Re: Any coordinate system in GR?

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Topic:	Science > Physics
User:	"Edward Green"
Date:	24 Aug 2006 06:00:20 PM
Object:	Re: Any coordinate system in GR?

Edward Green wrote:

The offhand claim is sometimes made that GR, which I take it means the
formal machinery of GR, treats all coordinate systems equally. Is this
true? Aside from questions about smoothness, it seems to me there is
at least some other implicit condition on permissible coordinate
systems.

Consider a 1-dimensional manifold, a real coordinate x, and the metric
ds = |dx|. Next, stretch this manifold by a postive factor a(x), also
stretching the coordinates (so that points on the manifold are still
labeled by the same numbers). Calling the new, stretched, coordinates
u, the most natural transformation law for the metric would be

ds = a(u)|du|

Now, instead of stretching the manifold, compress the coordinates by
this same function a( ), so that dx = a(u)du. The most natural
transformation law for the metric would again be

ds = a(u)|du|

So, if we allow "any coordinate system", we would apparently be
unable to distinguish the (physical) case of the stretched manifold
from the (unphysical) case of the compressed coordinates.

It may be objected that by compressing the coordinates, we have
implicitly changed physical units, and this is not allowed. But that's
a new rule, isn't it? What's a "unit"? We need additional structure
to specify what we mean by this; we must equip our manifold with some
physics -- a local property establishing the natural scale of the
coordinates. This will insure that we don't arbitrarily distort the
coordinates, and any distortion can be attributed to the manifold.

Igor wrote:

Stretching or compressing the coordinate system just redefines it
mathematically. There's nothing physical there.

Of course.

You can define a
coordinate transformation any way you want to provided that it is
invertible. But that doesn't result in any new physics.

Obviously, my dear man.

It may result
in so-called fictictious forces arising in the new coordinate system
that weren't there previously, but those are never really physical
either.

We are straying from the question.

And changing units can indeed be allowed under coordinate
transformations. Think of going from rectangular to polar or vice
versa. Rectangular has both coordinates with length units, and polar
has one with length and the other with angular units.

Ok. Good point.

The metric will always compensate to make up for the difference.

So you claim. I believe I have found the answer to my own question (as
usual), and while I did not state the requirement 100% correctly, I was
on the right track. I happened upon the correct formulation in
Schwarzschild's 1916 paper on the field of a mass point:
"The field equations ... have the fundamental property that
they preserve their form under the substitution of other
arbitrary variables in lieu of x1,x2, x3, x4, as long as the
determinant of the substitution is equal to 1."
http://arxiv.org/PS_cache/physics/pdf/9905/9905030.pdf
If the old coordinates were orthogonal and the new coordinates remained
orthogonal (which requirement I did not state) and we kept the same
units, mine would be a sufficient condition to meet the above
requirement on the named determinate. In general we have more latitude
than that, _but_ we are not free to chose the new coordinates
arbitrarily, even if they are invertible in a neighborhood. The unit of
_volume_ at least must be preserved. In the case of a one-dimensional
manifold this reduces to the requirement to keep the same units.
Are we in agreement now?
.

User: "Koobee Wublee"
Title: Re: Any coordinate system in GR?	24 Aug 2006 07:25:41 PM

Edward Green wrote:

Igor wrote:

The metric will always compensate to make up for the difference.

This is an excellent article. It is very easy to read and understand.
Thanks to Schwarzschild's ingenuity.

If the old coordinates were orthogonal and the new coordinates remained
orthogonal (which requirement I did not state) and we kept the same
units, mine would be a sufficient condition to meet the above
requirement on the named determinate. In general we have more latitude
than that, _but_ we are not free to chose the new coordinates
arbitrarily, even if they are invertible in a neighborhood. The unit of
_volume_ at least must be preserved. In the case of a one-dimensional
manifold this reduces to the requirement to keep the same units.

Schwarzschild found a unique solution to the differential equations of
Einstein Field Equations in free space. Hilbert found another one that
he called it Schwarzschild Metric. Recently, Mr. Rahman also a
contributor of this newsgroup presented another solution. The
spacetime with this metric is
ds^2 = c^2 dt^2 / (1 + K / r) - (1 + K / r) dr^2 - (r + K)^2 dO^2
The metric above indeed is another solution which anyone can easily
verify because its simplicity. Notice Rahman's metric and
Schwarzschild's original metric do not manifest black holes.
However, since Schwarzschild Metric is much simpler than
Schwarzschild's original solution, Schwarzschild Metric is embraced by
the physics communities today. Mr. Bielawski and Igor have not
understood Schwarzschild's original paper and choose to blindly reject
Schwarzschild's original solution and others.
As multiple solutions to the vacuum field equations are discovered,
there are actually an infinite number of them. With infinite number of
solutions, it is shaking the very foundation of GR and SR. The house
of cards will soon inevitably collapse. However, refusing to give up
GR and to comfort themselves in false sense of security, they choose to
embrace Voodoo Mathematics. In doing so, they blindly claim all
solutions are indeed the same regardless manifesting black holes,
constant expanding universe, accelerated expanding universe. VOODOO
MATHEMATICS REPRESENTS THE ACHIEVEMENT IN PHYSICS DURING THE LAST 100
YEARS. It is very sad that these clowns are regarded as experts in
their field.
.

User: "JanPB"
Title: Re: Any coordinate system in GR?	24 Aug 2006 07:53:39 PM

Koobee Wublee wrote:

No, the metric above is equal to Schwarzschild's metric. The form above
is obtained by a coordinate change from the original one, hence the
metric remains the same (tensors do not change under coordinate
changes).

However, since Schwarzschild Metric is much simpler than
Schwarzschild's original solution, Schwarzschild Metric is embraced by
the physics communities today.

It's embraced because it's the same.

Mr. Bielawski and Igor have not
understood Schwarzschild's original paper and choose to blindly reject
Schwarzschild's original solution and others.

There is nothing to reject. One can _prove_ Schwarzschild's metric is
unique. Off the horizon it follows immediately from the particular form
of the Einstein equation in the spherically symmetric case (which is
what we have) and the uniqueness of the extension over the horizon is
slightly more involved but it follows from a similar argument.

As multiple solutions to the vacuum field equations are discovered,
there are actually an infinite number of them.

Yes, and 2+2=5.

With infinite number of
solutions, it is shaking the very foundation of GR and SR.

Sure. My boots are all torn already.
Give us a break.
--
Jan Bielawski
.

User: "I.Vecchi"
Title: Re: Any coordinate system in GR?	25 Aug 2006 07:35:50 AM

JanPB ha scritto:

Koobee Wublee wrote:

However, since Schwarzschild Metric is much simpler than
Schwarzschild's original solution, Schwarzschild Metric is embraced by
the physics communities today.

It's embraced because it's the same.

Mr. Bielawski and Igor have not
understood Schwarzschild's original paper and choose to blindly reject
Schwarzschild's original solution and others.

I would say that there are at least two distinct ways to prolong the
solution across the horizon, which yield respectively the black hole
and the white hole solution, describing two different physical
phenomena. They correspond to two distinct choices of the
Eddington-Finkelstein coordinates at the horizon, i.e. to crossing the
horizon on future-directed or on past-directed curves. The extended
solution depends on how space-time is extended beyond the horizon.
IV
.

User: "Tom Roberts"
Title: Re: Any coordinate system in GR?	25 Aug 2006 10:35:27 AM

I.Vecchi wrote:

The Kruskal-Szerkes coordinates show that these two extensions are
merely two aspects of the complete (inextensible) manifold.
Tom Roberts
.

User: "I.Vecchi"
Title: Re: Any coordinate system in GR?	25 Aug 2006 03:14:20 PM

Tom Roberts ha scritto:

I.Vecchi wrote:

The Kruskal-Szerkes coordinates show that these two extensions are
merely two aspects of the complete (inextensible) manifold.

Your construct relies on the arbitrary duplication of the the horizon,
inventing two copies of what is actually a single
physical/observational domain. It remains a fact that the extension
across the horizon is not unique and that each extension yields a
distinct physical object.
A black hole is not a white hole. Period.
IV
.

User: "Tom Roberts"
Title: Re: Any coordinate system in GR?	25 Aug 2006 05:17:26 PM

I.Vecchi wrote:

Tom Roberts ha scritto:

I.Vecchi wrote:

The Kruskal-Szerkes coordinates show that these two extensions are
merely two aspects of the complete (inextensible) manifold.

Your construct relies on the arbitrary duplication of the the horizon,

Sure. As is easily seen in a Kruskal diagram, and as I said. Here you
seem to be in violent agreement with what I said (but later you get it
wrong).
It is arbitrary which portion of the horizon you consider, but as both
are contained in the manifold it is only the analyst who is subject to
this choice, not the manifold itself (or the physical system for which
it is a model).

inventing two copies of what is actually a single
physical/observational domain.

Not really. Those regions of the complete (inextensible) manifold are
DIFFERENT. That is, considered from a given point in spacetime near but
outside the horizon, the past and future horizons are NOT "the same".
This difference between past and future is true in ANY spacetime, of course.
That is, for any point in any manifold of GR, every locus
in the past lightcone of the point is disjoint from every
locus in the future lightcone of the point. This is true
in everyday life -- just think about how different is your
ability to observe events in the past from events in the
future.

It remains a fact that the extension
across the horizon is not unique and that each extension yields a
distinct physical object.

Not "physical object" but rather region of the manifold. Basically you
happened to choose E-F coordinates that do not cover the manifold, but
both sets of E-F coordinates cover the exterior region; which set of E-F
coordinates you choose will determine into which region of the manifold
you can extend. BTW neither choice includes other regions of the
complete (inextensible) manifold, but Kruskal-Szerkes coordinates
include them all.

A black hole is not a white hole. Period.

Sure. And both are contained in the Kruskal diagram, and in the complete
(inextensible) manifold.
Have you never looked at a Kruskal diagram? -- you seem rather
unknowledgeable about basic aspects of this manifold (yes, singular --
you are discussing different regions of a single manifold). Any
reasonably modern textbook on GR will have it.
Tom Roberts
.

User: "I.Vecchi"
Title: Re: Any coordinate system in GR?	26 Aug 2006 08:04:54 AM

Tom Roberts ha scritto:

I.Vecchi wrote:

Your construct relies on the arbitrary duplication of the the horizon,

Sure. As is easily seen in a Kruskal diagram, and as I said. Here you
seem to be in violent agreement with what I said (but later you get it
wrong).

It is arbitrary which portion of the horizon you consider, but as both
are contained in the manifold it is only the analyst who is subject to
this choice, not the manifold itself (or the physical system for which
it is a model).

Are you saying that for every black hole in the universe there is a
corresponding white hole?

inventing two copies of what is actually a single
physical/observational domain.

Not really. Those regions of the complete (inextensible) manifold are
DIFFERENT. That is, considered from a given point in spacetime near but
outside the horizon, the past and future horizons are NOT "the same".
This difference between past and future is true in ANY spacetime, of course.

That is, for any point in any manifold of GR, every locus
in the past lightcone of the point is disjoint from every
locus in the future lightcone of the point. This is true
in everyday life -- just think about how different is your
ability to observe events in the past from events in the
future.

It remains a fact that the extension
across the horizon is not unique and that each extension yields a
distinct physical object.

Not "physical object" but rather region of the manifold.

Which corresponds to a physical/observational object. Otherwise we are
talking about nothing.

Basically you
happened to choose E-F coordinates that do not cover the manifold, but
both sets of E-F coordinates cover the exterior region; which set of E-F
coordinates you choose will determine into which region of the manifold
you can extend.

Yes, which set of coordinates I choose will determine it . And
correspondily yield the black hole or the white hole solution.

BTW neither choice includes other regions of the
complete (inextensible) manifold, but Kruskal-Szerkes coordinates
include them all.

Yes, it's mathematical construct useful for didactic purposes.

A black hole is not a white hole. Period.

Sure. And both are contained in the Kruskal diagram, and in the complete
(inextensible) manifold.

Have you never looked at a Kruskal diagram? -- you seem rather
unknowledgeable about basic aspects of this manifold (yes, singular --
you are discussing different regions of a single manifold). Any
reasonably modern textbook on GR will have it.

The point here is its relevance to the topic of this discussion. As for
extending my knowledge, which may indeed be limited, vigorous (and,
ideally, polite) online discussion appears to be a fruitful approach.
Beside the above, I surmise that there are other space-time extensions
across the horizon corresponding to hybrid white hole/black hole
solutions (*). This would be impossible according to your argument,
right?
Cheers,
IV
(*) In the corresponding charts the metric gets singular at one point
on the horizon, but that does not make it more unphysical than, say,
the Schwarzschild solution.
.

User: "Tom Roberts"
Title: Re: Any coordinate system in GR?	26 Aug 2006 11:21:08 PM

I.Vecchi wrote:

Tom Roberts ha scritto:

It is arbitrary which portion of the horizon you consider, but as both
are contained in the manifold it is only the analyst who is subject to
this choice, not the manifold itself (or the physical system for which
it is a model).

Are you saying that for every black hole in the universe there is a
corresponding white hole?

Not at all! But I am pointing out that in the Schwarzschild manifold
there is both a black hole and a white hole. That manifold _IS_ what we
are discussing. It is quite clear that the universe we inhabit is not
described by the Schw. manifold.

It remains a fact that the extension
across the horizon is not unique and that each extension yields a
distinct physical object.

Not "physical object" but rather region of the manifold.

Which corresponds to a physical/observational object. Otherwise we are
talking about nothing.

I don't know what you mean by "a physical/observational object". But it
IS clear that a region of the manifold is not any sort of "object" at
all. <shrug>
One might call it an "observational domain" I suppose, but "object" is
quite definitely not applicable.

Beside the above, I surmise that there are other space-time extensions
across the horizon corresponding to hybrid white hole/black hole
solutions (*). This would be impossible according to your argument,
right?

The geodesically complete extension of the Schwarzschild charts is
unique, given by the Kruskal chart.
Tom Roberts
.

User: "I.Vecchi"
Title: Re: Any coordinate system in GR?	27 Aug 2006 04:05:34 AM

Tom Roberts ha scritto:

I.Vecchi wrote:

....

Are you saying that for every black hole in the universe there is a
corresponding white hole?

Not at all! But I am pointing out that in the Schwarzschild manifold
there is both a black hole and a white hole. That manifold _IS_ what we
are discussing.

What we are discussing is "the uniqueness of the extension over the
horizon" and the relevant selection criteria . What I am saying is that
there are at least two ways to extend the solution over the horizon,
correspoinding to two different physical situations, the black hole and
the white hole.

It is quite clear that the universe we inhabit is not
described by the Schw. manifold.

Yes, as I said, it is a physically irrelevant mathematical construct,
whose physical irrelevance reflects the irrelevance of the criteria it
fulfills.

It remains a fact that the extension
across the horizon is not unique and that each extension yields a
distinct physical object.

Not "physical object" but rather region of the manifold.

Which corresponds to a physical/observational object. Otherwise we are
talking about nothing.

I don't know what you mean by "a physical/observational object". But it
IS clear that a region of the manifold is not any sort of "object" at
all. <shrug>

One might call it an "observational domain" I suppose, but "object" is
quite definitely not applicable.

You may call it as you like.

The geodesically complete extension of the Schwarzschild charts is
unique, given by the Kruskal chart.

The question is whether the requirement of geodesic completeness (which
in this setting is obtained by shifting trouble to infinity) is
appriopriate, i.e. physically relevant.
I surmise that there are geodesically incomplete extensions,
corresponding to hybrid solutions which may be physically relevant.
I am not keen on throwing away interesting extensions/solutions just
because they do not not comply with some arbitrary uniqueness
criterion.
Cheers.
IV
.

User: "Daryl McCullough"
Title: Re: Any coordinate system in GR?	27 Aug 2006 07:02:45 AM

I.Vecchi says...

The question is whether the requirement of geodesic completeness (which
in this setting is obtained by shifting trouble to infinity) is
appriopriate, i.e. physically relevant.

The answer is definitely "yes". Consider an observer in freefall near the
event horizon. Using his local coordinates, there is *nothing* to prevent
him from reaching and passing the event horizon in a finite amount of
proper time, because for him, spacetime near the event horizon is
approximately *flat*. Anything other than demanding geodesic completeness
would violate the equivalence principle, I think.

I surmise that there are geodesically incomplete extensions,
corresponding to hybrid solutions which may be physically relevant.

I am not keen on throwing away interesting extensions/solutions just
because they do not not comply with some arbitrary uniqueness
criterion.

I don't think that there is anything arbitrary about the criteria.
--
Daryl McCullough
Ithaca, NY
.

User: "I.Vecchi"
Title: Re: Any coordinate system in GR?	27 Aug 2006 01:49:36 PM

Daryl McCullough wrote:

I.Vecchi says...

The question is whether the requirement of geodesic completeness (which
in this setting is obtained by shifting trouble to infinity) is
appriopriate, i.e. physically relevant.

Yes, but the problem is to define what is the event horizont for him.
What does it mean PHYSICALLY that he has crossed the event horizon?
Let me go back for a moment to the Agata and Bruno example ([1]). For
Agata, Bruno never crosses the horizon . She cannot associate a time
T_o to Bruno stepping over the horizon. For an external observer Bruno
NEVER crosses the horizon.
Let's consider Bruno perspective then. You are right that nothing
prevents him to reach the horizon, the problem is the the horizon's
position in his perspective becomes arbitrary. If his proper time at
the horizon had a physically determined value he should be able to
look at his clock just before talking the plunge (e.g. switching off
the rocket that keeps him over the horizon) calculate and say: "In ten
minutes I will have crossed the horizont", and that would mean that in
his proper time 10 minutes would pass and then he'd look at his clock
and know he's over . The problem is that in order to calculate that
proper time he needs to know on which KS chart he is. Now, as we have
seen ([2]), the correspondence between the KS chart and the
Schwarzschild chart is undetermined, so he will not be able to do
that. The question is then, is there any chart independent way (i.e.
independent from the space-time measurement model that defines a chart)
to calculate Bruno's proper time across the horizon? And my tentative
answer is, no there is not. Actually, it seems to me that different
charts correspond to different proper times that may be attributed to
Bruno and that there is no way to associate a unique chart to Bruno
based on the Schwarzschild chart alone (i.e. based on information that
is available this side of the horizont).

Anything other than demanding geodesic completeness
would violate the equivalence principle, I think.

Would it? I would like to inspect a rigorous proof that it would, and
examine the role that proper time plays in it. I am not saying that no
such proof exists, though indeed i doubt it, but it seems to me that
the arguments being thrown around are riddled with implicit assumption
that may or may not be warranted. It may be worthwhile to make them
explicit and scrutinise them.
Besides, note that neither KS is geodesically complete. It's just that
in it the only obstruction to geodesic completeness is the curvature
singularity at the origin. That does not violate the equivalence
principle , as far as I know.

I don't think that there is anything arbitrary about the criteria.

Good for you.
Cheers,
IV
PS I am reading a paper by Christian Fronsdal
(http://arxiv.org/abs/gr-qc/0508048) built around "an observer who is
aware of a limited portion of space", where "the horizon recedes as it
is approached and has no physical reality". It's good deconstructivist
stuff, highlighting some of the issues we are discussing.
[1]
http://groups.google.com/group/sci.physics.relativity/msg/9ca04b30b02cc1ef
[2]
http://groups.google.com/group/sci.physics.relativity/msg/8a5ffdd9c5944461
.

User: "Daryl McCullough"
Title: Re: Any coordinate system in GR?	27 Aug 2006 07:26:53 PM

I.Vecchi says...

Daryl McCullough wrote:

I.Vecchi says...

The question is whether the requirement of geodesic completeness (which
in this setting is obtained by shifting trouble to infinity) is
appriopriate, i.e. physically relevant.

Yes, but the problem is to define what is the event horizont for him.
What does it mean PHYSICALLY that he has crossed the event horizon?

Using Schwarzchild coordinates, you can compute the proper time for
a freefalling observer as a function of his radius r. The exact
computation depends on the initial conditions, but for one particular
geodesic, the relationship is simple:
r(tau) = 3/2 (2m)^{1/3} (tau_0 - tau))^{2/3}
where tau_0 is a constant depending on the initial conditions.
For this geodesic, the observer passes the event horizon
at a proper time tau_1 given by
tau_1 = tau_0 - (2/3)^{3/2} 2m
and reaches r=0 at proper time
tau = tau_0
So what it means physically for the observer to pass the
event horizon is just that his proper time is greater than
tau_1. As I said, from the point of view of the freefalling
observer, there is no physical reason for his proper time
to stop before reaching tau_1 (when he crosses the event
horizon) and tau_0 (when he reaches r=0).

Let me go back for a moment to the Agata and Bruno example ([1]). For
Agata, Bruno never crosses the horizon . She cannot associate a time
T_o to Bruno stepping over the horizon. For an external observer Bruno
NEVER crosses the horizon.

Yes, that shows that Agata's coordinate system is incomplete. There
are physically meaningful events that are given no coordinate whatsoever.
The same thing happens in flat spacetime with accelerated observers.
If you are on board a rocket ship with constant proper acceleration g,
then times and distances as measured by clocks and rulers on the ship
will be related to times and distances as measured by inertial clocks
that are at rest relative to the ship's initial reference frame as
follows:
x = X cosh(gT)
t = X sinh(gT)
where (x,t) are the coordinates as measured by the inertial observers,
and (X,T) are the coordinates as measured by the accelerated observers.
Note that the point (x=1 light year, t=1 year) is given no coordinates
*at* *all* in the accelerated coordinate system: there is no values of
X and T such that X cosh(gT) = 1 and X sinh(gT) = 1.
From the point of view of the accelerated
observer, the point (x=1, t=1) can be approached as T--> infinity,
but it can never be reached. That just means that the accelerated
coordinates are incomplete.
The same thing is true of the Schwarzchild coordinates. They are
incomplete in exactly the same way. There are valid points on
the manifold that are given no coordinates whatsoever in the
Schwarzchild coordinate system.

Let's consider Bruno perspective then. You are right that nothing
prevents him to reach the horizon, the problem is the the horizon's
position in his perspective becomes arbitrary.

Yes, from his point of view, there is nothing special about the
horizon. However, Agata can calculate the proper time tau_1 at
which Bruno will hit the event horizon, and she can make the
prediction that Bruno's clock will never advance past tau_1.
Bruno can watch his clock and prove that Agata is wrong (although
his message gloating about it will never reach Agata).

If his proper time at
the horizon had a physically determined value

It does. Agata can calculate it.

he should be able to look at his clock just before talking the
plunge (e.g. switching off the rocket that keeps him over the
horizon) calculate and say: "In ten
minutes I will have crossed the horizon",

Yes, and at 1 second past ten minutes, assuming he hasn't
hit the singularity, he'll know that he has crossed the
event horizon.

and that would mean that in his proper time 10 minutes would
pass and then he'd look at his clock and know he's over. The
problem is that in order to calculate that proper time he
needs to know on which KS chart he is.

The calculation can be done using KS, or using Schwarzchild
coordinates. It's not hard using Schwarzchild coordinates.
From the metric and the geodesic equation, you can prove
that there are two constants of motion for radial geodesics:
(1-2m/r) V^t = P
(1-2m/r) (V^t)^2 - 1/(1-2m/r) (V^r)^2 = E
where V^t = dt/dtau and V^r = dr/dtau. Using
the first equation in the second, we find
P/(1-2m/r) - (V^r)^2/(1-2m/r) = E
So
V^r = - square-root(P - E(1-2m/r))
dr/dtau = - square-root(P - E(1-2m/r))
(where I chose the minus sign because the observer is falling, so
dr/dtau is negative). This gives tau as a function of r:
dtau/dr = - 1/square-root(P - E(1-2m/r))
tau = tau_0 - integral of 1/square-root(P - E(1-2m/r)) dr
The integral is a perfectly ordinary integral, with no singularities
except at r=0. So the value of tau at which r=2m is computable.

Now, as we have seen ([2]), the correspondence between the KS chart
and the Schwarzschild chart is undetermined, so he will not be able
to do that.

The relationship between tau and r is perfectly well determined.
The relationship between tau and t is not completely determined
but that's not relevant to the question of what the value of tau
is when the infalling observer hits the event horizon.

The question is then, is there any chart independent way (i.e.
independent from the space-time measurement model that defines a chart)
to calculate Bruno's proper time across the horizon?

Yes. I just gave it to you. If you know Bruno's initial radius
and the initial time on Bruno's clock, then you can easily
compute the time on Bruno's clock when he reaches the event
horizon.

Anything other than demanding geodesic completeness
would violate the equivalence principle, I think.

Would it? I would like to inspect a rigorous proof that it would,

I don't know what a rigorous proof would look like, but suppose
that we compute (as shown above) that Bruno will hit the event
horizon at proper time tau = 10 minutes. Bruno, working in his
local coordinate system can see no physical reason that his clock
would stop running before reaching tau=10 minutes.

Besides, note that neither KS is geodesically complete. It's just that
in it the only obstruction to geodesic completeness is the curvature
singularity at the origin. That does not violate the equivalence
principle, as far as I know.

The singularity at r=0 is present in *all* coordinate systems,
including the coordinate system of a freefalling observer. In
contrast, the "singularity" at the event horizon vanishes
completely when you change coordinate systems. So there is
a big difference there.
--
Daryl McCullough
Ithaca, NY
.

User: "I.Vecchi"
Title: Re: Any coordinate system in GR?	01 Sep 2006 02:28:16 AM

Daryl McCullough ha scritto:

I.Vecchi says...

Daryl McCullough wrote:

I.Vecchi says...

The question is whether the requirement of geodesic completeness (which
in this setting is obtained by shifting trouble to infinity) is
appriopriate, i.e. physically relevant.

Yes, but the problem is to define what is the event horizont for him.
What does it mean PHYSICALLY that he has crossed the event horizon?

Using Schwarzchild coordinates, you can compute the proper time for
a freefalling observer as a function of his radius r. The exact
computation depends on the initial conditions, but for one particular
geodesic, the relationship is simple:

r(tau) = 3/2 (2m)^{1/3} (tau_0 - tau))^{2/3}

where tau_0 is a constant depending on the initial conditions.

For this geodesic, the observer passes the event horizon
at a proper time tau_1 given by

tau_1 = tau_0 - (2/3)^{3/2} 2m

and reaches r=0 at proper time

tau = tau_0

So what it means physically for the observer to pass the
event horizon is just that his proper time is greater than
tau_1. As I said, from the point of view of the freefalling
observer, there is no physical reason for his proper time
to stop before reaching tau_1 (when he crosses the event
horizon) and tau_0 (when he reaches r=0).

Yes, that shows that Agata's coordinate system is incomplete. There
are physically meaningful events that are given no coordinate whatsoever.

The same thing happens in flat spacetime with accelerated observers.

If you are on board a rocket ship with constant proper acceleration g,
then times and distances as measured by clocks and rulers on the ship
will be related to times and distances as measured by inertial clocks
that are at rest relative to the ship's initial reference frame as
follows:

x = X cosh(gT)
t = X sinh(gT)

where (x,t) are the coordinates as measured by the inertial observers,
and (X,T) are the coordinates as measured by the accelerated observers.
Note that the point (x=1 light year, t=1 year) is given no coordinates
*at* *all* in the accelerated coordinate system: there is no values of
X and T such that X cosh(gT) = 1 and X sinh(gT) = 1.
From the point of view of the accelerated
observer, the point (x=1, t=1) can be approached as T--> infinity,
but it can never be reached. That just means that the accelerated
coordinates are incomplete.

Yes, very nice.

The same thing is true of the Schwarzchild coordinates. They are
incomplete in exactly the same way. There are valid points on
the manifold that are given no coordinates whatsoever in the
Schwarzchild coordinate system.

Let's consider Bruno perspective then. You are right that nothing
prevents him to reach the horizon, the problem is the the horizon's
position in his perspective becomes arbitrary.

If his proper time at
the horizon had a physically determined value

It does. Agata can calculate it.

Yes, and at 1 second past ten minutes, assuming he hasn't
hit the singularity, he'll know that he has crossed the
event horizon.

The calculation can be done using KS, or using Schwarzchild
coordinates. It's not hard using Schwarzchild coordinates.
From the metric and the geodesic equation, you can prove
that there are two constants of motion for radial geodesics:

(1-2m/r) V^t = P
(1-2m/r) (V^t)^2 - 1/(1-2m/r) (V^r)^2 = E

where V^t = dt/dtau and V^r = dr/dtau. Using
the first equation in the second, we find

P/(1-2m/r) - (V^r)^2/(1-2m/r) = E

So

V^r = - square-root(P - E(1-2m/r))
dr/dtau = - square-root(P - E(1-2m/r))

(where I chose the minus sign because the observer is falling, so
dr/dtau is negative). This gives tau as a function of r:

dtau/dr = - 1/square-root(P - E(1-2m/r))
tau = tau_0 - integral of 1/square-root(P - E(1-2m/r)) dr

The integral is a perfectly ordinary integral, with no singularities
except at r=0. So the value of tau at which r=2m is computable.

Yes, but this assumes that the parameter r corresponds to to the the
radius of the black hole, right? When you write that tau_0 "depends on
the initial conditons" you are making some assumptions on the time it
will take to reach the singularity at r=0. This sounds circular, but
you are welcome to expand on your argument and/or to provide a
reference. I don't think that an infalling observer can measure the
inner radius, since in order to do that he would have to reach the
singularity at r=0, where the metric blows up for good. More in
general, apparently your argument assumes that to an infalling observer
the inner domain appears as a sphere of radius r, as it appears from
the outside to external observers, who however have no access to
space-time events beyond the horizon. Is there any reason to assume
that it must be a sphere of radius r for infalling observers too?
Consider the following alternative scenario. Take the complete Kruskal
solution and paste together the horizons of the white and of the black
hole. This is still a well-behaved extension. The infalling
observers, raising his eyes after checking on his clock that he's
passed the horizon of the black hole, would just find himself gushing
out of a white hole horizon. After browsing around I think this is
called an Einstein-Rosen bridge. I surmise this may be the same as a
blackhole were the singularity at r=0 is blown up to infinity
(together with tau_0) through a coordinate change for the interior
domain 2m<r<0. Repeating the above question, how do we know that this
is not the "right" chart (i.e, the chart corresponding to space-time
measurements of an infalling observer)?
Cheers,
IV
NB In this setting different charts do not correspond to different
"physical" situations, but to different space-time measurement models.
.

User: "Daryl McCullough"
Title: Re: Any coordinate system in GR?	01 Sep 2006 11:57:46 AM

I.Vecchi says...

No, not really. Let me switch from tau to s, because s is
easier to write. The geodesic equation implies the following
constraints on the motion
1. (1-2m/r) (dt/ds)^2 - 1/(1-2m/r) (dr/ds)^2 = 1
2. (1-2m/r) (dt/ds) = some constant k
Using 2, we can simplify 1 to get
k^2/(1-2m/r) - (dr/ds)^2/(1-2m/r) = 1
or
1'. (dr/ds)^2 + 2m/r + (1-k^2) = 0
To compute k, you need to know the initial conditions.
In particular, let's suppose that the observer is initially
at rest according to the Schwarzchild coordinates. That
means that when s=0, we have: dr/ds = 0.
In that case, letting r0 be the value of r when s=0
Then we have, at proper time s=0:
(1-2m/r0) (dt/ds)^2 = 1
from equation 1 and
(1-2m/r0) (dt/ds) = k
from equation 2. These two together imply
k = square-root(1-2m/r0)
So our equation 1' above becomes
(dr/ds)^2 + 2m/r + 2m/r0 = 0
This equation gives us r as a function of s and r0.

I don't think that an infalling observer can measure the
inner radius, since in order to do that he would have to reach the
singularity at r=0, where the metric blows up for good.

Yes, you're right that r is not directly measurable. However,
using the above equation, you can *compute* r from your proper
time s.

More in general, apparently your argument assumes that to an
infalling observer the inner domain appears as a sphere of radius r,

No, I didn't assume that. I just assumed that r and t are two
coordinate such that the metric expressed in terms of r and t
is
ds^2 = -(1-2m/r) dt^2 + 1/(1-2m/r) dr^2 + angular part

Consider the following alternative scenario. Take the complete Kruskal
solution and paste together the horizons of the white and of the black
hole.

I'm not sure what this means. Look at the Kruskal diagram as
shown in
http://io.uwinnipeg.ca/~vincent/4500.6-001/Cosmology/Black_Holes_files/image011.gif
In that picture, the Kruskal "time" coordinate is v, whose axis is shown
oriented vertically. The Kruskal "space" coordinate is u, whose axis is
shown oriented horizontally.
In terms of compass directions: The black hole singularity is to the
north, the white hole singularity is to the south, the black hole
exterior is to the east, and the mirror exterior is to the west.
The black hole event horizon is the V-shaped boundary of the northern
quadrant. It runs diagonally from the northwest corner to the center
(v=0, u=0), and then turns a right angle and runs diagonally to the
northeast corner.
The white hole event horizon is the inverted V-shaped boundary of
the southern quadrant. It runs from diagonally from the southwest
corner to the center, and then diagonally from the center to the
southeast corner.
So which portions of the horizon are you proposing to paste together?

This is still a well-behaved extension. The infalling
observers, raising his eyes after checking on his clock that he's
passed the horizon of the black hole, would just find himself gushing
out of a white hole horizon. After browsing around I think this is
called an Einstein-Rosen bridge.

As I understand it, the bridge connects the exterior
in the east to the mirror exterior in the west. I don't
see how it is accurate to describe someone crossing the
bridge as "gushing out of the white hole horizon".
--
Daryl McCullough
Ithaca, NY
.

User: "I.Vecchi"
Title: Re: Any coordinate system in GR?	05 Sep 2006 06:57:29 AM

Daryl McCullough ha scritto:

I.Vecchi says...

Objection 1.
Both equations may hold for r!=2m, otherwise they are meaningless as t
becomes infinite. I also do not understand what may be the meaning of
dt/ds and hence of the second equation inside the horizon, where t
becomes space-like.

Using 2, we can simplify 1 to get

k^2/(1-2m/r) - (dr/ds)^2/(1-2m/r) = 1

or

1'. (dr/ds)^2 + 2m/r + (1-k^2) = 0

Again, only if r>2m. There is no reason to assume that this holds on
both sides of the horizon. The matching at the horizon appears
arbitrary and it constitutes an "ad hoc" assumption. I think that it
entails assuming that the radius seen from inside the blackhole matches
the radius as seen from the outside.

To compute k, you need to know the initial conditions.
In particular, let's suppose that the observer is initially
at rest according to the Schwarzchild coordinates. That
means that when s=0, we have: dr/ds = 0.

objection 2.
This means that observer has yet to switch his rockets off, right?
I see another issue here. When he switches the rockets off there is an
abrupt change in dr/ds. How does it get encoded in the equation, if at
all? The problem here is that you must establish a map between the time
of a stationary (hence accelerated) observer and the time of an
inertial observer in free-fall. In order to establish such a
correspondence you must track proper time as it diverges from the time
of a stationary observer. I'd rather plot dr/ds as a Heaviside
function.

In that case, letting r0 be the value of r when s=0
Then we have, at proper time s=0:

(1-2m/r0) (dt/ds)^2 = 1

from equation 1 and

(1-2m/r0) (dt/ds) = k

from equation 2. These two together imply

k = square-root(1-2m/r0)

So our equation 1' above becomes

(dr/ds)^2 + 2m/r + 2m/r0 = 0

This equation gives us r as a function of s and r0.

I don't think that an infalling observer can measure the
inner radius, since in order to do that he would have to reach the
singularity at r=0, where the metric blows up for good.

Yes, you're right that r is not directly measurable. However,
using the above equation, you can *compute* r from your proper
time s.

More in general, apparently your argument assumes that to an
infalling observer the inner domain appears as a sphere of radius r,

No, I didn't assume that. I just assumed that r and t are two
coordinate such that the metric expressed in terms of r and t
is

ds^2 = -(1-2m/r) dt^2 + 1/(1-2m/r) dr^2 + angular part

Consider the following alternative scenario. Take the complete Kruskal
solution and paste together the horizons of the white and of the black
hole.

I'm not sure what this means. Look at the Kruskal diagram as
shown in
http://io.uwinnipeg.ca/~vincent/4500.6-001/Cosmology/Black_Holes_files/image011.gif

In that picture, the Kruskal "time" coordinate is v, whose axis is shown
oriented vertically. The Kruskal "space" coordinate is u, whose axis is
shown oriented horizontally.

In terms of compass directions: The black hole singularity is to the
north, the white hole singularity is to the south, the black hole
exterior is to the east, and the mirror exterior is to the west.
The black hole event horizon is the V-shaped boundary of the northern
quadrant. It runs diagonally from the northwest corner to the center
(v=0, u=0), and then turns a right angle and runs diagonally to the
northeast corner.

The white hole event horizon is the inverted V-shaped boundary of
the southern quadrant. It runs from diagonally from the southwest
corner to the center, and then diagonally from the center to the
southeast corner.

So which portions of the horizon are you proposing to paste together?

I am folding and pasting the line r=2m symmetrically with respect to
the origin, i.e, along the red circle in
http://casa.colorado.edu/~ajsh/wormbig_gif.html . See below.
..

As I understand it, the bridge connects the exterior
in the east to the mirror exterior in the west.

Yes. Einstein-Rosen originally suggested casting the metric into the
coordinate rho^2=r-2m and considering only real values of rho, with
rho>0 corresponding to the black hole exterior and rho<0 corresponding
to the white hole exterior. They are connected at rho=0 where the
determinant of the metric vanishes, i.e. on the white and black
horizons that are thus identified. I read that the Einstein-Rosen
construct is considered erroneous/arbitrary, but, as I explained
already, it's not clear to me yet that similar objections can't be
raised for the Schwarzschild interior solution.

I don't
see how it is accurate to describe someone crossing the
bridge as "gushing out of the white hole horizon".

Well, you cross r=2m and you are on the other side.
IV
.

User: "I.Vecchi"
Title: Re: Any coordinate system in GR?	05 Sep 2006 08:28:38 AM

I.Vecchi ha scritto:

Daryl McCullough ha scritto:

I.Vecchi says...

objection 2.

This means that observer has yet to switch his rockets off, right?
I see another issue here. When he switches the rockets off there is an
abrupt change in dr/ds. How does it get encoded in the equation, if at
all? The problem here is that you must establish a map between the time
of a stationary (hence accelerated) observer and the time of an
inertial observer in free-fall. In order to establish such a
correspondence you must track proper time as it diverges from the time
of a stationary observer. I'd rather plot dr/ds as a Heaviside
function.

Forget objection 2. It's dr^2/ds^2 that changes abruptly.
IV
.

User: "Daryl McCullough"
Title: Re: Any coordinate system in GR?	05 Sep 2006 09:55:00 AM

I.Vecchi says...

Daryl McCullough ha scritto:

Objection 1.
Both equations may hold for r!=2m, otherwise they are meaningless as t
becomes infinite.

Right, they are valid in the region r > 2m.
They are *also* valid in the region r < 2m.

I also do not understand what may be the meaning of dt/ds and
hence of the second equation inside the horizon, where t
becomes space-like.

The meaning of dt/ds is the rate of change of the coordinate
t as a function of proper time s. The fact that t is spacelike
makes dt/ds *less* mysterious, it would seem to me. dt/ds is
just proper velocity along the t-axis. It's no more mysterious
than dx/ds or dy/ds in Minkowsky spacetime.

Using 2, we can simplify 1 to get

k^2/(1-2m/r) - (dr/ds)^2/(1-2m/r) = 1

or

1'. (dr/ds)^2 + 2m/r + (1-k^2) = 0

Again, only if r>2m. There is no reason to assume that this holds on
both sides of the horizon.

There is nothing in the derivation that makes any assumption
about the sign of (1-2m/r), so the derivation is just as valid
when r < 2m as when r > 2m.

The matching at the horizon appears arbitrary and it constitutes
an "ad hoc" assumption.

There is nothing arbitrary about it. It's the unique answer,
as far as I see.
To put together a Lorentzian spacetime manifold, you need
to do the following:
You have to split spacetime up into a collection of overlapping
"patches" such that each patch can be described using a coordinate
system for which the metric components are nonsingular. The patches
have to describe an *open* subset of points (this insures that
overlapping patches overlap in a 4-D *region*). If two patches
overlap, then there must be a 2-way mapping function
between the two coordinate systems that is valid (differentiable,
nonsingular) in the region of overlap.
So following these guidelines, we can pick one patch to cover
the region infinity > r > 2m,
infinity > t > -infinity.
The Schwarzchild coordinates are perfectly adequate for describing
this region. But they don't cover the points on r=2m. So you have
to come up with a second patch that covers this region. But since
patches have to be *open* sets, you need a patch covering a little
bit on both sides of r=2m.
One possibility is to use a Kruskal patch. You have coordinates
R, T with the metric
32m^3/u exp(-u/2m) (-dT^2 + dR^2)
where u is an implicit function of R and T given by
T^2 - R^2 = (1-u/2m) exp(u/2m)
Using the mapping
r_schwarzchild = u(R,T)
t_schwarzchild = 4m arctan(T/R)
we find that the metrics agree in the region of their overlap,
and that the Kruskal patch covers the region r=2m, as well.
You claim that choosing a Kruskal patch is arbitrary. Then
show me a different choice: Show me another patch that extends
Schwarzchild spacetime past r=2m which (1) agrees with the
Schwarzchild patch in the region r > 2m, and (2) *disagrees*
with the Kruskal patch in the region r < 2m. I don't think
that there is such a patch. So the Kruskal extension is *unique*,
not arbitrary.

I think that it entails assuming that the radius seen from
inside the blackhole matches the radius as seen from the outside.

No, it doesn't make any assumptions about what the infalling
observer *sees*.

objection 2.

This means that observer has yet to switch his rockets off, right?

It's a geodesic. The particle is in freefall.
All motion is without the help of rockets. I'm
assuming an initial condition in which the test
particle has dr/ds = 0.

So which portions of the horizon are you proposing to paste together?

I am folding and pasting the line r=2m symmetrically with respect to
the origin, i.e, along the red circle in
http://casa.colorado.edu/~ajsh/wormbig_gif.html . See below.

I don't understand what that diagram means. However, if you identify
the two regions labelled r=2m, then you are identifying the black
hole with the white hole. That doesn't affect the event horizon.
--
Daryl McCullough
Ithaca, NY
.

User: "I.Vecchi"
Title: Re: Any coordinate system in GR?	07 Sep 2006 01:13:18 AM

Daryl McCullough wrote:
....

You have to split spacetime up into a collection of overlapping
"patches" such that each patch can be described using a coordinate
system for which the metric components are nonsingular. The patches
have to describe an *open* subset of points (this insures that
overlapping patches overlap in a 4-D *region*). If two patches
overlap, then there must be a 2-way mapping function
between the two coordinate systems that is valid (differentiable,
nonsingular) in the region of overlap.

So following these guidelines, we can pick one patch to cover
the region infinity > r > 2m,
infinity > t > -infinity.

The Schwarzchild coordinates are perfectly adequate for describing
this region. But they don't cover the points on r=2m. So you have
to come up with a second patch that covers this region. But since
patches have to be *open* sets, you need a patch covering a little
bit on both sides of r=2m.

One possibility is to use a Kruskal patch. You have coordinates
R, T with the metric

32m^3/u exp(-u/2m) (-dT^2 + dR^2)

where u is an implicit function of R and T given by

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

Using the mapping

r_schwarzchild = u(R,T)
t_schwarzchild = 4m arctan(T/R)

we find that the metrics agree in the region of their overlap,
and that the Kruskal patch covers the region r=2m, as well.

You claim that choosing a Kruskal patch is arbitrary. Then
show me a different choice: Show me another patch that extends
Schwarzchild spacetime past r=2m which (1) agrees with the
Schwarzchild patch in the region r > 2m, and (2) *disagrees*
with the Kruskal patch in the region r < 2m. I don't think
that there is such a patch. So the Kruskal extension is *unique*,
not arbitrary.

You did not use the KS chart to show that an infalling observer reaches
the singularity at r=0 in finite proper time. Your argument, as well as
any relevant argument I have spotted so far, relies on the interior
Schwarzschild chart. I have a problem with that.
Both the interior and the exterior Schwarzschild chart correspond to
stationary observers (i.e. they map measurement outcomes of space-time
events performed by observers that are stationary).
For the exterior chart this poses no problem since stationary observers
are well defined. However, in the interior domain, according to the
current theory, there are no stationary observers. This implies that
the interior Scwharzschild chart and the interior soultion defined on
it are meaningless, since they refer to measurements that cannot be
performed . The t and r coordinate and the metric for r<2m do not
correspond to any physically measurable/meaningful quantity. They are
just blots on a sheet of paper.
I see two mutually exclusive possibilities here.
a) It is true that there are no stationary observers in the interior
domain. In this case the interior Schwarzschild solution is unphysical
and hence meaningless. It should be simply discarded.
b) There are stationary obsevers in the interior domain. The arguments
purporting to show that there are no stationary observers in the domain
of the interior chart are flawed ( e.g. they relie on conservation of
quantities that become unobservable/unphysical across the horizon).
The Schwarzschild interior solution is valid but it has been grossly
misinterpreted.
Cheers and thanks for your patient feedback,
IV
.

User: "Daryl McCullough"
Title: Re: Any coordinate system in GR?	08 Sep 2006 07:54:17 AM

I.Vecchi says...

You did not use the KS chart to show that an infalling observer reaches
the singularity at r=0 in finite proper time.

Well, the argument goes through the same for Kruskal coordinates.
The metric (or line element, as Tom Roberts insists) for KS coordinates
is:
ds^2 = = f(r) (dT^2 - dR^2) + angular part
where f(r) = 32m^3/r exp(-r/2m) and where r is
an implicit function of R and T defined by
R^2 - T^2 = (r/2m - 1) exp(r/2m)
The geodesic equations of motion for radial motion
(dtheta = dphi = 0) are
2 d/ds(f(r) (dT/ds)) = df/dr @r/@T ((dT/ds)^2 - (dR/ds)^2)
2 d/ds(f(r) (dR/ds)) = df/dr @r/@R ((dT/ds)^2 - (dR/ds)^2)
with the constraint
f(r) ((dT/ds)^2 - (dR/ds)^2) = 1
These equations can be solved for T and R as a function of
proper time s, and you can compute the proper time at which
r(T,S) = 0.

Both the interior and the exterior Schwarzschild chart correspond to
stationary observers (i.e. they map measurement outcomes of space-time
events performed by observers that are stationary).

To be more precise, according to relativity, there is no
absolute notion of "stationary". What's special about the
Schwarzchild chart is that for any observer following
a worldline dr/ds = 0, the metric is unchanging. Of course,
inside the event horizon (r < 2m), it is impossible to have
such a worldline (it represents faster-than-light travel
there).

For the exterior chart this poses no problem since stationary observers
are well defined. However, in the interior domain, according to the
current theory, there are no stationary observers. This implies that
the interior Scwharzschild chart and the interior soultion defined on
it are meaningless, since they refer to measurements that cannot be
performed.

How does it follow that they are meaningless? Coordinates are useful
in *analyzing* a physical situation. They rarely correspond to anything
that is locally observable. For example, in good old Euclidean space,
we set up coordinates x,y,z. But for an alien observer that is billions
of miles from Earth, and who has never heard of Earth, can he measure
his value of x? No, he can't. He can set up his own local coordinates,
and if the time ever comes when that alien meets up with an Earthling,
the two can figure out mappings between their coordinate systems.
The same is true of Schwarzchild coordinates. There is no way an
observer inside the event horizon can directly measure the
Schwarzchild coordinates r and t. However, he can certainly set
up his own local coordinate system, and *we* can relate that
coordinate system to the coordinate system of an external observer.

The t and r coordinate and the metric for r<2m do not
correspond to any physically measurable/meaningful quantity.
They are just blots on a sheet of paper.

Yes, that's what coordinates are. They are just labels for
spacetime points. They are nothing physically meaningful.
If I call a distant star "Polaris", that has no physical
meaning---it is just a label. Similarly, if I identify a
spacetime event with coordinates (r,t,theta,phi), that's
just a label---it has no physical significance other than
for the purposes of saying which event you are talking
about.

a) It is true that there are no stationary observers in the interior
domain. In this case the interior Schwarzschild solution is unphysical
and hence meaningless. It should be simply discarded.

How does that make any sense? The only meaningful, coordinate-independent
way to say that an observer is "stationary" is to say that he is following
a worldline such that the metric is constant along that worldline. In
the *real* universe, there *are* no stationary observers by that
criterion. So does that mean that *all* coordinate systems in the
real universe are unphysical and meaningless?
Alternatively, you can let your notion of "stationary" be
coordinate-dependent, and just say that an observer is "stationary"
in coordinate system x^u if x^0 is timelike, and x^1, x^2, and x^3
are all constant for that observer. In that case, we can perfectly
well have "stationary" observers in the interior. Just let
x^0 = -r, x^1 = t, x^2 = theta, x^3 = phi. Inside the event
horizon, it is possible to have a timelike worldline with
t, theta, and phi all constant.

b) There are stationary obsevers in the interior domain. The arguments
purporting to show that there are no stationary observers in the domain
of the interior chart are flawed ( e.g. they relie on conservation of
quantities that become unobservable/unphysical across the horizon).

In the Kruskal coordinates, nothing traumatic happens at the horizon,
so there is no reason for conserved quantities to stop being conserved
as the observer crosses the horizon.

The Schwarzschild interior solution is valid but it has been grossly
misinterpreted.

That's probably true, but I think that it has been studied so extensively
that most of the misconceptions about it have been uncovered.
--
Daryl McCullough
Ithaca, NY
.

User: "I.Vecchi"
Title: Re: Any coordinate system in GR?	10 Sep 2006 07:49:52 AM

Daryl McCullough wrote:

I.Vecchi says...

....

How does it follow that they are meaningless? Coordinates are useful
in *analyzing* a physical situation. They rarely correspond to anything
that is locally observable. For example, in good old Euclidean space,
we set up coordinates x,y,z. But for an alien observer that is billions
of miles from Earth, and who has never heard of Earth, can he measure
his value of x? No, he can't. He can set up his own local coordinates,
and if the time ever comes when that alien meets up with an Earthling,
the two can figure out mappings between their coordinate systems.

The same is true of Schwarzchild coordinates. There is no way an
observer inside the event horizon can directly measure the
Schwarzchild coordinates r and t. However, he can certainly set
up his own local coordinate system, and *we* can relate that
coordinate system to the coordinate system of an external observer.

The t and r coordinate and the metric for r<2m do not
correspond to any physically measurable/meaningful quantity.
They are just blots on a sheet of paper.

Yes, that's what coordinates are. They are just labels for
spacetime points.

Space-time events (or points as you call them) correspond to
measurement outcomes relative to an observer. Coordinates encode
measurement outcomes relative to an observer's reference frame.

They are nothing physically meaningful.
If I call a distant star "Polaris", that has no physical
meaning---it is just a label. Similarly, if I identify a
spacetime event with coordinates (r,t,theta,phi), that's
just a label---it has no physical significance other than
for the purposes of saying which event you are talking
about.

Any physical event (including space-time events) must be measured by
an observer. A physical event corresponds to a measurement outcome.
The question here is what measurement outcome corresponds to r inside
the horizon. In order words, for r<2m, what does it mean to say that A
is at r=r_0? What measurement outcome corresponds to such a statement?

a) It is true that there are no stationary observers in the interior
domain. In this case the interior Schwarzschild solution is unphysical
and hence meaningless. It should be simply discarded.

See my parallel reply to Steve Carlip.
Cheers,
IV
.

User: "Daryl McCullough"
Title: Re: Any coordinate system in GR?	10 Sep 2006 08:16:43 AM

I.Vecchi says...

See my parallel reply to Steve Carlip.

I looked over your reply again, and I certainly didn't see anything
addressing this point. Here are your comments:
1. A stationary (hence accelerated) observer (say, on a space-ship) can
determine her r coordinate by measuring her proper acceleration a,
weighing objects or herself with a dynamometer. Normalising everything
in sight a= -m/(r^2*sqrt(1-2m/r)) for r>2m.
2. The problem arises because there is no such thing as an interior
observer at fixed r. There are no stationary observers in the interior
domain, hence the above measurement of r is impossible.
3. I am saying that "proper measurements" of r in the interior domain are
physically impossible.
4. As I wrote, since no observer can hover "at constant r" in the interior
domain, r is not a measurable quantity.
None of those address the question of what it means for a coordinate
to be measurable in a realistic universe in which there are no "stationary"
observers.
--
Daryl McCullough
Ithaca, NY
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User: "Daryl McCullough"
Title: Re: Any coordinate system in GR?	10 Sep 2006 08:13:18 AM

I.Vecchi says...

Yes, that's what coordinates are. They are just labels for
spacetime points.

Space-time events (or points as you call them) correspond to
measurement outcomes relative to an observer.

No, they really don't.

Any physical event (including space-time events) must be measured by
an observer.

No, that's not true. As I said, coordinates are just *labels*.
They don't need to correspond to anything measurable.
Figuring out what is going on in the universe is a difficult
inverse problem. We take the observations, and we try to work
backwards to what caused those observations. The model of the
universe that we use is *not* determined by our observations,
but is only *constrained* by observations. Observations can
prove a model to be wrong, but observations cannot prove a model
right.
Coordinates are part of our *model*. Observations can show that
our model is wrong, but the model is not uniquely determined by
the observations.

See my parallel reply to Steve Carlip.

I didn't see you address this point.
--
Daryl McCullough
Ithaca, NY
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User: "I.Vecchi"
Title: Re: Any coordinate system in GR?	12 Sep 2006 12:20:41 AM

Daryl McCullough wrote:

I.Vecchi says...

....

Space-time events (or points as you call them) correspond to
measurement outcomes relative to an observer.

No, they really don't.

Any physical event (including space-time events) must be measured by
an observer.

No, that's not true. As I said, coordinates are just *labels*.
They don't need to correspond to anything measurable.

Labels for what? What's written on them? Who writes it?
According to Wigner ([1]), "in relativity theory, the state is
described by a metric which consists of a network of points in
space-time, that is, a network of events, and the distances between
these events. If we wish to translate these general statements into
something concrete, we must decide what events are, and how we measure
the distance between events" ([1]).
This is basic. If you cannot define your space-time events in terms of
measurement outcomes you are talking about nothing and/or building
circular arguments such as yours.
Specifically, when you write the Schwarzschild metric as
ds^2 = -(1-2m/r) dt^2 + 1/(1-2m/r) dr^2 + ...
you must be able to specify which measurement procedure determines (or
labels, as you prefer) r and t, i.e. who is measuring them and how.
Otherwise, as I said, you are talking about nothing. In the exterior
domain, local procedures to determine r and t can be implemented.
However, as long as a measurement protocol for r and t in the interior
domain cannot be defined, the so-called interior solution is physically
meaningless.
Cheers,
IV
[1] E. Wigner "Relativistic Invariance and Quantum Phenomena" Rev. Mod.
Phys. 29, 255 (1957)
PS I add a further quote from [1], which I find quite relevant here,
albeit indirectly:
"... the measurement of position, that is, of the space coordinates, is
certainly not a significant measurement if the postulates of of the
general theory are adopted: the coordinates can be given any value one
wants. ... Most of us have struggled with the problem of how, under
these premises, the general theory of GR can make meaningful statements
and predictions at all. ... This is a point that which cannot be
emphasised strongly enough and it is the basis of a much deeper dilemma
.... . It pervades the general theory, and to some degrees we mislead
both our students and ourselves when we calculate , for instance, the
mercury perihelon without explaining how our coordinate system is fixed
in space, what defines it in such and such a way that it cannot be
rotated, by a few seconds a year, to follow the perihelion apparent
motion. ... . There must be some assumption on the nature of the
coordinate system that keeps it from following the perihelion. ... . A
difference in the tacit assumptions which fix the coordinate system is
increasingly recognized to be at the bottom of the many conflicting
results arrived at in calculations based on the general theory of
relativity."
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User: "Daryl McCullough"
Title: Re: Any coordinate system in GR?	12 Sep 2006 06:31:28 AM

I.Vecchi says...

Daryl McCullough wrote:

No, that's not true. As I said, coordinates are just *labels*.
They don't need to correspond to anything measurable.

Labels for what?

Coordinates are labels for points on the manifold.
The manifold is part of the *model* of what is going on.
It's a mathematical object, and we use coordinates to describe
points on that mathematical object.

According to Wigner ([1]), "in relativity theory, the state is
described by a metric which consists of a network of points in
space-time, that is, a network of events, and the distances between
these events. If we wish to translate these general statements into
something concrete, we must decide what events are, and how we measure
the distance between events" ([1]).

This is basic.

Yes, I think that you are missing the basics here. The measurements
he is talking about are all *local* measurements. Those are the only
things that are physically significant. Global coordinate systems
are not (in general) measurable.

If you cannot define your space-time events in terms of
measurement outcomes you are talking about nothing and/or building
circular arguments such as yours.

What are you saying is circular? We're *starting* with a particular
solution of Einstein's field equations, and then we're exploring
the features of that solution. What's circular about it?

Specifically, when you write the Schwarzschild metric as

ds^2 = -(1-2m/r) dt^2 + 1/(1-2m/r) dr^2 + ...

you must be able to specify which measurement procedure determines (or
labels, as you prefer) r and t, i.e. who is measuring them and how.

No, I don't. It's a mathematical model. Mathematical models are
not reducible to a finite set of observations. What's important
is that you can use the model to *describe* observations. If
you want to know what an observer sees at various points on the
manifold, the metric above allows you to figure that out. But
there is no requirement that observations allow you to figure
out what manifold you are on, or that observations allow you
to figure out what your coordinates are.
You are coming at this from a particular philosophy of science
which I certainly don't agree with, and I'm not interested in.
To me, science is about building models that *predict* what
observers see, not building models *from* what observers see.
The latter problem is impossible---observations never uniquely
pin down the model.

Otherwise, as I said, you are talking about nothing.

I don't agre with your philosophy of science here, but it
doesn't matter. We're talking here about a particular *model*,
namely the Schwarzchild solution to Einstein's field equations.
This model was developed by physicists using mathematics. It was
not developed from any actual observations---nobody we know has
actually fallen into a black hole, and the universe is *not* described
by the Schwarzchild metric.

In the exterior domain, local procedures to determine
r and t can be implemented.

You are confused. The only way that r and t are measurable
in the exterior is by the *same* methods as can be used
in the interior.

However, as long as a measurement protocol for r and t in the interior
domain cannot be defined, the so-called interior solution is physically
meaningless.

There is no difference between the interior and the exterior
as far as the measurability of t and r is concerned.

[1] E. Wigner "Relativistic Invariance and Quantum Phenomena" Rev. Mod.
Phys. 29, 255 (1957)

PS I add a further quote from [1], which I find quite relevant here,
albeit indirectly:
"... the measurement of position, that is, of the space coordinates, is
certainly not a significant measurement if the postulates of of the
general theory are adopted: the coordinates can be given any value one
wants. ... Most of us have struggled with the problem of how, under
these premises, the general theory of GR can make meaningful statements
and predictions at all. ...

He's talking here about philosophical issues that I don't think have
been resolved. But they are issues about GR in *general*, not about
the Schwarzchild interior versus Schwarzchild exterior solutions.
--
Daryl McCullough
Ithaca, NY
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User: "Tom Roberts"
Title: Re: Any coordinate system in GR?	12 Sep 2006 09:23:55 AM