
Black holes do not exist where space and time do not exist, says new theory - lelf
http://phys.org/news/2015-01-black-holes-space-theory.html
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nilkn
I've never heard of "gravity's rainbow". It seems that it's what happens when
one derives general relativity using "doubly special relativity" instead of
normal special relativity. I hadn't heard of "doubly special relativity"
either.

[http://en.wikipedia.org/wiki/Doubly_special_relativity](http://en.wikipedia.org/wiki/Doubly_special_relativity)

It seems to have been around for at least ten years but perhaps never gained
much traction. In short, this seems like it may be an interesting result that
only applies in a fringe variation of accepted modern theory. Perhaps someone
with actual training in physics can comment in more detail.

~~~
tjradcliffe
This isn't a "fringe" variation in any but the sociological sense. I hadn't
heard of either of these ideas until today, but my first thought was, "Yeah,
that makes sense" (I'm a physicist, but not a gravity person.)

If you believe in the Planck scale (and almost everyone does) then
incorporating it into SR as an "observer independent maximum energy" that is
comparable to the usual "observer independent maximum velocity" is a very
natural thing to do.

The theory might not have got much play, but this result will definitely
increase interest because the black hole information paradox is a pain.
However: the amount of interest a theory gets does not reflect the likelihood
that it is correct. Caloric was once a very popular idea. Currently string
theory is. Tomorrow maybe "extra special relativity" will be.

~~~
otakucode
>If you believe in the Planck scale (and almost everyone does)

What? The idea that spacetime is quantized is most definitely not widely
accepted. Spacetime is presumed to be continuous by basically every mainstream
physicist (though most will admit that we don't have conclusive evidence
either way, we only know that spacetime is not quantized down to a certain
scale (which is short of the planck length by a good deal)). I personally
think spacetime is quantized, but the only physicists pursuing the
consequences of such an idea tend to be the fringe ones...

~~~
tjradcliffe
I should have said "take seriously" rather than "believe in". Most people take
the notion seriously, for some value of "seriously". It's not a binary state.

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PaulHoule
As a physics PhD I feel like an idiot for accepting the theory of the
classical black hole.

If you write the equations for the space around a black hole in the
straightforward way you see a singularity at the event horizon. You can make a
coordinate change that makes this singularity go away, but then you get the
classical black hole singularity where there is either an "end of time" or an
infinitely dense point or ring.

Physicists believed that this coordinate transformation was valid but it is
not because it infinitely stretches space/time which at some point will blow
up the Planck scale to be visible, at which point something completely
indescribable happens -- it might be something like DSR or it might be
different, but you void any warranty on the space time continuation, so the
classical black hole is science fiction.

The obvious semiclasicalization is for the infinitely dense singularity to
become a highly dense object which has some dimensions on the length of the
Planck scale, but no model of quantum black hole using actual quantum gravity
looks like that at all.

We don't have a "complete" theory of quantum gravity but we do have a number
of competing partial theories, and we can already make some strong
conclusions, such as holography, that are generic to many or any QG models.
Underlying this are principles such as unitarity (unitarity is pretty much the
one thing you need to make QM work) that means no information is lost when
stuff falls in a black hole and that can't be reconciled with the mass being
concentrated all in one place.

~~~
fargolime
As _not_ a physics PhD it seems so much easier to realize that GR's
equivalence principle fails at the event horizon, in which case black holes
are a mistake of GR. (Let downvotes commence since I must be a crackpot.)

Let a particle be above the horizon and escaping to infinity, as GR allows. By
definition of a black hole, a signal can't be sent from below the horizon to
the escaping particle. In an inertial frame, signals can be sent between any
two points in the frame. Then an inertial frame relative to which the escaping
particle is at rest can't extend below the horizon. That's a violation of the
equivalence principle, because (we'll call this law of physics K:) inertial
frames can wholly contain other inertial frames, and the inner frames can be
extended to fill all of the outer frame. But as we've proven, a frame falling
through the horizon of a black hole violates law K. An inertial frame relative
to which the escaping particle is at rest _cannot_ be extended to fill all of
a frame falling through the horizon. Texts on GR tell us that an inertial
frame can fall through the horizon, but we've just proven that such frame
cannot be inertial without violating the equivalence principle. That principle
is the core of GR, so black holes need be rejected as a mistake of the theory.

Black holes are predicted by the Schwarzschild metric. That metric can be
tweaked to not predict black holes, while still agreeing with all physical
experiments of Schwarzschild geometry to date. In the tweaked metric the
escape velocity is always less than the speed of light (including at r < 2M),
so that escape is always possible in principle. Singularities vanish (with
escape always possible, no body need implode to one), so no quantum gravity is
needed to make GR compatible with QM, and much better agreement to Occam's
razor. Also solves the black hole information loss paradox.

Please downvote this major advance of physics if you haven't already.

~~~
antognini
I don't completely understand the argument you're making, so please correct me
if I'm misunderstanding you, but I think I see the problem.

It is true that in an inertial frame you can send a signal between two points.
But if you have an inertial frame which is falling into a black hole any
signal sent from beneath the event horizon will not reach any other point
until that point has _also_ fallen beneath the event horizon.

Remember that the inertial frame is in free fall, so points which are fixed in
the inertial frame are moving according to a distant observer.

~~~
fargolime
> But if you have an inertial frame which is falling into a black hole any
> signal sent from beneath the event horizon will not reach any other point
> until that point has also fallen beneath the event horizon.

Problem is, the escaping particle never falls beneath the horizon. So an
inertial frame in which that particle is at rest cannot extend below the
horizon _at all_. That's a violation of law K for a frame falling into a black
hole. By violating law K, the latter frame cannot be inertial. Note that law K
is about allowing inner frames be extended to fill all of an outer frame.

Let a cloud of particles straddle the horizon. Let all the particles above the
horizon be escaping to infinity. According to GR, such cloud must be splitting
apart. The particles below the horizon must move inexorably inward, toward the
singularity at the center of the black hole, whilst the particles above the
horizon move ever outward, away from the black hole. Then a frame falling
through the horizon of a black hole cannot be inertial, for in an inertial
frame the cloud needn't be splitting apart (since law K applies). For example,
in an inertial frame in Earth's atmosphere you can have a cloud of particles
in which half the particles are given to be escaping to infinity, and the
cloud needn't be splitting apart (just let all the particles escape in
formation).

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RickHull
Where is it that space does not exist? The headline seems like a contradiction
in terms.

~~~
Florin_Andrei
Your question, actually, is self-contradictory. "Where" already assumes space.

~~~
0942v8653
I believe that was the point. The article title makes the same mistake.

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filearts

       "In gravity's rainbow, space does not exist below a certain minimum length, and time does not exist below a certain minimum time interval"
    

Does this imply that neither space nor time can go below the given threshold
independently? I.e.: could a situation be imagined where time remains above
the minimal interval, but space is below? Also, would those answers be
different depending on the observer's time / position?

~~~
phn
I had never head of anything like this. Does that mean that space and time are
somehow discrete instead of continuous?

~~~
tjradcliffe
They have a scale. They are not "discrete" in the sense that pixels are
discrete because there is no grid (no orientation, no preferred direction,
etc).

Think of it like the resolution of a microscope: you can only see objects down
to scales of 1 micron, say. That doesn't make what you are looking at discrete
at the 1 micron level. It makes it blurry (invisible, or in the case of space
and time, non-existent) below that level.

~~~
PaulHoule
How it shows up depends on the model. Superstrings are objects of (usually)
Planckian size, so if the particles are the size of the "grain" then that is
one way a scale could manifest.

In other models like loops quantum gravity you can only ask questions about
loops in space and not about points so that is another way the scale
manigests.

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givan
Having so little evidence and information about the phenomena
[http://en.wikipedia.org/wiki/Black_hole#Observational_eviden...](http://en.wikipedia.org/wiki/Black_hole#Observational_evidence)
the entire theory around black holes is a mathematical model that probably has
nothing to do with the real phenomena.

~~~
tjradcliffe
Absence of information in Wikipedia isn't very good evidence that there isn't
very much evidence.

Black holes can be observed in multiple ways:

1) radiation from infalling matter 2) gravitational lensing 3) dynamics of
orbiting objects

Remember: we not see the moon by its own light. We see the moon by the light
it reflects from the sun. That reflected light is a direct causal consequence
of the moon's existence, and we quite properly take it as such... even though
the moon emits no light of its own.

No one, to the best of my knowledge, has ever suggested we ought to treat the
evidence for the Moon as scanty or inadequate on this basis. So the fact that
the radiation we see from black holes originates from the heat of infalling
matter, which is a direct causal consequence of the black hole's existence,
should constitute similarly strong evidence.

Lensing and dynamical measurements are also quite direct, strong and
compelling. There exist compact bodies with more than 1.4 solar masses, which
is the limit for nuclear matter (we would have to be wrong about quite a lot
of nuclear physics for this to be otherwise.)

The question is: what is the best theory to describe black holes, given we
know based on this deep and broad evidence that they exist? Currently we have
a rough combination of GR and quantum theory, and while the theory has issues,
the claim it "probably has nothing to do with the real phenomenon" is
minimally plausible.

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mrfusion
Wouldn't something like the uncertaintity principle mean information is
destroyed all the time?

I don't see how you can find an earlier state from knowing the current state
since you can't know position and momentum, right?

~~~
Retra
The uncertainty principle doesn't imply that any information is being
destroyed, it implies that there is a limit to the amount of information you
have. Position and momentum are things that you derive from the available
information. You just can't use the same limited pool of information to derive
the both in a way that would use more information than you have.

>I don't see how you can find an earlier state from knowing the current state
since you can't know position and momentum, right?

State has little to do with position and momentum directly, but the underlying
information. We use those concepts to indirectly refer to state, but state is
more fundamental. Information arises from a correlation between states, and
position and momentum are different measures of this correlation.

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memracom
Could it be that when space and time get stretched beyond the Planck interval,
then the black hole ceases to exist at all beyond that point? In other words,
the center of a black hole is non-existence itself.

Looking at it from the time angle, we all blink in and out of existence
constantly. But when time and space gets stretched too far, it becomes
impossible to blink back into existence, as if the process runs out of gas if
the out-of-existence time period is too long.

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mrfusion
Would this imply hawking radiation doesn't exist? Can we no longer use black
holes as a potential energy source?

~~~
tjradcliffe
On a quick read before I've entirely finished my morning coffee, Hawking
radiation will still exist.

The argument seems to go like this:

1) the event horizon in GR is _infinitely sharp_ 2) gravity's rainbow implies
nothing can be infinitely sharp (because that requires space and time to exist
on scales below the Planck scale) 3) therefore gravity's rainbow implies the
event horizon does not exist _at those scales_

Above those scales, something that looks very much like the event horizon
still exists. The time taken to fall into a black hole will still be very
long, just not infinite. Presumably the details of the theory show how this
happens, which is the interesting bit. It's one thing to say "Hey, maybe the
horizon is smooth or doesn't exist at small scales!" but to be useful the
theory has to show how reality will actually behave if that is the case.

With regard to using black holes as power sources: if we have the technology
to do that we'd be unlikely to need black holes as power sources.

~~~
mrfusion
I didn't see what about your explanation says why hawking radiation would
still exist? It's caused by particle-anti particle pairs being created with
one on each side of the event horizon, no?

~~~
tjradcliffe
When I first posted that response I was thinking that the scale that mattered
was the compton wavelength of the particles being emitted, but having actually
finished my coffee now I'm no longer so sure. My point was that so long as the
horizon is sufficiently fine on the scale of the particle's compton wavelength
then hawking radiation can still be emitted, even granted the horizon is not
infinitely fine. This may be correct, or not.

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mattbgates
This article looks like one big paradox that shouldn't actually exist.

