[ExI] Dark Energy and Causal Cells
johnkclark at gmail.com
Sat Jan 20 17:03:29 UTC 2018
On Fri, Jan 19, 2018 at 10:21 PM, Stuart LaForge <avant at sollegro.com> wrote:
> Ok, let me a rephrase that again: A causal cell is a finite volume of
> space-time enclosed by an event
> horizon wherein all inertial observers should agree on the temporal
> ordering of time-like or light-like separated events, thus sharing an arrow
> of time. In other words, if there was any possible way A could have caused
> B, then we would all agree that A happened before B.
If I say A caused me to do B nobody within my Hubble volume would disagree
and say my doing B caused A to happen, although somebody outside my Hubble
>> If your "causal cell" is the Hubble volume then it is case #2 that I
>> mentioned above, it is the volume of the universe we can see now. If they
>> mean the same thing I think it would be wise to use the more standard term
>> rather than one you made up, it would help avoid confusion.
> >Firstly, "now" is a word fraught with peril in General Relativity. Whose
> now? My now? Your now?
If we’re talking about my Hubble volume then the answer to your question is
my now, if you’re standing a foot from me your Hubble volume will be
slightly different than mine, very very very slightly.
>> If L=A*D then L can not be a constant because, due to the expansion and
>> acceleration of the universe, the area A of the Hubble volume is shrinking,
>> and so is the density of the Hubble volume.
> > I know that equation looks ridiculously simple but it is derived from
> the Schwartzschild metric and as counter intuitive as it sounds, a
> white-hole's density should *increase* as it loses matter and energy and
> its event horizon shrinks.
Then we can’t be living in a white hole because there is no experimental
evidence the density of the universe is increasing and plenty of evidence
it is decreasing.
>>> Yes, causal cells are not invariant. What is invariant is the product
>>> of the density and event horizon surface area of a casual cell.
>> I don't see how that could be if both are shrinking.
> That's just it. It is not possible for both of them to be shrinking.
> The Friedmann Equations are wrong because they are based on
> the Robertson-Walker metric.
Forget the mathematics, in physics as in all science empirical evidence
always outranks theory, if what a theory predicts and the facts about the
universe are inconsistent the universe doesn’t care because facts remain
facts. Experiment outranks theory and the facts say both the density and
the area of our Hubble volume are decreasing, so the product of the two
can’t be constant.
> The R-W metric takes it as an assumption the universe is isotropic and
> homogeneous. While observations clearly indicate it is neither.
Obviously on a small scale that is true, but if you look at boxes a few
hundred million light years on a side it is pretty homogeneous, and without
simplifying assumptions none of the fundamental laws of physics would have
ever been discovered because things are just too complicated.
> the so-called-universe cannot possibly be homogeneous because there is an
> actual bona-fide event horizon 14 billion light years away from us.
You’re using the same word for both the surface area of a Black Hole and of
a Hubble volume, and I think that’s a bad idea. If I’m anywhere inside a
Black Hole I will come into contact with things at the event horizon when
we both reach the singularity, but I will never come into contact with
things on the surface of my Hubble volume.
>> I think if you're going to talk about reversing the arrow of time you
>> can't just stick with General Relativity, you're going to have to get into
>> Quantum Mechanics because CPT Symmetry says a observer couldn't tell if
>> time reversed direction,...well..., that is to say an observer couldn’t
>> tell if positive and negative electrical charge was also reversed and
>> things were viewed in a mirror.
>If I want to bring electrical charges into it, I have to switch from
> Schwarzschild metric to Reissner–Nordström metric. Maybe later, after I
> have the fundamentals down.
If you want to preserve CPT symmetry I think you’ll need to switch to the
Kerr–Newman metric because although Reissner–Nordström understands
electrical charge it ignores spin, and a key element of spin is parity. And
parity is the P in CPT.
John K Clark
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