Re: Misinterpretation of the radial parameter in the Schwarzschild solution?


Tom Roberts wrote:
LEJ Brouwer wrote:
(i) Is time going backwards or forwards as the particle falls towards
the singularity, and how do we know?

The proper time of an infalling timelike particle never goes backwards.
We know this because the path of the particle is parameterized by its
proper time, and an infalling geodesic starting at some point with r>2M
simply goes through the horizon and intersects the singularity at r=0,
all the while with increasing proper time; indeed the geodesic ends
after a finite proper time from its initial point, so we _know_ there is
a singularity at r=0 and a boundary of the manifold.

This is not particularly convincing. If the direction of time is
reversed in the interior, then your picture of a particular falling in
to the singularity would be incorrect - rather we would have what
appears to be a particle-antiparticle annihilation event, with the
antiparticle emerging from the singularity and annihilating with an
infalling particle at the event horizon. From the metric, I do not see
how you can tell whether dt is positive or negative. Clearly we need
t^2 = 4m^2 at the event horizon, and t^2 = 0 at the singularity, but
how do you know whether t evolves from -2m to 0 (i.e. time is going
forwards) or from +2m to 0 (i.e. time is going backwards?). Simply
stating your belief that the former is true is not sufficient.

As regards the singularity at r = 0, this means that the solution is
not geodesically complete, and is hardly satisfactory. Particles cannot
simply disappear into never-never-land. Having to reply "I don't know"
to what happens to the particle suggests strongly that you have gotten
it wrong.

As I said before, in the region r<2M the vector -d/dr is the
future-pointing timelike basis vector of the Schw. coordinates of the
interior region.

How do you know the future-pointing time-like basis vector is not
+d/dr? Both are consistent with the boundary conditions.

This is geometry, and one can assign coordinates in _ANY_ way
whatsoever, including using the symbol "r" for the time coordinate, and
arranging it so the value of that coordinate decreases as time elapses
into the future. Indeed the interior Schw. coordinates are just like
that. <shrug>

Whether times goes backwards or forwards gives two completely different
physical scenarios as to what is happening at the event horizon. You
are now admitting that you are *arbitrarily* choosing one of these and
discarding the other.

You are totally hung up on the symbols, and they are _meaningless_. Just
because the geometry does not behave the way you expect it to behave
does not mean the geometry is wrong. Indeed, thorough analysis over many
decades by many gifted people shows your naive ideas and claims are
wrong. <shrug>

Yeah, yeah, blah blah blah... the usual statement of faith. You might
as well say 'I am a sheep' and throw progress out of the window.

And besides, the geometry you describe of the interior region is
totally incompatible with the exterior region, which means that the
event horizon is no ordinary surface.

(ii) If we do not call it a radius, do you still agree that the area of
the 2-spheres is decreasing as t increases/decreases?

If one considers an infalling test particle, and looks at the areas of
successive spheres through its location at successive values of its
proper time, then the areas of those 2-spheres decrease with time, all
the way down to asymptotically zero area at the singularity at r=0.

Or do you mean '...the areas of those 2-spheres _increase_ with time',
with time starting of negative at the event horizon and increasing to
zero?

Since we know
that the 'radius' of the 2-sphere is 2m when the particle hits the
event horizon, does this not mean that the timelike coordinate (let's
call it 'r') just inside the EH must have r = +/- 2m, depending on
whether r is increasing (-2m) or decreasing (+2m)? If so, which of the
latter is correct and why?

Using the Schw. interior chart, immediately after a particle enters this
region from outside, it has a value of r infinitesimally smaller than
2M. For a timelike object the value of r is necessarily decreasing,
because -d/dr is the future-pointing timelike basis vector of this
coordinate chart.

Wrong. The value of r^2, not r is infinitesimally smaller. This does
not allow you to conclude the direction of time in the interior.

No. It's just that -d/dr is the future-pointing timelike basis vector of
these coordinates. That is, the singularity at r=0 is in the _future_ of
every interior point (every object entering this region from the
exterior enters at r=2M). This is why in a Kruskal diagram the
singularity is not a point but rather a hyperbola (which makes it a 3-d
locus); a timelike object located in the interior can no more avoid the
singularity than you can avoid tomorrow.

Or... a timelike object can no more have avoided emerging from the
singularity than you can avoid emerging from the past. Which one will
it be?

(iii) Bearing in mind the above, how do we know that the singularity is
in the future of every interior point, rather than its past?

Because for an infalling test particle, the trajectory can be
parameterized by proper time, which monotonically increases until the
particle reaches the singularity at r=0.

Evidence, please?

Just _LOOK_ at the Kruskal diagram -- every timelike object must move
upward within +-45 degrees of vertical.

The Kruskal diagram as drawn is highly misleading. The whole of r=2m is
physically at the same point. Sectors I and IV are adjacent. As I said,
to an external observer, the motion of the particle appears to change
direction in time as the particle moves smoothly from sector I to
sector IV.

The singularity is a hyperbola
at the top of the diagram, and hence is in the future of every point in
the region r<2M. Of course there are actually two singularities, and the
other one is in the past of a similar region with r<2M.

The existence of singularities means that the solution is physically
invalid and must be rejected.

(iv) If the singularity is a hyperbola, what actually happens to a
particle hitting the singularity?

The singularity is a hyperbola only in the Kruskal-Szerkes chart. The
singularity is outside the manifold in which GR applies, and GR can make
no statements about what happens there. Here there be dragons.

And we all know that dragons don't really exist. Don't we?

Does it re-emerge from the white hole
solution?

GR can make no statements about what happens. We have no other theory
that applies, either.

That is just plain stupid. Stop fooling yourself. Discard the interior
solutions and stick to what makes physical sense.

(v) If the claim that the region around the event horizon is perfectly
ordinary is true, why is the description of what happens in the
interior so different from the exterior,

Because the geometry is different. <shrug>

Well, in the locality of the event horizon the geometry should be the
same if this claim about the event horizon being an ordinary point is
true. Conclusion: the event horizon is not ordinary as usually claimed.


e.g., 't' is free and need not
be continuous across the horizon?

t cannot possibly be continuous across the horizon for an infalling test
particle, because t->infinity as r->2M for such a trajectory.

It doesn't just do that - it does a double backflip somersault and
becomes spacelike across the event horizon. So impressive as to be
physically impossible.

[As I have said before, this does NOT mean the particle never
gets to the horizon, it merely means that the Schw. exterior
coordinates are ill behaved there. The particle's proper time
advances as usual right through the horizon and down to the
singularity.]

Sure. I have no qualms about that.

Why are the spacelike and timelike
coordinates so different in character here?

Because both the geometry and the coordinates are different. <shrug>

The geometry shouldn't be different on one side of the EH from the
other.

I repeat: Don't let the similarity of the symbols confuse you -- these
are _completely_ different regions of the manifold.

I agree. So let's stop pretending that nothing special happens at the
event horizon.

If one looks in a naive way at the wormhole-like picture of the
Schwarzscild spacetime (e.g. figure 6.10 of Wald), it seems very clear
that there should be a coordinate patch smoothly connecting sectors I
and IV connecting the black hole and white hole exterior.

Of course there is! It is the very Kruskal-Szerkes coordinates used to
make Fig. 6.9. But this is a highly-curved manifold, and embedding it in
a flat space (Fig 6.10) does not show all the relationships inherent in
the geometry. For instance, Fig. 6.10 is a spacelike 3-surface at t=0,
but no timelike object can ever be at r=2M at t=0.

BTW that is not at all a "wormhole-like" diagram -- the manifold is the
2-d surface, not the "hole" one sees in the embedding 3-space.


This is
obscured by the usual Kruskal diagram (figure 6.9) which makes it
appear that sector I an IV are physically separated, so that a particle
hitting the event horizon appears as if it must head towards the
singularity, despite the complete mismatch in the interpretations of
spacelike and timelike coordinates should such a thing occur.

You are very confused, because you keep trying to think in terms of
Schw. coordinates. Just look at Wald's Fig. 6.9: any timelike object
must move upward within +-45 degrees of vertical (because the light
cones are drawn at +-45 degrees).

They got the gluing wrong. Sector I should be glued to sector IV and
sector II should be glued to sector III.

A object with r=const (>2M) follows a
hyperbola in region I. An infalling object moves generally upward,
crossing the horizon (the big X) and intersecting the singularity at the
top (r=0). No timelike path can connect Regions I and IV because that
would require it to move greater than 45 degrees from vertical, and no
timelike path can do that.

It depends which frame of reference you use. It is perfectly possible
for the infalling particle to be travelling forwards in time but an
external observer to see his trajectory gradually reverse in time - in
fact this is precisely what happens. The infalling particle emerges on
sector IV where the direction of time is reversed relative to sector I,
so the external observer and the (in)falling particle will at that
point disagree with regards to the direction of time.

Instead of one misleading Kruskal diagram, there should be two separate
diagrams - one with just sectors I and IV, and another with only
sectors II and III.

No. One manifold, one diagram.

Yes, we should throw away sectors II & III as physically irrelevant.

Only the sector I & IV diagram is physically
relevant,

Not true at all! What justification do you have for ignoring regions II
and III?? -- they are demonstrably part of the manifold. Just "because I
want to" is not sufficient for the rest of us. <shrug>

Ditto. I prefer my solutions to be physically reasonable and
singularity free.

with geodesics going directly from sector I to sector IV,

No timelike trajectory can do that. <shrug>

Yes it can. It depends whose watching.

and
no mental gymnastics required to patch basically incompatible
coordinate systems when going from sector I to sector III.

Just _LOOK_ at the Kruskal diagram -- no such patching is necessary. It
is only you, who seem personally wedded to the Schw. coordinates, that
is confused. You need to learn how to think _geometrically_ not via
coordinates. Those coordinates are ill behaved and are the source of
your confusion. <shrug>

The Kruskal diagram is a misleading reflection of the geometry. Just
_LOOK_ at the wormhole-like surface depiction of the Kruskal
coordinates. The interior and the exterior are completely different in
nature, and only the exterior solutions are physically compatible with
each other.

Tom Roberts

- Sabbir.

.

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