This web page was inspired by a discussion on the Usenet group **sci.physics.research**
entitled “The Black Fishing Hole”, in which Edward Green asked for an
account of precisely what would happen if someone lowered an object through a black
hole’s horizon on a fishing line.
I gave a reply describing a slightly different, but related, scenario: that of a
constantly accelerating observer in flat spacetime trailing an object behind
them.
Darryl McCullough made the connection explicit, pointing out that a first-order
approximation of the Schwarzschild metric near a black hole’s horizon gives a coordinate
system for flat spacetime describing just such a class of observers.

The purpose of this web page, then, is to analyse in detail (using only special relativity) some interesting thought experiments that can be carried out by a constantly accelerating observer, who sees a “Rindler horizon” in spacetime that is very similar to the event horizon of a black hole. This is certainly not a perfect substitute for a full, general-relativistic analysis of the same experiments carried out near a black hole’s horizon. Rather, what it represents is an interesting limiting case: a black hole so massive that the spacetime curvature at its horizon is negligible.

My thanks to Chris Hillman for many helpful comments and suggestions.

- Preliminaries
- Static situations
- Free fall
- Dynamic situations
- String unreeled at a constant rate
- Conclusions

Imagine that it’s possible to build a spaceship that constantly accelerates.
We don’t really care how it does this; this is a thought experiment in physics, not
an exercise in interstellar spacecraft engineering. Because we’re dealing with
relativistic velocities, we need to be precise and say that the spaceship undergoes
constant **proper acceleration**: passengers will measure a constant inertial
force pushing them towards the back of the ship, whereas observers in
motion relative to the ship will necessarily describe its acceleration differently.

If we start out by treating the spaceship as a single point, we can describe its
world line in an inertial reference frame with coordinates (*t*, *x*) as:

x(τ) = (t(τ),x(τ)) = (sinh(aτ)/a, cosh(aτ)/a)

where τ is proper time measured by a clock in the spaceship (set to zero when *t*=0),
*a* is the acceleration the passengers feel, and we have chosen coordinates so that
*x*=1/*a* when *t*=τ=0. (We are using units where the
speed of light is set equal to 1, and this means that whenever we talk about
an inverse acceleration being equal to a distance, or vice versa,
their product will equal *c*^{2} in conventional units.
For example 1/*g*, where *g* is the gravitional
acceleration at the Earth’s surface, corresponds to a distance of 9.18 × 10^{15}m,
or about a light-year, while 1/*x*, where *x* is one metre, corresponds to an
acceleration of 9.18 × 10^{15}g.)

The world line of the spaceship is a hyperbola:

This hyperbola can also be described by the equation:

a^{2}x^{2}–a^{2}t^{2}– 1 = 0, or

x·x= 1/a^{2}

where in the second line we’re using the Lorentzian dot product:

(t_{1},x_{1}) · (t_{2},x_{2}) = –t_{1}t_{2}+x_{1}x_{2}

Now, the dashed line in the figure above shows the world line of a flash of light
emitted from *x*=0 at *t*=0. The spaceship’s world line is asymptotic to that of the flash
of light, so the light never catches up with the spaceship. In (*t*, *x*) coordinates the
two draw ever closer together, but that’s not the case in the spaceship’s own
coordinates. If we take a family of inertial reference frames that are
instantaneously co-moving with the accelerating spacecraft, each frame will have an
orthonormal basis {**e**_{τ}, **e**_{s}}, where:

e_{τ}= cosh(aτ)e_{t}+ sinh(aτ)e_{x}

e_{s}= sinh(aτ)e_{t}+ cosh(aτ)e_{x}

For any value of τ, we then have:

(0, 0) =x(τ) – (1/a)e_{s}

In other words, the event of the emission of the flash of light is
always in the present, and always a distance
of 1/*a* behind the ship. As far as the passengers on the ship are concerned, the
flash of light isn’t getting any closer.

Any signal sent from an event that lies to the left of the dashed line in the figure
will be unable to
reach the spaceship, because that would mean out-racing the pulse of light
that itself fails to catch the spaceship. So, as far as the passengers on the ship
are concerned, the dashed line constitutes a kind of event horizon, known as the
**Rindler horizon**. Assuming that the spaceship continues to accelerate forever,
anything that ends up behind the Rindler horizon will be lost to
the passengers forever.

Of course, this is not the same as a black hole’s event horizon in two very important respects. Firstly, it’s always possible to stop the spaceship accelerating, so this horizon’s persistence is a matter of choice, not physical law. Secondly, there is nothing corresponding to a black hole’s singularity to do any actual damage to anything passing through the horizon.

However, the analogy extends this far: a constant force is required to keep the spacecraft a constant distance (by its own measurements) from the Rindler horizon, and, apart from the effects of spacetime curvature (which can be made as small as we like by imagining a black hole with a large enough mass), all the other phenomena (red shift, time dilation and so on) that occur in the vicinity of a black hole’s horizon can be found in the vicinity of a Rindler horizon. There is even an analogue of Hawking radiation, known as Unruh radiation.

We can construct a useful coordinate system that covers the region of spacetime to the right of the horizon. Every straight line of the form:

cosh(q)t– sinh(q)x= 0

will intersect the spaceship’s
world line orthogonally to its velocity 4-vector **e**_{τ} and parallel to
the vector **e**_{s}. Now, consider a family of hyperbolas of the form:

x^{2}–t^{2}–s^{2}=x·x–s^{2}= 0, or

(t(q),x(q)) = (ssinh(q),scosh(q))

These will intersect all of the straight lines of constant *q* orthogonally.
So, in the region to the right of the Rindler horizon, we can use *q* and *s*
as coordinates:

Along the *x*-axis, *s* agrees with *x*, and *q*=0. Although *q* can be used to parameterise
each of the hyperbolas of constant *s*, it does not correspond to elapsed proper time
along these curves; rather, that is given by τ=*q**s*. Along lines of constant *q*,
*s* measures proper distance.

We can extend our field of orthonormal frames {**e**_{τ}, **e**_{s}} away
from the spaceship itself to arbitrary values of *s*:

e_{τ}= cosh(q)e_{t}+ sinh(q)e_{x}

e_{s}= sinh(q)e_{t}+ cosh(q)e_{x}

We can imagine a flotilla of spaceships, each remaining at a fixed value of *s* by
accelerating at 1/*s*. In principle, these ships could be physically connected together
by ladders, allowing passengers to move between them. Although each ship would have
a different proper acceleration, the spacing between them would remain constant as far
as each of them was concerned. The easiest way to see this is to note that
increasing *q* is equivalent to performing a boost on everything, leaving
intervals in spacetime invariant. Another way to describe this is to say that
the vector field ∂_{q} generates
an **isometry** of spacetime.

When an object is accelerated in such a way as to preserve the proper distances
between its different parts, this is referred to as **Born rigid**
motion (after the physicist Max Born).

The easiest results to calculate involve imagining various observers and objects
at fixed positions in terms of our *s* coordinate, so these are the first things
we’ll consider; more interesting possibilities will follow. We will allow light
signals to cross between different values of *s*, but nothing else will move.

Suppose a flash of light is emitted from *s*_{1} at *q*=*q*_{1},
in the forwards direction. Then if an adjacent observer considers the energy of the
pulse to be *E*_{1}, its energy-momentum vector will be:

p=E_{1}(e_{τ}+e_{s})

=E_{1}(cosh(q_{1}) + sinh(q_{1})) (e_{t}+e_{x})

=E_{1}exp(q_{1}) (e_{t}+e_{x})

By setting *q*_{1} to 0 and solving for the value of *q*_{2} where
the world line for the flash of light hits the hyperbola *s*=*s*_{2}, we find:

exp(q_{2}) =s_{2}/s_{1}

and since we can write an expression for **p** at this second
point that takes the same form as the one at (*q*_{1},*s*_{1}),
it follows that:

E_{1}s_{1}=E_{2}s_{2}

where *E*_{2} is the energy of the flash measured at *s*_{2}.
This will hold regardless of the particular value of *q*_{1}. So as we’d
expect, light that travels forward away from the Rindler horizon is red-shifted.
Similarly, light that travels towards the horizon is blue-shifted.

Another way to
obtain this result is to note that the vector field
∂_{q} = *s***e**_{τ}
is a Killing field; that is, as we’ve already noted, it
generates an isometry of spacetime. The result then follows from the fact that
any geodesic has a constant dot product with a Killing field, and so:

E_{1}s_{1}=s_{1}p·e_{τ}=s_{2}p·e_{τ}=E_{2}s_{2}

Because observers with fixed *s* coordinates are constantly accelerating,
light from distant stars will suffer constantly increasing Doppler shifts,
so (unlike the case with black holes) it’s meaningless to talk about a blue
shift for light arriving from infinity.

Suppose we fix the *s* coordinate of one end of a string and let it dangle
above the Rindler horizon. We’ll call the *s* coordinate of the
fixed end *s*_{2}, and that of the free end *s*_{1}.
Because the tension in the string becomes zero at the free end, it’s easier to
deal with the boundary conditions if we describe everything in terms of *s*_{1}.

We will assume that the string obeys Hooke’s law exactly (that is, each part
of it stretches in proportion to the force applied)
even as the tension it experiences grows arbitrarily large; this is unrealistic, but
it is the simplest assumption we can make that’s compatible with special relativity,
which does not permit perfectly inelastic bodies.
(There are some further restrictions needed, which we’ll describe shortly.)
We’ll call the constant of proportionality *k*, and we will assume that the string has a constant proper mass density of
ρ
per unit of unstretched length. We’ll use a parameter
λ
to describe the unstretched distance along the string, starting from the free end.

Rather than working explicitly with the tension in the string, we will compute
everything in terms of pressure, since this is a more common starting point for
the stress-energy tensor that we’ll need to write down; we simply need to remember
that the pressure will be negative. We start by expressing the quantities we need
in terms of *n*(*s*), the factor by which the string has been *compressed*
(which will be less than 1 when the string has been stretched):

n(s) = dλ(s)/ds

p(s) =k(1 – 1/n(s))

u(s) = (k/2) (1 – 1/n(s))^{2}

Here *p*(*s*) is the pressure, and *u*(*s*) is the density of potential energy stored in the
string, per unit of unstretched length. We can rewrite *n*(*s*) and *u*(*s*) in terms of the
pressure:

n(s) =k/ (k–p(s))

u(s) =p(s)^{2}/ (2k)

Now, if we work in just one dimension of space and one of time, the stress-energy tensor for the string will be:

T=n(s) (ρ +u(s))e_{τ}⊗e_{τ}+p(s)e_{s}⊗e_{s}

where the first term comes from the total energy density of the string, and the
second from the pressure. In order for this stress-energy tensor to satisfy
**the weak energy condition** — the requirement that the energy density measured
by any observer is non-negative — the tension in the string can not be greater
than the energy density.

If we compute the divergence of *T*, which must be
zero to conserve energy-momentum, the
τ component is identically zero,
while the *s* component gives us the equation:

(1/s)n(s) (ρ +u(s)) + (1/s)p(s) +p'(s) = 0

Since the proper acceleration of any object with a fixed *s* coordinate is 1/*s*,
this equation is *almost* just *F* = *m* *a* for an element of the string occupying
one unit of *s*, with *p*'(*s*), the rate of change of pressure in the string,
being the opposite of the force on the element, and
*n*(*s*) (ρ + *u*(*s*))
giving its total inertial mass. However, the extra term (1/*s*) *p*(*s*) is necessary
to conserve momentum; this arises because the flow of momentum associated with the
pressure is changing its direction in spacetime as the element of the string
accelerates, turning it into a density of momentum. (Earlier versions of
this page, prior to 11 December 2006, neglected this term.)

If we express everything in terms of *p*(*s*), we obtain the
differential equation:

s(p(s) –k)p'(s) + (1/2)p(s)^{2}–kp(s) –kρ = 0

This is solved by:

p(s) =k(1 – (1/K) √(1 +K^{2}–s_{1}/s))

where *K*^{2}=*k*/(2ρ), and
*p*(*s*_{1}) = 0 as required by our boundary conditions. Because
√(*k*/ρ)
is the speed of sound in the string (in the small-amplitude limit, for the
relaxed material), we should
impose the restriction *K*^{2} < 1/2,
in order that the speed of sound does not exceed 1, the speed of light.

Here are some plots of –*p*(*s*)/*k* versus *s*/*s*_{1} for *K* ranging from 0.1 to 0.7.

Note that for given values of *K* and *k*, the tension in the string is bounded,
regardless of its length:

–p(s) <k(Q/K– 1)

where we’ve defined *Q*^{2}=1+*K*^{2}. It turns out that
this bound is precisely what’s needed to guarantee that the stress-energy
tensor satisfies the weak energy condition! The same condition can be
expressed more concisely as:

ds/dλ = 1/n(s) <Q/K

In other words, **the string can not be stretched by a factor greater
than Q/K**. For

**N.B.** We have already restricted the value of *K* so that the speed of sound
*in the relaxed material* does not exceed the speed of light. However, as shown on
this page, the speed of sound will vary with
the tension, and for a string under uniform tension, the speed
of sound will reach the speed of light when it is stretched by a factor of
*Q*/(√3 *K*), or just 57% of the stretching that
would violate the weak energy condition. We can’t expect the same result to hold
exactly for our suspended string (which will have a complicated dispersion
relation, with different frequencies of sound travelling at different speeds), but
it’s highly likely that the hyperelastic model will become physically impossible
at some point significantly *below* the stretch factor of *Q*/*K*. Computing
that point exactly is beyond the scope of this analysis, so we will present solutions
that obey the hyperelastic model all the way up to *Q*/*K*, but these results
need to be viewed with the caveat that even idealised materials will necessarily
depart from them at sufficiently high tension.

We can find
dλ(*s*)/d*s* from our solution for the
pressure, and then integrate to obtain
λ(*s*):

dλ(s)/ds=K√(s/ [Q^{2}s–s_{1}])

λ(s)/s_{1}= (K/Q^{3}) [log[Q√(s/s_{1}) + √(Q^{2}s/s_{1}– 1)] – log[Q+K] +Q[√(s/s_{1}(Q^{2}s/s_{1}– 1)) –K]]

Below are some plots of the solution, for *K* ranging from 0.1 to 0.7.

Note that as *s*/*s*_{1} →
∞,
λ(*s*) is asymptotic to
(*K*/*Q*) *s*, in the sense that the ratio of the two expressions tends to 1
(the difference between them does not tend to zero, but grows smaller
as a proportion of the expressions themselves).

We can numerically invert this solution to see how the locations of the
top (*s*_{2}) and bottom (*s*_{1}) ends of the string are related,
for a string of unstretched length *L*:

Each curve starts at *s*_{2}/*L* = *Q*/*K*, then descends to a minimum value for *s*_{2},
below which there can be no equilibrium. Depending on how easily stretched the string
is, this can occur when the top of the string is much further from the horizon than the
unstretched length *L*, and even for the maximum value of *K* the minimum for
*s*_{2}/*L* is approximately 1.72.
For values of *s*_{2} greater than this minimum but less than *Q*/*K*,
there are two equilibrium positions for *s*_{1}: a higher one, which
will be stable, and a lower one, which will not.

It follows that, for a given value of *K* (which sets the stiffness of the string)
and a given value of *s*_{2} (the height of the top of the string), there is
a maximum unstretched length *L* for a string suspended in stable equilibrium.
For strings with the maximum possible stiffness (i.e. with a speed
of sound equal to the speed of light), this is approximately 0.58 *s*_{2}.

**The picture on the left** shows strings of the maximum possible stiffness
of various unstretched lengths *L*, all suspended from
the same height above the Rindler horizon, in stable equilibrium.
Segments of equal unstretched length are marked in bands of alternating colour.

Although we haven’t studied any dynamic processes here, it seems safe to conclude that:

- if you suspend a given string from too low a point, it will be subject to unbounded stretching (and any real string would subsequently break);
- if you anchor the string from a safe point but with
*s*_{2}/*L*<*Q*/*K*, and then apply an extra force to drag the bottom end down past the point of unstable equilibrium, the string will be unable to snap back up, and will suffer the same unbounded stretching.

If these results seem hard to believe, don’t forget that we’re talking about
extremely powerful accelerations.
The acceleration associated with *s*_{2} = 10 metres is 9.18 × 10^{14}g,
and although it’s obvious that you can’t trail anything longer than 10 metres behind
you at that acceleration, we’ve found that even a string just 6 metres long
would be doomed.

Suppose a passenger named Adam on our flotilla of spaceships, travelling at an
*s* coordinate of *s*_{0}, steps off the ship. Without loss of generality we
can choose coordinates so that he does this at (*t*, *x*) = (0, *s*_{0}).
His free-falling world line is then just *x*=*s*_{0}, and his proper time will
correspond to the *t* coordinate of our global inertial frame.
At *t*=*s*_{0}, he will pass behind the Rindler horizon, after which no signal
from him can ever catch up with any part of the flotilla.

It’s worth noting Adam’s rates of change for the coordinate
*s* = √(*x*^{2} – *t*^{2}), which
are just the rates of change with respect to *t*:

ds/dt= –t/s

d^{2}s/dt^{2}= –(s^{2}+t^{2}) /s^{3}

Now suppose that Adam’s
friend Eve was travelling with him at *s*=*s*_{0}, and remains on the same ship.
There is no finite *q* coordinate for any point on the Rindler horizon,
which means there is no time for Eve when, in her co-moving inertial reference frame,
Adam passes through the horizon. In that sense,
Eve could claim that Adam never reaches the horizon as far as she’s concerned. However,
not only is it clear that Adam really does cross the horizon, there is a time
τ_{crit}
for Eve after which any signal she sends to Adam will reach him only after he’s on the
other side.

It’s easy to compute this time. The event of Adam reaching the horizon is (*t*, *x*) =
(*s*_{0}, *s*_{0}), and the null line passing through this event for light
coming back towards Adam from the direction of the ships consists of points of the
form (*t*, *x*) = *s*_{0}(1–*z*, 1+*z*). Eve’s world line is *s*_{0}(sinh(*q*), cosh(*q*)),
and so it intersects that null line when:

sinh(q) + cosh(q) = 1–z+ 1+z

exp(q) = 2

q= log(2)

τ_{crit}=s_{0}log(2)

After a time of τ_{crit}
has passed for Eve, she must concede that it’s too late for her to send
Adam a message asking him to hitch a ride and catch up with the ship, since every signal
she now sends will be received by him on the other side of the horizon.

Suppose Adam decides to tie a rope around his waist when he steps off the ship, but
Eve agrees to feed out the rope in such a way that Adam remains in free fall. Is this
possible? Clearly it is, because we can imagine a rope of arbitrary length sitting
motionless in our (*t*, *x*) coordinates, and all Eve has to do to keep her and Adam’s
rope slack is to feed it out in such a way that it matches that reference rope.
This will require Eve to give the section of rope she is dispensing
a velocity equal and opposite to her own ordinary
velocity in the (*t*, *x*) frame, which is tanh(τ/*s*_{0}).
If Eve sticks to her notion of simultaneity then she’ll never admit that Adam has
passed through the horizon, so her task is endless (and the velocity she needs to
give the rope will asymptotically approach the speed of light),
but if she takes a more sensible approach and concedes that
after a time of τ_{crit}
has elapsed there’s no hope of hauling him back on to the ship, she will have fed out
a length of just
*s*_{0} [cosh(τ_{crit}/*s*_{0}) – 1] = *s*_{0}/4 before reaching that point.
The velocity at which she will be dispensing the rope at
τ_{crit} will be
tanh(τ_{crit}/*s*_{0}) = 3/5.

We should note that the velocity Eve must give the rope at the point
where she is unreeling it does *not* equal the rate at which the total proper
length of rope she has dispensed is increasing.
The latter is equal to *v*/√(1–*v*^{2}),
or in this case, sinh(τ/*s*_{0}).
At τ_{crit}, this rate is
sinh(τ_{crit}/*s*_{0}) = 3/4.

If Eve clamps down on the rope after
τ_{crit}
(or in fact somewhat earlier, depending on the speed of sound in the rope), then
the wave of tension passing down the rope will not reach Adam until he is behind
the horizon, and however powerful an acceleration this gives him, he will not be able
to pass back through the horizon, because that would mean travelling faster than light.

When will the rope break? There is no particular time when this *must* happen;
it will depend on the rope’s actual properties
(though as we’ve noted earlier, the weak energy condition does put an upper limit
on the amount by which any given rope can be stretched).
The tension in the rope does not
become infinite just because it crosses the horizon. What a clamped rope
spanning the horizon does imply, though, is that it must *eventually* break,
because the portion of it in the neighbourhood of the horizon will always be moving
backwards relative to the ship.

Suppose that Eve opens a box of dust and lets it waft freely out of the ship at a constant rate; in other words, by her clock the time between one particle and the next leaving the ship is constant. The world lines of these dust particles will be tangents to the ship’s world line. They are the straight grey lines, labelled by values of d, in the spacetime diagram below.

The curves on the diagram that cross between the dust’s world lines,
labelled by values of *w*, have been chosen to be orthogonal everywhere to those world
lines. Someone like Adam, who jumped ship, would consider these curves (locally)
to be surfaces of simultaneity. It’s interesting to ask what he would consider
the density of the dust trail to be.

To do this, we need to describe these tangent lines and the curves orthogonal to
them in more detail. We’ve chosen to label the world lines of the dust
particles by the ship time at which they were emitted, so where these lines
commence at the ship’s world line, *d*=τ.
We have also labelled the curves of constant *w* by the ship time when they meet the
ship’s world line.

If we wish to find the *d* coordinate of some event **x**, suppose that **x**
lies on the world line of the dust particle that left the ship at the event
**p** = *s*_{0}(sinh(*d*(**x**)/*s*_{0}), cosh(*d*(**x**)/*s*_{0})). Then
we know that **x**–**p** is the tangent to the ship’s world line at
**p**, and since the ship’s world line is orthogonal to any line
that connects it to the origin (the lines of constant *q* in our (*q*, *s*) coordinates),
**x**–**p** must be orthogonal to **p** itself:

p· (x–p) = 0

p·x=p·p=s_{0}^{2}

Define *k*(**x**) to be the proper time that has elapsed for the dust since it
left the ship, i.e.

k(x)^{2}= –(x–p) · (x–p) = –(x·x– 2p·x+p·p) =s_{0}^{2}–x·x

k(x) = √(s_{0}^{2}–x·x)

If we define:

d(x) =s_{0}[ arctanh(t/x) – arctanh(k(x)/s_{0}) ]

=s_{0}log[(x+t) / (s_{0}+k(x))]

=s_{0}log[(s_{0}–k(x)) / (x–t)]

then it’s not hard to show that
**p** = *s*_{0}(sinh(*d*(**x**)/*s*_{0}), cosh(*d*(**x**)/*s*_{0}))
satisfies **p** · **x** = *s*_{0}^{2}.

Note that on the ship’s world line, **x** · **x** = *s*_{0}^{2}
and *k*(**x**) = 0, so:

d(x) =s_{0}log[(x+t) /s_{0}] =s_{0}log[exp(q)] =s_{0}q= τ

If we define:

w(x) =k(x) +d(x)

then on the ship’s world line we will have *w*(**x**) = *d*(**x**) =
τ. For generic events **x**,
since *k*(**x**) is the proper time for the dust particle since it left the
ship, and *d*(**x**) specifies the value of τ
when it left, *w*(**x**) is the total proper time for the dust since it passed
through the event (*t*, *x*) = (0, *s*_{0}), including both its shipboard time
and its time in free fall.

Now, if we compute the derivatives of *d*(**x**) and *w*(**x**) with respect to
*t* and *x*, we get:

∂_{t}d(x) =s_{0}(x–s_{0}t/k(x)) /x·x

∂_{x}d(x) =s_{0}(s_{0}x/k(x) –t) /x·x

∂_{t}w(x) = (s_{0}x–k(x)t) /x·x

∂_{x}w(x) = (k(x)x–s_{0}t) /x·x

It’s easy to see that the gradients of these functions are orthogonal.

The magnitude of the gradient of *d*(**x**) turns out to be *s*_{0}/*k*(**x**).
Since *k*(**x**) is the proper time that has elapsed since the dust left the ship
this is singular at the instant the dust leaves the ship, which makes sense because
we’re modelling the source of the dust trail as a single point.
Since the magnitude of the gradient corresponds to the particle density of the dust
trail, our result shows that the spacing between the particles grows linearly with time.

We’ve already seen that the proper time for any free-falling object to reach the
Rindler horizon is *s*_{0}, so as the dust trail crosses the horizon, *k*(**x**)
= *s*_{0} and the local particle density becomes equal to 1. In other words,
if particles were leaving the ship at a rate of 1 every second, they would now be spaced at
one every light-second! If we imagine that this is some kind of smart dust
made of particles that can signal to each other, then if a particle reaches the horizon
and sends a light signal back to the one coming after it, the
receiving particle will get the message at precisely the moment when it, too, arrives
at the horizon.

The dust-trail scenario can easily be generalised to the case where the dust is
thrown out backwards with a speed of –*v*, by replacing
*s*_{0} in the preceding analysis with *s*_{0} /
√(1 – *v*^{2}). Then by the time the
dust has reached *s*=*s*_{0}, it will have a velocity relative to the flotilla of
–*v*.

We’ve seen that it requires an acceleration of 1/*s* to maintain a fixed distance of *s*
from the horizon, an acceleration that obviously grows without bounds as you approach the horizon
itself. We’ve also seen that there is no problem with free-falling right through the
horizon (apart from the fact that you can’t cross back).

This leads to the question of exactly what kinds of motion, apart from free fall, can take you through the horizon. To what degree can you control your fall, without requiring infinite acceleration?

Suppose Adam decides that, rather than free-falling through the horizon,
he wants to use a rocket backpack to control his descent.
His goal is to maintain a constant ordinary velocity of –*v* relative to each ship
in the flotilla that he passes, all the way down to the horizon itself.
Of course any real flotilla must end at some finite distance above the horizon,
but Adam plans to follow the same criterion whether there’s an actual ship beside
him or not.

At any event (*t*, *x*) with *x* > *t*, the *q* coordinate
is arctanh(*t*/*x*), so the 4-velocity of the flotilla ship passing through this event
is:

u_{flotilla}= (cosh(q), sinh(q)) = 1/√(x^{2}–t^{2}) (x,t)

What we want is a world line for Adam that has a constant dot product with
this vector field of **u**_{Adam} · **u**_{flotilla} =
–1/√(1–*v*^{2}).
It turns out that we can describe such a world line
in (*t*, *x*) coordinates by the equation:

x+t=A^{2}(x–t)^{–(1–v)/(1+v)}

where *A*^{2} is a constant that sets the position of the world line without
changing its properties of interest. We can write the (*t*, *x*) coordinates of this
world line in terms of a parameter *w* = *x*–*t*, while being careful to keep in
mind that this parameter is *not* proper time along the world line:

(t,x) = (1/2) (A^{2}w^{–(1–v)/(1+v)}–w,A^{2}w^{–(1–v)/(1+v)}+w)

The tangent vector with respect to *w* is then:

∂_{w}(t,x) = –(1/2) ((1–v)/(1+v)b(w)^{2}+ 1, (1–v)/(1+v)b(w)^{2}– 1)

whereb(w) =Aw^{–1/(1+v)}

We can normalise this vector to a squared length of –1 to get Adam’s 4-velocity,
also taking account of the fact that ∂_{w}(*t*, *x*)
points backwards in time so we need to reverse it:

u_{Adam}= 1/[2√(1–v^{2})] ((1–v)b(w) + (1+v) /b(w), (1–v)b(w) – (1+v) /b(w))

If we note that, along Adam’s world line:

x+t=wb(w)^{2}, and

x^{2}–t^{2}=w^{2}b(w)^{2}

then it’s not hard to show that:

u_{Adam}·u_{flotilla}= –1/√(1–v^{2})

This verifies our claim that the ordinary velocity Adam and the flotilla ship will
measure for each other has a magnitude of *v*.

Finding Adam’s proper acceleration is then just a matter of calculating the rate of change of his 4-velocity along his world line with respect to his proper time, and taking the magnitude of the resulting vector. This turns out to be:

a_{Adam}= 1/[s√(1–v^{2})]

In other words, his acceleration is just as singular in *s* as that required to remain
stationary with respect to the horizon. In fact the acceleration for any given value
of *s* is actually greater when *v* is non-zero!

So Adam’s goal of passing right through the horizon in this manner is physically impossible; even though he’s not trying to remain stationary, his rocket would still have to provide an infinite acceleration as he crossed the horizon.

What does this tell us about lowering an object on a string?
We can’t say that we’ve ruled out the possibility of
feeding out string from a reel at a constant rate, because the unavoidable elasticity
of the string means the object being lowered need not end up moving at a constant
velocity relative to the flotilla.
There is also the further complication of the difference between clocks at the top and
bottom of the string.
So we can only say that any scheme that aims to adjust the rate at which the string
is fed out in such a way as to cause *the object itself* to be lowered at a
constant rate compared to the flotilla (according to its own clock)
will necessarily fail as the object approaches the horizon.

Now that we can see that Adam must accelerate with respect to the flotilla as he passes through the horizon, let’s try to estimate the kind of acceleration that’s necessary.

The world line:

(t,x) = (1/a)(1 + sinh(aτ), cosh(aτ))

is a world line of constant proper acceleration *a*, and it crosses the horizon
*s* = √(*x*^{2} – *t*^{2}) = 0
at τ=0. To lowest order in τ, we can
approximate it as:

(t,x) = (1/a)(1 +aτ, 1 +a^{2}τ^{2}/ 2)

and then to lowest order in τ:

s= √(–2τ/a)

τ = –as^{2}/2

The sign inside the square root here reminds us that we’ll have to consider
everything for small negative values of τ,
not positive ones, because the *s* coordinate is not defined on the far side of the
horizon.

The derivatives of *s* with respect to τ are
then:

ds/dτ = –1/(as)

d^{2}s/dτ^{2}= –1/(a^{2}s^{3}).

So in order to achieve a finite proper acceleration at the horizon, Adam must
accept that he will see the *s* coordinate of the flotilla changing with respect to
his proper time in this fashion. Note the inverse square relationship: the lower
Adam’s proper acceleration, *a*, the greater the coefficient in the leading term for
the second derivative of *s*.

We can get our hands on one class of exact solutions for elastic strings descending
through the horizon by the simple expedient of taking a string that is static in
one flotilla’s (*q*, *s*) coordinates, and then viewing it from the perspective of a
different flotilla. We can obtain the world lines for the ships in the new flotilla
simply by shifting the origin of the (*t*, *x*) coordinates. So if the world lines in our
“reference” flotilla are:

(t,x) = (ssinh(q),scosh(q))

then those in the second flotilla, which we’ll call the “primed” flotilla, will be:

(t,x) = (A+s' sinh(q'),B+s' cosh(q'))

for some constants *A* and *B*. Of course, if the event (*A*, *B*) lies above (or to the
right of, in a conventional spacetime diagram) the Rindler horizon of the reference
flotilla, the entire primed flotilla will also remain above the horizon,
so those values of *A* and *B* will be of no use to us.

We will suppose that there is a stationary string suspended above the horizon
of the primed flotilla, with the bottom of the string at a coordinate of
*s*'_{1}. If we wish to make use of this solution in the reference flotilla,
we can pick a location *s*_{2} and imagine that we are feeding out the string
from that point. Suppose the bottom of the string passes *s*_{2} at
τ=0, and its ordinary
velocity is measured by Eve, riding in the reference flotilla at *s*_{2}, to
be –*v*_{0}, heading towards the horizon. We can then solve for *A* and *B*:

A=s'_{1}γ_{0}v_{0}

B=s_{2}– γ_{0}s'_{1}

where γ_{0}= 1 / √(1–v_{0}^{2})

Of course the velocity of the string adjacent to Eve (or being unreeled by her, if she is matching the solution by her own actions) will not remain constant. The expression for the velocity is:

v(τ) = – √[g(τ)^{2}–s'_{1}^{2}–s_{2}^{2}+ 2s'_{1}s_{2}γ_{0}] / [g(τ) +s_{2}]

whereg(τ) =s'_{1}cosh(τ/s_{2}– log(Γ_{0})) –s_{2}cosh(τ/s_{2})

and Γ_{0}= γ_{0}(1–v_{0}) = √[(1–v_{0}) / (1+v_{0})]

The bottom of the string passes through the reference flotilla’s horizon at:

t=x= [Γ_{0}s_{2}(2s'_{1}γ_{0}–s_{2})] / [2 (s'_{1}– Γ_{0}s_{2})]

and it follows that after

τ_{crit}=s_{2}log([Γ_{0}(2s'_{1}γ_{0}–s_{2})] / [s'_{1}– Γ_{0}s_{2}])

no influence from Eve can reach the bottom of the string before it goes behind
her Rindler horizon. At τ
= τ_{crit}
(as at all finite values for τ)
everything Eve measures about the string —
the tension, the rate at which she is feeding it out, and the
total length she has unreeled — will all be finite.
However, if Eve tries reeling the string back up after this time,
the bottom will remain on the far side of the horizon, and the string will
eventually break.

The function *v*(τ) which describes the
velocity at which Eve must unreel the string is plotted below for the case
*s*'_{1} = *s*_{2} = 1, for various values of *v*_{0}.

Also plotted is τ_{crit} as a function
of *v*_{0} (the red curve). Naturally the slower the string is being unwound,
the longer it is until the bottom of the string has effectively been lost behind the
horizon.

The velocity at which the string is being unwound by Eve when this happens,
*v*(τ_{crit}), is shown on
the graph for a few different values of *v*_{0}.
For *s*'_{1} = *s*_{2}, in the limit of *v*_{0} → 0,
*v*(τ_{crit}) → –1/3. In other
words, for a string that starts out being unreeled slowly, there is a ceiling
of roughly one third lightspeed on the speed at the point of unreeling at
τ_{crit}.

Although the descending string described in the previous section has the advantage of an exact, analytical solution, it’s a rather artificial set-up. What we’d really like to understand is what happens when a string is unreeled at a constant rate.

What we’ll take this to mean is that a constant unstretched length of string,
*R*, is unreeled per unit of proper time from some fixed point on the flotilla.
To understand what this entails, it’s worth looking first at the Newtonian solution
for a string
being unreeled at a constant rate by an observer undergoing constant acceleration.
The *x* coordinate of an element of string then obeys the linear wave equation:

∂_{t,t}x(λ,t) =w^{2}∂_{λ,λ}x(λ,t)

where *w* is the speed of sound in the string; in terms of our previous
parameters:

w^{2}= 2K^{2}=k/ ρ

The appropriate solution is a quadratic in
λ and *t*:

x(λ,t) =x_{0}+ λ –Rt+ (a/2) (λ^{2}+w^{2}t^{2}) / (w^{2}+R^{2})

If we put
λ = *R* *t*, we get:

x(Rt,t) =x_{0}+ (a/2)t^{2}

which is just the *x* coordinate of the constantly accelerating observer who is
unreeling the string.

The derivatives of this solution are easily computed:

∂_{λ}x(λ,t) = 1 +aλ / (w^{2}+R^{2})

∂_{λ,λ}x(λ,t) =a/ (w^{2}+R^{2})

∂_{t}x(λ,t) = –R+aw^{2}t/ (w^{2}+R^{2})

∂_{t,t}x(λ,t) =aw^{2}/ (w^{2}+R^{2})

The second derivatives make it plain that the wave equation is satisfied, while the
first derivative with respect to λ
shows that at the bottom of the string, where
λ = 0, stretched and unstretched
length change at the same rate, i.e. there is no tension in the string.
The tension depends only on λ
and grows linearly along the string. Explicitly, if
*p*(λ) is pressure:

–p(λ)/k=aλ / (w^{2}+R^{2})

The first derivative of *x* with respect to *t* shows us that the ordinary velocity
of the string, relative to the constantly accelerating observer
dispensing the string (whose own velocity at time *t* is *a* *t*), will be:

v= –R+aw^{2}t/ (w^{2}+R^{2}) –at = –R–aR^{2}t/ (w^{2}+R^{2})

This is independent of λ,
but if we want the relative velocity of the string as it is being
dispensed, we set *t* = λ / *R*, and
then:

v_{dispensed}(λ) = –R–aRλ / (w^{2}+R^{2}) = –R∂_{λ}x(λ,t)

It might seem odd to think of the string’s velocity relative to the person unwinding it not being constant, despite the fact that a constant amount of unstretched length is being unreeled. However, unless some mechanism completely isolated the portion of the string wrapped around the reel from the tension in the rest of the string, the string on the reel would be stretched to some degree, so the reel would be rotating ever faster to dispense unstretched length at a constant rate. If a tension-isolating mechanism was used, the string would be accelerated within that mechanism, as it underwent the transition from zero tension to full tension.

The fact that the velocity of the string is independent of
λ and the tension is
independent of *t* shows us that, in the Newtonian case,
the “artificial” trick of using a static string suspended by
one accelerating observer and then viewing it in a different accelerating
frame gives us *exactly the same result* as unreeling unstretched
length at a constant rate! The tension in a string
that is static in a frame with acceleration *a*' is given by:

–p(λ)/k=a' λ /w^{2}

So we can match the tension exactly if we set:

a' =aw^{2}/ (w^{2}+R^{2})

Our “unreeling observer” with acceleration a will overtake the frame
with the static string and acceleration *a*', but if they start out with the bottoms of
their strings coinciding — and with the “unreeling observer”
already moving with speed *R* relative to the static string —
then the tension and velocity of the two strings will match all
along their length.

The plot below shows the trajectories of elements of the string separated by equal increments of unstretched length λ, along with the trajectory of the unreeling point.

Unfortunately, the Newtonian relationship between static solutions and solutions with constant unreeling does not extend to the relativistic case. The right choice of parameters can give a rate of unreeling that is constant to first order, but then the rate necessarily increases. Specifically, we can set:

v_{0}=R/ √(1 +R^{2})

s'_{1}=s_{2}(2K^{2}+v_{0}^{2}) / [2K^{2}√(1 –v_{0}^{2})]

Here *s*'_{1} is the location of the bottom of the string in the
flotilla in which the string is static, *s*_{2} is the point of
unreeling in the flotilla in which the string is being lowered, and
v_{0} is the initial velocity of the bottom of the string measured by
the observer who is lowering it.
Then to second order in τ,
the rate of unreeling can be shown to be:

dλ/dτ = (dλ/ds') (ds'/dτ) =R(1 + (4K^{4}+ 2K^{2}v_{0}^{2}+v_{0}^{4}) τ^{2}/ [2s_{2}^{2}(2K^{2}+v_{0}^{2})^{2}] )

Here we’ve made use of the expression for
dλ/d*s*'
derived in our treatment of a static string suspended above a horizon, and
d*s*'/dτ comes from the change of coordinates
between different flotillas discussed in the previous section.
We’ve given the second-order Taylor series, to make it clear that the
first-order term is zero, but taking account of higher order terms doesn’t help;
the exact expression is unbounded as τ
increases. In fact, it’s asymptotically proportional to
exp(τ/(2 *s*_{2})). This is slower than,
but not all that different from, the free fall rate of dispensing string,
sinh(τ/*s*_{2}).

So, when viewed from a flotilla whose horizon the static string eventually crosses, the “rate of unreeling” of the string will be unbounded. This in turn suggests that a string being unreeled at a constant rate will experience greater tension than the static string, because unreeling the string more slowly will generally increase the tension. But the tension in a static string is already asymptotic to the limit set by the weak energy condition, so it seems likely that a string being unreeled at a constant rate will eventually exceed that limit and break.

Below is a spacetime diagram showing numerical results ^{[1]} for the case
*K*=0.1 (i.e. the speed of sound, *K*√2,
is 0.1414 times the speed of light), and *R*=0.2.
The black curves are world lines of elements of the string, separated by
equal increments of λ.
The coloured lines are spacelike slices, everywhere orthogonal to those
world lines, separated by equal increments of proper time at the point of
unreeling. (The topmost of these lines is incomplete because the numerical data
was not accurate along the full slice; the lines are shown where the solution
is estimated to be accurate to within 1%.)

The graph below shows the tension as a function of λ along a succession of spacelike slices.

The results are quite close to the Newtonian values, for which the tension would
simply be directly proportional to λ
everywhere. For comparison, the tension in a static string
(which matches the rate of unreeling to first order) is shown;
this does not increase as rapidly with
λ.
Note that the limit on the tension set by the weak energy condition,
(*Q*/*K* – 1), is 9.05.

Where the results depart most clearly from both the Newtonian case and the relativistic static string is in the acceleration of elements of the string. This is plotted below, for the same spacelike slices as the tension.

Whereas the acceleration in the Newtonian case is both constant in time and identical for all elements of the string, and for a static string each element has a constant acceleration which is smaller for higher values of λ, in the case of constant unreeling, the acceleration of each element increases over time as well as decreasing along each spacelike slice.

Although a string lowered in this fashion appears doomed to break *eventually*,
this example shows that there’s no reason it can’t survive intact
not only after τ_{crit},
but also after the bottom of the string has passed through the horizon
according to the string’s own definition of simultaneity.
In other words, a spacelike slice orthogonal to the world lines
of the string’s elements can run all the way from the bottom of the string, on the
far side of the Rindler horizon, to the point of unreeling, with the tension everywhere
remaining well below the limit set by the weak energy condition.

**The animation on the left** shows a string being unreeled at a constant rate,
with *R*=0.2 and *K*=0.1.
The distance shown vertically here is measured along the spacelike slices,
so it is “proper distance” in the sense that these slices are everywhere
orthogonal to the world lines of the elements of the string.
The bands along the string are segments of 0.02 unstretched length; the animation
shows the unreeling of a total unstretched length of 0.36.
The red line that appears during the descent marks the Rindler horizon;
the distance from the point of unreeling to the horizon, measured along these spacelike
slices, is not fixed.

What happens when an object is lowered across a horizon
on the end of an elastic string?
We have only been able to obtain *exact* solutions for two special cases:
a perfectly slack string in free fall, and a string which
is lowered in such a way as to be static in its own “flotilla” of
accelerating frames.
However, it seems likely that whenever an object is lowered across a horizon
this way, the main qualitative features of those solutions will always be present:

- Given some function of proper time which expresses the rate at which
the string is being unreeled,
there will be a time τ
_{crit}after which no signal from the point of unreeling can reach the object at the end of the string before it passes through the horizon. - After τ
_{crit}, the only way to keep the string from eventually breaking (assuming it hasn’t broken already) will be to continue to feed it out at an increasing rate. Clamping the string, attempting to pull the object back up, or even just feeding out the string at a constant rate, will all ultimately cause the string to break, though the exact time when that happens will depend on the physical properties of the string.

The Rindler metric for flat spacetime and its similarity to the Schwarzschild
metric for black holes is discussed in Section 6.4 of *General Relativity*
by Robert M. Wald.

[1] A partial differential equation was constructed by describing the
world lines of the elements of the string with functions
*T*(τ, λ)
and *X*(τ, λ);
here λ identifies the particular
element, τ varies along
each world line (but is not necessarily proper time for the element), and *T* and *X*
are the spacetime coordinates in some inertial frame. Given the assumption of
a hyperelastic string — that is, a string obeying Hooke’s Law exactly —
the stress-energy tensor was expressed in terms of the functions *T* and *X*,
and the requirement that it be divergence-free produced a
single partial differential equation in *T* and *X*:

To eliminate *T*,
it was necessary to make a “choice of gauge” for
τ. A convenient choice in this
case was to define it to be the lapse in coordinate time since the
unreeling of each element, making *T* simply:

T(τ, λ) =s_{2}sinh(λ / (Rs_{2})) + τ

An approximate solution was
found by setting *X* equal to a 32-order polynomial *P* in
λ and
τ, plus an initial offset equal to the *x*-coordinate
of the point of unreeling:

X(τ, λ) =s_{2}cosh(λ / (Rs_{2})) +P(τ, λ)

This made the boundary condition describing the way the string was unreeled very simple:

P(0, λ) = 0

The coefficients of the polynomial *P* were then found
by requiring that all derivatives of the PDE up to order 32 were zero (at the point
λ=0,
τ=0), and that the tension at the bottom of the
string was zero to the same order in τ.
The solution was validated by comparing the purely geometrical proper acceleration
of each world line with the acceleration required by the PDE; for the data displayed,
these two values differ by less than 1%.

More details on the derivation of the PDE can be found in this supplementary page on relativistic elasticity.