Would proton decay allow for black holes?
Jarek Duda
positron and neutral pion, which quickly decays into two photons.
Such decays would allow standard matter to completely change into EM
waves (proton + electron -> ~4 photons).
So this decay allow to get to more stable state and temperatures in
collapsing neutron stars should make it easier - it suggests that
neutron star instead of creating a mysterious matter state (black
holes), should 'evaporate' - turn its core into photons ...
I've looked at a few papers and I haven't found any considered this
type of consequences?
Maybe they could explain extremely high energetic cosmic rays?
What do you think about proton decay?
If it would be true - would black holes be created?
Sam Wormley
Not been observed.
Jarek Duda
Would it contradict black holes creation? Be a nature's failsafe to
avoid infinite densities?
If not, does black holes have infinite density?
If they have a finite density - what state of matter is it?
Sam Wormley
What, in your opinion would proton decay have to do with black holes.
Background:
http://en.wikipedia.org/wiki/Black_hole
Jarek Duda
creating infinitely energy density state in a black hole, should decay
into nothing, emitting its energy in form of photons - while e.g.
neutron star collapse in it's center would start
neutron - > ~ 4 photons
decay, heating up the star and not allowing it to create infinite
density ...
Particles are some local energy minimals, but there is always lower
energy state than any particle - no particles.
States prefer to deexcitate to lower energy state, radiating the
difference in form of photons (the deeper local energy minimum, the
higher energy barrier and so the more difficult this deexcitation is).
The higher the average energy (temperature) the easier these
deexcitations/decays are (lower expected life time) .. and the most
extreme conditions for standard matter we consider are those in the
center of neutron star ...
Autymn D. C.
> If the baryon number doesn't have to be conserved, matter instead of
> creating infinitely energy density state in a black hole, should decay
> into nothing, emitting its energy in form of photons - while e.g.
> neutron star collapse in it's center would start
> neutron - > ~ 4 photons
> decay, heating up the star and not allowing it to create infinite
> density ...
Neutròns are more than fotònic, so that can't happen. Infinite
density is a blunder where relativity feard to tread:
http://twitter.com/alysdexia/status/4323682209.
> Particles are some local energy minimals, but there is always lower
> energy state than any particle - no particles.
> States prefer to deexcitate to lower energy state, radiating the
> difference in form of photons (the deeper local energy minimum, the
There are never no motes. Matter is always. When no other matter's
around, positronium will spring back and forth without annihilare; its
entropy hits a deck, and its dissociation constand is 1. Coinomatter
has no lower minimal unless the first WIMP is found to start
catalýsis--WIMPs are well beyond dipiòn-stars. Look for the
pentaquark-star first.
-Aut
Uncle Al
Super-Kamiokande and its 50 kilotonnes of H2O put limits on proton
decay half-life. It doesn't happen. SUSY was then rewitten to add a
couple of orders of magnitude to the number. That is also crap.
<http://www-sk.icrr.u-tokyo.ac.jp/sk/physics/pdecay-e.html>
(5x10^10 g H2O)(2)(6.022x10^23)/18.01528 = 3.34x10^33 protons
Zero candidate decays detected, 1996-2009. Proton decay half-life
must exceed 8.2x10^33 years to a big pile of sigmas.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz4.htm
Jarek Duda
decay on
the way) ... but generally in the core of neutron star there is
thermodynamically still a lot of protons and electrons ...
> When no other matter's around, positronium will spring back and forth without annihilare
When it's happening in the core of neutron star, wanting or not - this
positron would quickly meet with some electron ... but the more
important are photons created directly while proton decay - they would
carry most of proton's GeV ...
But there appeared interesting question:
If in the whole universe there would be only two particles: electron
and positron, would they finally meet?
Electromagnetically they should create periodic stable motion ...
But there are also other forces, 'dark' energy of vacuum - I
personally believe that they would finally meet.
Uncle Al - and so they've showed that this eventual decay time isn't
smaller than 10^32 years ... but what if it is let say 10^40 years?
Extreme temperature in neutron star core should make crossing this
energy barrier much easier - decrease this decay time ...
Can we be sure that it's not possible?
Jarek
john
Each proton is a volume of
space taken up by a 'black hole'.
The tiny black hole which is the proton
is identical to every other proton
in the galaxy in terms of spin. They are
all exactly opposite in terms of spin to
the large black hole at the galaxy's
center. Together the one large hole and the
myriad small holes are equal and opposite,
if they could be shoved back together, there would
be nothing. But they can't be- it's like shoving
bubbles back into champagne, or popped
popcorn back into its shell.
john
galaxy model for the atom
http://users.accesscomm.ca/john
Uncle Al
The aren't any protons in neutron star, bozo, absolutely none a mile
below its surface. You don't know what you are talking about. You
propose no experiments to justify what you spew.
Start reading relevant papars. Ignorance is educable, stupidity is
forever. Make your choice.
Jarek Duda
> The aren't any protons in neutron star, bozo, absolutely none a mile
> below its surface. You don't know what you are talking about. You
> propose no experiments to justify what you spew.
about some kind of thermodynamical beta equilibrium (proton<->neutron)
Please show me some paper in which they are sure that there are only
neutrons there, with exactly no protons?
Even if it's true - do You believe that if proton would be able to
decay, neutron would be ... stable?
These theories usually says that inner core is made of some different
state of matter, like being one huge nucleon - why can we be sure that
in such extreme temperatures, proton decay doesn't start to be
statistically important (instead?)?
Uncle Al
Neutrons within a neutron star cannot decay - they are gravitationally
bound by the overlying pressure. Do more reading.
The half-life of a free neutron is 885.7 seconds. What is the
half-life of deuterium?
Jarek Duda
I've seen greater estimations for neutron stars, but let's take 'only'
10^12K - it means each degree of freedom has at average almost 90MeV -
it's muuuch more than energy difference between proton and neutron -
in any pressure, they would be almost indistinguishable ...
Quite a lot of these degrees of freedom would exceed neutron's mass -
if there is more stable state it can decay to (like nothing) - it
should just do it ...
Could You cite some concrete paper?
Jarek Duda
http://www.astro.umd.edu/~miller/nstar.html
"If the core is composed of only "ordinary" matter (neutrons, protons,
and electrons), then when the temperature drops below about 10^9 K all
particles are degenerate and there are so many more neutrons than
protons or electrons that the URCA processes don't conserve momentum
(...) "
http://nrumiano.free.fr/Estars/neutrons.html
"Fluid core - mainly neutrons with other particles"
http://abyss.uoregon.edu/~js/ast122/lectures/lec19.html
"superconducting protons plus superfluid neutrons core"
I agree - mainly neutrons, but why You are so sure that there are
completely no protons there?
john
IMHO a neutron is a proton bound to an electron
Uncle Al
In physics' non-humble opinion, a neutron is three quarks as is a
proton. The lepton is bookkeeping. Do some reading.
Jarek Duda
by so called quark-strings ... so probably they have topological
nature and change of this topological configuration in extreme
temperature would make it decay ...
john
> john wrote:
>
> > On Dec 26, 1:38 pm, Jarek Duda <duda...@gmail.com> wrote:
> > > First three pages google shown:http://www.astro.umd.edu/~miller/nstar.html
> > > "If the core is composed of only "ordinary" matter (neutrons, protons,
> > > and electrons), then when the temperature drops below about 10^9 K all
> > > particles are degenerate and there are so many more neutrons than
> > > protons or electrons that the URCA processes don't conserve momentum
> > > (...) "http://nrumiano.free.fr/Estars/neutrons.html
> > > "Fluid core - mainly neutrons with other particles"http://abyss.uoregon.edu/~js/ast122/lectures/lec19.html
> > > "superconducting protons plus superfluid neutrons core"
>
> > > I agree - mainly neutrons, but why You are so sure that there are
> > > completely no protons there?
>
> > IMHO a neutron is a proton bound to an electron
>
> In physics' non-humble opinion, a neutron is three quarks as is a
> proton. The lepton is bookkeeping. Do some reading.
>
> --
> (Toxic URL! Unsafe for children and most mammals)http://www.mazepath.com/uncleal/qz4.htm- Hide quoted text -
>
> - Show quoted text -
First thing I read said a neutron
decays into a proton and an
electron and an emr- whatever.
john
Ace0f_5pades
Its all a fabrication, but I recall the foto with a far greater
scale...
just numbers...
The idea though, there's the greatness. and its not even moday... so
who knows. perhaps in order of magnatude "get real"
Jarek Duda
matter? :)
Another question: if baryon number is always conserved, how nonzero
total baryon number in observed universe (matter-antimatter asymmetry)
could be created?
Autymn D. C.
On Dec 26 2009, 11:17 am, Jarek Duda <duda...@gmail.com> wrote:
> Great ... but what temperatures we are talking about?
> I've seen greater estimations for neutron stars, but let's take 'only'
> 10^12K - it means each degree of freedom has at average almost 90MeV -
almost 45 MeV
> it's muuuch more than energy difference between proton and neutron -
> in any pressure, they would be almost indistinguishable ...
Wrong, they must keep orbital momentum/magnètic moment under pressure.
> Quite a lot of these degrees of freedom would exceed neutron's mass -
no
> if there is more stable state it can decay to (like nothing) - it
> should just do it ...
There is no nothing! A few days ago I learnd a new interpretation for
the Planck limit/barrier, and therefrom how two elèctròns could
gravitally fuse--would there now be a awkward predicament where
inmassive charges could bode and also fall toward a singularity as a
"Planck black hole"? Not--for black holes still break conservation of
momentum: A dielèctròn at Planck* width** is at equilibrium with two
elèctròns at infinity, so Planck width is /not/ the smallest
meaningful width; they can still fall in but only insofar as they end
up as cold as their background. Such dielèctròns [or dipositròns]
could be everywhere after a big bang had reached Planck temperature--
there would be a huge thunderclap between potential maximum at √3 R_0
(Read subscript "_0" as "nouht¿".) and Planck width at R_0 (It's not
Planck energhy, a Coulomb gauge extrapolation alone; how about
Einstein energhy***? By the way, look up Carrington event of 1859.),
and a little one (Dirac energhy spare background)--and now are in a
helium-2-like six-body set at millielectròn-volts (whereas the
neutrino is a deuterium-like three-body set). A good name for my
tetraleptòn is.. collapsium, and for neutrino-hood is.. nucleoghen.
*Newton is a better name than Planck for its gravital range.
**Standard term for scale is Planck length, but see
http://google.com/groups?q=%22Comparisons+for+the+illiterate%22. It's
for a wavespan--which they call wavelength--whereas the wriht scale is
the arm R and not the span S (or λ); multiply by Alfa.
***It's very easy to find Planck units--okay, forget Planck; from now
on I shall call them Newton. Set inverse Newton and Coulomb
potentials of two elèctròns equal:
-(-Gm^2/r) = KQ^2/r. (In my short'hand, uppercase are constands and
lowercase are variabils.) In other words, 0 = -Gm^2/r + KQ^2/r and
there's no potential.
|K| = |C^2 10^-7|. (Go SI!)
Solve for m, the lone variabil:
m = √K/G Q := 5·88)9 kg. (Read ")" as "in", a floating point mark; and
":=" as "is", or three bars. Read "=" as "likens" if you like English
over bad Latin.)
Now Newton mass M_0 := 5·88)9 kg. Newton energhy is easily MC^2 := √K/
G QC^2 := 529 MJ! := 3·30(27 eV.
Scientists held identity E := MC^2 = HC/λ to find λ alone, but E :=
MC^2 = KQ^2/r = HC/λ is more meaningful, and r/λ := Α := KQ^2/HC ~
1/137. I'll forget wavespan and wavestint unless told otherwise.
Find Newton arm:
E_0 := √K/G QC^2 = KQ^2/r;
r = R_0 := √GK Q/C^2 (~ |√G10^-7 Q/C|; read "~" as "about".) := 1·38)
36 m.
Find Newton length:
t = T_0 := R/C := √GK Q/C^3 := 4·60)45 s.
Now find maxima for huge thunderclap:
Set your elèctròns' whole potential E_W := U + V := -Gm^2/r + KQ^2/r.
But m = V/C^2 for bodies at rest, (There's a huge snag if I set m =
E_W/C^2 or U/C^2--gravita potential then becomes recursive, and r ends
up in the numerator! Maybe it's everything's self-gravity thas is why
it expands, and why clear matter's concentration grows farther out?)
so E_W := -GK^2Q^4/C^4r^3 + KQ^2/r. Slope E'_p = 0 := 3GK^2Q^4/C^4r^4
+ -KQ^2/r^2 := 3GK^2Q^4/C^4r^2 - KQ^2.
Uh-oh. E_p has a substitute mass but E_0 doesn't; Newton didn't know
about relativity anyway. But mass rests on charge so Newton units are
sound. It's time to make Einstein's dream of tien gravity and
elèctrism come true. Behold, the Einstein units!:
Einstein arm r = R_V := √3GK^2Q^4/C^4KQ^2 := √3GK Q/C^2 := √3 R_0 :=
2·39)36 m.
Einstein length t = T_V := R_V/~C := 7·97)45 s.
Einstein energhy e = E_V := -GK^2Q^4/C^4(√3GK Q/C^2)^3 + KQ^2/(√3GK Q/
C^2) := -KQC^2/3√3GK + KQC^2/√3GK := 2/3√3 √K/G QC^2 := 2/3√3 E_0 :=
204 MJ := 1·27(27 eV. (No more e for Euler constand; it's Υ now,
Ýpsilon.)
Einstein wavespan and wavestint are 137 2·39)36 m = 3·27)34 m and 137
7·97)45 s = 1·09)42 s, each.
Lastly find loca minima for little thunderclap:
Cosmic microwave background is at 2·725 K. A covalent or
semidegenerate two-body (such as N2) has three translative and two
vibrative--and two degenerate rotative--degrees of freedom. (A two-
body is of "dissociate state"--by the way, some months back I'd found
there were ten states of matter and six stages of matter, therefore 60
stage-states of matter. It's under my About.com profile at Apr 2009:
wetstuff matrix. I'd reared to post something here and sci.chem, and
a primer, but couldn't find the time!) However, intergalactic
hydrogen is at megakelvins! It's clear the far-field and near-field
are at equilibrium when they are cold and hot, each. Matter is
usually much hotter than background, by gravity, and is coldest when
there are many degrees of freedom in the [hot] convective background
sink--such as for thin clouds. So we'll need to settle on 30 K.
Hydrogen is a fine conductor of heat, whereas insulative Oort cloud is
somewhere at 4–9 K.
Castleton warmth e = E_B := 5K_B (30 K)/2 := 1)21 J := 6 meV. (B is
for Boltzmann, or background--whatever.)
Castleton heat e = E_F := E_D - E_B ~ E_D := 1·02210986 MeV - 6 meV
(Oh well.) := 1·63760034)13 J := 0 - (-GK^2Q^4/C^4r^3 + KQ^2/r) :=
GK^2Q^4/C^4r^3 - KQ^2/r. (The fat term's coefficiend is a googolth!
Grapher couldn't find the root at all. It's a poor calculator too.
On to Magic Number Machine...)
Newton's method:
f(r) := E_Fr^3 + KQ^2r^2 - GK^2Q^4/C^4 :~ 1·63760034)13r^3 + 2·3071)
28r^2 - 4·395)100 = 0;
f'(r) := 3E_Fr^2 + 2KQ^2r :~ 4·91280128)13r^2 + 4·6142)28r.
0 - f(0)/f'(0) = GK^2Q^4/C^4 = O;
O - f(O)/f'(O) = O - (E_FO^3 + KQ^2O^2 - O)/(3E_FO^2 + 2KQ^2O);
O - (E_FO^3 + KQ^2O^2 - O)/(3E_FO^2 + 2KQ^2O) - (E_F(O - (E_FO^3 +
KQ^2O^2 - O)/(3E_FO^2 + 2KQ^2O))^3 + KQ^2(O - (E_FO^3 + KQ^2O^2 - O)/
(3E_FO^2 + 2KQ^2O))^2 - (O - (E_FO^3 + KQ^2O^2 - O)/(3E_FO^2 +
2KQ^2O))/3E_F(O - (E_FO^3 + KQ^2O^2 - O)/(3E_FO^2 + 2KQ^2O))^2 + 2KQ^2
(O - (E_FO^3 + KQ^2O^2 - O)/(3E_FO^2 + 2KQ^2O)). (Argh. Backtrack.)
-4·395)100 - (1·63760034)13 4·395)100^3 + 2·3071)28 4·395)100^2 -
4·395)100)/(4·91280128)13 4·395)100^2 + 4·6142)28 4·395)100) = -4·395)
100 - (1·390)311 + 4·4564)227 - 4·395)100)/2·0279)127 = 2·1672(27;
(Shit. Smaller exponends fall out, so it's not that much work.)
2·1672(27 - (1·63760034)13 2·1672(27^3 + 2·3071)28 2·1672(27^2 - 2·1672
(27)/(4·91280128)13 2·1672(27^2 + 4·6142)28 2·1672(27) = 2·1672(27 -
(1·6669(53 + 1·0836(27 - 2·1672(27)/2·3074(42 = 2·1672(27. (Argh!)
Okay, new strataghèm: Plug in Newton arm:
1·38)36 - (1·63760034)13 1·38)36^3 + 2·3071)28 1·38)36^2 - 4·395)100)/
(4·91280128)13 1·38)36^2 + 4·91280128)13 1·38)36) = 1·38)36 - (4·30)
121 + 4·394)100 - 4·395)100)/(9·356)85 + 6·780)49) = 1·38)36 (Whew.
Okay, E_D is so lyt and R_0 so small, and E_W so steep, binding stays
near R_0. However motes can be boostt and smasht at even-smaller
sizes.)
Castleton arm r = R_U :~ R_0 := 1·38)36 m.
When two motes bind and shed mass, their mutual hulls blow up and
overlap until their [inner] size (Outter size is still at their causal
horizons.) holds up the newer inner potential. So whiles they may be
next together at big energhies (E_V±E_V+E_F), their rim will be at lyt
energhies (E_B). This is the width at which a dielèctròn and
dipositròn may meet. Now get rid of U; it's atchsom:
Castleton rim r = R_B := KQ^2/E_B := 2·3)7 m := 230 nm.
Castleton wavespan λ = Λ_B := R_B/2Α := 15·8 μm.
I thus forsee dileptòns (±±) and trileptòns (±; ±±±) as "grey matter"
if not clear matter; they readily take on the background temperature
and bear more weiht than expected of normal stuff. Think of them as
ectoplasm[a].
Collapsium however is not dielèctronium (dielèctròn dipositronide) or
dielèctrinium (dielèctròn dipositride); former would match up
perfectly and become infrared heat, after a very long lifetime, and
latter is dihydrogen-cation-like. In order for bound motes to be very
leiht, their potentials must nearly overlap, and to be very still (or
inert), their moments and momenta must overlap; this is why neutrinos
lose so much weiht over their free halves and why hydrogens don't.
Mesòns, unlik barýòns, can power up a leptòn with their valent quarks
so thas the former may shrink. Therefore, collapsium is a dineutrino,
or a dinucleoghen.
-Aut
Autymn D. C.
R -> R_0
E_p -> E_W
Autymn D. C.
Autymn D. C.
> There is no nothing! A few days ago I learnd a new interpretation for
> the Planck limit/barrier, and therefrom how two elèctròns could
> gravitally fuse--would there now be a awkward predicament where
> inmassive charges could bode and also fall toward a singularity as a
> "Planck black hole"? Not--for black holes still break conservation of
> momentum: A dielèctròn at Planck* width** is at equilibrium with two
> elèctròns at infinity, so Planck width is /not/ the smallest
> meaningful width; they can still fall in but only insofar as they end
> up as cold as their background. Such dielèctròns [or dipositròns]
And even if elèctròns could lose all of their mass, their mass is
finite, and so would their last potential and size be finite. All
black holes are a sham.
Brian Davis
> Dammit I don't proofread.
Don't worry about it. Nobody else would have caught it either. It
would have required bothering to read this stuff, and most of us have
*far* better thing to do with our time. Sleeping, for instance.
--
Brian "grouchy today" Davis
Erik Max Francis
Followed shortly by standing next to a nondescript wall and staring at
it in silence.
--
Erik Max Francis && m...@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 18 N 121 57 W && AIM/Y!M/Skype erikmaxfrancis
Yes I'm / Learning from falling / Hard lessons
-- Lamya
Mike Ash
Erik Max Francis <m...@alcyone.com> wrote:
> Brian Davis wrote:
> > On Jan 7, 11:02 pm, "Autymn D. C." <lysde...@sbcglobal.net> wrote:
> >
> >> Dammit I don't proofread.
> >
> > Don't worry about it. Nobody else would have caught it either. It
> > would have required bothering to read this stuff, and most of us have
> > *far* better thing to do with our time. Sleeping, for instance.
>
> Followed shortly by standing next to a nondescript wall and staring at
> it in silence.
I would read Autymn's posts, but I haven't finished my daily quota of
stabbing myself in the leg repeatedly with a sharpened pencil.
--
Mike Ash
Radio Free Earth
Broadcasting from our climate-controlled studios deep inside the Moon