Schwarzschild questions
I.Vecchi
and in general about solutions to the Einstein equation in GR.
Some terminology first, so everyone can decide if we agree on what we
mean by the words I'll use.
The Schwarzschild solution is given by a 4-dimensional
pseudo-Riemannian manifold , i.e by a metric defined on a 4-dimensional
differentiable manifold.
The manifold's elements are space-time events. The differentiable
structure is provided by a set of charts that associate space-time
events with points in R^4. The charts cover the manifold and provide an
atlas, hence they define the manifold.
My first, rhetorical question is the following, which I will motivate
below. Do different atlases, i.e, different sets of charts, possbly
define different differentiable structures? In other words, does the
manifold depend on our choice of charts?
The obvious answer is "No! The manifold does not depend on your choice
of coordinates" . However, this is potentially circular, since it
implicitly assumes that the manifold has already been defined, i.e. is
already there. If we start from the set of space-time events, it is not
immediately clear to me that any set of charts will define the same
manifold. I will now explain why I think that the above question may
not amount to pure quibbling.
The overall motivation for the question is the following. Observers
measure space-time events and label them according to the measurement
outcomes. Such labelling provides charts from the set of space time
events into R^4, which may be used to define an atlas and hence a
manifold, our manifold. Different observers may label the same
space-time event differently. An observer may not be able to measure
all space-time events. For example, a space time-event beyond the
horizon of the Schwarzschild solution cannot be measured by an observer
this side of the horizon. The charts of external observers may extend
only up to the horizont without reaching it , since no information
about space-time events can be extracted from the interior. Observers
inside the horizon on the other hand may measure events inside the
horizon and they define charts there.
Actually the charts defined inside the horizon and over it correspond
exclusively to inside observers. However, they stretch backwards over
the horizon, where they overlap with the charts of external observers.
Now, as far as I understand, Eddington-Finkelstein and
Kruskal-Szekeres provide such charts, but it's not clear to me what
characterises them. Suppose I want to define a chart for an inside
observer. What conditions must be fulfilled at horizon so as to provide
a valid overlap (i.e. to identify univocally space-time events) with
the external observers' charts on the other side?
Is the condition that the value of the metric tensor be non-singular
there (when we evaluate it there in the chart's coordinates) the only
requirement?
Are the Kruskal-Szekeres and Eddington-Finkelstein charts smoothly
defined over the horizon?
Cheers,
IV
Daryl McCullough
[stuff deleted]
>Now, as far as I understand, Eddington-Finkelstein and
>Kruskal-Szekeres provide such charts, but it's not clear to me what
>characterises them. Suppose I want to define a chart for an inside
>observer. What conditions must be fulfilled at horizon so as to provide
>a valid overlap (i.e. to identify univocally space-time events) with
>the external observers' charts on the other side?
>Is the condition that the value of the metric tensor be non-singular
>there (when we evaluate it there in the chart's coordinates) the only
>requirement?
>Are the Kruskal-Szekeres and Eddington-Finkelstein charts smoothly
>defined over the horizon?
I think that the metric being nonsingular is the only requirement.
As you say, you have to have a manifold *first* before you can
say whether a particular coordinate system is smooth on that
manifold. But for General Relativity, what is physically meaningful
is an atlas of *local* charts that could be measured by a
(slower-than-light) observer using only local means (clocks and
rulers). The metric (or rather, the line element computed from
the metric) contains *exactly* this information. If you know the
metric in a region, then you can figure out what local observers
would measure.
The coordinate system is smooth if nearby points, as measured
by local observers are mapped by the coordinate system to nearby
points on R^4, and vice-versa. The coordinate system is nonsmooth
if (1) points that are close together, as measured by local observers,
are mapped to distant points, as measured by the coordinate system, or
(2) points that are close together, as measured by the coordinate
system, are mapped to points that are distant, as measured by local
observers.
The Schwarzchild coordinates are nonmooth in this sense. For
example, let A be a point with coordinates
r=2m, t=0, theta=0, phi=0 and let B be a point with coordinates
r=2m, t=1 billion years, theta=0, phi=0. These are far apart, according
to the coordinates, but the temporal separation between them, as
measured by freefalling observers, is *zero*.
For another example, let A be the point with coordinates
(r=2m, t=0, theta=0, phi=0), and let B be the point with
coordinates (r=2m + x, theta=0, phi=0), where x is some
tiny fraction of the Schwarzchild radius, say 8m * 10^{-12}.
These two points are very close together in Schwarzchild
coordinates. However, the proper separation between these
points is 8m * 10^{-6}, a *million* times the coordinate
difference.
In Schwarzchild coordinates, small proper separations do not
imply small coordinate differences, and small coordinate
differences do not imply small proper separations. That's
the sense in which Schwarzchild coordinates are nonsmooth.
--
Daryl McCullough
Ithaca, NY
JanPB
>
> The coordinate system is smooth if nearby points, as measured
> by local observers are mapped by the coordinate system to nearby
> points on R^4, and vice-versa. The coordinate system is nonsmooth
> if (1) points that are close together, as measured by local observers,
> are mapped to distant points, as measured by the coordinate system, or
> (2) points that are close together, as measured by the coordinate
> system, are mapped to points that are distant, as measured by local
> observers.
Just to be a bit anal: what you described is more like continuity of
coordinate systems. Smoothness I guess would be this continuity plus
"smooth curves get mapped to smooth curves", i.e. if a curve tangent
exists at a point according to one observer, it exists according to the
other, or paths cannot develop observer-dependent "kinks". In real life
this sort of criterion is not really measurable though... But it makes
the mathematics of the situation consistent.
--
Jan Bielawski
I.Vecchi
Daryl McCullough ha scritto:
> I.Vecchi says...
>
> [stuff deleted]
>
> >Now, as far as I understand, Eddington-Finkelstein and
> >Kruskal-Szekeres provide such charts, but it's not clear to me what
> >characterises them. Suppose I want to define a chart for an inside
> >observer. What conditions must be fulfilled at horizon so as to provide
> >a valid overlap (i.e. to identify univocally space-time events) with
> >the external observers' charts on the other side?
>
> >Is the condition that the value of the metric tensor be non-singular
> >there (when we evaluate it there in the chart's coordinates) the only
> >requirement?
> >Are the Kruskal-Szekeres and Eddington-Finkelstein charts smoothly
> >defined over the horizon?
>
> I think that the metric being nonsingular is the only requirement.
> As you say, you have to have a manifold *first* before you can
> say whether a particular coordinate system is smooth on that
> manifold. But for General Relativity, what is physically meaningful
> is an atlas of *local* charts that could be measured by a
> (slower-than-light) observer using only local means (clocks and
> rulers). The metric (or rather, the line element computed from
> the metric) contains *exactly* this information. If you know the
> metric in a region, then you can figure out what local observers
> would measure.
...
>
> In Schwarzchild coordinates, small proper separations do not
> imply small coordinate differences, and small coordinate
> differences do not imply small proper separations. That's
> the sense in which Schwarzchild coordinates are nonsmooth.
It is now my understanding that the apparent metric singularity at the
horizont is induced by the Schwarzschild chart not being defined there.
Essentially, one is taking a limit of the metric's value to the charts
boundary, where the chart is not defined and this results on the
apparent singularity. This is well illustrated by JanPB's example, the
physical interpretation being that an external observer cannot measure
and chart space-time events at the horizon and beyond. The chart blows
up as the distant external observer measures space-time events closer
and closer to the horizon.
I will mull over your remarks, and but if I understand you correctly,
you are suggesting that once you have a chart that extends the metric
smoothly to the boundary and beyond, you are home. Basically
Eddington-Finkelstein and Kruskal-Szekeres define the "right" metric
over the horizon through such a smooth extension.
A natural question for me is whether any such smooth extrapolation
process yields the same non-singular metric over the horizon. It's not
clear that it should, at least not to me.
IV
JanPB
> Here some thoughts and some questions about the Scwharzschild solution
> and in general about solutions to the Einstein equation in GR.
>
> Some terminology first, so everyone can decide if we agree on what we
> mean by the words I'll use.
>
> The Schwarzschild solution is given by a 4-dimensional
> pseudo-Riemannian manifold , i.e by a metric defined on a 4-dimensional
> differentiable manifold.
>
> The manifold's elements are space-time events. The differentiable
> structure is provided by a set of charts that associate space-time
> events with points in R^4. The charts cover the manifold and provide an
> atlas, hence they define the manifold.
>
> My first, rhetorical question is the following, which I will motivate
> below. Do different atlases, i.e, different sets of charts, possbly
> define different differentiable structures?
Yes. The result is manifolds which can be homeomorphic but not
diffeomorphic. One well-known example is the 7-dimensional sphere which
has 28 different smooth structures (so there are 28 different smooth
7-dimensional manifolds, all topologically equivalent to S^7). Such
non-standard smooth structures are called "exotic" and the
corresponding manifolds "fake <such-and-such>". So there are e.g. 27
exotic structures on S^7, or there exist 27 fake S^7's.
The problem of determining how many distinct smooth structures a given
topological manifold can support is quite deep and holds a major
surprise in dimension four (as luck would have it). Not surprisingly,
the first attack on this problem was to look at the transition
functions. In a topological manifold they are all homeomorphisms. In
order to "smooth the manifold out" one might try to alter these
transition maps somehow to make them diffeomorphisms while preserving
the original topology. It turns out that this problem in dimensions >=5
can be solved by "classifying spaces" methods where a smoothing
essentially corresponds to lifting maps in a certain "fibration". So in
the case of S^7 for example one can establish that its smooth
structures form the group Z/28 (the cyclic group of 28 elements).
Manifolds in dimension >=5 have finitely many exotic structures.
In dimension 4 the situation is very different. It was actually quite
mysterious until the mid-80s when Simon Donaldson created a bunch of
totally new invariants of smooth 4-manifolds. These invariants are
based (unexpectedly) on gauge theory - one looks at spaces of solutions
of Yang-Mills-like (but elliptic) PDEs which depend very strongly on
the smooth structure of the original manifold. These spaces of
solutions turn out to be finite dimensional manifolds themselves (a
point in such manifold is an equivalence class of connections on
certain SU(2) bundle over the original manifold) and are very hard to
investigate but it's possible to get just enough information from them
to prove, for example, that there exist uncountably many(!) exotic
structures on the plain R^4. What's even more bizarre, some of those
fake R^4 can be embedded into the standard R^4. In the '90s the
Seiberg-Witten invariants were constructed which are easier to
calculate and seem to retain the full power of Donaldson's approach.
> In other words, does the manifold depend on our choice of charts?
> The obvious answer is "No! The manifold does not depend on your choice
> of coordinates" . However, this is potentially circular, since it
> implicitly assumes that the manifold has already been defined, i.e. is
> already there. If we start from the set of space-time events, it is not
> immediately clear to me that any set of charts will define the same
> manifold.
Right, it's not clear at all. It's a very deep business.
> I will now explain why I think that the above question may
> not amount to pure quibbling.
>
> The overall motivation for the question is the following. Observers
> measure space-time events and label them according to the measurement
> outcomes. Such labelling provides charts from the set of space time
> events into R^4, which may be used to define an atlas and hence a
> manifold, our manifold. Different observers may label the same
> space-time event differently. An observer may not be able to measure
> all space-time events. For example, a space time-event beyond the
> horizon of the Schwarzschild solution cannot be measured by an observer
> this side of the horizon. The charts of external observers may extend
> only up to the horizont without reaching it , since no information
> about space-time events can be extracted from the interior. Observers
> inside the horizon on the other hand may measure events inside the
> horizon and they define charts there.
I'm not sure if I'll sound relevant here but I think it's important to
not identify a chart domain with a single particular observer. A chart
domain is just an area where certain metrical parametrisation is valid.
It registers/quantifies spatial and temporal relationships and the fact
that a portion of it would be inaccessible to an observer sitting in
some other portion of that domain is not invalidating any porton of it.
> Actually the charts defined inside the horizon and over it correspond
> exclusively to inside observers.
Well, I don't know what it means. I mean, I do but I'm pretending I
don't because the notion of chart has no concept that would make
precise what you are saying.
> However, they stretch backwards over
> the horizon, where they overlap with the charts of external observers.
> Now, as far as I understand, Eddington-Finkelstein and
> Kruskal-Szekeres provide such charts, but it's not clear to me what
> characterises them. Suppose I want to define a chart for an inside
> observer. What conditions must be fulfilled at horizon so as to provide
> a valid overlap (i.e. to identify univocally space-time events) with
> the external observers' charts on the other side?
As long as there exists a solution expressed in terms of a compatible
chart that extends across the horizon. (Not sure if I'm answering your
question.)
> Is the condition that the value of the metric tensor be non-singular
> there (when we evaluate it there in the chart's coordinates) the only
> requirement?
It still has to be a solution of Einstein's equation there.
> Are the Kruskal-Szekeres and Eddington-Finkelstein charts smoothly
> defined over the horizon?
Yes, you can see it right in the components.
--
Jan Bielawski
I.Vecchi
...
>
> > Actually the charts defined inside the horizon and over it correspond
> > exclusively to inside observers.
>
> Well, I don't know what it means. I mean, I do but I'm pretending I
> don't because the notion of chart has no concept that would make
> precise what you are saying.
My argument is that the manifold's elements are space-time events. The
manifold is determined by the charts that establish a correspondence
between space-time events ( i.e. measurement outcomes relative to an
observer) and points in R^4. An outside observer cannot measure
space-time events beyond the horizont. Hence the measurement outcomes
that chart the space-time events inside the horizont correspond only to
inside observers.
>
> > However, they stretch backwards over
> > the horizon, where they overlap with the charts of external observers.
> > Now, as far as I understand, Eddington-Finkelstein and
> > Kruskal-Szekeres provide such charts, but it's not clear to me what
> > characterises them. Suppose I want to define a chart for an inside
> > observer. What conditions must be fulfilled at horizon so as to provide
> > a valid overlap (i.e. to identify univocally space-time events) with
> > the external observers' charts on the other side?
>
> As long as there exists a solution expressed in terms of a compatible
> chart that extends across the horizon. (Not sure if I'm answering your
> question.)
Yes, but see below.
> > Is the condition that the value of the metric tensor be non-singular
> > there (when we evaluate it there in the chart's coordinates) the only
> > requirement?
>
> It still has to be a solution of Einstein's equation there.
Of course.
>
> > Are the Kruskal-Szekeres and Eddington-Finkelstein charts smoothly
> > defined over the horizon?
>
> Yes, you can see it right in the components.
Well , I am thinking about this. Looking at the derivation of the KS
and EF charts I gather that at the horizont there is some some matching
of the coordinates u and v at the horizon that is essentially
hand-made. There appears to be an implicit selection criterion that
reflects some handy implicit assumptions on the insider observers.
Different matching criteria may work as well., but as far as I can
guess now, they would yield the same results from the point of view of
an outside observer, who will never get to talk to the insiders and ask
them "How did it go? What did you see?". Experimental verification is
an intrinsecally tricky business as far as blackholes are concerned,
due to the epistemic structure of the problem.
Anyways, I find it helpful to look at the issue from an epistemic
perspective, i.e. in terms of information exchanges between observers.
It provides some explanatory juice. I don't know whether epistemic
models here may turn out to be physically relevant, i.e. be
experimentally testable, but I would not rule it out. Looking at the
historical record, as soon as an epistemic model is successfully
verified experimentally , it is declared physical.
Let me add what I regard as an intriguing analogy from the theory of
nonlinear hyperbolic equations. Take the Hopf equation and start with
smooth initial data. When characteristics cross you may recover
existence by admitting distributional solutions (satisfying
Rankine-Hugoniot conditions) but you lose uniqueness. In order to
recover well-posedness you may impose an entropy condition, which
essentially requires that no new information is created at the
singularities. It's basically a causality condition.
I think the problem here is somehow similar. The question of defining
the "right path" across the Scwharzschild surface may have been solved
"ad hoc" with devices such as Eddington-Schwarzschild and
Kruskal-Szekeres. I am not entirely sure that they are the end of the
story.
Cheers,
IV
-------------------------------
"Wovon man nicht sprechen kann, darueber muss man schwaetzen"
Wittgenstein&Vecchi
Daryl McCullough
>Well , I am thinking about this. Looking at the derivation of the KS
>and EF charts I gather that at the horizont there is some some matching
>of the coordinates u and v at the horizon that is essentially
>hand-made.
No, the relationship between Kruskal coordinates and Schwarzchild
coordinates seems ad hoc, but there is nothing funny going on if
we just look at the Kruskal coordinates (I always say "Kruskal"
instead of "Kruskal-Szekeres" because I can never remember how
to spell or pronounce "Szekeres").
The KS metric looks like this:
http://en.wikipedia.org/wiki/Kruskal_coordinates
ds^2 = 32 m^3/r exp(-r/2m) (-dT^2 + dR^2) + r^2 dOmega^2
where r is a well-defined function of R and T, implicitly
defined by
R^2 - T^2 = (r/2m - 1) exp(r/2m)
The event horizon in KS coordinates is nothing special:
near r=2m, the metric just looks like this:
ds^2 = 16/e m^2 (-dT^2 + dR^2) + 4m^2 dOmega^2
Absolutely nothing weird happens as you cross the event
horizon. (The event horizon in these coordinates is
the locus of points where |R| = |T|).
So, if we look at the manifold as described using KS coordinates,
there is nothing weird going on anywhere except at r=0 (which
in KS coordinates is the locus of points with T^2 - R^2 = 1.
>There appears to be an implicit selection criterion that
>reflects some handy implicit assumptions on the insider
>observers.
>Different matching criteria may work as well, but as far as I can
>guess now, they would yield the same results from the point of view of
>an outside observer, who will never get to talk to the insiders and ask
>them "How did it go? What did you see?".
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)
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. 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.
>I think the problem here is somehow similar. The question of defining
>the "right path" across the Scwharzschild surface may have been solved
>"ad hoc" with devices such as Eddington-Schwarzschild and
>Kruskal-Szekeres. I am not entirely sure that they are the end of the
>story.
As I said before, it seems to me that the "right path" is defined
by looking at what a local observer sees in the neighborhood of the
event horizon. A smooth coordinate system should be able to
map smoothly to the local coordinate system of a freefalling
observer. KS can do this just fine: near the event horizon,
KS coordinates look like
ds^2 = 16/e m^2 (-dT^2 + dR^2) + angular part
which is just a rescaling of the local coordinate system
of a freefalling observer.
I.Vecchi
Daryl McCullough ha scritto:
...
> I.Vecchi says...
...
>
> >Different matching criteria may work as well, but as far as I can
> >guess now, they would yield the same results from the point of view of
> >an outside observer, who will never get to talk to the insiders and ask
> >them "How did it go? What did you see?".
>
> Yes, the relationship between KS coordinates and Schwarzchild
> coordinates is not completely specified.
That's the key point. See below.
> 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)
>
> 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. 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.
Beautiful.
>
> >I think the problem here is somehow similar. The question of defining
> >the "right path" across the Scwharzschild surface may have been solved
> >"ad hoc" with devices such as Eddington-Schwarzschild and
> >Kruskal-Szekeres. I am not entirely sure that they are the end of the
> >story.
>
> As I said before, it seems to me that the "right path" is defined
> by looking at what a local observer sees in the neighborhood of the
> event horizon. A smooth coordinate system should be able to
> map smoothly to the local coordinate system of a freefalling
> observer. KS can do this just fine
Yes, in the perspective of an infalling observer yes, but in the
perspective of of an outside observer his path is undetermined, short
of "ad hoc" assumptions fixing a correspondence between KS and
Schwartschild coordinates.
Isn't this perplexing? What's the model telling us?
Here is a , er, speculative guess.
The indeterminacy of the correspondence between A and B might be
interpreted as the model telling us that we should NOT fabricate such a
correspondence, i.e. that all matchings should be given equal weight.
In a relational setting "alla Rovelli" , one may say that, in the
perspective of on outside observer, the infalling observer gets
scattered at the horizon.
Scattering does not affect the infalling observer. Cats and Wigner's
friends get scattered and don't notice it. We all get scattered.
Scattering happens only in the perspective of the outside observer who
cannot track the infalling observer's path over the horizon.
OK, we were talking GTR and not QG, but I think that the model might be
telling us that GTR is not enough here, that the the horizon is an
epistemic boundary and as such needs to be considered in a QG setting.
Cheers,
IV
JanPB
> Daryl McCullough ha scritto:
>
> > As I said before, it seems to me that the "right path" is defined
> > by looking at what a local observer sees in the neighborhood of the
> > event horizon. A smooth coordinate system should be able to
> > map smoothly to the local coordinate system of a freefalling
> > observer. KS can do this just fine
>
> Yes, in the perspective of an infalling observer yes, but in the
> perspective of of an outside observer his path is undetermined, short
> of "ad hoc" assumptions fixing a correspondence between KS and
> Schwartschild coordinates.
> Isn't this perplexing? What's the model telling us?
This is an interesting question which - naturally - textbooks tend to
gloss over. Let me paraphrase it slightly:
We have the Schwarzschild solution on two disjoint domains: 0<r<2m and
2m<r<infty (plus the usual other variables). An extension of this
solution is obtained by changing these two charts by the
diffeomorphism:
t' = t + 2m ln|r/2m - 1|
(other variables unchanged)
...which yields a new coordinate expression for the metric
(Eddington-Finkelstein) the coefficients of which are not singular at
r=2m anymore. So we extend our domain to the full range 0<r<infty by
"filling in" the missing value r=2m. This keeps the extended metric
both smooth and satisfying the Einstein equation.
Now your question is: how do we know this procedure is unique? IOW, how
do we know that there doesn't exist some *other* diffeomorphism on
0<r<2m and 2m<r<infty (plus the usual other variables) with the
resulting charts and metric coefficients extending over the gap
*differently* (i.e. perhaps across a horizon of nonzero thickness? So
it's more like a thick "wall"? Etc.)
This seems to hinge on constraints imposed by the Einstein equations
themselves. Without them it is certainly possible to smoothly extend
the manifold and the metric over the "gap" between the two
Schwarzschild chart domains in infinitely many distinct ways. I don't
see anything besides the Einstein equation that would prohibit this.
For example, in the Eddington-Finkelstein case - before "filling in"
the horizon - we have the three coeficient functions: -(1-2m/r), 4m/r,
(1+2m/r), and the angular part. Instead of the usual "filling in" by
inserting the point r=2m, we might try to insert a thick "wall" between
the two domains and define the metric coefficients to be constant
across (equal to the above coefficients with r=2m substituted). This
would define a different metric satisfying the Einstein equations
(trivially: derivatives of constants are zero) except the metric would
not be smooth at the "wall" boundaries where we sharply switch to
constant functions. We might try to smooth those kinks out but then
we'll loose the Einstein equations at the "rounded corners" as they
wouldn't be guaranteed to satisfy them.
The above paragraph is only intended to indicate that the claim that it
is the Einstein equations that force the extension uniqueness is a
plausible one. I suspect an airtight argument could be constructed from
uniqueness theorems for solutions to ODEs (to which Einstein's PDEs
tend to reduce here) or from some special uniqueness for PDEs that
perhaps applies to Einstein's equation in this case.
> Here is a , er, speculative guess.
> The indeterminacy of the correspondence between A and B might be
> interpreted as the model telling us that we should NOT fabricate such a
> correspondence, i.e. that all matchings should be given equal weight.
> In a relational setting "alla Rovelli" , one may say that, in the
> perspective of on outside observer, the infalling observer gets
> scattered at the horizon.
Before we get into all that perhaps we could ask Steve Carlip about
this. It may all simply follow from some standard PDE/ODE theory.
--
Jan Bielawski
I.Vecchi
My question is motivated by the considerations about charts and
observers in my previous posts, so I would not rephrase the problem as
you do, since I think your formulation blurs the underlying
physical/epistemic argument. I would not say that metric is different,
but that that the correspondence between the Schwarzschild and KS
charts is underdetermined. However with some goodwill we should be
able to agree on the mathematical structure of the problem no matter
how we formulate it.
>
> Now your question is: how do we know this procedure is unique? IOW, how
> do we know that there doesn't exist some *other* diffeomorphism on
> 0<r<2m and 2m<r<infty (plus the usual other variables) with the
> resulting charts and metric coefficients extending over the gap
> *differently* (i.e. perhaps across a horizon of nonzero thickness? So
> it's more like a thick "wall"? Etc.)
I think Daryl McCullough's previous post provides an answer to this
question. The T/R correspondence at the boundary is underdetermined,
since any choice of tha A and B parameters will do. Tthe underlying
issue is the lack of intrinsic uniqueness for the correspondence
between the proper time of an outside observer and that of an infalling
observer at the horizon.
> This seems to hinge on constraints imposed by the Einstein equations
> themselves. Without them it is certainly possible to smoothly extend
> the manifold and the metric over the "gap" between the two
> Schwarzschild chart domains in infinitely many distinct ways. I don't
> see anything besides the Einstein equation that would prohibit this.
>
> For example, in the Eddington-Finkelstein case - before "filling in"
> the horizon - we have the three coeficient functions: -(1-2m/r), 4m/r,
> (1+2m/r), and the angular part. Instead of the usual "filling in" by
> inserting the point r=2m, we might try to insert a thick "wall" between
> the two domains and define the metric coefficients to be constant
> across (equal to the above coefficients with r=2m substituted). This
> would define a different metric satisfying the Einstein equations
> (trivially: derivatives of constants are zero) except the metric would
> not be smooth at the "wall" boundaries where we sharply switch to
> constant functions. We might try to smooth those kinks out but then
> we'll loose the Einstein equations at the "rounded corners" as they
> wouldn't be guaranteed to satisfy them.
In your argument you seem to assume that there is a matching (i.e. a
choice of A,B) that is "better than the others" to start with. I can't
make sense of what you mean.
> The above paragraph is only intended to indicate that the claim that it
> is the Einstein equations that force the extension uniqueness is a
> plausible one. I suspect an airtight argument could be constructed from
> uniqueness theorems for solutions to ODEs (to which Einstein's PDEs
> tend to reduce here) or from some special uniqueness for PDEs that
> perhaps applies to Einstein's equation in this case.
If the underlying structure is that of nonlinear hyperbolic equation
like the Hopf equation, you do not get uniqueness without additional
constraints, which in this case appear unwarranted.
> > Here is a , er, speculative guess.
> > The indeterminacy of the correspondence between A and B might be
> > interpreted as the model telling us that we should NOT fabricate such a
> > correspondence, i.e. that all matchings should be given equal weight.
> > In a relational setting "alla Rovelli" , one may say that, in the
> > perspective of on outside observer, the infalling observer gets
> > scattered at the horizon.
>
> Before we get into all that perhaps we could ask Steve Carlip about
> this. It may all simply follow from some standard PDE/ODE theory.
You are obviously free to ask anyone you want.
Cheers,
IV
JanPB
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
I.Vecchi
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?