The wall of silence behind a rocket that never stops

A rocket sets off and holds one gravity for ever. The passengers feel their ordinary weight — which no experiment inside can tell from gravity, the engine never falters, and the speed measured from the launch pad climbs toward c without reaching it. Nothing in that description sounds like a boundary of any kind, and it produces one.

The dividing line is the asymptote itself. Everything at or behind it is permanently out of reach, and at one gravity that boundary sits 0.97 light years behind the rocket’s starting point.

Why the worldline is a hyperbola #

Constant proper acceleration is what an accelerometer on board reads and what a passenger feels as weight. It is not a constant rate of increase of speed in the launch frame, because that would take the rocket past c in a year.

The condition is that the acceleration measured in the rocket’s own instantaneous rest frame is always the same a. Writing the worldline in terms of proper time τ, the result is

$$ct = \frac{c^2}{a}\sinh\frac{a\tau}{c}, \qquad x = \frac{c^2}{a}\cosh\frac{a\tau}{c},$$

and since $\cosh^2 - \sinh^2 = 1$, these satisfy

$$x^2 - c^2t^2 = \left(\frac{c^2}{a}\right)^2.$$

A hyperbola, and specifically a curve of constant interval from the origin — which is why it is the same curve the calibration of a spacetime diagram is built on.

The two hyperbolic functions are worth reading as a pair. $\cosh$ is the Lorentz factor γ and $\sinh$ is γβ, so the worldline’s two coordinates are exactly the two components of the four-velocity multiplied by $c^2/a$. What is being said is that a uniformly accelerating body’s four-velocity rotates through a hyperbolic angle at a constant rate, which is the flat-spacetime version of “turning at a constant rate” — and the reason the trajectory is a conic section is the same reason a circle is.

The quantity that rises steadily is the rapidity. $a\tau/c$ is the hyperbolic angle, it grows linearly with the traveller’s own clock, and speeds are its hyperbolic tangent. That is the sense in which the motion is uniform: a quantity does increase at a constant rate, and it is not the velocity.

Rapidity is the natural variable because velocities do not add and rapidities do, exactly. That is what makes “hold one gravity for a year, then another year” a statement about a sum rather than about a composition rule: after a proper year at one gravity the rapidity is 1.03 and the speed is 0.77c; after ten proper years the rapidity is 10.3 and the Lorentz factor is fifteen thousand. The speed has gone almost nowhere and the rapidity has gone ten times as far.

And the acceleration never changes the interval. A body on one of these hyperbolas stays at a fixed invariant distance from the origin for ever, which is the tidiest statement of what “uniform acceleration” means: not a constant velocity, not a constant coordinate acceleration, but a worldline that keeps its interval from one event. Every quantity that seems to change — speed, Lorentz factor, coordinate distance — is a reading of one curve on which nothing invariant changes at all.

Where the horizon is, and what is at it #

The horizon is a fact about a family of observers rather than a place. Stop accelerating — coast for a moment — and the worldline becomes a straight line, the asymptote is crossed, and everything that had been cut off arrives. Nothing has changed anywhere in space; what changed is the motion of the observer, and with it which events can ever reach them.

The asymmetry is worth stating carefully. The rocket can still send signals into the region behind the horizon — its own light cone opens backwards without limit — so the boundary cuts off what can be received and not what can be sent. That one-way property is what makes it a horizon rather than a wall, and it is the same one-way property a black hole’s has, with the direction reversed.

And it is not a distance to somewhere. An observer 0.97 light years behind the rocket, at rest in the launch frame, sees nothing unusual at all: their own light cone is complete, their own future is unbounded, and signals from them are travelling toward the rocket at c the whole time. The horizon exists only in the rocket’s accounting, and it is exactly the boundary of the set of events the rocket can ever learn about.

An observer in uniform motion has no horizon at all, and the contrast is the whole point. Their light cone reaches the whole of spacetime given enough time — every event in the past of any point on their worldline is a signal they could receive — so no part of the diagram is permanently out of reach. Inertial motion is what has no horizon. Acceleration is what makes one, and stopping is what removes it again.

The redshift that goes with it. A source at a fixed position in the launch frame, sending pulses at a steady rate, is received by the rocket at longer and longer intervals — and for a source right at the horizon the received interval grows without limit. So the last thing the rocket sees of anything approaching that boundary is an ever-slower, ever-redder image that never quite goes out, which is what a distant observer actually receives from a fall. That is the second signature of a horizon, and it arrives here from a hyperbola rather than from a metric.

Every observer in the family has a different gravity #

The horizon was introduced as a fact about one rocket. Adding the rest of the ship turns it into a statement about clocks, and the statement is the one general relativity is usually credited with.

Consider a whole fleet of observers, each on their own hyperbola, arranged so that the distances between them stay fixed as measured by themselves. Each hyperbola has its own $c^2/a$, and every one of them is asymptotic to the same light ray — the fleet shares a single horizon. An observer whose hyperbola passes closer to that ray has a smaller $c^2/a$, which means a larger proper acceleration:

$$a(X) = \frac{c^2}{X},$$

with $X$ the distance from the horizon. So the observer nearest the horizon has to push hardest, the one furthest has to push least, and the required thrust runs to infinity as the horizon is approached. Holding a rigid formation under acceleration is not a matter of everyone doing the same thing.

The clocks then follow. Proper time along each hyperbola runs at a rate proportional to $X$, so the ratio of any two observers’ clock rates is the ratio of their distances from the horizon — fixed for ever, since the distances are. A clock at the rear of the fleet runs slow compared with one at the front, permanently, by a factor that is exactly one plus the acceleration times the separation over $c^2$ when the separation is small.

That is the gravitational redshift, derived with no gravity anywhere. Two clocks at different heights in a uniformly accelerating cabin disagree, by $gh/c^2$, and by the equivalence principle two clocks at different heights in a field must disagree by the same amount. The rear of the fleet is “lower” in exactly the sense that matters, and the horizon is the floor.

From this horizon to a black hole’s temperature #

The Unruh temperature and the Hawking temperature look like two results and are very nearly one, and the bridge is the fleet above.

Take an observer hovering at a fixed radius just outside a black hole, firing their engine to stay put. Over a small enough region they are simply an accelerating observer in flat spacetime, so what they detect is the Unruh bath at $\hbar a/2\pi ck_B$ with $a$ their own local proper acceleration. Close to the horizon that acceleration is enormous, so the local temperature is high.

Now take that radiation out to a distant observer. It climbs out of the well and is redshifted by exactly the factor that made the local acceleration large in the first place, and the two effects cancel: the temperature measured far away is finite, the same for every hovering observer whatever radius they hover at, and it comes out as the surface gravity divided by $2\pi c$ in the same units. Which is Hawking’s result.

So the two temperatures are the same phenomenon read at two places, and the reason a black hole has a single well-defined temperature is that the divergence of the local one is exactly compensated by the divergence of the redshift. Neither statement is derivable from this essay’s figures — both need a field theory — but the relation between them is pure kinematics, and it is the strongest form of the claim that the resemblance between the two horizons is structural.

The same shape as a black hole’s, and the differences #

A black hole’s horizon is at a place. Its factor $\sqrt{1 - r_s/r}$ goes to zero at a definite radius; it is there for every observer; it has an area; and nobody can stop accelerating to be rid of it. The rocket’s horizon is at $c^2/a$ behind whoever is accelerating, so two rockets at different accelerations have different ones and a rocket that cuts its engines has none. Same geometry, entirely different status.

What the two share is the structure that matters, which is why the comparison is worth making rather than merely noting. In both cases there is a family of observers who hold station, both are describing a region they cannot receive from, and in both cases a signal from a fixed source falls to zero frequency and fades away rather than cutting off — which is the same infinite redshift, arrived at from an acceleration in one case and a field in the other.

And the equivalence principle says the resemblance is not a coincidence. A sealed cabin holding one gravity of thrust is indistinguishable from a sealed cabin standing on a planet, so anything the accelerating cabin has must have a counterpart for the stationary one, and the argument runs both ways. The horizon is the counterpart of a horizon, and it is the cleanest example of a general-relativistic feature derived with no general relativity in it.

The principle doing the work is the simplest form of the equivalence principle. Light crossing an accelerating cabin follows a curve in the cabin’s own frame, and the curvature is exactly what a cabin at rest in a field of the same strength must see. Every feature of the accelerated frame that survives being made local is a statement about gravity — that is the whole method — and the horizon is what happens when the acceleration is not made local but held for ever.

What a journey looks like from on board #

The hyperbola has a consequence that is more cheerful than the horizon, and it comes from the same two lines of algebra.

Distance covered in the launch frame is $x = (c^2/a)(\cosh(a\tau/c) - 1)$, which grows exponentially in the traveller’s own time once $a\tau/c$ is past one. At one gravity, $c/a$ is 0.97 years, so a proper year gets 0.56 light years, five proper years get 84, and ten get 15,000. A trip to the centre of the galaxy — 26,000 light years — takes about twenty years on board and 26,000 years at home.

Nothing about that violates anything. The traveller never exceeds c in any frame, the elapsed time at the destination is longer than the distance in light years, and the asymmetry is the ordinary one: the traveller’s clock has run slow because the traveller accelerated and the Earth did not. What makes the arithmetic feel wrong is the exponential, and the exponential is a hyperbolic cosine that has been there since the first equation.

And the fuel is what makes it impossible. A rapidity of ten needs a mass ratio of $e^{10}$ even at the theoretical maximum exhaust speed, which is c. That is twenty-two thousand tonnes of fuel per tonne of rocket, perfectly converted, with the exhaust perfectly collimated. The kinematics is unremarkable and the energetics is the barrier.

The temperature, and why it is not detectable #

An accelerating detector in empty space registers a thermal bath at $T = \hbar a/2\pi c k_B$. At one gravity that is 4 × 10⁻²⁰ kelvin, and the arithmetic of why nobody has measured it is the same arithmetic as the horizon distance: reaching a millikelvin requires 2.5 × 10¹⁷ gravities.

The corresponding temperature for a black hole is the same expression with the surface gravity in place of the acceleration, and the numbers are what make both undetectable. A solar-mass hole sits at 60 nanokelvin, four orders of magnitude colder than the cosmic microwave background, which is why such a hole absorbs rather than evaporates. Both temperatures are absurdly small for one reason: $\hbar/c$ is a very small number of joule-seconds per metre.

The interesting part is not the number but the statement. The vacuum an inertial observer finds empty is not empty to an accelerating one — the two disagree about how many particles are present, which means “how many particles are there” is not a question with an observer-independent answer. That is a strong conclusion drawn from a picture with one hyperbola in it.

A sharper version of the same statement. The number of particles in a field is defined by splitting the field into positive- and negative-frequency parts, and “frequency” needs a time. An inertial observer’s time and an accelerating observer’s differ by more than a shift or a scaling — they mix the two parts — so a mode that is purely positive-frequency for one is a mixture for the other, and a state with no quanta for the first has quanta for the second. That is the whole derivation, and it contains no thermodynamics: the thermal spectrum falls out of the mixing rather than being assumed.

What the picture does not show #

The rocket is a point. A real rocket has length, and the front and back cannot both hold the same proper acceleration if the length is to stay fixed — the rear must push harder, by a factor that grows as the rear approaches the horizon, which is what a rigid body costs in relativity. At one gravity and a hundred-metre rocket the difference is one part in 10¹⁴ and is invisible; at a rear end near c²/a it is unbounded.

Nothing here is quantum until the last section. The horizon, the hyperbola, the signals that never arrive and the coordinate wedge are all classical kinematics in flat spacetime, derivable with a ruler and a light ray. The temperature is the one statement that needs a field theory, and it is separable from the rest: an observer who never accelerated could be told everything above and would find nothing to disagree with.

The acceleration is held for ever. Every conclusion above depends on that, and no rocket has that much fuel: reaching a rapidity of ten at one gravity takes ten years of proper time and a mass ratio, at the best exhaust speed there is, of $e^{10}$ — twenty-two thousand to one. The horizon is a feature of an idealised worldline, and the idealisation is the part that is impossible rather than the conclusion.

Drawn against time rather than as a worldline, the same motion reads three ways at once. Speed climbs toward $c$ and never reaches it; distance covered in the launch frame becomes linear in time; and the proper time on board falls further and further behind the coordinate time. The third is why a journey of any distance takes a finite time on board, and it is the clock the acceleration is held constant with respect to — the proper time along the worldline, not anything the launch frame measures.

The horizon moves with the rocket. It is always c²/a behind wherever the rocket currently is, which sounds like it is being carried along and is not: the horizon is a fixed light ray in the launch frame, and the rocket is receding from it at an ever-decreasing rate in its own reckoning. That the proper distance from rocket to horizon stays at c²/a while the coordinate gap grows without limit is the same kind of statement as a length contraction, and it is why the rocket’s crew, measuring with their own rulers, find the boundary at the same distance for ever.

And the coordinates it suggests do not cover everything. Building a coordinate system from the family of uniformly accelerating observers — one for each distance behind the leader — covers only the wedge to the right of the asymptote. The rest of spacetime is not described by those coordinates at all, which is a defect of the coordinates rather than of spacetime, and is the exact analogue of what a coordinate system anchored outside a black hole does at its horizon.

The same coordinate failure was found first in the gravitational case. Signals from something falling through a horizon arrive at longer and longer intervals and never stop arriving, so the outside observer’s coordinate never records the crossing — while the faller’s own clock passes through in a finite and unremarkable time. Both statements are true, they are about different clocks, and the rocket’s horizon has the identical structure with no gravity in it anywhere.

A horizon that ends is not a horizon #

One qualification deserves to be stated on its own, because the word “for ever” is carrying the whole argument and it is easy to miss how much it carries.

Every claim above requires the acceleration to be eternal. If the rocket burns for a century and then coasts, its worldline is a piece of hyperbola followed by a straight line, and that straight line crosses the asymptote almost immediately. Every signal that had been chasing it arrives, in order, late. Nothing was ever cut off; the delivery was delayed.

So a horizon of this kind is not a property of a stretch of motion but of a whole infinite future, and no finite observation can establish that one exists. A crew a year into their burn cannot tell whether they have a horizon behind them, because the answer depends on what their engine does in a thousand years. That is an unusual kind of physical statement and it is worth naming as such: the horizon is defined by the global structure of a worldline, not by anything local along it.

The same is true of a black hole’s horizon, and it is the reason the definition is so awkward. What makes a surface an event horizon is that nothing crossing it ever escapes — a statement about the entire future of the spacetime, which no observation inside any finite region can settle. Which is why the practical definitions used in calculations are local approximations to it rather than the thing itself.

Where this ladder goes next #

This rung has one observer accelerating for ever and a horizon behind them. The next asks about the region those observers describe: the wedge, the coordinates that fit it, and what happens at the point where they fail. What comes out is that the surface where the coordinates break is not a surface where anything physical happens — an inertial observer sails through it noticing nothing — and that the whole apparatus of a horizon can therefore be a property of a description rather than of a place, which is the single most useful thing to have understood before meeting a black hole.