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This thread is open to all.

I posed these questions to Thornton here, which he can't seem to answer. Can you help him out?

Thornton:

Uranium-thorium decay from the surrounding rocks is one known way.

No it is not for a sufficient concentration. So again, you failed to specify the concentration of Uranium necessary to create an indicated C14 age of 40,000 years or for that matter any C14 age. Bertche just hand waved. You haven't provided one peer-reviewed article that gives the requisite concentration. I asked earlier is it:

99%, 98%......1%....0.1%.....

Just a single number would suffice. You're strangely silent on this substantive matter, but abundant in loud meaningless ridicule.

Because we have lots of coal samples with zero C14 which means the C14 isn't intrinsic to coal.

That is a non-sequitur. Absence of C14 in some samples doesn't mean the samples that have C14 are due to contamination. You're lack of logic is showing. Btw, you've again failed to defend your claim that most carboniferous coal has no C14. So what is the percentage without C14?

99%, 98%....1%.......

You had your chance to shine, but so far, you're just kind of not too bright.
stcordova

Posts: 447
Joined: Wed Mar 05, 2014 1:41 am

Sal,

Taking the second question first, a mistake has been made. Some samples of coal have no currently measured carbon-14, but that does not mean they have no carbon-14 whatsoever. In any measurement there is the possibility of alpha error, believing the null hypothesis when the measurement is too inaccurate to make a determination that the null hypothesis is wrong, but in fact the null hypothesis is wrong.

We don't really have a handle on what proportion of coal has measurable carbon-14 dates. The closest thing to an unbiased sample is the data from Baumgardner. That suggests that virtually all coal has measurable carbon-14. The number is 100%, although the 95% confidence limits extends to around 70%. More measurements in this area might be helpful.

Taking your first question, it can be split up into 2 parts. Second, and easiset, carbon-14 can be a rare decay product of radium. Rotta wrote an article in Creation Research Society Quarterly which argued that for this mechanism to work, uranium would have to be in a 99+% concentration, in which case we wouldn't call the material coal, but uranium with carbon impurities. Wickipedia disagrees (my emphasis):
Most man-made chemicals are made of fossil fuels, such as petroleum or coal, in which the carbon-14 should have long since decayed. However, such deposits often contain trace amounts of carbon-14 (varying significantly, but ranging up to 1% the ratio found in living organisms, a concentration comparable to an apparent age of 40,000).[24] This may indicate possible contamination by small amounts of bacteria, underground sources of radiation causing the 14N(n,p) 14C reaction, direct uranium decay (although reported measured ratios of 14C/U in uranium-bearing ores[25] would imply roughly 1 uranium atom for every two carbon atoms in order to cause the 14C/12C ratio, measured to be on the order of 10[^]−15), or other unknown secondary sources of carbon-14 production.

So if you believe them it would only take 90% uranium, which still seems unrealistic. (Can anyone say Durrett and Schmidt?)

As far as making neutrons underground is concerned, there are varying number-based estimates. The most optimistic one I have seen is in Baumgardner, and implies that the usual concentration of uranium must be increased by some 13,000 times in order to supply the requisite neutrons. This is one place where more independent data and calculations would be welcome.
Paul Giem

Posts: 17
Joined: Sun Mar 09, 2014 12:23 am

Sal, can you enable superscript and subscript?
Paul Giem

Posts: 17
Joined: Sun Mar 09, 2014 12:23 am

I just figured out the correct formula for the 95th percent confidence limit if one finds carbon-14 in 10 out of samples. It is the 10th root of 0.05, or 0.74, multiplied by 10. That is, if we expect 7.4 out of the 10 samples, or more, to have measurable carbon-14, then equal to or greater than 5% of the time we will find that all 10 of our samples have carbon-14.
Paul Giem

Posts: 17
Joined: Sun Mar 09, 2014 12:23 am

Dr. Giem,

You are so kind to visit the few of us in the world that are interested in this discussion.

I should caution you posted on a thread where anyone can comment, and thus we may get spammed.

I've provided other discussions here where they cannot spam the discussion such as:
C14 in fossils is credible evidence fossils they are young. I did this so as to archive such information as you have just provided for posterity in a way that is free of troll spam.

I organized the forum this way to allow for individuals like Thornton to be contrasted with your reasoned discourse while also allowing for the compartmentalization and archiving of actually good discussions. But on occasion, engaging in debate sharpens our understanding of the issues, and it is good that we recognize substantive criticisms like the question of Uranium contamination.

I would highly recommend for your own sanity and the sake of those here like myself that enjoy reading what you have to say to feel free to archive some of your best thoughts in threads like C14 in fossils is credible evidence fossils they are young. I'll probably create a thread in the admin section that will link to quality discussions for posterity.

Ok, I've made some minor tweak to the software. The original color scheme from 2 weeks ago was changed and I hope the forum is easier on the eyes than when you first posted to the homochirality thread. Sometimes browsers need to have their cache cleared so that the changes can be seen. Strange bug, but I don't know how to fix it.

Now you can do super and subscripts, and if you really need it, there is modest latex capability. I provide examples here, and I'll post in the Admin forum some BBCode and Latex tips:

Code: Select all
H[sub]2[/sub]O

yields

H2O

Code: Select all
C[sup]14[/sup]

yields
C14

Code: Select all
[sub]7[/sub]C[sup]14[/sup]

yields
7C14

Code: Select all
C[sub]7[/sub][sup]14[/sup]

yields
C714

Code: Select all
$${ _{6} ^{14} C}$$

yields
${ _{6} ^{14} C}$

Code: Select all
$${SO_{4}^{2-}}$$

yields

${SO_{4}^{2-}}$

Code: Select all
$${^{227}_{90}Th^{+}}$$

yields
${^{227}_{90}Th^{+}}$

Code: Select all
$${CO_{2} + C \rightarrow {2CO}}$$

yields
${CO_{2} + C \rightarrow {2CO}}$

Code: Select all
$${{SO_{4}^{2-} + Ba^{2+} \rightarrow BaSO_{4} \downarrow}$$

yields
${{SO_{4}^{2-} + Ba^{2+} \rightarrow BaSO_{4} \downarrow}$

Code: Select all
$$\Intz \left( 1 \ +\ \sqrt{\omega_{i+1} + \zeta -\frac{x+1}{\Theta +1} y + 1} \right) = 1$$

$\Intz \left( 1 \ +\ \sqrt{\omega_{i+1} + \zeta -\frac{x+1}{\Theta +1} y + 1} \right) = 1$

Code: Select all
$$\left[ {\bf X} + {\rm a} \ \geq\ \underline{\hat a} \sum_i^N \lim_{x \rightarrow k} \delta C \right]$$

$\left[ {\bf X} + {\rm a} \ \geq\ \underline{\hat a} \sum_i^N \lim_{x \rightarrow k} \delta C \right]$

Code: Select all
$$\int_a^b x^2 dx=2$$

yields
$\int_a^b x^2 dx=2$
stcordova

Posts: 447
Joined: Wed Mar 05, 2014 1:41 am

I am a member of CRS and have access to Rotta's paper. Incredible. Public domain portions I will post here if I have time.
stcordova

Posts: 447
Joined: Wed Mar 05, 2014 1:41 am

From Wikipedia a Curie

The curie (symbol Ci) is a non-SI unit of radioactivity, named after Marie and Pierre Curie.[1][2] It is defined as
1 Ci = 3.7 × 1010 decays per second.

Here is a sketch of the proof, and it will take some fleshing out to understand.

Uranium is composed of the following isotopes in the following propotions:
U-238 99.2745%
U-235 0.720%
U-234 0.0055%

Since the half-life of C14 is much less than Uranium, we have what is known as secular equilibrium where the C14 amount is related to the Uranium decay rate based on Uranium concentration.

1 mole U results in equilibrium concentration of 4.9 x 10 -9 microcuries of C14

I presume Rotta means that this will result in C14 having that many emissions based on an equilibrium Uranium generation process.

let
NU = number Uranium atoms
NTh = number Thorium atoms [correction, edited later]
NC = number Carbon 14 atoms
RC14/C = ratio of C14 atoms to C atoms

Rotta claims:

$\frac{N_{U}}{N_{C}} = \frac{1}{8 \cdot 10^{-17} } \cdot R_{C^{14}/C}$

$\frac{N_{Th}}{N_{C}} = \frac{1}{1.6 \cdot 10^{-17} } \cdot R_{C^{14}/C}$

for a C14 age of 40,000 years we need

RC14/C = 1 x 10-14

plugging into the above formula we need 125 atoms uranium per carbon atom, which is a concentration by weight of 99.96%.

Rotta's paper was not exactly clear how he got that formula so that may take some fleshing out. It would be reassuring to see the derivation in detail.
Last edited by stcordova on Mon Mar 17, 2014 7:22 pm, edited 1 time in total.
stcordova

Posts: 447
Joined: Wed Mar 05, 2014 1:41 am

Stcordova,

I am confused by NTh = number Uranium atoms in your prior entry. What you posted seems complete with only references to NU. Does NTh refer to Thorium as another, albeit weaker, source of C14?

Stephen
sterusjon

Posts: 20
Joined: Thu Mar 13, 2014 1:37 pm

SteRusJon,

Thank you, I made a mistake, and corrected it to Thorium. I'm still working through the paper and it seems this section is a bit confusing in the paper. I'm trying my best to summarize the salient points. I can e-mail you the paper if you think you can help me uncoil it. I might try to see if some of the nuclear chemistry students at JMU can help me work the problem out.

Sal
stcordova

Posts: 447
Joined: Wed Mar 05, 2014 1:41 am

Sal,

Please forward the paper to me. I do not know how much help I can be since the cobwebs are rather dense in the my mind's physics archive. But, right now, I am in the mood for a little house cleaning in that section.

Stephen
sterusjon

Posts: 20
Joined: Thu Mar 13, 2014 1:37 pm

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