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

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

RAZD

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Message 47 of 142 (487666)
11-03-2008 7:13 AM

Reply to: Message 46 by peaceharris
11-03-2008 4:06 AM

Re: Time taken for equilibrium

Thanks peaceharris,

This is a reply to isotope equilibriums and decay probabilities (Message 200 of Thread polonium halos in Forum Dates and Dating), since I am replying regarding U-halos.

Good idea, thanks.

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

The mistake here is that this calculation is not applicable to Uranium halos found in coal. Uranium inclusions in coal did not start as a system in equilibrium.

So we are in the process of reaching equilibrium, and the equilibrium equations are not applicable yet.

To me it is like a string of buckets with holes that leak into the next bucket, and the flow rate is proportional to the depth in each bucket and the size of the hole. The smallest hole is the 238U hole. We know a billion drops have flowed into the 234Th bucket because we have a 238U halo. We don't know how long this has taken. The time "t₁" for this to occur is unknown.

The flow from the 234Th bucket to the 234Pa bucket and from the 234Pa to the 234U buckets is virtually instantaneous due to their short half-lives (0.065984 and 0.000076479 years respectively). We can assume that this all occurs in the time t₁ for the 238U decay, and in essence treat this as one decay event from 238U to 234U.

The 234U Bucket hole (245,500 yr half-life) is a smaller than the holes in the 234Th and 234Pa buckets, but still bigger than the 238U hole (4,4680,000,000 yr half-life), so there will be some backup in this bucket but not much over the time t₁, and determining how much will mean integrating the outflow against the inflow taking into account the concurrent outflow.

Okay so far?

Enjoy.

Edited by RAZD, : added detail

Edited by RAZD, : added more

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This message is a reply to:
	Message 46 by peaceharris, posted 11-03-2008 4:06 AM		peaceharris has replied

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	Message 48 by peaceharris, posted 11-03-2008 8:41 PM		RAZD has replied

RAZD

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Message 49 of 142 (488331)
11-10-2008 12:06 AM

Reply to: Message 48 by peaceharris
11-03-2008 8:41 PM

Re: Time taken for equilibrium

Hey peaceharris, I've not forgotten.

The string of buckets is a good analogy... continue.

Thanks, but I don't believe it is original.

I've been working on the math, and the biggest problem I see is how to present it without going ballistic on post length. I also see no easy solution, as I have one too many variables for solution, thus some assumptions are needed. I've been playing with some of the variables to see how the spreadsheet reacts, and I've learned a few things.

Message 46

It takes a few hundred thousand years for equilibrium among all the daughter products of 238U to be reached. Assume that Uranium is soluble in water, but Thorium isn’t. Then the tree would suck groundwater that contains Uranium but not Thorium. As the Uranium decays, the uranium halo forms. But since there is no Thorium, the halo due to 230Th will not start forming immediately, but will have to wait until a significant number of Uranium atoms have decayed.

The 234U decay is the bottleneck here, as you say, however, starting with 100% of 10^{^9} decay events for the first ring formation from 238U decay, and running through the buckets, at ~490,000 years I am getting ~45% of 10^{^9} 230Th production from 234U decay, and ~32% of 10^{^9} 226Ra production from 230Th decay, and ~32% of 10^{^9} production of 222Rn from 226Ra decay, adding up to ~110% of the 10^{^9} decay events required to form a ring, assuming their decay impacts combine. This would be about the earliest second ring you could observe.

At this point I am also getting ~32% of 10^{^9} 222Rn production from 226Ra, and similar for the rest of the decay chain down to ~32% of 10^{^9} for 210Po. I would expect the same combination of decay events that combine 234U, 230Th and 226Ra decays into one visible ring to also combine the 222Rn and 210Po events into ~65% of the required 10^{^9} decay events to form a ring.

Certainly this is much more than we see in the 238U halo that Gentry labels as "embryonic" ...

... leaving me, still, with my original conclusion: that by the time these two rings have formed to this extent, that you would have a much more visible third ring than seen here.

There are 2 ways to prove that the time taken for equilibrium has not elapsed for embryonic U halos.
1. The ratio of 238U:206Pb is some of these Uranium radiocenters is very high (~27000).
2. The 214Po halo is not visible, but the Uranium halo is visible.

Both of these would also be true if 222Rn leaves the inclusion site before decaying, so no, this does not prove that the time needed for equilibrium has not occurred.

Conversely the existence of more than ~32% of 10^{^9} 206Pb in the inclusion could mean that equilibrium had been reached in the inclusion before it came to be embedded in the crystal lattice, and thus you should have much more 222Rn decay (and other daughter products) than is visible here.

The problem I have with the term "embryonic" is that you can have a partially formed halo that will never become fully formed due to 222Rn departure.

Of course this also means that this partial halo is half a million years old ....

Enjoy.

Edited by RAZD, : Of course ...

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	Message 50 by peaceharris, posted 11-13-2008 12:55 AM		RAZD has replied

RAZD

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Message 51 of 142 (488641)
11-13-2008 11:30 PM

Reply to: Message 50 by peaceharris
11-13-2008 12:55 AM

Formulas and calculations

Thanks, peaceharris,

I don't think this calculation is right. Maybe if you show us your formulas we might understand what you are trying to say and correct you.
I don't think you are doing the math correctly.

Gladly, however I'll take it in stages to make sure we are on the same footing.

Assumptions:

that we start with pure 238U rather than with a piece of ore that is already along the way or at equilibrium.
that the half-lives are as listed in Wikipedia or similar source/s
that "X" decay events are needed to form a ring of density "Y" and that "X" and "Y" are the same (or nearly so) for all rings in the decay chain sequence (ie - number of decay events to form a ring does not change with the isotope that decays)

Initial Conditions:

N_o = amount at time t_o, = unknown
N = amount at time t, varies with t
H = half-life, known
A = time for 1 ring amount of 238U decay events to occur, = unknown (because N_o unknown)
at time A, N_o-N_A = N_R = enough decay events to form 1 ring

The Basic Decay Formula/s:

We can write the standard radioactive decay formula as a dimensionless formula with:
(N/N_o) = (1/2)^{^(t/H)}
or
%238U = 100(N/N_o) = 100(1/2)^{^(t/H)}
Now we have one formula and two unknowns, so no mathematical solution is possible, but we can express A in terms of N (or N/N_o) or N in terms of A (or N/N_o in terms of A):
at 100%Ring the %238U = 100(N_A/N_o) = 100(1/2)^{^(A/H)}
Then we can vary A or N_A (or N_A/N_o) and see what the effects are on the results. I'll save this for later.

Now if we start with assuming that (N_A/N_o) = 99.9% (for example), we see that the production of 234Th is fairly linear:

       Time                   238U%   234Th%  Ring% 
t = 0    0%   N(0.0A)/No =   100.00%   0.00%    0.0%
10%   N(0.1A)/No =    99.99%   0.01%   10.0%
20%   N(0.2A)/No =    99.98%   0.02%   20.0%
30%   N(0.3A)/No =    99.97%   0.03%   30.0%
40%   N(0.4A)/No =    99.96%   0.04%   40.0%
50%   N(0.5A)/No =    99.95%   0.05%   50.0%
60%   N(0.6A)/No =    99.94%   0.06%   60.0%
70%   N(0.7A)/No =    99.93%   0.07%   70.0%
80%   N(0.8A)/No =    99.92%   0.08%   80.0%
90%   N(0.9A)/No =    99.91%   0.09%   90.0%
t = A  100%   N(1.0A)/No =    99.90%   0.10%  100.0%

Ignoring the decay of 234Th for now, the decay of one atom of 238U produces one atom of 234Th and one α particle (that goes out into the wilderness to build a ring):
N_(234Th) = N_o(238U) - N_(238U)
at t = A
N_(234Th) = N_R
and
%Th = 100(N_(234Th)/N_o(238U)) = 100[1 - (1/2)^{^(t/H_238U))}]
or, in terms of ring decay events:
%Ring = 100(N_234Th/N_R) = 100{1 - (1/2)^{^(t/H_238U))}} x {N_o(238U)/N_R}

%Ring =100{1 - (1/2)^{^(t/H_238U))}} x {N_o(238U)/N_R}

Enough for now?

If the ring is due to U238 is visible, then the ring due to U234 should also be visible. This is because the activity ratio of U234: U238 in many natural systems is close or greater than unity (http://cat.inist.fr/?aModele=afficheN&cpsidt=16322329)

This means that for a Uranium inclusion in a solid that just formed, the U238 and U234 ring will form simultaneously. But if we start with the assumption that the Uranium radiocenter did not contain any Thorium and other isotopes initially, other rings due to other isotopes such as Radon and Polonium will have to wait until significant amounts of U234 have decayed.

IFF you have both 238U and 234U in the inclusion. You would likely form a strong 234U ring (and daughter product rings) before you got a 238U ring.

In K-Ar dating, the assumption used is that Argon being a gas gets evaporated before lava solidifies. After solidification, any Argon produced remains in the rock. If you are right in saying gases can continue to escape after a rock has been solidified, you are basically saying that every geologist that uses K-Ar and U-Pb dating could be grossly mistaken. Perhaps you should do some experiments to prove your point, and get some fame. You might want to consider asking creationist organizations for funding since they too might be delighted to see mainstream scientists making serious mistakes.

Always a possibility eh?

But not all rocks are the same. Some are porous and some are dense and solid. What we are talking about here are rocks that went through a period of deformation after cooling that fractured the rock, allowing gas and fluid to move through the structure, and a replacement process that recrystallizes the rock afterward (see linked abstract re Collins of article in Lithos).

Enjoy.

Edited by RAZD, : subtitle

Edited by RAZD, : clarity

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This message is a reply to:
	Message 50 by peaceharris, posted 11-13-2008 12:55 AM		peaceharris has replied

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	Message 54 by peaceharris, posted 11-18-2008 10:59 PM		RAZD has replied

RAZD

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Message 52 of 142 (488714)
11-15-2008 6:09 PM

Reply to: Message 50 by peaceharris
11-13-2008 12:55 AM

Thorium Production

Continuing to the next stage of calculations:

Maybe if you show us your formulas we might understand what you are trying to say and correct you.

We see from the previous post (Message 51), that a dimensionless formula can be used:
%238U = 100(N/N_o) = 100(1/2)^{^(t/H)}

where

N_o = amount at time t_o, = unknown

N = amount at time t, varies with t

H = half-life, known

And we define "A" as the time needed for 238U decay to form the first (innermost) ring. From this we develop forumlas for the production of 234Th and %ring formation:

%Th = 100[1 - (N/N_o)_(238U)] = 100[1 - (1/2)^{^(t/H_238U)}]

%Ring =100[1 - (1/2)^{^(t/H_238U)}]•[N_o(238U)/(N_o(238U)-N_A(238U))]
or
%Ring =100R[1 - (1/2)^{^(t/H_238U)}]

where

N_A(238U) = amount of 238U remaining at time t_A, and

(N_o(238U)-N_A(238U)) is the amount of 238U decay needed to form a ring at time A, and

R = [N_o(238U)/(N_o(238U)-N_A(238U))] == [1/(1- (1/2)^{^(A/H_238U)})]

With (N_A/N_o)_(238U) = 0.25 → A = 2H → R = 4/3, we would get the following graphs of %238U decay (squares), %234Th production (triangles) and %Ring formation (diamonds):

while with (N_A/N_o)_(238U) = 0.90 → A = 15.2%H → R = 10, we would get the following graphs of %238U decay (squares), %234Th production (triangles) and %Ring formation (diamonds):

Thus however we vary A or N/No we still get a 100% ring formation at time A. Either way you measure it, we have produced enough decay events for ...238U (H = 4.468x10^9 yr) → 234Th + α (at 4.27 MeV)
... to cause the formation of the first ring at the ring density that is visible.

This sets the initial conditions to then look at the amount of decay for other isotopes/rings on an equal basis.

Enjoy.

Edited by RAZD, : finishing ...

Edited by RAZD, : stopped at better stopping point

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RAZD

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Message 53 of 142 (488733)
11-16-2008 1:40 AM

Reply to: Message 50 by peaceharris
11-13-2008 12:55 AM

Thorium Decay (edited)

Now the next stage:

Message 51

Ignoring the decay of 234Th for now, the decay of one atom of 238U produces one atom of 234Th and one α particle ...

The problem, of course, is that 234Th does decay and rather rapidly by comparison:
234Th (H = 24.10 days) → 234Pa + β (at 0.27 MeV)
Let H1 = the half-life of 238U = 4.468x10^9 yr
and H2 = the half-life of 234Th = 24.10 days = 0.06598 yr

The ratio of half lives, H1:H2, is 6.7713x10^{^10}:1

The next stage is to put the decay of 234Th on the same time-scale as the decay of 238U. Going back to the original decay formula:
(N/N_o) = (1/2)^{^(t/H)}
For 234Th this is
%234Th = (N/N_o)_234Th = (1/2)^{^(t/H2)}
or
%234Th = (1/2)^{^(t/H2)(H1/H1)}
or
%234Th = (1/2)^{^(t/H1)(H1/H2)}

with a^{^b•c} == [a^{^b}]^{^c} (or [a^{^c}]^{^b})
then
%234Th = [(1/2)^{^(t/H1)}]^(H1/H2)
and
%234Th = [(1/2)^{^(t/H1)}]^{^6.7713x10^{^10}}

Thus even though we do not know what the time scale is from 0 to A we can compare different decay rates. The results for "X" "Y" and "Z" decay, where Hx = 2Hy = 4Hz is:

where you can compare values to see that:

%X (at t = 20%) = %Y (at t = 10%) = %Z (at t = 5%),

%X (at t = 40%) = Y (at t = 20%) = Z (at t = 10%),

X (at t = 80%) = Y (at t = 40%) = Z (at t = 20%),

at t = 50%, X = 50%, Y = 25% (50%^{^2}), Z = 6.25% (50%^{^4})

at t = 100%, X = 25%, Y = 6.25% (25%^{^2}), Z = 0.39% (25%^{^4})
for some examples, thus validating the formula relationships at this point.

With this approach we can calculate the decay of 234Th over the same time-frame as we have for the production of 234Th. The difficulty with just applying the decay formula though, is that 234Th is being produced during this period, and we have a situation similar to this:

Where the blue circles represent the parent decay curve (here the N/N_o curve for 238U) from t = 0 to A, the red diamonds represent the total daughter production from t = 0 to A (in terms of (N_o-N)/N_R where N_R = the number of alpha particle decays needed to form a ring). Let's look at in generic terms first, using parent, daughter terms and assume that the parent decays for ~2 half-lives (as shown above) and the daughter decays at a half-life 1/5th of parent (as shown above).

In this case I have divided the daughter production into 10 time periods (0 to 10%, 10% to 20%, ... , 80% to 90%, and 90% to 100%), and considered the production only in that time period, so the value for the green square is the delta between the end and beginning values for the daughter total production values.

To figure out what level of daughter isotope we have remaining at 10% intervals from t = 0 to 100%A we then need to calculate the decay of each green square value within that time period and add them up to get the total for that time period.

For a rough approximation we can assume that the decay from the initial portion of each section is the same as the decay for the delta amount as a fixed amount at the average time point of the interval, thus we set the Noi₁ value = DY₁ at 5% for the first interval, the Noi₂ value = DY₂ at 15%, the Noi₃ value = DY₃ at 25%, etc.

The orange circle curve in the above graph is the daughter decay curve started at 5%A and then showing values at 10%, 20%, ... 90%, 100% so that we can use these for our summation process. Note that in this case, with the relative decay rates used so far, that the initial decay in the first 5%A is ~30% (actual number is 29.29%) so we can start each interval with ~70% of the green square curve values for this example (the value changes with the ratio of decay rates).

Where the red diamonds are the total daughter isotope 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 it's daughter isotope (the next bucket). In this (theoretical) instance the result is ~92% of the amount of decay needed to form a ring of the same density as the first ring ... except that this is theoretical.

Of course the real half-life of 234Th is much less than 1/5 of the half-life for 238U and the blue points will all be close to 0.0 while the orange points will be close to the red ones, and virtually all the 234Th will decay to 234Pa. Even with A = 1,000 years (unlikely) the graph for 234Th looks like this:

With complete decay of all 234Th (H = 24.10 days) → 234Pa + β (at 0.27 MeV).

The decay of 234Pa would be similar (H = 6.70 hr) → 234U + β (at 2.20 MeV), and thus we come to the next alpha decay event:

234U (H = 245,500 yr) → 230Th + α (at 4.86 MeV)

Let me know if you concur with this approach so far, or whether I have missed something in the process.

Enjoy.

Edited by RAZD, : fixed

Edited by RAZD, : updated

Edited by RAZD, : 98

Edited by RAZD, : clarity

Edited by RAZD, : simplified (I hope) previous hidden

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RAZD

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Message 55 of 142 (488940)
11-19-2008 11:05 PM

Reply to: Message 54 by peaceharris
11-18-2008 10:59 PM

Re: Formulas and calculations

Thanks peaceharris,

So what if a rock is porous? Does rock porosity in any way prove your point that radon gas can sublime out of a rock? To explain Polonium halos you were saying the opposite thing, deposition inside the rock after the rock has solidifed. To explain embryonic Uranium halos you are saying that radon escapes from inside the Uranium inclusion.

It has nothing to do with sublimation - just the movement of gas through fissures, movement that is documented by decay damage along fissures. Both polonium\radon and partially formed uranium halos can be explained by the same mechanism - gas flows through fissures.

Regarding your radiometric formulas, I still haven’t seen the formula you used to come up with your statement that “starting with 100% of 10^9 decay events for the first ring formation from 238U decay, and running through the buckets, at ~490,000 years I am getting ~45% of 10^9 230Th production from 234U decay”

I take it that you see no errors in the methodology, formulas and calculations so far then.

Enjoy.

Edited by RAZD, : sig

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	Message 56 by peaceharris, posted 11-20-2008 5:34 AM		RAZD has replied

RAZD

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Message 57 of 142 (489007)
11-20-2008 10:03 PM

Reply to: Message 56 by peaceharris
11-20-2008 5:34 AM

Re: Formulas and calculations

Thanks peaceharris,

You are trying to argue that gases escapes from solid rock that has already solidified, ...

Not really, for the rock that IS solidified does not have halos. Halos appeara only in rocks that have undergone secondary processes involving stress that causes fractures.

The method you are using is very lengthy, and I find it hard to understand what you are saying. Please use ”standard’ formulas that everyone uses.

The standard formula for decay is
N = N_o • e^{^-λt} where λ = the decay constant
The version of this formula I use (and prefer to use) is:
N = N_o • (1/2)^{^t/H} where H = the half-life
and H == ln(2)/λ (or λ = ln(2)/H)

They are the same formula, but the latter version is more intuitive and easy to understand:

at t = H, N = N_o/2;
at t = 2H, N = N_o/4 and
at t = 3H, N = N_o/8

If we divide by N_o then we have the formula in Message 51:
(N/N_o) = (1/2)^{^(t/H)}

Let U238(0) be the initial number of U238 atoms present in an Uranium inclusion. Let U238(T) denote the number of U238 atoms remaining after time (T) has elapsed,. Then the number of alpha particles emitted during that time interval T from the decay of U238 can be calculated as

U238(0)-U238(T)

Correct, and the number of daughter isotopes produced is the same as well:
238U → 234Th + α

We also know (all things being equal) that if the amount of decay at time "T" (or "A" as I used) is sufficient to form a ring, that then that is the amount of decay needed to form other rings, and that this time from t = 0 to t = "T" is the same for the whole decay chain.

You may neglect Th234 and Pa234, assuming that U238 decays directly U234. You will get an accurate result by making this assumption.

Thanks, but rather than just make this assumption I confirmed it (Message 53).

You may assume anything you want for U238(0) and U234(0). To calculate U238(T) and U234(T),

Seeing as we started out with the equilibrium position, and this showed that all the rings should be visible once you have the first two rings, it seems logical to go to the other extreme and assume no existing daughter isotopes in the inclusion: if the same conclusion holds, then we know that it holds for all the conditions in between.

please use the formulas at http://www.soes.soton.ac.uk/staff/pmrp/GY309/Module6/m6.html

I looked at your site, and the impression I got was that it was very well for people that know what we are talking about, but that it gets lost in the jargon for most people.

I also have a problem with just accepting the blanket statement and the derived formula:

quote:
This equation is difficult to solve unless you know quite a lot of maths, so I will just present the solution here;

As I wouldn't feel right to tell other people this is so without verifying it. Feel free to use it to check my calculations if you like.

It seems much easier and more practical to use a simple approximation to achieve the same basic results and be able to explain it simply at the same time.

So we start with a simple decay curve from t=0 to t=A:
(N/N_o) = (1/2)^{^(t/H)}
We calculate the number of α decay events in the same period:
N_α = N_o - N

N_α/N_o = (N_o - N)/N_o = 1 - N/N_o = 1 - (1/2)^{^(t/H)}

And we note that at t=A, N_α is 100% of the amount needed to form the innermost ring (238U decay ring), N_α(A) = N_o - N_A, and so we have these three curves:

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

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

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

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.

As you can see this is the same approach you are suggesting, just presented graphically.

Whatever assumption you use, your end result should explain the following observations:
1. There are 2 rings that can be seen with the same clarity, this implies that the number of alpha particles that formed these 2 rings are approximately equal.

Agreed, however we have 3 isotopes (234U, 230Th and 226Ra) where their decay contributes to the second ring formation instead of just one.

2. There is a 3rd ring slightly visible, so the number of alpha particles that created this barely visible ring should be much less than the number of alpha particles for the 2 inner rings.

Agreed, and if we find more decay calculated than what we see, then logically we should assume that the difference is due to 222Rn leaving the site of the inclusion.

Given the half-lives of the isotopes after 226Rn are on the same order of magnitude as 234Th and 234Pa, compared to the half-lives of 234U and 238U, we should see the same virtually instantaneous decay of those isotopes once the 226Ra has decayed.

3. Other rings are not visible, so the number of alpha particles that have been emitted to form the other invisible rings should be less than the 3rd slightly visible ring.

We also have two isotopes (222Rn and 210Po) where their decay contributes to the third ring formation, while the remaining two have only one isotope, so they should be at a ratio of 2:1.

Enjoy.

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	Message 56 by peaceharris, posted 11-20-2008 5:34 AM		peaceharris has replied

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	Message 58 by peaceharris, posted 11-21-2008 4:30 AM		RAZD has replied

RAZD

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Message 60 of 142 (489015)
11-21-2008 7:42 AM

Reply to: Message 59 by cavediver
11-21-2008 6:46 AM

technique and materials chose the layer?

How do you know that the plane of the photo is not the fracture? By virtue that we are looking at a photo of a radiohalo, this suggests that the rock was fractured in this very plane. Or are we just taking lucky slices of random rocks???

You're dealing with mica, and Gentry's technique is to use scotch tape to lift layers.

Another clue is that several halos appear on the same plane, rather than partially on those where one is at a complete center.

And any damage along the fissure would not be seen from above, except as a general discoloration on the whole layer.

Enjoy.

Edited by RAZD, : added

Edited by RAZD, : more

Edited by RAZD, : No reason given.

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RAZD

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Message 61 of 142 (489037)
11-21-2008 10:38 PM

Reply to: Message 58 by peaceharris
11-21-2008 4:30 AM

Re: Formulas and calculations

What say peaceharris?

Are we in agreement that this is correct so far (Message 57):

quote:

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

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

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

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.

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-lives

H1 = half life of the parent isotope
H2 = half life of the daughter isotope

The 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)/H

and e^{^-λt} == (1/2)^{^t/H}

so we can write it as:
N₂ = [(ln(2)/H1)/(ln(2)/H2-ln(2)/H1)]•N₁_o•[(1/2)^{^t/H1} - (1/2)^{^t/H2}]
and neglect the N₂_o (assume = 0)

Multiplying the top and bottom of the first bracket fraction by H1•H2/ln(2), and dividing both sides by N₁_o we get:

N₂/N₁_o = [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 == (N₁_o - N_A) = N_{α_238U} (at 4.27 MeV) and N_R - N₂ = N_{α_234U} (at 4.86 MeV).

and N_A = N₁_o•(1/2)^{^A/H1} so N_R = N₁_o•(1-(1/2)^{^A/H1}) or N₁_o = 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.

Message 49 ... at ~490,000 years I am getting ~45% of 10^9 230Th production from 234U decay, ...

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 clarity

Edited by RAZD, : added more

Edited by RAZD, : added end

Edited by RAZD, : clarity

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RAZD

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Message 62 of 142 (489055)
11-22-2008 3:15 PM

Reply to: Message 59 by cavediver
11-21-2008 6:46 AM

added comment, Cavediver

filling in the evidence

How do you know that the plane of the photo is not the fracture? By virtue that we are looking at a photo of a radiohalo, this suggests that the rock was fractured in this very plane.

From Message 188

quote:
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.

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?

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.

Seems pretty clear to me.

Enjoy.

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	Message 59 by cavediver, posted 11-21-2008 6:46 AM		cavediver has replied

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	Message 64 by cavediver, posted 11-22-2008 4:43 PM	RAZD has seen this message but not replied
	Message 65 by peaceharris, posted 11-24-2008 12:44 AM	RAZD has replied

RAZD

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Message 63 of 142 (489057)
11-22-2008 4:26 PM

Reply to: Message 58 by peaceharris
11-21-2008 4:30 AM

The results please ...

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 clarity

Edited by RAZD, : complete

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RAZD

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Message 71 of 142 (489578)
11-28-2008 10:35 AM

Reply to: Message 70 by peaceharris
11-24-2008 11:22 PM

Equilibrium or not, the picture does not match the pattern that should exist

Thanks, peaceharris,

Message 62

OK, I understand your method of explaining the embryonic halo and your math is correct.

Good.

The problem with your explanation is it doesn’t follow the principle of Uniformitarianism.

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).

You are assuming that the amount of U234 in the radiocenter was 0 initially and then showing that the embryonic halo is ~500000 years.

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

Can you tell me what I have wrong here?
Your calculation of isotopes that have decayed only applies to systems in equilibrium.

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).

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 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.

So activity of U238 is 99.27*7.04e8 / 0.72/ 4.463e9 (=21.7) times greater than that of U235

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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RAZD

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Message 72 of 142 (489732)
11-29-2008 3:45 PM

Reply to: Message 39 by peaceharris
09-30-2008 5:01 AM

Re: Not about Polonium, Not about a Young Earth.

getting back to this comment, peaceharris,

Could you give us a photo of a fully developed Uranium halo. A halo where the 238U and the 214Po ring can be seen?

These are provided by Gentry:

These are all well past the 500,000 year stage where the third ring would be visible

Anyway, if accelerated radioactive decay did not take place, how old are the Uranium halos in Gentry's paper? Pls tell me how you arrive at your answer.

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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RAZD

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Message 73 of 142 (489737)
11-29-2008 4:25 PM

Reply to: Message 65 by peaceharris
11-24-2008 12:44 AM

More complete view

Thanks peaceharris,

The second problem with your method is it gives a lot of freedom to interpret data the way you believe it to be.

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.

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.

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).

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.

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?

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.

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 208

Enjoy.

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	Message 65 by peaceharris, posted 11-24-2008 12:44 AM		peaceharris has replied

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	Message 74 by peaceharris, posted 12-01-2008 5:04 AM		RAZD has replied

RAZD

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Posts: 20714

From: the other end of the sidewalk

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Message 75 of 142 (490050)
12-01-2008 10:09 PM

Reply to: Message 74 by peaceharris
12-01-2008 5:04 AM

Re: More complete view

Thanks again peaceharris for your input.

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.

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.

Dude, you don’t understand anything!

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?

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.

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.

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.

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.

Try doing a google search on metamictization.

Curiously I do not find where this process creates weak fracture planes rather than general and directionally nonspecific damage.

Are there scientists who share your view that fissures and Uranium radiocenters go together?

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.

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.

Strangely this just seems to be another dodge rather than discussing the issue.

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 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.

I agree with you, there’s nothing I can do to convince you that embryonic Uranium halos are young.

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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