Excuse my slow responses, a deadline at work on Thursday and all that.

Daryl McCullough wrote:

Yes, the relationship between KS coordinates and Schwarzchild
coordinates is not completely specified. The usual relationship
is this:

R^2 - T^2 = (r/2m - 1) exp(r/2m)

T/R = arctanh(t/4m) (in the exterior region)
= arccoth(t/4m) (in the interior region)

Typo: tanh and coth?

However, since only *differences* of t are observable,
we could just as well say

T/R = arctanh(t/4m + A) (in the exterior region)
= arccoth(t/4m + B) (in the interior region)

where A and B are arbitrary constants. There is really
no way to fix the relationship between A and B.

Let me write A and B in the numerator (makes no difference since
4m=const.):

T/R = tanh((t + A)/4m) (exterior)
= coth((t + B)/4m) (interior)

Altering A and B then corresponds to shifting the exterior and interior
Schwarzschild regions up and down by A and B units, resp.:

t' = t - A (exterior)
t' = t - B (interior)

....which are obviously diffeomorphisms. So nothing really changes. The
metric stays the same and we still have two disconnected regions. Fine.
But I thought IV's question was (paraphrasing) "can we have distinct
ways of patching over the horizon?" IV, am I understanding your
question right?

Continuity
doesn't do it, because there is no way to go from the
exterior region to the interior region without passing
through t=infinity. Infinity + a constant is still infinity.

Right, it's still two disconnected regions so you can move them around
independently and diffeomorphically without messing up anything (or
changing anything physically/tensorially).

Just trying to get IV and me on the same page...

--
Jan Bielawski

JanPB ha scritto:

Excuse my slow responses, a deadline at work on Thursday and all that.

Daryl McCullough wrote:

Yes, the relationship between KS coordinates and Schwarzchild
coordinates is not completely specified. The usual relationship
is this:

R^2 - T^2 = (r/2m - 1) exp(r/2m)

T/R = arctanh(t/4m) (in the exterior region)
= arccoth(t/4m) (in the interior region)

Typo: tanh and coth?

However, since only *differences* of t are observable,
we could just as well say

T/R = arctanh(t/4m + A) (in the exterior region)
= arccoth(t/4m + B) (in the interior region)

where A and B are arbitrary constants. There is really
no way to fix the relationship between A and B.

Let me write A and B in the numerator (makes no difference since
4m=const.):

T/R = tanh((t + A)/4m) (exterior)
= coth((t + B)/4m) (interior)

Altering A and B then corresponds to shifting the exterior and interior
Schwarzschild regions up and down by A and B units, resp.:

t' = t - A (exterior)
t' = t - B (interior)

...which are obviously diffeomorphisms. So nothing really changes. The
metric stays the same and we still have two disconnected regions. Fine.
But I thought IV's question was (paraphrasing) "can we have distinct
ways of patching over the horizon?" IV, am I understanding your
question right?

Continuity
doesn't do it, because there is no way to go from the
exterior region to the interior region without passing
through t=infinity. Infinity + a constant is still infinity.

Right, it's still two disconnected regions so you can move them around
independently and diffeomorphically without messing up anything (or
changing anything physically/tensorially).

Just trying to get IV and me on the same page...

Here is what I've been mulling over lately. There are two orders of
questions.


1) The change of coordinates.

At the horizon where T=-+R it's not clear to me how smooth the
resulting map is (i.e. whether one can differentiate up to the order
required by the Einstein equations.). The assumptions in the sign
choice at the horizon are discussed by Carroll in [1]. The latter
corresponds to a deliberate choice of the future-oriented direction.
In general, the question of precisely characterising the charts that
extend the Schwarzschild solution across the horizon seems worth
pondering, at least to me. Diffeent charts mean different space-time
measurement models. It's not clear to me whether the KS chart
correponds to the usual meeasurement of space-time events with clocks
and rods.

2) The A and B parameters.

Consider two observers, Agata e Bruno, who are initially outside the
horizont . Agata dumps Bruno and he plunges into a blackhole. I was
saying the following. Since the matching between Schwarzschild and KS
coordinates is undetermined (any A and B can be chosen), Bruno's path
over the horizon (i.e, across the disconnected components of the
Schwarzchild chart's domain, where KS however is well defined) is
undetermined in Agata's perspective. (*)

I then surmised that , from Agata's perspective, without arbitrary
matching at the horizon Bruno gets scattered into a continuum of Brunos
with different clock readings.

Now, what's the safe reply? That the whole argument is void, because in
Agata's perspective Bruno never reaches the horizon ("According to the
Schwarzschild metric, nothing crosses the event horizon in finite
coordinate time").

In other words, can Agata look at her clock at time T_o and rightfully
say "I still see Bruno's faint image flickering over the horizon, as it
will forever , getting dimmer and dimmer, but I know that by now in
reality he's crossed the horizon".

No, she cannot. Nothing ever crosses the horizon in Agata's "reality".

However this raises obvious questions about the blackhole's
gravitational field as it extends/swallows stuff. The Schwarzchild
stationary model becomes inadequate, but the epistemic boundary at the
horizon is still there. What does that mean from the perspective of an
outside observer?
Will Agata be able to say something like "See, my clock indicates T_o
and the gravitational field of the blackhole has changed. Bruno's 90kg
are inside now"?

Cheers,

IV

(*) Since the horizon is a null-surface Bruno can "wait" over there
there without increasing his Schwarzschild time. I suppose this is the
origin of the the indeterminacy in the correspondence between
Schwarzschild and KS coordinates. Besides, since the length of curves
that switch from time-like to space-like is not defined, it is unclear
to me what we mean by proper time of Bruno beyond the horizon. Can
Bruno still read his proper time simply by looking at his clock and
comparing his reading to what he got when he decided to take the jump?

[1] http://arxiv.org/abs/gr-qc/9712019