Are Uranium Halos the best evidence of (a) an old earth AND (b) constant physics?

What say peaceharris?Are we in agreement that this is correct so far (Message 57):

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Where the brown square cuve is the parent decay curve
(N/N_o) = (1/2)^{^(t/H)}

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The red triangle curve is the daughter production curve
N_α/N_o = 1 - (1/2)^{^(t/H)}

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And the blue diamond curve is the ring production curve
N_α/N_R = [N_o/N_R]•[1 - (1/2)^{^(t/H)}]

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Where N_R = (N_o - N_A) = the number of decay events needed to form a ring, and this is the same for all rings.

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It doesn't really matter how many decay events are needed to cause a ring, we just use the observation that at time A we have a ring, and we use that to set the amount of decay needed for subsequent isotopes.Thus we need only concern ourselves with the blue diamond curve and how that decays.Because we don't know the actual time scale or the actual starting quantity we have to deal with decay of the daughter isotope in relation to the time scale of the parent decay.As noted in Message 53, this is done by the ratio of the half-livesH1 = half life of the parent isotope
H2 = half life of the daughter isotopeThe decay curve for the daughter isotope would be:
N/N_o = (1/2)^{^(t/H2)}And to convert this to measure t along the same parent time scale is a simple matter:
N/N_o = (1/2)^{^(t/H2)(H1/H1)}
or
N/N_o = (1/2)^{^(t/H1)(H1/H2)}
with a^{^b•c} == [a^{^b}]^{^c} (or [a^{^c}]^{^b}), then
N/N_o = [(1/2)^{^(t/H1)}]^(H1/H2)For the curves above and a hypothetical (H1/H2) = 5 we get the following:Where the brown square curve is the parent decay curve and the blue diamond curve is the ring production curve, as before, and the red triangle curve is the daughter decay curve on the parent decay curve time scale.The next step is to model how the blue diamond curve decays by the red triangle curve.Of course you can calculate it by your link site formula where λ == ln(2)/Hand e^{^-λt} == (1/2)^{^t/H}so we can write it as:
N₂ = [(ln(2)/H1)/(ln(2)/H2-ln(2)/H1)]•N_1o•[(1/2)^{^t/H1} - (1/2)^{^t/H2}]
and neglect the N_2o (assume = 0)Multiplying the top and bottom of the first bracket fraction by H1•H2/ln(2), and dividing both sides by N_1o we get:N₂/N_1o = [H2/(H1-H2)]•[(1/2)^{^t/H1} - (1/2)^{^t/H2}]But we need to know how N₂ (net remaining daughter at t=A) relates to N_R (total daughter produced from t=0 to t=A to make the first ring) ...
... where N_R == (N_1o - N_A) = N_α238U (at 4.27 MeV) and N_R - N₂ = N_α234U (at 4.86 MeV).and N_A = N_1o•(1/2)^{^A/H1} so N_R = N_1o•(1-(1/2)^{^A/H1}) or N_1o = N_R/(1-(1/2)^{^A/H1})and at t=A we would have:N₂/N_R = [H2/(H1-H2)]•[(1/2)^{^A/H1} - (1/2)^{^A/H2}]/(1-(1/2)^{^A/H1})And with A = 490,000 years I get N₂/N_R = 54.16%, which means ~45.84% decayed into 230Th. This is what the previous graph (see end of Message 53 note: edited to simplify) looks like with the values for 234U and A = 490,000 years:Where the red diamonds are the total daughter isotope (U234) production, the green squares are the net daughter isotope production at each time increment and the orange circles are the decay curve for the daughter isotope starting at 5%A, as before, and the blue circles are the remaining daughter isotope after decay, summed at each interval, and the grey diamonds are the total daughter isotope decay = total production of the next generation (230Th, the next bucket). In this (actual) instance the result is ~45.88% of the amount of decay needed to form a ring of the same density as the first ring ... except that this isn't the whole story: we still have the remaining isotopes to go, and the decay of 230Th and 226Ra have very similar decay energies and combine with the 234U decay to make the second ring.Cumbersome? Perhaps, but with the advantage of being able to use the same process for the next generation/s. Perhaps you'd like to check my numbers?Enjoy.Edited by RAZD, : == means defined as equal, added some for clarityEdited by RAZD, : added moreEdited by RAZD, : added endEdited by RAZD, : clarity

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filling in the evidence From Message 188

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Key word "visible" -- what you have are sample after sample after sample taken by cleaving the rocks along convenient fissures in the rocks. Mica is lifted with normal everyday cellophane tape, demonstrating (a) weak bonds, and (b) ready fissure planes. Once you have removed the layer you have removed the evidence of the fissure.

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Look at the pictures by Gentry where you have a number of halos all on the same wafer thin sample that he has removed from mica by the tape method: why are they all on the same plane and not distributed up and down from it?

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Do you suppose they all formed along a single fissure plane in a easy to split crystal for some arcane purpose or because it was a fissure that allowed the radon gas to penetrate the crystal.

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Seems pretty clear to me.Enjoy.

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Well, peaceharris,Having validated my approach by getting virtually the same value for the decay of 234U to 230Th, we can now proceed to discuss the rest of the ring formation.Up to now we have discussed damage density as if it were constant for each ring, however in fact it should vary inversely with the radius of the ring. If we had single rings with the radii discussed previously (see polonium thread) with the same number of decay events, then (neglecting the duplications) we would expect to see:ring 1 = 238U decay = 4.27Mev = 14.1μm = 100.00%
ring 2 = 234U decay = 4.86MeV = 17.0μm = 82.94%
ring 3 = 222Rn decay = 5.59MeV = 20.4μm = 69.12%
ring 4 = 218Po decay = 6.12MeV = 23.5μm = 60.00%
ring 5 = 214Po decay = 7.88MeV = 34.6μm = 40.75%And to compensate for this decline with radius, to have the same density per unit area of ring we would need to see:ring 1 = 238U decay = 100.00%
ring 2 = 234U decay = 120.57%
ring 3 = 222Rn decay = 144.68%
ring 4 = 218Po decay = 166.67%
ring 5 = 214Po decay = 245.39%While there is no physical way to increase decay in daughter isotopes to more than the original parent decay (unless it is pre-existing), rings 2 and 3 are formed by multiple isotopes with similar decay energies.Ring 2 is formed from decay of 234U, 230Th and 226Ra, so if we set the time "A" such that these total 120% of 238U decay then we should have the same density ring visible.This occurs at ~550,000 years where I have:234U ~50%230Th ~35%226Ra ~35%And because all remaining half-lives are less than 226Ra they will all be at ~35%Ring 3 is also composed of decay from multiple isotopes, 222Rn and 210Po, so the decay produced would be ~2x35% = 70% where we need 144.68% for a full density ring, so we should see 70/145 = ~48% of the density seen in the two inner rings.Ring 4 should be 35/166.67 = ~21% of the density of the two inner rings, and Ring 5 should be 35/245.39 = ~14% of the density seen in the inner two rings.This would give us an "embryonic" halo with ring density per unit ring area (compensated for radii) of:ring 1 = 238U decay = 100%
ring 2 = 234U/230Th/226Ra decay = 100%
ring 3 = 222Rn/210Po decay = 48%
ring 4 = 218Po decay = 21%
ring 5 = 214Po decay = 14% It seems to me that the amount of ring damage that we see in these outer areas is substantially, significantly, under the amount calculated based on the appearance of the two inner rings, and thus the only logical conclusion is that 222Rn, as a gas, left the sight of the inclusion in significant proportion and quantity.Lower density outer rings would have to show lower density for the second ring compared to the first for this to be an "embryonic" halo, and it seems to me that the second ring is slightly denser than the first.What we see is a partially formed halo due to 222Rn departure.Enjoy.Edited by RAZD, : added for clarityEdited by RAZD, : complete

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Yep, I'm forced to agree

OK, I understand your method of explaining the embryonic halo and your math is correct.The problem with your explanation is it doesn’t follow the principle of Uniformitarianism. You are assuming that the amount of U234 in the radiocenter was 0 initially and then showing that the embryonic halo is ~500000 years. If you start with the assumption that the initial activity of U234 was approximately the same as U238, (since this is what is observed in all natural waters and magmas today) you will be forced to reduce your estimate by a factor of ~10. If you want to insist that your assumption is reasonable, you should find a source where the amount of U234 is 0, then you can argue that possibly 500000 years ago we could find a similar source having no U234.The second problem with your method is it gives a lot of freedom to interpret data the way you believe it to be. While discussing Po-halos, you were arguing that the daughter products get accumulated in the radiocenter, and while discussing U-halos you argue that Uranium daughter products leave the radiocenter. Crystal rocks have cleavage planes that are weak so any crack would probably occur along the plane. Planes having more radiocenters will be weaker since radiation weakens bonding.Any radiometric method can be modified using your method to ”explain’ results different from the theory. You are basically doing what many geologists do to explain discordance of U-Pb dating of zircon. Whenever data plots below the Concordia curve, they say there has been Pb loss from the zircon. Whenever data plots above the Concordia curve, they say there has been Pb gain to the zircon.The difference between you and them is that you are claiming that there has been radon loss/gain to justify your theory, while they are claiming there has been Pb loss/gain to justify their theory.

Say what? U234 is a daughter of U238. What do you mean by "activity?"

What it means is that for every U238 decay there is a U234 decay i.e. the isotopes are in equilibrium.Thus is seems that we have the following situation: Message 46 states that RAZD made an error in assuming that U234 and U238 would be in equilibrium. Message 65 states that the new calculation is not applicable because U234 and U238 WILL be in equilibrium.(The intermediate stages in the decay chain are so short that they won't have any significant effect).

Activity is the number of decays per second.

This is often not the case. Waters are known in which the U-234 is highly enriched. The decays leading to U-234 can leave it in a quite different chemical state from the parent U-238, through what are called hot-atom effects. Further, the intermediate nuclides have quite different chemistries. For example, thorium is often insoluble where uranium is soluble. If you are actually thinking of U-235, which is primordial, your comments are correct.

There are indeed many sources enriched in U234, but let’s be generous to RAZD. Even if he wanted to assume that the initial activity ratio of U234: U238 was as low as 0.8, we should allow him (see Florida Geological Survey | Florida Department of Environmental Protection )The isotope percentage of U238 is 99.27%
The isotope percentage of U235 is 0.72%
Half life of U238=4.463e9 years
Half life of U235=7.04e8 yearsSo activity of U238 is 99.27*7.04e8 / 0.72/ 4.463e9 (=21.7) times greater than that of U235This explains why people usually match only rings of daughter products of u238 while determining the origin of each halo.

Thanks, peaceharris, Message 62
Good. I don't like using that word because it is so misused that most people misunderstand what is being said. What I have done is assume that the basic physics that we know today were in operation during the formation of the rings, as (a) there is no evidence of any change in physics, and (b) the evidence of the rings says that the decay rates remained constant during the formation of the rings (decay rate being tied directly to alpha particle energy). Correct. That is the starting point for this exercise. We already looked at what happens if we assume the inclusion to be in equilibrium at the time of the rock formation: in Message 46 your said:

quote:

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This is a reply to Message 200, since I am replying regarding U-halos.

RAZD writes:

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Can you tell me what I have wrong here?

Your calculation of isotopes that have decayed only applies to systems in equilibrium.

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So now we look at the other extreme condition, with no daughter isotopes, to see what it would look like, and in this case we see that there should still be sufficient decay production of daughter isotopes to be visible once we have enough decay of 234U + 230Th + 226Ra to form a second ring at least as dense as the first ring (as shown in the picture). The problem is yours, for you are boxed in by both sides now. As PaulK noted, we started this with the assumption of equilibrium, and the equations showed that the rings would not be as seen in the picture. Then we went to the other extreme with no daughter isotopes, and again the equations showed that the rings would not be as seen in the picture, that we should see more ring formation in the third ring area that is seen.There is no intermediate stage that would result in removal of daughter isotope ring production, unless one of the isotopes is bodily removed from the vicinity, ergo the only logical assumption is that 222Rn (gas) has left the site of the inclusion in the picture. And then the second ring would not be of the same visible density as the first ring, which is the condition of the halo in the picture. You need to explain the second ring to explain the picture.Enjoy.

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getting back to this comment, peaceharris, These are provided by Gentry: These are all well past the 500,000 year stage where the third ring would be visible Old enough to form the rings, which can be a younger date than the rock, if the rock is fissured and allowed the entry of 238U particles into the structure after the rock was formed.Enjoy.

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Thanks peaceharris, Can you show how this affects the calculations? If there is "freedom to interpret data" then you should be able to show a different calculation with a different result. The basic argument is that 222Rn is a gas, and as a gas it behaves like a gas: it leaves the original inclusion just as any gas would, and it behaves like a gas in attenuating to equalize partial pressures throughout it's available volume. This would mean over time that there would be a higher percentage of probability that the gas would decay at a location where there would be slightly more net molecules -- ie small voids that, while small, are many time larger than the fine fissures that permeate these kinds of crystals (those cleavage planes are there because there is not a perfect fit from one crystal layer to another). Do you have any evidence of the radiation damage to chemical bonds? or is this just an ad hoc concept to ignore the fact that these fissures are characteristic of this kind of rock, and they were there before the halos? I know U/Pb calculations are a bete noir with you, however I don't think you can compare these. In this case we have:

• Uranium halos with insufficient decay damage after 226Ra in the decay chain ...
• the very next isotope, 222Rn, is an inert gas, with nothing to bond it to the original inclusion ...
• evidence of lots of "free" decay events (not tied to any inclusion particle or specific location) along fissures throughout these rocks ...
• a rock crystal lattice that can chemically absorb certain atoms into it's structure, including daughter isotopes below 222Rn ...
• places where later generation isotopes are observed and that produce halos (222Rn, 218Po, 214Po and 210Po), and ...
• halos for 222Rn, 218Po, 214Po and 210Po in the same relative proportion as their net equilibrium levels ...
• higher than normal end product of this one decay chain, 206Pb, levels in all those later generation halos than would be normal for a natural inclusion (but normal for an isotope deposition process),
• levels of the end product of this one decay chain, 206Pb, much lower in the 238U inclusions (indicating a different process formed the different inclusion particles) ...
• halos only in rocks that show evidence of secondary formation processes that occur at lower temperatures and that cause opening in the crystal lattice if not wholesale replacement of atoms throughout the crystal.

The uranium halos can be considerably younger than the base rock, as we do not know when the secondary formation occurred, or how many times such processes occurred.And we are still stuck with a long time for the formation of uranium halos, especially the more complete ones that Gentry shows. There is no need to invoke special decay rates, no evidence for any change in the basic behavior of physics, and evidence in the rings of consistent decay energies for each of the alpha decay stages in the 238U decay chain.As such I think we are done with the 238U decay process, and any further discussion should be redirected back to the polonium thread at: Message 208Enjoy.

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You first started assuming that all isotopes of the U238 decay chain were in equilibrium initially . If all isotopes of the U238 decay chain in the radiocenter start off being in equilibrium, then all rings would be visible. The very definition of ”embryonic Uranium halos’ is that the halos due to the decay of Uranium can be seen, but the halos due to the decay of other products of the U238 decay chain cannot be seen.Then you used the assumption that there is only U238 without U234, and explained that it could be 500000 years. No scientist would take your assumption seriously that 500000 years ago you could find a source with U238, but without U234. Dude, you don’t understand anything! I don’t need to explain it, all I need to do is assume that others who read my posts understand that if U234 is in equilibrium with U238 when the rock is formed, the U234 ring will form simultaneously with the U238 ring. There are people who are able to understand this without me explaining it. No, Radon trapped in a solid doesn’t behave like a gas. It is known that helium and Argon stay inside the rock that has cooled down.People have experimented and found out that by heating a rock you can cause these inert gases to escape from the rock. The Radon atoms is bigger than both Helium and Argon, so it follows that it is easier for He and Argon to escape from a rock compared to Radon.Furthermore Radon -222, the inert gas being discussed has a half life of a few days compared to Argon and Helium that are stable. You are trying to convince people who know that Argon cannot inspite of being a stable atom that Radon-222 can escape within a few days! Good luck to you.Furthermore, even expert evolutionists have understood inert gases stay embedded in a rock that has cooled down. Joe Meert wrote, “The temperatures at which minerals close is easily verifiable through experimentation and this has been conducted numerous times including the famous experiments of Bowen” - One of the main objections to radiometric dating Creationists also have done experiments checking whether helium can escape from rocks and they also know that most helium is retained in the rock at low temperatures. Try doing a google search on metamictization. There are scientists, (other than Gentry) who have concluded that Uranium radiocenters formed during crystallization. - refer Geochemical Journal Vol 10, pg 188. Are there scientists who share your view that fissures and Uranium radiocenters go together? On the contrary U/Pb is one of my favorite radiometric methods. I have understood that U-Pb data can provide an upper bound for the age of the sample and I have used U/Pb data to show others that coal beds are few orders of magnitude younger than what evolutionists say.U/Pb calculations based on reasonable assumptions are a bete noir with you. Understanding embryonic U halos is a prerequisite to understand the more complete U halos, and I have no intentions to explain the more complete U- halos to someone who either doesn’t understand the embryonic ones or is in a state of denial. We are in agreement here. But you need to invoke the unrealistic assumption that there was a source having U238, but without U234, 500000 years ago. I agree with you, there’s nothing I can do to convince you that embryonic Uranium halos are young. I am wasting my time reading your posts and replying to you.

Thanks again peaceharris for your input. The problem is that this can also be due to the loss of 222Rn from the site of the inclusion, so the existence of two rings does not necessarily mean that the halo is embryonic. Given that you have check my math and my method and have not pointed out any errors in them, this seems to be more denial than anything else. Can you show how 234U can form the second ring without any decay of other daughter isotopes? Have you calculated what the resulting daughter decay would be? And yet you seem to be unable to explain it to me, and just repeat an assertion rather than offer any new evidence\argument.Would not starting with any initial 234U mean you are starting from a position between the two cases I have delineated? Both of these cases show that there should be more outer rings seen that in the picture. Again the rocks in question have undergone secondary kinematic processes that have cracked the rocks. Comparing these to rocks that have not undergone the same process is faulty at best. Curiously I do not find where this process creates weak fracture planes rather than general and directionally nonspecific damage. This is where we started.There are scientists that have shown complete secondary replacement processes occurring in rocks like the ones with halos, resulting in less dense rock structure and plenty of opportunity for both the influx of the 238U inclusions and for the transport of the much smaller 222Rn atoms. See the Lithos (Collins et al) article on the polonium thread. Strangely this just seems to be another dodge rather than discussing the issue. I think you are missing the point. The case of 238U without any other daughterr isotopes is taken as an extreme case to show the minimum case for developing subsequent rings. It just happens to work out to ~500,000 years as a minimum age. I'm perfectly happy to consider that the inclusions had several daughter isotopes when the inclusions were incorporated into the rock during the secondary (low temperature) reformation process. And yet, interestingly, all you have offered to date is repeated assertion of young age. You have not provided any kind of explanation at all, nor have you demonstrated that my math and method is erroneous.You are correct that just repeated assertion of a position without any substantiation will not convince me of it's validity.Enjoy.

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