rho = 0$ belongs to `aq-sr-rindler-horizon`. Both are linked by slug and neither is claimed here. Hyperbolic motion is recalled from `aq-sr-relativistic-uniform-acceleration`, the Born-rigid bound from `aq-sr-born-rigidity`; neither is re-derived. The Unruh effect is gestured at, not derived — it is `aq-sr-unruh-effect`'s, and no claim is made about its contents.

Builds on. Relativistic Uniform Acceleration and the Rindler Hyperbola · Proper Acceleration and the Comoving Frame · The Equivalence Principle. Leads to. The Rindler Metric and Position-Dependent Clock Rates · The Rindler Horizon · Born Rigidity · The Unruh Effect.

If the Rindler metric has a horizon and gravitational time dilation, in what sense is this not gravity?

In the only sense that counts: the Riemann tensor is identically zero. Curvature is what cannot be transformed away — it is a tensor, so zero in one chart means zero in all of them, and it is what produces tidal forces and geodesic deviation. Rindler spacetime has none, because it is Minkowski with a different grid drawn on it. What it does have is a nonzero connection, $\Gamma^{\eta}{}_{\eta\rho} = 1/\rho$ , and the connection is not a tensor: it can be set to zero at any point by better coordinates — which is exactly what "go to the freely falling frame" means. Real gravity is what survives after you kill $\Gamma$ at a point. Here, nothing survives.

Why does the observer nearer the horizon accelerate harder? Should the rigid array not accelerate uniformly?

"Uniformly" is the trap: it means two things. A Born-rigid array keeps every pair of neighbours at constant proper distance — that is rigidity — and the price is that they cannot all share one proper acceleration. Equation (4) is exact: $\alpha(\rho) = c^2/\rho$ , so smaller $\rho$ pushes harder. Geometrically it is forced — the worldlines must share one vertex to have a common simultaneity fan, and a hyperbola with a nearer vertex is more sharply curved. An array in which everyone accelerates equally is a different program: its worldlines do not share a vertex, the proper distances stretch, and that is Bell's spaceship paradox (`aq-sr-bell-spaceship-paradox`). Rigid and equally accelerated are mutually exclusive.

Why do Rindler coordinates only cover a quarter of Minkowski spacetime?

Because the map with $\rho > 0$ has range exactly $X > |cT|$ : $\cosh$ is never smaller than $|\sinh|$ . Reaching the other wedges would need $\rho^2 < 0$ or complex $\eta$ . This is not a defect to be patched — it reflects something true. An eternally accelerating observer genuinely cannot be reached by a signal from beyond $X = -cT$ , since catching them would mean outrunning light. The chart's coverage is the observer's causal reach. Continuing analytically past $\rho = 0$ is how one shows the horizon is a coordinate, not a physical, boundary — `aq-sr-rindler-horizon`'s business.

Is the Rindler time coordinate eta the same as anyone's proper time?

Only for one observer, and only if you scale it to make it so. From the metric, $d\tau = (\rho/c)\,d\eta$ — the lapse depends on $\rho$ , so $\eta$ ticks at a different rate against every observer's wristwatch. It is a coordinate, not a clock. In the scaled convention $\eta_a = (c/a)\eta$ one has $d\tau/d\eta_a = a\rho/c^2$ , which equals exactly $1$ at $\rho = c^2/a$ — so $\eta_a$ is the proper time of the single reference observer whose proper acceleration is $a$ , and nobody else's.

Does the diverging acceleration at rho equals zero mean the horizon is physically dangerous?

No, and this is the crispest illustration of what a horizon is. $\alpha = c^2/\rho \to \infty$ is a statement about what it costs to hover at fixed $\rho$ — not about what is there. Nothing is there. Curvature is zero at the horizon exactly as everywhere else; an inertial probe crosses $X = cT$ smoothly and notices no boundary. What diverges is the thrust required to refuse to cross. The identical structure appears at a black-hole horizon: surface gravity finite, hovering acceleration divergent, infalling observers crossing uneventfully.

If the Christoffel symbols are nonzero, how can a free particle move in a straight line here?

It does, and the geodesic equation says so. A free particle obeys $\ddot{x}^\mu + \Gamma^{\mu}{}_{\nu\lambda}\dot{x}^\nu\dot{x}^\lambda = 0$ , and the $\Gamma$ terms are precisely what turn the straight Minkowski worldline into curved-looking $(\eta,\rho)$ motion. To the Rindler observer the free particle appears to accelerate toward the horizon — it looks like it falls — and that appearance is entirely the connection. Transform back to $(T,X)$ and it is a straight line at constant velocity, as it was all along. The equivalence principle read backwards: what looks like a gravitational field in the accelerated frame is the absence of one in the inertial frame.