Register | Sign In


Understanding through Discussion


EvC Forum active members: 65 (9162 total)
8 online now:
Newest Member: popoi
Post Volume: Total: 915,815 Year: 3,072/9,624 Month: 917/1,588 Week: 100/223 Day: 11/17 Hour: 7/1


Thread  Details

Email This Thread
Newer Topic | Older Topic
  
Author Topic:   Are Uranium Halos the best evidence of (a) an old earth AND (b) constant physics?
peaceharris
Member (Idle past 5596 days)
Posts: 128
Joined: 03-28-2005


Message 46 of 142 (487658)
11-03-2008 4:06 AM


Time taken for equilibrium
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.
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.
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.
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.

Replies to this message:
 Message 47 by RAZD, posted 11-03-2008 7:13 AM peaceharris has replied

  
RAZD
Member (Idle past 1404 days)
Posts: 20714
From: the other end of the sidewalk
Joined: 03-14-2004


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

we are limited in our ability to understand
by our ability to understand
Rebel American Zen Deist
... to learn ... to think ... to live ... to laugh ...
to share.


• • • Join the effort to solve medical problems, AIDS/HIV, Cancer and more with Team EvC! (click) • • •

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

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

  
peaceharris
Member (Idle past 5596 days)
Posts: 128
Joined: 03-28-2005


Message 48 of 142 (487712)
11-03-2008 8:41 PM
Reply to: Message 47 by RAZD
11-03-2008 7:13 AM


Re: Time taken for equilibrium
The string of buckets is a good analogy... continue.

This message is a reply to:
 Message 47 by RAZD, posted 11-03-2008 7:13 AM RAZD has replied

Replies to this message:
 Message 49 by RAZD, posted 11-10-2008 12:06 AM peaceharris has replied

  
RAZD
Member (Idle past 1404 days)
Posts: 20714
From: the other end of the sidewalk
Joined: 03-14-2004


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 ...

we are limited in our ability to understand
by our ability to understand
Rebel American Zen Deist
... to learn ... to think ... to live ... to laugh ...
to share.


• • • Join the effort to solve medical problems, AIDS/HIV, Cancer and more with Team EvC! (click) • • •

This message is a reply to:
 Message 48 by peaceharris, posted 11-03-2008 8:41 PM peaceharris has replied

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

  
peaceharris
Member (Idle past 5596 days)
Posts: 128
Joined: 03-28-2005


Message 50 of 142 (488565)
11-13-2008 12:55 AM
Reply to: Message 49 by RAZD
11-10-2008 12:06 AM


Re: Time taken for equilibrium
RAZD writes:
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 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.
RAZD writes:
This would be about the earliest second ring you could observe.
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.
In other words 'the earliest second ring you could observe' is the same time as the first ring formation
RAZD writes:
Certainly this is much more than we see in the 238U halo that Gentry labels as "embryonic"
I don't think you are doing the math correctly.
RAZD writes:
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.
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.
Edited by peaceharris, : typo

This message is a reply to:
 Message 49 by RAZD, posted 11-10-2008 12:06 AM RAZD has replied

Replies to this message:
 Message 51 by RAZD, posted 11-13-2008 11:30 PM peaceharris has replied
 Message 52 by RAZD, posted 11-15-2008 6:09 PM peaceharris has not replied
 Message 53 by RAZD, posted 11-16-2008 1:40 AM peaceharris has not replied

  
RAZD
Member (Idle past 1404 days)
Posts: 20714
From: the other end of the sidewalk
Joined: 03-14-2004


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:
  1. that we start with pure 238U rather than with a piece of ore that is already along the way or at equilibrium.
  2. that the half-lives are as listed in Wikipedia or similar source/s
  3. 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:
  • No = amount at time to, = 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 No unknown)
  • at time A, No-NA = NR = 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/No) = (1/2)^(t/H)
or
%238U = 100(N/No) = 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/No) or N in terms of A (or N/No in terms of A):
at 100%Ring the %238U = 100(NA/No) = 100(1/2)^(A/H)

Then we can vary A or NA (or NA/No) and see what the effects are on the results. I'll save this for later.
Now if we start with assuming that (NA/No) = 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) = No(238U) - N(238U)
at t = A
N(234Th) = NR
and
%Th = 100(N(234Th)/No(238U)) = 100[1 - (1/2)^(t/H238U))]

or, in terms of ring decay events:
%Ring = 100(N234Th/NR) = 100{1 - (1/2)^(t/H238U))} x {No(238U)/NR}
%Ring =100{1 - (1/2)^(t/H238U))} x {No(238U)/NR}
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

we are limited in our ability to understand
by our ability to understand
Rebel American Zen Deist
... to learn ... to think ... to live ... to laugh ...
to share.


• • • Join the effort to solve medical problems, AIDS/HIV, Cancer and more with Team EvC! (click) • • •

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

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

  
RAZD
Member (Idle past 1404 days)
Posts: 20714
From: the other end of the sidewalk
Joined: 03-14-2004


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/No) = 100(1/2)^(t/H)
where
  • No = amount at time to, = 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/No)(238U)] = 100[1 - (1/2)^(t/H238U)]
    %Ring =100[1 - (1/2)^(t/H238U)]•[No(238U)/(No(238U)-NA(238U))]
    or
    %Ring =100R[1 - (1/2)^(t/H238U)]
    where
  • NA(238U) = amount of 238U remaining at time tA, and
  • (No(238U)-NA(238U)) is the amount of 238U decay needed to form a ring at time A, and
  • R = [No(238U)/(No(238U)-NA(238U))] == [1/(1- (1/2)^(A/H238U))]
  • With (NA/No)(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 (NA/No)(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

    we are limited in our ability to understand
    by our ability to understand
    Rebel American Zen Deist
    ... to learn ... to think ... to live ... to laugh ...
    to share.


    • • • Join the effort to solve medical problems, AIDS/HIV, Cancer and more with Team EvC! (click) • • •

    This message is a reply to:
     Message 50 by peaceharris, posted 11-13-2008 12:55 AM peaceharris has not replied

      
    RAZD
    Member (Idle past 1404 days)
    Posts: 20714
    From: the other end of the sidewalk
    Joined: 03-14-2004


    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/No) = (1/2)^(t/H)

    For 234Th this is
    %234Th = (N/No)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/No curve for 238U) from t = 0 to A, the red diamonds represent the total daughter production from t = 0 to A (in terms of (No-N)/NR where NR = 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

    we are limited in our ability to understand
    by our ability to understand
    Rebel American Zen Deist
    ... to learn ... to think ... to live ... to laugh ...
    to share.


    • • • Join the effort to solve medical problems, AIDS/HIV, Cancer and more with Team EvC! (click) • • •

  • This message is a reply to:
     Message 50 by peaceharris, posted 11-13-2008 12:55 AM peaceharris has not replied

      
    peaceharris
    Member (Idle past 5596 days)
    Posts: 128
    Joined: 03-28-2005


    Message 54 of 142 (488876)
    11-18-2008 10:59 PM
    Reply to: Message 51 by RAZD
    11-13-2008 11:30 PM


    Re: Formulas and calculations
    RAZD writes:
    Always a possibility eh?
    But not all rocks are the same. Some are porous and some are dense and solid.
    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.
    Gentry had a fine way to excuse himself of answering his critiques. He asked his critiques to create Polonium halos on a granite that initially had none, since his critiques were of the opinion that Polonium halos formed inside the granite after the granite solidified.
    RAZD, if you want your theories to become established fact, I suggest you perform an experiment where Noble gases can be removed from the rock without heating the rock or an experiment where Polonium halos can be formed in the rock after the rock has solidified.
    Experiments have already been done which prove that a lot of Helium is retained in a rock. Helium is much smaller than Radon atoms so it should be easier for He to escape (yet significant amounts of He is retained). Not only that, you are saying that Radon which has a half-life of a few days (compared to stable Helium) can actually escape from the Uranium inclusion. I do not want to waste any more time arguing against your theories which I think are extremely unlikely, if not impossible.
    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”

    This message is a reply to:
     Message 51 by RAZD, posted 11-13-2008 11:30 PM RAZD has replied

    Replies to this message:
     Message 55 by RAZD, posted 11-19-2008 11:05 PM peaceharris has replied

      
    RAZD
    Member (Idle past 1404 days)
    Posts: 20714
    From: the other end of the sidewalk
    Joined: 03-14-2004


    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

    we are limited in our ability to understand
    by our ability to understand
    Rebel American Zen Deist
    ... to learn ... to think ... to live ... to laugh ...
    to share.


    • • • Join the effort to solve medical problems, AIDS/HIV, Cancer and more with Team EvC! (click) • • •

    This message is a reply to:
     Message 54 by peaceharris, posted 11-18-2008 10:59 PM peaceharris has replied

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

      
    peaceharris
    Member (Idle past 5596 days)
    Posts: 128
    Joined: 03-28-2005


    Message 56 of 142 (488951)
    11-20-2008 5:34 AM
    Reply to: Message 55 by RAZD
    11-19-2008 11:05 PM


    Re: Formulas and calculations
    RAZD writes:
    It has nothing to do with sublimation - just the movement of gas through fissures,....
    You are saying that Radon can escape from the Uranium inclusion. This is different from 'movement of gas through fissures'. Deep beneath the earth, molten rock will release gases produced from radioactive decay. This gas may find its way to the earth's surface because soil and rocks are porous.
    You are trying to argue that gases escapes from solid rock that has already solidified, and by saying that you are implying that the vast majority of mainstream scientists are making serious mistakes by trusting results obtained through K-Ar and U-Pb dating since you are saying noble gases escape from solid rock.
    RAZD writes:
    I take it that you see no errors in the methodology, formulas and calculations so far then.
    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.
    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)
    To calculate the number of alpha particles emitted during that time interval T from the decay of U234, the formula is
    U238(0)-U238(T) + U234(0) - U234(T)
    You may assume anything you want for U238(0) and U234(0). To calculate U238(T) and U234(T), please use the formulas at http://www.soes.soton.ac.uk/staff/pmrp/GY309/Module6/m6.html
    You may neglect Th234 and Pa234, assuming that U238 decays directly U234. You will get an accurate result by making this assumption.
    To explain the embryonic Uranium halo found at http://www.halos.com/images/ctm-rc-6-a.jpg tell me what assumption you are using for the initial amount of U238(0), U234(0) and T.
    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.
    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.
    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.
    In my opinion if you assume the value of T is ~500000 years as you stated earlier, you will not be able to explain those 3 observations, anyway maybe I am mistaken.

    This message is a reply to:
     Message 55 by RAZD, posted 11-19-2008 11:05 PM RAZD has replied

    Replies to this message:
     Message 57 by RAZD, posted 11-20-2008 10:03 PM peaceharris has replied

      
    RAZD
    Member (Idle past 1404 days)
    Posts: 20714
    From: the other end of the sidewalk
    Joined: 03-14-2004


    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 = No • e^-λt where λ = the decay constant

    The version of this formula I use (and prefer to use) is:
    N = No • (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 = No/2;
    • at t = 2H, N = No/4 and
    • at t = 3H, N = No/8
    If we divide by No then we have the formula in Message 51:
    (N/No) = (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.
    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/No) = (1/2)^(t/H)

    We calculate the number of α decay events in the same period:
    Nα = No - N
    Nα/No = (No - N)/No = 1 - N/No = 1 - (1/2)^(t/H)
    And we note that at t=A, Nα is 100% of the amount needed to form the innermost ring (238U decay ring), Nα(A) = No - NA, and so we have these three curves:
    Where the brown square cuve is the parent decay curve
    (N/No) = (1/2)^(t/H)
    The red triangle curve is the daughter production curve
    Nα/No = 1 - (1/2)^(t/H)
    And the blue diamond curve is the ring production curve
    Nα/NR = [No/NR]•[1 - (1/2)^(t/H)]
    Where NR = (No - NA) = 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.

    we are limited in our ability to understand
    by our ability to understand
    Rebel American Zen Deist
    ... to learn ... to think ... to live ... to laugh ...
    to share.


    • • • Join the effort to solve medical problems, AIDS/HIV, Cancer and more with Team EvC! (click) • • •

    This message is a reply to:
     Message 56 by peaceharris, posted 11-20-2008 5:34 AM peaceharris has replied

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

      
    peaceharris
    Member (Idle past 5596 days)
    Posts: 128
    Joined: 03-28-2005


    Message 58 of 142 (489012)
    11-21-2008 4:30 AM
    Reply to: Message 57 by RAZD
    11-20-2008 10:03 PM


    Re: Formulas and calculations
    razd writes:
    Halos appeara only in rocks that have undergone secondary processes involving stress that causes fractures.
    No. There are many photos of halos (including the embryonic halo being discussed at http://www.halos.com/images/ctm-rc-6-a.jpg ) where the radiocenter is not along a fracture. Without any visible conduit for Radon, you are claiming that Radon escapes the radiocenter.

    This message is a reply to:
     Message 57 by RAZD, posted 11-20-2008 10:03 PM RAZD has replied

    Replies to this message:
     Message 59 by cavediver, posted 11-21-2008 6:46 AM peaceharris has not replied
     Message 61 by RAZD, posted 11-21-2008 10:38 PM peaceharris has not replied
     Message 63 by RAZD, posted 11-22-2008 4:26 PM peaceharris has not replied

      
    cavediver
    Member (Idle past 3643 days)
    Posts: 4129
    From: UK
    Joined: 06-16-2005


    Message 59 of 142 (489013)
    11-21-2008 6:46 AM
    Reply to: Message 58 by peaceharris
    11-21-2008 4:30 AM


    Re: Formulas and calculations
    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???

    This message is a reply to:
     Message 58 by peaceharris, posted 11-21-2008 4:30 AM peaceharris has not replied

    Replies to this message:
     Message 60 by RAZD, posted 11-21-2008 7:42 AM cavediver has not replied
     Message 62 by RAZD, posted 11-22-2008 3:15 PM cavediver has replied

      
    RAZD
    Member (Idle past 1404 days)
    Posts: 20714
    From: the other end of the sidewalk
    Joined: 03-14-2004


    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.

    we are limited in our ability to understand
    by our ability to understand
    Rebel American Zen Deist
    ... to learn ... to think ... to live ... to laugh ...
    to share.


    • • • Join the effort to solve medical problems, AIDS/HIV, Cancer and more with Team EvC! (click) • • •

    This message is a reply to:
     Message 59 by cavediver, posted 11-21-2008 6:46 AM cavediver has not replied

      
    Newer Topic | Older Topic
    Jump to:


    Copyright 2001-2023 by EvC Forum, All Rights Reserved

    ™ Version 4.2
    Innovative software from Qwixotic © 2024