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Author Topic:   Are Uranium Halos the best evidence of (a) an old earth AND (b) constant physics?
RAZD
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Message 1 of 142 (478523)
08-16-2008 6:30 PM


Not about Polonium, Not about a Young Earth.
Many people in this debate are familiar with the "problem" of Polonium halos, however this thread is not about Polonium halos, and any further mention is off topic (see PRATT CF201: Polonium Halos). Anyone wanting to talk about Polonium halos is free to start their own thread, and not clutter this one up, thanks.
Where I am starting is from Dr Wiens:
Radiometric Dating
quote:
13. "Radiation halos" in rocks prove that the Earth was young.
This refers to tiny halos of crystal damage surrounding spots where radioactive elements are concentrated in certain rocks. Halos thought to be from polonium, a short-lived element produced from the decay of uranium, have been found in some rocks. ...
At any rate, halos from uranium inclusions are far more common. Because of uranium's long half-lives, these halos take at least several hundred million years to form. Because of this, most people agree that halos provide compelling evidence for a very old Earth.
(bold added for empHASis, part deleted not about uranium halos)
The stock YEcreationist response is that the decay rates changed, and thus all you are seeing is the result of fast decay rather than long time.
Aside from this being another PRATT (see CF210: Constancy of Radioactive Decay) which is also off topic and not part of this thread, it struck me that Uranium halos are evidence that this did not occur.
The basic radiohalo principle is simple: radioactivity produces alpha decay, and the alpha particle have a certain energy (usually measured in million electron volts, MeV) based on the familiar e=mc² formula and the conservation of energy/mass (see ref):
M1 = M2 + mp + e/c²
Thus when you have isotopes decaying into other isotopes by alpha decay, the energy of the alpha particle is unique to that decay stage because of the unique before and after mass of the decaying isotope and the constant mass of the alpha particle.
This unique energy then determines how far (on average) an alpha particle will travel before it gets stopped and absorbed into the surrounding material (and causes the ring pattern to be visible) and the result is a halo or a number of halos around decaying inclusions that look like rings, but are actually spherical, and something like this:
The halos require more than one particle to form as each one only makes a point on the ring. Thus uranium, with it's long half-life, takes "several hundred million years to form."
Now the fun part: this is based on our knowledge of physics and the physical constants that tell us how things behave in the universe, so what happens if you have fast decay instead of old time?
Not being a physicist, I am not familiar with the equations that link decay rate to decay energy, so I am going on memory, but I found this interesting tid-bit in Alpha Decay, Alpha detectors and identification:
quote:
However, if the alpha has enough energy to surmount this barrier then it will regain that energy as electrostatic repulsion once it gets outside the range of the attractive strong nuclear force. One important consequence of this is that all alpha emissions have at least ~5 MeV energy. Furthermore, half-life is inversely related to decay energy.
(bold for empHASis)
Very simply put, if you change the decay rate, you change the decay energy, and the diameter of the halo changes.
There should be no characteristic uranium halos with the unique energy of uranium alpha decay from fast decay.
The existence of (common) uranium halos then is evidence that shows the physical constants have not changed while they were formed, and their formation in turn is evidence that the earth is old, at least several hundred million years old.
Enjoy.
ps - I would appreciate any of the physics mavens supplying confirmation of my rambling, and possibly provide the equation/s relating decay rate to decay energy. Thanks.
ppss - anyone that can tell what goes on in the ring formation would also help.
DATES AND DATING forum please (I am thinking this is another correlation item)
Edited by RAZD, : was "an alpha particle will travel before it decays" changed decays into "stopped and absorbed"
Edited by RAZD, : added bold and subtitle
Edited by RAZD, : .
Edited by Admin, : Fix title.

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RAZD
Member (Idle past 1432 days)
Posts: 20714
From: the other end of the sidewalk
Joined: 03-14-2004


Message 3 of 142 (478533)
08-16-2008 9:03 PM
Reply to: Message 1 by RAZD
08-16-2008 6:30 PM


What is that ring stuff anyway?
I found this article abstract The nature of radiohaloes in biotite: Experimental studies and modeling:
quote:
Several micro-techniques (confocal laser-Raman microprobe, optical absorption micro-spectroscopy, high-resolution transmission electron microscopy, electron microprobe analysis) were employed in the detailed characterization of radiohaloes in biotites from two Variscan rocks from Germany. The studied biotites are intermediate members of the phlogopite-annite series with Mg/Fe2+ ratios in the range 1.6 -1.0. Radiohaloes in biotite resulted from the impact of 4He cores (α-particles) emitted from actinide-bearing inclusions. Monte Carlo simulations yielded α (238U, 235U, and 232Th series) penetration ranges in biotite between 12.5 and 37.3 μm, which are in reasonable agreement with the observed radii of radiohaloes in natural biotites. The coloration pattern of a radiohalo closely correlates with the calculated distribution pattern of point defects generated in displacive events. Calculated point defect densities in the range from < 10-5 to at most 10-2 dpa (displacements per lattice atom) suggest that there are only scattered point defects in a mainly preserved biotite lattice. This is consistent with HRTEM studies that did not reveal any indication for initial volume amorphization in the haloes. However, general Raman band broadening and intensity loss suggest that the short-range order in radiohaloes is significantly disturbed. The darkened color of radiohaloes, when compared with the un-irradiated host biotite, is caused by increased light absorption over the complete visible range due to increased point defect density. No additional color centers were found, and the absorbances of the VIFe2+, Fe2+< sup> - Fe3+, and Fe2+ - Ti4+ centers seem hardly to be changed. Both Raman and optical absorption spectra obtained from radiohaloes retain a clear orientational dependence. The results suggest that the formation of point defects rather than ionization is the main process causing the coloration of radiohaloes in natural biotites. The haloes represent an early stage of structural radiation damage, characterized by significantly disturbed short-range order but still widely preserved long-range order of the structure.
(bold for empHASis)
SO the ring would be caused by the alpha particle causing a "point defect" in the surrounding material, interrupting the normal light pattern
Edited by RAZD, : .
Edited by RAZD, : corrected symbols

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RAZD
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Posts: 20714
From: the other end of the sidewalk
Joined: 03-14-2004


Message 5 of 142 (478535)
08-16-2008 9:13 PM
Reply to: Message 1 by RAZD
08-16-2008 6:30 PM


Mechanics of alpha decay?
I also found this simple article on the forces in a nucleus and how that affects decay:
The Strong Nuclear Force, Alpha Decay and Fission
quote:
Unlike the electric forces, whose strengths are given by the simple Coulomb force law, there is no simple formula for how the strong nuclear force depends on distance. Roughly speaking, it is effective over ranges of ~1 fm, but falls off extremely quickly at larger distances (much faster than 1/r2). Since the radius of a neutron or proton is about 1 fm, that means that when a bunch of neutrons and protons are packed together to form a nucleus, the strong nuclear force is effective only between neighbors.
In a very heavy nucleus, (c), a proton that finds itself near the edge has only a few neighbors close enough to attract it significantly via the strong nuclear force, but every other proton in the nucleus exerts a repulsive electrical force on it. If the nucleus is large enough, the total electrical repulsion may be sufficient to overcome the attraction of the strong force, and the nucleus may spit out a proton. Proton emission is fairly rare, however; a more common type of radioactive decay in heavy nuclei is alpha decay, shown in (d). The imbalance of the forces is similar, but the chunk that is ejected is an alpha particle (two protons and two neutrons) rather than a single proton.
It is also possible for the nucleus to split into two pieces of roughly equal size, (e), a process known as fission.
This is just background information, not for discussion yet.
Enjoy.
Edited by RAZD, : .
Edited by RAZD, : fixed

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RAZD
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Posts: 20714
From: the other end of the sidewalk
Joined: 03-14-2004


Message 6 of 142 (478536)
08-16-2008 9:17 PM
Reply to: Message 4 by Coragyps
08-16-2008 9:06 PM


Thanks Coragyps,
I'm betting the YEC response will sound like, "Yeah, now alpha energy and half-life correlate inversely, but back during Teh Fludde......"
That's why it would be nice to have the formulas.
Surely it's chemical/mechanical damage.
And it looks to be more mechanical than chemical (ionic).
Thanks
Edited by RAZD, : .

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RAZD
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From: the other end of the sidewalk
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Message 7 of 142 (478537)
08-16-2008 9:42 PM
Reply to: Message 5 by RAZD
08-16-2008 9:13 PM


More on the Mechanics of alpha decay?
From Alpha Barrier Penetration
quote:
The energy of emitted alpha particles was a mystery to early investigators because it was evident that they did not have enough energy, according to classical physics, to escape the nucleus. Once an approximate size of the nucleus was obtained by Rutherford scattering, one could calculate the height of the Coulomb barrier at the radius of the nucleus. It was evident that this energy was several times higher than the observed alpha particle energies. There was also an incredible range of half lives for the alpha particle which could not be explained by anything in classical physics.
The resolution of this dilemma came with the realization that there was a finite probability that the alpha particle could penetrate the wall by quantum mechanical tunneling. Using tunneling, Gamow was able to calculate a dependence for the half-life as a function of alpha particle energy which was in agreement with experimental observations.
Why an alpha particle and not a proton?
No webpage found at provided URL: Alpha Binding Energy
quote:
The nuclear binding energy of the alpha particle is extremely high, 28.3 MeV. It is an exceptionally stable collection of nucleons, and those heavier nuclei which can be viewed as collections of alpha particles (carbon-12, oxygen-16, etc.) are also exceptionally stable. This contrasts with a binding energy of only 8 MeV for helium-3, which forms an intermediate step in the proton-proton fusion cycle.
I envisage it as a pyramid with each particle in contact with the other, and therefore bound by the strong force.
Edited by RAZD, : size of image
Edited by RAZD, : .

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RAZD
Member (Idle past 1432 days)
Posts: 20714
From: the other end of the sidewalk
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Message 8 of 142 (478538)
08-16-2008 9:49 PM
Reply to: Message 7 by RAZD
08-16-2008 9:42 PM


Relation found
This, I believe is the crux of the issue:
Alpha Tunneling Model
quote:
The illustration represents an attempt to model the alpha decay characteristics of polonium-212, which emits an 8.78 MeV alpha particle with a half-life of 0.3 microseconds. The Coulomb barrier faced by an alpha particle with this energy is about 26 MeV, so by classical physics it cannot escape at all. Quantum mechanical tunneling gives a small probability that the alpha can penetrate the barrier. To evaluate this probability, the alpha particle inside the nucleus is represented by a free-particle wavefunction subject to the nuclear potential. Inside the barrier, the solution to the Schrodinger equation becomes a decaying exponential. Calculating the ratio of the wavefunction outside the barrier and inside and squaring that ratio gives the probability of alpha emission.
Change the decay rate, and you change the energy of the alpha particle. Not a strict inverse relationship (exponential?)
Enjoy.
Edited by RAZD, : .

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RAZD
Member (Idle past 1432 days)
Posts: 20714
From: the other end of the sidewalk
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Message 9 of 142 (478556)
08-17-2008 1:49 PM
Reply to: Message 8 by RAZD
08-16-2008 9:49 PM


Another piece of the relationship ... but ?
Digging deeper along the coulomb wall, I found another article that goes into equations:
PHYS 490/891 - Winter 2007, 2.8 Alpha-Decay
quote:
Qα = Eb(Z, A) - Eα(Z - 2, A - 4) - Eα(2,4) .
Now the we can translate the above condition for the possibility of an α-decay to Qα > 0. Fig. 14 shows a diagram with Qα-values for β-stable nuclei. Positive values start to appear for A > 150.
While the energy range of alpha-decays observed in nature is relatively narrow (~ 2 - 12 MeV) the lifetimes span a range from 10 ns to more than 10^19 years. To better understand this behaviour we will investigate the mechanism of this decay a little closer.
...
So we find for the decay constant:
λ = w(α)vαe-G/2R

Okay, I'm having trouble getting from Qα to λ
Translation needed.
Enjoy.
Edited by RAZD, : 10^19 not 1019
Edited by RAZD, : darn smilies
Edited by RAZD, : .
Edited by RAZD, : formula restored
Edited by Zen Deist, : fixed formula again. tried to fix link but it is subject to annual revision
Edited by Zen Deist, : disable smilie

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RAZD
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Posts: 20714
From: the other end of the sidewalk
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Message 13 of 142 (478582)
08-18-2008 7:29 AM
Reply to: Message 11 by johnfolton
08-18-2008 5:15 AM


Re: Young Earth !!!!!!! - NOT THE TOPIC
Well, johnfulton et al et al et al,
Makes sense if you take genesis literally you could have 3,000 to 4,000 years of accelerated decay.
This thread is not about how you twist reality into your private mythology.
It is about Uranium halos, and why Uranium halos are (a) evidence that there has been no change to the rate of decay for hundreds of millions of years and (b) that the earth is at least several hundred million years old.
If the decay rates had changed we would not have Uranium halos, because the energy of the alpha particles would change WITH the change in decay rate, and the resulting rings would NOT be at the correct diameter, if they weren't smeared to much to see.
You don't discuss this at all, so your entire post is irrelevant, typical of your failure to deal with reality.
Enjoy.
Edited by RAZD, : .

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RAZD
Member (Idle past 1432 days)
Posts: 20714
From: the other end of the sidewalk
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Message 15 of 142 (478628)
08-18-2008 7:07 PM
Reply to: Message 14 by johnfolton
08-18-2008 11:35 AM


Re: Young Earth !!!!!!! - NOT THE TOPIC
Thanks johnfolton et al et al et al,
I will only address the portions related to the topic.
Where is your evidence that accelerated decay would change the energy of the alpha particle affecting ring diameter, etc...?
Everywhere.
But you can start here: Message 8
The decay rate is based on the probability of decay for an atom, change the probability (increase decay) and you change the decay rate of an isotope.
Change the decay rate, and you change the time a alpha particle takes to get through the Coulomb barrier (it follows the decay curve). This changes the energy absorbed in the process, which changes the energy left over. The energy left over is what sends the alpha particle out to make the ring.
Enjoy.
Edited by RAZD, : ]
Edited by RAZD, : .

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RAZD
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Posts: 20714
From: the other end of the sidewalk
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Message 25 of 142 (478698)
08-19-2008 7:31 PM
Reply to: Message 19 by peaceharris
08-19-2008 12:45 AM


Neither Young Earth OR Polonium Please - NOT THE TOPIC
Thanks peaceharris, for your help.
Please keep in mind that this thread is intended (see Message 1) to talk about uranium halos and exclude talk about polonium (feel free to start a topic if you wish).
In particular, the question is whether uranium halos:
(1) show that the energy of alpha particle decay has not changed while the halos are formed (otherwise ring would be blurred or have some different diameters).
(2) because the energy of alpha particles is a result of the decay rate of an isotope, this means that the decay rate has not changed while the halos are formed.
(3) therefore the halos are several hundred million years old.
Thanks, and
Enjoy.
Edited by RAZD, : .

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RAZD
Member (Idle past 1432 days)
Posts: 20714
From: the other end of the sidewalk
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Message 26 of 142 (478699)
08-19-2008 7:37 PM
Reply to: Message 24 by Joe Meert
08-19-2008 4:13 PM


Not Polonium Niether ... Uranium, and the age shown by Uranium halos etc ...
Hey Joe, nice to see you back and posting again. I've been reading your thread, and it is interesting stuff to this amateur geologist/naturalist.
So are decay rates constant, or not?
That is the big question here, and what I can see so far is that the decay rates are tied to the energy the alpha particle ends up with when it tunnels through the Coulomb wall\barrier, so that if one changes the other must also change.
Are there other variables? Would there be other effects (atoms falling apart)?
Thanks, and
Enjoy.
Edited by RAZD, : .

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RAZD
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From: the other end of the sidewalk
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Message 29 of 142 (478707)
08-19-2008 9:47 PM


THIS THREAD IS NOT ABOUT POLONIUM
SHEESH.

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RAZD
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Posts: 20714
From: the other end of the sidewalk
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Message 30 of 142 (478708)
08-19-2008 9:57 PM
Reply to: Message 27 by peaceharris
08-19-2008 8:31 PM


Re: Neither Young Earth NOR Polonium NOR Dating Methods
Thanks peaceharris.
So in either case, whether someone chooses to neglect U234 or whether he wants to be more accurate without neglecting U234, the conclusion is that U halos do not in any way prove an old earth.
Let's stick to the simple parts first: this is not about radiometric dating methods, but about the simple decay of uranium causing a steady stream of alpha particles that damage the surrounding rock at diameters directly proportional to their (present day) energy levels.
The consistency of these rings is not dependent on any correction or dating methodology, it is just a simple fact, bedded in the rocks.
There is no change in the alpha particle energies that form the uranium rings in all the time that it takes to form the rings.
We don't need to date the uranium halos to know that they are old: the decay rates for those rings unique to the long half-life isotopes in the uranium decay series, and the large number of impacts necessary to form a visible ring means that de facto the rings are several hundred million years old (a drop in the bucket for geological time, but too old for any YEC mythology).
It's simple evidence of an old earth, no rocket science corrections necessary.
Enjoy.
Edited by RAZD, : .

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RAZD
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Message 35 of 142 (484517)
09-29-2008 7:31 AM
Reply to: Message 34 by peaceharris
09-29-2008 4:56 AM


STILL Not about Polonium, STILL Not about a Young Earth.
Curiously you are still not on topic. Read the subtitle again, read Message 1 again.
Uranium halos. Focus.
Enjoy.

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RAZD
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Message 42 of 142 (484691)
09-30-2008 10:36 PM
Reply to: Message 37 by peaceharris
09-30-2008 12:30 AM


It;s about gas ... but the topic is uranium.
Thanks peaceharris
“Specifically, it was discovered that the halos (Fig. 1a) surrounding the -active sites are typically embryonic, that is, they do not generally exhibit the outer 214Po ring characteristic of fully developed U halos in minerals.”
He is basically saying that he can see the U halo but cannot see the 214Po ring, thus he has defined it as “embryonic”.
Curiously they may not have Radon-222 halos either -- the stage before Po214, which is also the first and only gas phase, whence the atoms could (and do) readily leave the parent inclusion.
In fact, finding Uranium Halos absent Po214 halos in the same rocks you find Po214 halos without Uranium halos kind of adds up to one logical cause: the one leads directly to the other by Radon gas diffusion.
The fact that not one of the uranium decaying atoms sufficient to form a uranium halo was able to decay to Radon-222 means either something was stopping decay OR that the Radon-222 diffused away from the parent inclusion. So far there is no known process to stop radioactive decay at any intermediate stage, so that leaves one valid conclusion: gas diffusion.
Do you agree that Uranium halos which do not have the 214Po ring are not significantly more than 245 thousand years? Please answer this question, so that I can try to explain this concept more clearly.
Not at all, they could be many times that age, as all that needs to happen is that the Radon-222 gas formed (just before the Polonium-214 stage) leaves the Uranium inclusion.
If you cannot see the 214Po ring, but can see the 238U ring, what does that mean? It means that there have been lots of 238U atoms that have decayed, but most of these decayed descendants have not yet become 214Po atoms. This implies that the sample is not significantly more than 245 thousand years.
Not at all, because Radon is a gas. The uranium decayed down to Radon-222 in one location, the radon left (being a gas), then it decayed to Po-214 in another location, which then continued the sedentary decay of non-gaseous radioactive isotopes ending with lead.
Elvis has left the building.
Or do you think Gentry is blind . ...
Yes, I think he is blind to the evidence that shows how radon can, and does, seep through the rocks he studied/s.
But this topic is not about Gentry and his perennial polonium poppycock problem, it is about uranium and uranium halos.
And curiously you still need a minimum of 245,000 years to form the uranium halo eh? You still need to explain how a young earth concept is compatible with that idea (unless you acknowledge the earth is old). You still cannot change the decay rates.
Enjoy.
Edited by RAZD, : added

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