The Rindler Horizon

Leave Earth today under a steady $1\,g$ — comfortable, Earth-normal weight — and never shut the engine off. You will never reach the speed of light, and you will never leave the flat, uncurved spacetime of special relativity. But you will lose Earth. Not by outrunning it: light is faster than you, always, and every photon Earth emits will chase you forever. It will simply never arrive. Everything that happens on Earth after about 354 days of ordinary Earth time sends its light after you, and that light falls behind at a gap that never closes. Your family's next New Year broadcast is not delayed. It is unreachable, for the rest of eternity, so long as the engine burns. The last images you do receive show Earth slowing, dimming and reddening, freezing on the final frame. That is not a black hole. There is no gravity in this story, no curvature, no singularity — the Riemann tensor is zero everywhere. It is a Rindler horizon: a causal boundary that acceleration alone puts behind you, in flat empty space, behaving in every qualitative respect like the event horizon of a black hole. It is why horizons can be studied without ever leaving special relativity.

Key takeaways

Signature $(+,-,-,-)$ : $\mathrm{d}s^2 = c^2\mathrm{d}t^2 - \mathrm{d}x^2$ , timelike intervals positive. Spacetime is exactly Minkowski's; $R^{\lambda}{}_{\mu\nu\kappa} = 0$ everywhere on this page.
A worldline of constant proper acceleration $a$ is the hyperbola $x^2 - c^2t^2 = (c^2/a)^2$ , asymptotic to the null line $x = ct$ . That asymptote is the Rindler horizon.
No signal emitted from $x \le ct$ ever reaches an observer who accelerates forever. The gap between a chasing photon and the observer tends to a negative constant — the photon closes on the observer's position but never on the observer.
The horizon sits a fixed proper distance $d_H = c^2/a$ behind you, permanently. At $a = g$ that is $9.16\times10^{15}\ \mathrm{m} = 0.97$ light-year.
Owned result: the horizon in home time. Depart Earth at rest and burn at constant $a$ forever, and the last Earth event you can ever see is at Earth time $t_{\text{last}} = c/a$ — $0.97$ years at $1\,g$ . The signal you receive from Earth-time $t_E$ arrives redshifted by exactly $\nu_{\text{rec}}/\nu_{\text{emit}} = 1 - a t_E/c$ : a linear fade to zero. Earth freezes and reddens, exactly as an infaller freezes at a black hole's horizon.
A horizon is a property of a worldline, not a place. Cut the engine and it vanishes: the hidden region becomes visible again. Two observers with different accelerations have different horizons.
The Rindler metric in comoving (Møller) coordinates is $\mathrm{d}s^2 = +\left(1 + \frac{a\xi}{c^2}\right)^2c^2\mathrm{d}\tau^2 - \mathrm{d}\xi^2$ ; the lapse vanishes at $\xi = -c^2/a$ , which is the horizon, and the redshift from there diverges.
Key idea: you do not need gravity to have a horizon. You need only to accelerate forever — and "forever" is the whole content of the word.

The core insight

A light ray cannot be outrun, and it is not outrun here. What happens is that an eternally accelerating observer's speed approaches $c$ fast enough that a photon launched from far enough behind spends its entire infinite future closing a gap that is asymptotically constant and strictly positive. The photon is always gaining on where you were; it is never gaining on where you are. Causal disconnection in flat spacetime is not a race that light loses. It is a race that never ends.

Where this lives today

Every black-hole horizon is locally a Rindler horizon. Expand the Schwarzschild metric about a point just outside $r = 2GM/c^2$ and to leading order it becomes the Rindler metric, with the black hole's surface gravity $\kappa$ in place of $a$ . A static observer hovering outside a black hole is, locally, a Rindler observer: they must burn their engines to stay put, their thrust diverges as they approach the horizon exactly as $a = c^2/\xi$ does, and they see infalling matter freeze and redden precisely as the ship sees Earth do. The flat-space calculation is not an analogy — it is the near-horizon limit.

It is the reason horizon thermodynamics is believable. Hawking's derivation of black-hole radiation is technically formidable and rests on field modes near the horizon. Unruh's flat-space version rests on nothing but a wedge, a hidden region and a trace — and gives the same structure. That a fictitious horizon, produced by an engine in empty space, already carries a temperature is what convinced the field that the result was about horizons in general and not about black holes in particular.

And it constrains relativistic mission design, in principle if not in practice. A craft under continuous thrust is causally severed from part of the space it left behind — at $1\,g$ , everything more than $0.97$ light-years astern. The remedy is trivially available and physically revealing: coast. Shut the engine and the horizon dissolves, the backlog arrives, and the ship rejoins the causal universe it had been outrunning.

The physical picture

Draw the light cone of the origin. An observer with constant proper acceleration $a$ , momentarily at rest at $x = c^2/a$ when $t = 0$ , follows

x(t) = \sqrt{(ct)^2 + \left(\tfrac{c^2}{a}\right)^2} \; > \; ct \qquad \text{for every } t,

so the worldline lies strictly to the right of the null line $x = ct$ , forever, while approaching it without bound. That null line is the future Rindler horizon. Now emit a photon from an event $(t_0, x_0)$ behind it — meaning $x_0 \le ct_0$ — and send it rightward, chasing. Its worldline is $x = ct + (x_0 - ct_0)$ : a line parallel to the horizon, displaced by the negative constant $x_0 - ct_0$ . The observer is asymptotic to $x = ct$ from above; the photon is asymptotic to it from below. Both converge on the same line and they never meet. The gap tends to $|x_0 - ct_0|$ and stays there. That is the whole theorem, and it is worth pausing on how little it needed: no gravity, no curvature, no quantum mechanics. Only the fact that the observer never stops.

Figure

PUNCHLINE — a causal horizon in flat, empty, uncurved spacetime. The blue hyperbola is an observer at constant proper acceleration a; the gold dashed line x = ct is the future Rindler horizon, the hyperbola's own asymptote. The grey region beneath it is HIDDEN: every event there is causally cut off from the observer forever. The red ray is a photon launched from inside the hidden region, chasing at exactly c — it closes on the horizon from below while the observer closes on it from above, and the gap between them tends to the constant |x₀ − ct₀| and never shuts. Light does not lose the race. The race never ends. The horizon sits a fixed proper distance c²/a behind the observer — 9.16×10¹⁵ m, or 0.97 light-year, at a = 1 gee

Two things about this horizon make it not a black hole. It is not a place: it is a feature of one worldline. A ship accelerating harder has its horizon closer, one accelerating less has it further away, and an inertial observer has none at all — they sail straight across $x = ct$ and notice precisely nothing, because there is nothing there to notice. And it is not permanent: cut the engine and the hyperbola straightens into an ordinary inertial worldline, the asymptote stops being an asymptote, and the hidden region floods back into view. The signals were never destroyed — they were merely never going to arrive while you kept accelerating.

The core equations

The worldline, and its asymptote:

x^{2} - c^{2}t^{2} = \left(\frac{c^{2}}{a}\right)^{2} \qquad\Longrightarrow\qquad x(t) \xrightarrow[t\to\infty]{} ct \qquad (1)

where: $a$ — the constant proper acceleration, the invariant an accelerometer reads. Physical reading: the hyperbola is owned by `aq-sr-relativistic-uniform-acceleration`; we use only the one fact that it never crosses its own asymptote. The horizon's distance, which is the article's central number:

\boxed{\;d_{H} = \frac{c^{2}}{a}\;} \qquad (2)

where: $d_H$ — the proper distance to the horizon on the observer's own simultaneity slice. Physical reading: it is constant along the whole worldline. You never get closer to your horizon and never pull away from it; it follows you at a fixed distance forever, like the horizon at sea. The metric in comoving (Møller) coordinates, with $\xi$ the proper distance from the reference observer:

\boxed{\;\mathrm{d}s^{2} = +\left(1 + \frac{a\xi}{c^{2}}\right)^{2}c^{2}\,\mathrm{d}\tau^{2} \;-\; \mathrm{d}\xi^{2}\;} \qquad (3)

where: the $+$ on time and $-$ on space are the locked signature $(+,-,-,-)$ . Physical reading: the lapse $\left(1 + a\xi/c^2\right)$ vanishes at $\xi = -c^2/a$ — the horizon, where a clock at rest in the accelerated frame stops. The base prints this metric with both signs flipped; see the warning below. Clock rates and redshift across the accelerated frame:

\frac{\mathrm{d}\tau_{2}}{\mathrm{d}\tau_{1}} = \frac{1 + a\xi_{2}/c^{2}}{1 + a\xi_{1}/c^{2}}, \qquad\qquad \boxed{\;1 + z = \frac{1}{1 + a\,\xi_{\text{emit}}/c^{2}} \;\xrightarrow[\;\xi_{\text{emit}}\to -c^{2}/a\;]{}\; \infty\;} \qquad (4)

where: the observer sits at $\xi = 0$ , the source at $\xi_{\text{emit}} < 0$ — behind them, toward the horizon. Physical reading: infinite redshift from the horizon. A source lowered toward $\xi = -c^2/a$ reddens and dims without limit and freezes, never quite crossing. This is the flat-space rehearsal of the frozen infaller at a black hole. And, as a signpost only, the temperature the horizon is later found to carry:

T_{U} = \frac{\hbar a}{2\pi c\,k_{B}} \qquad (5)

where: $T_U$ is the Unruh temperature. Physical reading: quoted, not derived — it is quantum field theory on the Rindler wedge and belongs to `aq-sr-unruh-effect`. It is here only to say where the door is. The coefficient $\hbar/2\pi ck_B = 4.06\times10^{-21}$ K per $\mathrm{m/s^2}$ is why nobody has measured it.

Why nothing behind the horizon ever arrives

Derivation ANo signal from beyond x equals ct ever reaches an eternally accelerated observer

The claim is causal and deserves a causal proof, not a picture. We show that a photon — the fastest signal there is — launched from any event on or behind the null line $x = ct$ and aimed at the fleeing observer, fails to arrive: not late, but never. The mechanism is not that the observer outruns light. It is that both worldlines are asymptotic to the same null line from opposite sides, and the offset between them is fixed at launch and never changes.

Step 1. Write the observer's position. From (1), taking the right branch, $x_{\text{obs}}(t) = \sqrt{(ct)^2 + (c^2/a)^2}$ . Since $(c^2/a)^2 > 0$ , we have $x_{\text{obs}}(t) > ct$ for every $t$ , including every negative $t$ . Why: this is the whole of the geometric content. The observer is confined to the wedge $x > |ct|$ for all eternity, and no continuous worldline can leave it without the acceleration stopping.

Step 2. Launch the photon. From an emission event $(t_0, x_0)$ with $x_0 \le c t_0$ — on or behind the horizon — send a rightward light ray: $x_{\text{ph}}(t) = ct + (x_0 - ct_0)$ , with offset $\Delta \equiv x_0 - ct_0 \le 0$ . Why: a rightward photon is the best the emitter can do. Anything slower, or aimed otherwise, does worse. If this fails, everything fails.

Step 3. Subtract and take the limit. The photon catches the observer only if $x_{\text{ph}}(t) \ge x_{\text{obs}}(t)$ for some $t$ . Now

x_{\text{ph}}(t) - x_{\text{obs}}(t) = ct + \Delta - \sqrt{(ct)^2 + (c^2/a)^2} \; \xrightarrow[t\to\infty]{} \; \Delta \;\le\; 0 ,

because $\sqrt{(ct)^2 + k^2} - ct \to 0$ as $t \to \infty$ . The difference is monotonically increasing in $t$ and its supremum is $\Delta \le 0$ , which it never attains. So $x_{\text{ph}} < x_{\text{obs}}$ for all $t$ , always. Why: the photon does close the gap — that is what the monotonicity says, and it is why "light is faster, so it must catch up" feels so compelling. But it closes onto a limit it cannot pass: the supremum is $\Delta$ , and $\Delta$ is not positive.

Step 4. Read the special case. For an event exactly on the horizon, $\Delta = 0$ : the photon converges asymptotically, arriving at $t = \infty$ , which is to say not arriving. Strictly behind it, $\Delta < 0$ , and the photon stays a fixed distance astern forever. Why: the boundary case is the definition. $x = ct$ is precisely the locus of events whose light arrives "at $t = \infty$ ", and that is what makes it a horizon rather than an inconvenience.

Step 5. Note what breaks it. Every step used $x_{\text{obs}}(t) = \sqrt{(ct)^2 + (c^2/a)^2}$ for all $t \to \infty$ . Cut the engine at any finite time and the worldline straightens; the photon then closes at a finite rate and arrives. Why: the horizon is made of the word "forever". It is not a property of the acceleration but of the acceleration being eternal — which is exactly what separates a Rindler horizon from a black hole's, whose eternity belongs to the spacetime and not to anybody's engine. *Verified in verify_aq-sr-rindler-horizon.py: sympy confirms $x_{\text{obs}}(t) - ct > 0$ for all real $t$ and that $\lim_{t\to\infty}[x_{\text{ph}}(t) - x_{\text{obs}}(t)] = \Delta$ ; the difference is checked monotone increasing; and the horizon distance $d_H = c^2/a$ is recovered from the $t=0$ slice, where the vertex sits at $c^2/a$ and the asymptote passes through the origin.*

Go deeper · Derivation B · 2 minThe Moller metric the lapse that vanishes and the redshift that diverges

Derivation BThe Moller metric the lapse that vanishes and the redshift that diverges

The causal argument above is exact but coordinate-free, and does not say what the horizon looks like from on board. For that we need the accelerated observer's own chart. The goal is the comoving metric with the correct signature — the AllQFT base prints it flipped, an artefact of $(-,+,+,+)$ — and then to read the horizon off it as the surface where the lapse vanishes, which is exactly where an infalling source appears to freeze.

Step 1. Set up the comoving chart: the reference observer, proper acceleration $a$ and proper time $\tau$ , sits at $\xi = 0$ , and $\xi$ is proper distance from them along the acceleration axis. The map is

cT = \left(\frac{c^{2}}{a} + \xi\right)\sinh\frac{a\tau}{c}, \qquad X = \left(\frac{c^{2}}{a} + \xi\right)\cosh\frac{a\tau}{c} .

Why: this is the Rindler map of `aq-sr-rindler-metric` with the origin shifted to the reference worldline, so that $\xi$ is measured from the observer rather than from the horizon. It is the chart a navigator would actually build.

Step 2. Substitute into $\mathrm{d}s^2 = \mathrm{d}(cT)^2 - \mathrm{d}X^2$ — time positive, space negative, as our signature demands. The cross terms cancel by $\cosh\sinh - \sinh\cosh = 0$ and the diagonal terms collapse by $\cosh^2 - \sinh^2 = 1$ , leaving

\mathrm{d}s^{2} = +\left(1 + \frac{a\xi}{c^{2}}\right)^{2}c^{2}\,\mathrm{d}\tau^{2} - \mathrm{d}\xi^{2} .

Why: check the sign the fast way. A clock held at rest in the frame has $\mathrm{d}\xi = 0$ , so $\mathrm{d}s^2 = (1 + a\xi/c^2)^2c^2\mathrm{d}\tau^2 > 0$ — positive, hence timelike, hence a real proper time. The base's flipped version makes a stationary clock spacelike and its proper time imaginary. Never transcribe a metric.

Step 3. Find the horizon in this chart. The lapse $\left(1 + a\xi/c^2\right)$ vanishes when

\xi_{H} = -\frac{c^{2}}{a} ,

and $g_{\tau\tau} \to 0$ there: a standard clock held at $\xi_H$ logs no proper time at all per unit of the observer's time. Why: this is the same surface the causal argument found, now wearing coordinates. And $|\xi_H| = c^2/a$ is exactly the proper distance (2), so the two derivations agree — as they must, since $g_{\xi\xi} = -1$ means $\xi$ is proper distance.

Step 4. Get the redshift. Frequency is inversely proportional to the emitter's local proper time, so for an observer at $\xi = 0$ receiving from a static source at $\xi_{\text{emit}}$ ,

1 + z = \frac{\nu_{\text{emit}}}{\nu_{\text{obs}}} = \frac{\mathrm{d}\tau_{\text{obs}}}{\mathrm{d}\tau_{\text{emit}}} = \frac{1}{1 + a\,\xi_{\text{emit}}/c^{2}} \;\xrightarrow[\;\xi_{\text{emit}} \to \xi_H\;]{}\; \infty .

Why: infinite redshift from the horizon. A source lowered toward $\xi_H$ reddens without bound, dims without bound, and its clock is seen to slow to a stop. The observer never sees it cross. This is, feature for feature, what an outside observer sees of matter falling into a black hole — and here there is no black hole, no mass, and no curvature. *Verified symbolically in verify_aq-sr-rindler-horizon.py: sympy substitutes the map into $\mathrm{d}(cT)^2 - \mathrm{d}X^2$ and returns $+(1 + a\xi/c^2)^2c^2\mathrm{d}\tau^2 - \mathrm{d}\xi^2$ exactly, with the cross term identically zero; the lapse is confirmed to vanish only at $\xi = -c^2/a$ ; and $1 + z$ is confirmed to diverge there.*

Go deeper · Derivation C · 2 minThe last thing you will ever see of home

Derivation CThe last thing you will ever see of home

Everything so far concerned an observer already on their hyperbola with the horizon already behind them. The question a reader actually wants answered is sharper: I leave home today, at rest, and burn at a constant $g$ forever. What, exactly, do I lose? This card answers it in home's own time, and the answer is a closed form that turns the abstraction into a date on a calendar. It is this article's owned structural result.

Step 1. Set up the departure. Home sits at $x = 0$ in the launch frame, forever. The ship starts there at rest at $t = 0$ and accelerates at constant proper $a$ , so with $X \equiv c^2/a$ and rapidity $\varphi = a\tau/c$ its worldline is

x_{\text{ship}} = X\big(\cosh\varphi - 1\big), \qquad ct_{\text{ship}} = X\sinh\varphi .

Why: the $-1$ shifts the standard hyperbola so that it passes through the origin — the ship starts at home, not at the vertex $c^2/a$ . That shift is the whole point: it puts home's worldline a distance $X$ in front of the ship's horizon, not behind it, so home is not hidden at the outset.

Step 2. Chase the ship with a photon from home. Home emits at time $t_E$ ; the photon travels $x = c(t - t_E)$ . It reaches the ship when $x_{\text{ship}} = c\,t_{\text{ship}} - ct_E$ , that is

X\cosh\varphi - X = X\sinh\varphi - ct_{E} \qquad\Longrightarrow\qquad X\big(\cosh\varphi - \sinh\varphi\big) = X - ct_{E} .

Why: writing it this way exposes the combination that matters. $\cosh\varphi - \sinh\varphi = e^{-\varphi}$ , and an exponential that only ever decays is about to do all the work.

Step 3. Solve, and find the barrier.

\boxed{\;e^{-\varphi} = 1 - \frac{c\,t_{E}}{X} = 1 - \frac{a\,t_{E}}{c}\;}

Since $e^{-\varphi} > 0$ strictly, a solution exists if and only if $a t_E/c < 1$ :

\boxed{\;t_{\text{last}} = \frac{c}{a}\;}

Why: this is the result. Every home event before $t_{\text{last}}$ is eventually seen; every home event after it is never seen at all. The boundary is not fuzzy and not a matter of patience — the equation has no solution. At $a = g$ , $c/g = 3.06\times10^{7}\ \mathrm{s} = \mathbf{0.97\ years} = \mathbf{354\ days}$ .

Step 4. Watch home fade. The ship recedes with rapidity $\varphi$ when the signal arrives, so the received frequency is Doppler-shifted by $\nu_{\text{rec}}/\nu_{\text{emit}} = \sqrt{(1-\beta)/(1+\beta)} = e^{-\varphi}$ . With Step 3:

\boxed{\;\frac{\nu_{\text{rec}}}{\nu_{\text{emit}}} = 1 - \frac{a\,t_{E}}{c}\;}

Why: the fade is exactly linear in home time — a small surprise, and easy to check. Departure light arrives unshifted; light from $t_E = \tfrac12\,c/a$ arrives at half frequency; light from $t_E \to c/a$ arrives at zero frequency, infinitely redshifted and infinitely delayed. Home does not vanish — it freezes and reddens, its last frame stretched over the ship's entire infinite future.

Step 5. Recognise what has just been described: a distant object seen to slow, redden and dim asymptotically, its final instant smeared across all of the observer's remaining eternity. That is the textbook description of matter falling into a black hole, seen from outside. Here it is a planet sitting perfectly still in flat empty space, watched by someone with a rocket. Why: no gravity was used, no curvature exists, and the Riemann tensor is zero at every event in the calculation. Horizon phenomenology is a phenomenology of acceleration, not of gravity. That is this chapter's thesis, and this is its sharpest single instance. *Verified in verify_aq-sr-rindler-horizon.py: sympy solves the intercept condition symbolically and returns $e^{-\varphi} = 1 - ct_E/X$ ; the existence condition $t_E < c/a$ follows; $\sqrt{(1-\tanh\varphi)/(1+\tanh\varphi)} = e^{-\varphi}$ is confirmed; and $c/g = 3.0570\times10^{7}\ \mathrm{s} = 0.9687\ \mathrm{yr} = 353.8\ \mathrm{days}$ is recomputed and grepped against this article.*

Figure

The owned result, on a spacetime diagram. Home sits still at x = 0 (green). The ship (blue) leaves home at the origin and accelerates at a constant 1 gee forever. The gold dashed line is the ship's Rindler horizon — the asymptote of its hyperbola, which here passes through home's worldline at ct = c²/g. Home events BELOW that crossing (green, solid) send light that catches the ship. Home events ABOVE it (grey, hatched) never do: their light rays run parallel to the horizon and stay astern forever. The boundary is at Earth time t = c/g = 0.97 years = 354 days. And the fade is linear: a signal sent at Earth time t_E arrives with ν_rec/ν_emit = 1 − g·t_E/c, so home reddens smoothly to zero frequency and freezes on its last frame

Limiting cases

Case	Result	Meaning
$a \to 0$	$d_H = c^2/a \to \infty$	an inertial observer has no horizon. Every event is eventually visible
$a \to \infty$	$d_H \to 0$	the horizon crowds up behind you, hiding almost everything astern
$a = g$	$d_H = 9.16\times10^{15}$ m	$0.97$ light-year. This is why we never notice Rindler horizons
$\xi_{\text{emit}} \to -c^2/a$	$1 + z \to \infty$	infinite redshift; the source freezes and reddens without limit
engine cut at finite $\tau$	horizon dissolves	the hidden region floods back into view. The horizon was made of the word "forever"
home departure at $a$	$t_{\text{last}} = c/a$	every home event after $c/a$ is lost. At $1\,g$ : $354$ days

Common trapThe signature — and this base gets it wrong twice over

In $(+,-,-,-)$ the comoving Rindler metric is $\mathrm{d}s^2 = +\left(1 + a\xi/c^2\right)^2c^2\mathrm{d}\tau^2 - \mathrm{d}\xi^2$ . The AllQFT base for this article prints $\mathrm{d}s^2 = -\left(1 + a\xi/c^2\right)^2c^2\mathrm{d}\tau^2 + \mathrm{d}\xi^2$ , which is the $(-,+,+,+)$ convention imported wholesale. It is not a cosmetic difference: with the base's signs, a clock held at rest in the accelerated frame has $\mathrm{d}s^2 < 0$ , is therefore spacelike, and reports an imaginary proper time. Ask what a stationary clock does — the check takes four seconds and it catches this every time. Three of this chapter's five bases carry the same flip. Re-derive the metric from the map. Never transcribe one.

Common trapNothing about this is gravity — and two of the base's numbers do not survive arithmetic

This whole article lives in flat spacetime with $R^{\lambda}{}_{\mu\nu\kappa} = 0$ : no mass, no curvature, no tidal force, no general relativity. The horizon, the infinite redshift, the frozen fading image of home — all of it is special relativity, and the folklore that accelerated frames require GR is simply false. What GR is for is the thing with no counterpart here: tidal effects, which no coordinate change removes, and which `aq-sr-equivalence-principle` shows to be the true boundary. Two base numbers also fail: the base's limits card gives the horizon distance at $a = g$ as " $\approx 9.5\times10^{15}$ m" (it is $9.16\times10^{15}$ , and the base's own checks card says $9.2\times10^{15}$ — it contradicts itself); and its provenance claims $a = 10^{20}\ \mathrm{m/s^2}$ gives an Unruh temperature of " $\approx 40$ K", when $\hbar a/2\pi ck_B$ at that acceleration is $\mathbf{0.41\ K}$ — wrong by a factor of about $100$ .

Worked examples

1 · How far behind you is the horizon, and what does it cost to move it?

A ship holds a steady $1\,g$ . Where is its horizon? What acceleration would bring the horizon to within one astronomical unit? And what would it take to warm the horizon to $1$ K?

Step 1 — identify. All three questions are equation (2), $d_H = c^2/a$ , read forwards and backwards, plus the signpost (5).

Step 2 — the $1\,g$ horizon. $d_H = \dfrac{(2.99792458\times10^{8})^{2}}{9.80665} = \mathbf{9.16\times10^{15}\ m}$ , which against a light-year of $9.4607\times10^{15}$ m is $\mathbf{0.97}$ light-years. Your causal horizon is just under a light-year astern — and stays exactly there, forever, however long you burn.

Step 3 — pull the horizon in to $1$ AU. Invert: $a = c^2/d_H = \dfrac{8.98755\times10^{16}}{1.49598\times10^{11}} = \mathbf{6.01\times10^{5}\ m/s^{2}}$ , which is $\mathbf{6.1\times10^{4}}\,g$ — sixty thousand gees. To hide everything more than one astronomical unit behind you, you must be crushed by sixty thousand times your own weight.

Step 4 — warm it to $1$ K. From (5), $a = 2\pi ck_B T/\hbar = \mathbf{2.47\times10^{20}\ m/s^{2}}$ , or about $2.5\times10^{19}\,g$ . This is why the Unruh effect has never been seen. At the $1\,g$ of everyday life the temperature is $T_U = \mathbf{4.0\times10^{-20}\ K}$ — twenty orders of magnitude below the cosmic microwave background.

Step 5 — the boundary number. Every horizon we can build is either a light-year away or needs an acceleration that would vaporise the apparatus. The Rindler horizon is not an engineering constraint; it is a structural fact, and its importance is that it is the only horizon we can compute exactly.

2 · You leave Earth at 1 g today. What do you never see?

The ship departs at rest from Earth and burns at a constant $1\,g$ , forever. Which Earth events are lost, and what does Earth look like on the way out?

Step 1 — identify. This is derivation 3. The barrier is $t_{\text{last}} = c/a$ in Earth's time, and the fade is $\nu_{\text{rec}}/\nu_{\text{emit}} = 1 - a t_E/c$ .

Step 2 — the barrier. $t_{\text{last}} = \dfrac{c}{g} = \dfrac{2.99792458\times10^{8}}{9.80665} = 3.06\times10^{7}\ \mathrm{s} = \mathbf{0.97\ years} = \mathbf{354\ days}$ . Everything that happens on Earth from that moment on — every subsequent year, every subsequent century, every subsequent aeon — sends light that never arrives.

Step 3 — the fade. A broadcast made at Earth time $t_E$ arrives with $\nu_{\text{rec}}/\nu_{\text{emit}} = 1 - t_E/(354\ \mathrm{days})$ : at half a year, redshifted to $\mathbf{48\%}$ of its frequency; at nine months, $\mathbf{23\%}$ ; at $353$ days, $\mathbf{0.23\%}$ — a radio broadcast arriving as a slow thump. It never quite stops; it just stretches.

Step 4 — check whether Earth is actually hidden. No — and this subtlety is worth flagging. Earth's worldline is not behind the horizon: the ship's horizon crosses $x = 0$ at $ct = c^2/g$ , so Earth's entire history before that moment is visible. What is hidden is not the place. It is the future.

Step 5 — the boundary number. $354$ days. That is a human number, produced by a comfortable acceleration and an infinite amount of patience, in perfectly flat spacetime. It is the closest thing special relativity has to a black hole, and you can compute it on the back of an envelope.

Physics of the portrait

The Rindler horizon is a concept, not a person, and that is the interesting thing about it: it was noticed several times, by people looking for different things, before anyone thought it deserved a name.

Origins · 2 minRindler horizon and how this was discovered

Rindler horizon

1956–1976 · Rindler names the event horizon · the uniformly accelerated frame gets its wedge · Unruh gives the horizon a temperature

The name. Wolfgang Rindler (1924–2019) — born in Vienna, sent to England on the Kindertransport in 1938, doctorate at Liverpool, then a long career at the University of Texas at Dallas — introduced the term "event horizon" in a 1956 paper on cosmological horizons (Mon. Not. R. Astron. Soc. 116, 662), carefully distinguishing it from the particle horizon with which it had been routinely confused. The word entered black-hole physics from there. A decade later, in a 1966 American Journal of Physics paper relating Kruskal's extension of the Schwarzschild solution to the uniformly accelerated frame, he laid out the flat-space wedge and its horizon so cleanly that both now carry his name — though he did not name them himself, and the coordinates go back at least to Møller and, in essence, to Einstein's 1907 accelerating box. Why it matters. Rindler's insistence was that the accelerated observer's causal boundary is a genuine horizon, with the same one-way character and the same infinite redshift as a black hole's, in a spacetime that is exactly flat. That converted a curiosity into a laboratory: every hard question about horizons — what is seen, what is hidden, what "static" means, whether information is lost — can be posed first in the Rindler wedge, where the geometry is Minkowski's and every calculation closes in a line or two. The door it opens. In 1976 William Unruh showed that quantising a field on this wedge and tracing over the hidden region leaves the accelerated observer in a thermal state at $T = \hbar a/2\pi ck_B$ . The horizon is what forces the trace: no hidden region, no trace, no temperature. That is quantum field theory, not special relativity, and this article does not derive it — but it is why the flat-space horizon is taken as seriously as it is.

History

1907

— Einstein's Jahrbuch review treats the uniformly accelerated frame and finds that clocks in it run at position-dependent rates — the first sighting of the structure whose boundary is the horizon.

1943

— Christian Møller develops the comoving coordinates of a uniformly accelerated observer, in which the lapse vanishes at a finite distance astern.

1956

— Rindler introduces the term "event horizon" in Mon. Not. R. Astron. Soc. 116, 662, distinguishing it from the particle horizon and giving the concept its name.

1960

— Kruskal and Szekeres extend the Schwarzschild solution across its horizon, showing the coordinate singularity at r = 2GM/c² to be no singularity at all.

1966

— Rindler relates the Kruskal extension to the uniformly accelerated frame in Am. J. Phys. 34, 1174; the flat-space wedge and its causal boundary acquire his name.

1976

— Unruh, in Phys. Rev. D 14, 870, shows that an observer with proper acceleration a sees the Minkowski vacuum as a thermal bath at ħa/2πck_B, and that the horizon is what makes it so.

Going further

Texts. Rindler, Relativity: Special, General and Cosmological, 2nd ed., ch. 12 · Misner, Thorne & Wheeler, Gravitation, §6.6 · Wald, General Relativity, §6.4 and ch. 14 · Rindler, "Visual horizons in world-models", Mon. Not. R. Astron. Soc. 116, 662 (1956) · Rindler, "Kruskal space and the uniformly accelerated frame", Am. J. Phys. 34, 1174 (1966) · Unruh, Phys. Rev. D 14, 870 (1976).

Provenance. Signature $(+,-,-,-)$ . Every equation and number is recomputed in verify_aq-sr-rindler-horizon.py, which reads this file and asserts each printed value against what it computed. Base defects repaired: (i) the base's comoving metric $\mathrm{d}s^2 = -(1 + a\xi/c^2)^2c^2\mathrm{d}\tau^2 + \mathrm{d}\xi^2$ is a $(-,+,+,+)$ artefact; corrected and re-derived from the map. (ii) the base's limits card gives the $1\,g$ horizon distance as $\approx 9.5\times10^{15}$ m; it is $9.16\times10^{15}$ m, and the base's own checks card says $9.2\times10^{15}$ — the base is internally inconsistent. (iii) the base's provenance claims $T_U \approx 40$ K at $a = 10^{20}\ \mathrm{m/s^2}$ ; the correct value is $\mathbf{0.41}$ K, low by a factor of about $100$ (the acceleration needed for $1$ K is $2.47\times10^{20}\ \mathrm{m/s^2}$ ).

Scope. The hyperbolic worldline is owned by `aq-sr-relativistic-uniform-acceleration`; the chart and the position-dependent clock rate by `aq-sr-rindler-metric`; the profile $a = c^2/x$ by `aq-sr-born-rigidity`. All are used, not re-derived. The Unruh temperature is a signpost, never derived — it is quantum field theory and belongs to `aq-sr-unruh-effect`. The tidal boundary with general relativity is `aq-sr-equivalence-principle`'s.

Builds on. Relativistic Uniform Acceleration · Rindler Coordinates · The Rindler Metric. Leads to. The Equivalence Principle · Unruh Effect · Gravitational Redshift.

Light is faster than the ship. How can it possibly fail to catch up?

Because "faster" settles a race with a finish line, and there is no finish line. The photon does gain: the gap shrinks monotonically, forever. But it shrinks toward a limit — the constant $x_0 - ct_0$ fixed at emission, which is negative if the emission was behind the horizon. The photon spends an infinite time closing an interval it never finishes closing. Nothing exceeds $c$ , nothing is outrun, and the observer's speed stays strictly below $c$ throughout. The only ingredient is that the acceleration is eternal: stop it at any finite moment and the photon arrives promptly.

Is the horizon a real thing or a coordinate artefact?

Both, and the pair of answers is the point. The surface $x = ct$ is a coordinate singularity of the Rindler chart — the lapse vanishes there, as $g_{\theta\theta}$ vanishes at the origin of a polar chart — and it is a place where nothing physical happens: the Riemann tensor is zero, an inertial observer crosses without noticing, and no instrument could locate it locally. But the causal structure is not an artefact. Events beyond it genuinely cannot signal the accelerated observer, ever, and no change of coordinates alters that. A coordinate singularity marking a genuine causal boundary is exactly the status of a black hole's event horizon, which is why the analogy is taken seriously rather than merely admired.

Can the accelerated observer send signals INTO the hidden region?

Yes — and this asymmetry is the whole character of a horizon. The observer's future light cone extends into the region behind $x = ct$ , so messages can be sent in freely; nothing can be received back. The horizon is a one-way membrane, one-way in exactly the direction a black hole's is: an outside observer can drop a probe in and can never hear from it again. Leave Earth at $1\,g$ forever and you can broadcast to Earth indefinitely, and Earth will keep receiving you. You simply cannot hear their reply after their day $354$ .

Do two ships with different accelerations share a horizon?

No, and the error is consequential. A horizon belongs to a worldline, not to a region of spacetime. A ship at $2g$ has its horizon at $c^2/2g$ , half as far astern as one at $1g$ ; a ship starting elsewhere has its horizon displaced with it. Two ships share a horizon only if their hyperbolae share a vertex — which is precisely the Born-rigid condition, and precisely what Bell's identically-accelerating ships fail to satisfy. That failure has a startling consequence, worked out in `aq-sr-bell-spaceship-paradox-s6`: each of Bell's ships has its own horizon, and one of them ends up behind the other's.

Why do I never notice a Rindler horizon when I accelerate my car?

Because $c^2$ is enormous and your car is not. At the $3\ \mathrm{m/s^2}$ of a brisk acceleration the horizon lies $c^2/a = 3\times10^{16}$ m astern — about three light-years, further than the nearest star — and it exists only while you keep accelerating, which is a few seconds. A horizon inside the solar system would need about sixty thousand gees. The structure is always there in principle; $c^2$ simply pushes it so far away that it bears on nothing you can do.

What exactly does the Unruh temperature have to do with this article?

It is the door at the end of the corridor, and this article stops at the threshold. The Rindler horizon hides a region of spacetime; quantum field theory says that a field's state in the visible region, obtained by tracing out the hidden one, is thermal, at $T = \hbar a/2\pi ck_B$ . The horizon is what forces the trace — no hidden region, no trace, no temperature — so the flat-space causal boundary derived above is the direct cause of the effect. But the effect itself is quantum field theory, not special relativity, and it is derived in `aq-sr-unruh-effect`, not here. The scale is worth carrying: $4.06\times10^{-21}$ K per $\mathrm{m/s^2}$ , so $1$ K needs $2.47\times10^{20}\ \mathrm{m/s^2}$ .