jem patiently waded through my screed, and wrote:

Edward Green wrote:

jem wrote:

Edward Green wrote:

On the table before me I have a Jack-in-the-Box. I have a physical
theory of this Box, and it tells me that Jack will spring out when my
clock reads "t = infinity". Hmm... I say. Does Jack ever spring out?
I'd say not... saying he springs out "at infinity" is just a funny way
of saying "never".

"Never", as measured by your clock.

Agreed.

But now Tom comes in the room, and accuses me of being a coordinate
chauvinist. For, he, claims, closer examination of my theory reveals
that local physical law is invariant in form under a transformation
which places me inside the box and maps my "infinity" to a finite time.
Jack pops out in finite time by _his_ watch, so only my prejudice leads
me to claim he "never" pops out. I am a coordinate-centric pig.

Should this argument sway me?

The argument that it'll pop out in a finite time if you'll just keep
time with a different clock? I don't think so.

Good. I don't think too much of it either.

It's partially a question of semantics, but I'd still argue that in my
universe, this transformation doesn't amount to a hill of beans. I and
all my descendants can watch the box until the stars wink out, and it
won't pop. What they will see is that "time" inside the box is running
down, and in fact running asymptotically to a finite duration. Not
only is the box never going to pop, its internal clock is never going
to pass midnight. This existence of the cute transformation symmetry
doesn't change anything operational for me.

That is, unless you're capable of adopting different "time perspectives"
(e.g. by moving differently).

So far I haven't said anything about movement. I'm abstracting this
"different observers" business to a case where the observer who seems
to slow down and stop is right there in front of us, so we don't have
any weasel room about simultaneity and distant regions of spacetime.

Suppose an object is receding from us in flat spacetime, undergoing
constant acceleration -- asymptotically approaching c, of course, in a
fixed Lorentzian frame. Suppose we sent a light pulse after it,
initially lagging it by a distance d (in the same frame). The closure
rate in this fixed frame is given by c - v_m, where v_m -> c. The
closure rate goes to zero, and the net distance closed (I emphasize,
measured in the fixed frame) is given by the integral of the closure
rate from initiation of the pulse to t = oo. The acceleration profile
may be chosen so that net_distance_closed < d : the light never catches
the particle.

Yes.

I suspect that if we integrated the proper time of the particle under
such an acceleration profile we would find that total elapsed proper
time following the passing of any fixed coordinate point is finite: the
particle runs down, and only takes finite proper time to get to
infinity!

No.

Let's see... I worked this out. I think in terms of my eta (==
1/gamma), the total closure of the pursuing light pulse went as eta^2,
and the total elapsed time went as eta. Therefore, it's a stronger
condition that elapsed time remain finite, but still achievable (eta
goes to zero as v -> c).

That is a strange conclusion, I agree -- stranger I think than the twin
paradox, to which it is a relative -- but was confirmed by Ben RG, one
of my authority figures. ;-)

Supposing an acceleration profile can be produced with this property,
we seem to have a close analogy of an infalling particle in a black
hole: after some finite delay, light can no longer catch it, and it
takes finite proper time to reach an "event" we would tend to label
with "t = infinity".

No, the coordinate clocks encountered by the accelerating object will
always read a finite time.

I don't think you followed what I said.

suggest the accelerating observer will never see a clock reading
"t = oo" (what would that mean?), but will nonetheless take less than a
strictly bounded amount of proper time to get any coordinate time you
wish.

You know, this is a real acid test case for the "infalling observer
sees the horizon in finite proper time, and sails right past" trope.
_If_ all possible proper times are somehow created equal, in a free,
just and egalitarian universe, just what happens to our strongly
accelerated observer in flat spacetime when his clock ticks out that
last second of proper time? If his clock is as good as everybody else's
clock, it must just keep on ticking, mustn't it?

read "t_crit + 1". But where is our friend then, since all times up to
t_crit have already been mapped to coordinate times and positions from
here to infinity? He must have gone _past_ infinity, and right out of
our universe!

"Ooooh...."

Well, the result of this gedanken is not hard to fathom... during the
infinite time in our universe in which his clock is ticking its last
mortal second, something is bound to happen to the fellow. A rogue
comet will annihilate him, or the universe will go and heat death or
something. From our point of view he will be placed in suspended
animation (merely an extreme version of the traveling twin), and
something will happen to him between now and infinity, extricating us
from "but what happens in the _next_ second". Infinity is a very long
time. :-)

Does that mean the particle really "crosses infinity"?

Project for a rainy day: devise such an explicit acceleration profile,
and examine its behavior in the proper time of the particle
extrapolated past the finite proper time corresponding to "t =
infinity". Is there a second anomaly at a later, finite, proper time,
suggestive of a "spacetime singularity"?

You won't find your "singularity" by looking in the direction of
acceleration, however if you look in the opposite direction, you'll see
a region (the so-called Rindler Horizon), which, in some respects,
behaves like a BH horizon. In particular, any object that crosses this
horizon does so at the accelerator's "t = infinity".

Well, it works both ways then. If the accelerating observer sees a
horizon behind him, the stationary observer sees one in front of him...
well, the "horizon" in this analogy is really infinity, but at least
the stationary observer finds that after some elapsed time, he can no
longer even catch the fleeing ship with a radio message, urging it back
from the edge. "Come back, my love, all is forgiven!" Too late.
Forever too late.

jem wrote:

Edward Green wrote:

How long can I wait before hitting the button so that the light pulse
will reach the rocket before it crosses the horizon in its own proper
time, signaling it to return? [Ignore how much thrust will be
required, and how blueshifted the signal will be when it reaches the
rocket.]

Ignoring the limited lifetime stumbling block, you can wait forever -
you'll always be able to retrieve your rocket (provided it's capable of
unlimited thrust).

It has already been mentioned by authoritative posters (no irony
involved, since I can't reproduce their calculations, but have little
reason to doubt them) that this is incorrect. I have nothing to add.

What you haven't recognized, is that, in Relativity,
your forever doesn't correspond to everyone else's forever.

But I have something to add here (besides more of the fifth of scotch I
just purchased to my glass). Let's start at the beginning...

In SR, given not terribly unfriendly accelerations, all observers are
equivalent in a strong sense: at close of business, tired out from a
long day of playing with clocks and meter sticks, they may all turn
their rockets around and head for the station, where they can swap
stories over a few drinks at the bar.

But in some spacetimes of GR, or in the event of sufficiently
unfriendly accelerations in SR, all observers lose equivalency, in the
sense that for a given observer, there are some others with whom he can
eventually no longer communicate. They may as well be "out of his
universe". We may still like to hold them morally equivalent to us, in
a sense of fair play, but we might also claim with some justice that
what can never be communicated to us is no longer part of our world.

Here's a related scenario:

On the table before me I have a Jack-in-the-Box. I have a physical
theory of this Box, and it tells me that Jack will spring out when my
clock reads "t = infinity". Hmm... I say. Does Jack ever spring out?
I'd say not... saying he springs out "at infinity" is just a funny way
of saying "never".

But now Tom comes in the room, and accuses me of being a coordinate
chauvinist. For, he, claims, closer examination of my theory reveals
that local physical law is invariant in form under a transformation
which places me inside the box and maps my "infinity" to a finite time.
Jack pops out in finite time by _his_ watch, so only my prejudice leads
me to claim he "never" pops out. I am a coordinate-centric pig.

Should this argument sway me?

It's partially a question of semantics, but I'd still argue that in my
universe, this transformation doesn't amount to a hill of beans. I and
all my descendants can watch the box until the stars wink out, and it
won't pop. What they will see is that "time" inside the box is running
down, and in fact running asymptotically to a finite duration. Not
only is the box never going to pop, its internal clock is never going
to pass midnight. This existence of the cute transformation symmetry
doesn't change anything operational for me.

It is, or was, my supposition that black holes (the objects described
by the theory of GR) in some respects act like this box, but with a
twist. There is the added complication that not only will we never see
the box pop, but it is moved a far away from us, so that our mapping of
"simultaneous" inside and outside times becomes ambiguous! Damn
nuisance. This makes it harder to dismiss the aforementioned
transformation as a mathematical existent but operationally
inconsequent oddity. If we _could_ always get a signal to the box
before it popped (contact the infalling probe), ordering it to put on
the latch, then we would have support for the idea that it still never
popped, though we hid it far away. But we can't always get the signal
to it in time.

Now, like a persistent defense lawyer, I see this result as a setback,
but do not admit defeat. There is something funny going on here. Can
the following mapping be made mathematically precise?

Suppose an object is receding from us in flat spacetime, undergoing
constant acceleration -- asymptotically approaching c, of course, in a
fixed Lorentzian frame. Suppose we sent a light pulse after it,
initially lagging it by a distance d (in the same frame). The closure
rate in this fixed frame is given by c - v_m, where v_m -> c. The
closure rate goes to zero, and the net distance closed (I emphasize,
measured in the fixed frame) is given by the integral of the closure
rate from initiation of the pulse to t = oo. The acceleration profile
may be chosen so that net_distance_closed < d : the light never catches
the particle.

I suspect that if we integrated the proper time of the particle under
such an acceleration profile we would find that total elapsed proper
time following the passing of any fixed coordinate point is finite: the
particle runs down, and only takes finite proper time to get to
infinity!

Supposing an acceleration profile can be produced with this property,
we seem to have a close analogy of an infalling particle in a black
hole: after some finite delay, light can no longer catch it, and it
takes finite proper time to reach an "event" we would tend to label
with "t = infinity".

Does that mean the particle really "crosses infinity"?

Project for a rainy day: devise such an explicit acceleration profile,
and examine its behavior in the proper time of the particle
extrapolated past the finite proper time corresponding to "t =
infinity". Is there a second anomaly at a later, finite, proper time,
suggestive of a "spacetime singularity"?