Talk:Radiometric dating/Archive 1

This is an archive of past discussions about Radiometric dating. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page.

Archive 1

Archive 2

Invitation

Work is currently in progress on a page entitled Views of Creationists and mainstream scientists compared. Also currently being worked upon is Wikipedia: NPOV (Comparison of views in science) giving guidelines for this type of page. It is meant to be a set of guidelines for NPOV in this setting. People knowledgable in many areas of science and the philosophy of science are greatly needed here. And all are needed to ensure the guidelines correctly represent NPOV in this setting. :) Barnaby dawson 21:32, 27 Dec 2004 (UTC)

Does this mean every geology and archaeology article will have to contain creationist explanations as well? adamsan 23:06, 27 Dec 2004 (UTC)

No of course not. This page is only meant to refer to pages who's principle aim is a comparison of scientific positions (or in the case above positions where there is dispute over the scientific nature of a position). Barnaby dawson 11:47, 28 Dec 2004 (UTC)

They should contain creationist explanations. Everone on here uses the excuse "pseudoscience" to keep creationism from "leaking" onto the articles. Scorpionman 14:37, 12 April 2006 (UTC)

This is because creationism is not science, it is a religious belief. Radiometric dating establishes facts through objective, verifiable methods. Slrubenstein | Talk 17:13, 12 April 2006 (UTC)

Scorpionman is an avowed creationist, so we'll just ignore his blathering because he's pissed that a bunch of scientist-friendly folk over on Age of the Earth aren't letting him and others turn it into a bullet-point bitchfest comparison of science and creation "science". Rolinator 01:43, 13 April 2006 (UTC)

Needed additions

We seem to be missing information on U238/U234 dating--nixie 01:21, 10 Apr 2005 (UTC)

The use of those two U isotopes alone for dating is quite rare, perhaps best discussed under 238U-234U-230Th dating in which they play an essential part. Actinide 11:40, 20 Jun 2005 (UTC)

There is a uranium-uranium dating article now. The glaring omission is uranium-lead dating. Actinide 03:17, 11 July 2005 (UTC)

Radiocarbon dating is mentioned here, but not cosmogenic nuclide dating. Is this simply accidental omission of a relatively obscure technique, or are cosmogenic techniques seen as somehow less "absolute"? The list of techniques in this article seems to emphasize "hard rock" formation-decay type techniques (exceptions are radiocarbon, o. I'll be the first to admit there are non-absolute dating applications of cosmogenic nuclides (erosion rates, non-unique solutions, etc), but they are certainly not the rule. I'll add something on cosmo techniques if no one objects. Also, can we organize the long list of techniques at the beginning of the article? --BlueCanoe 01:41, 23 February 2006 (UTC)

Strictly speaking, cosmogenic nuclide dating is a calibrated radiometric technique rather than an absolute radiometric technique (the only constants requied for the latter are half-lives). But this article is about "radiometric dating," not "absolute radiometric dating," so of course cosmogenic dating should be mentioned here. And yes, this article needs a lot of work. Actinide 13:15, 23 February 2006 (UTC)

Merge from POV fork

The POV fork Controversy on Radiometric dating is inappropriate, as there is plenty of room here for anything that could be contained there. I propose merging that article back here and deleting/redirecting it. Peyna 15:55, 17 February 2006 (UTC)

Just a note I've listed that article for AfD. Peyna 02:29, 23 February 2006 (UTC)

New article: religious controversy

Controversy on Radiometric dating inappropriately handled religious reaction to radiometric dating with a POV slant. However, the existence of a separate article discussing the religious and sociological impact of radiometric dating results (as opposed to the scientific practice of radiometric dating) should be preferred to the alternative of squeezing both topics into one article. This was recently discussed in the Talk page of the radiocarbon dating article: Talk:Radiocarbon_dating#Controversy. I'm currently drafting a new article (not a recreation of the rightly deleted Controversy on... article) to handle this separate non-science topic. However, before I post the new article, I would like to hear from contributors to this article: are you agreement with the outcome of the debate in the radiocarbon dating article's Talk page? Thanks. jdbartlett 16:24, 23 June 2006 (UTC)

Another merge suggestion

The radiogeology article appears to be repetition of material found here. I suggest merging radiogeology into radiometric dating. --BlueCanoe 01:41, 23 February 2006 (UTC)

I wrote the "radiogeology" article as a broader and more simplified article than this one. I tried to make it as easy as possible to understand and as somewhat of an introduction to the matter. This article is more for those who already understand the topic and need further reading, in my opinion. Radiogeology is a broader topic than radiometric dating, and thus, the radiogeology page serves to direct readers to other pages--one of which may be the article on radiometric dating. --Primetime 02:42, 23 February 2006 (UTC)

I'd never heard of "radiogeology" before, and a quick Google search returned a miniscule 236 hits indicating usage of this word is essentially nonexistent (c.f. the phrase "Radiometric Dating" which returns 302,000 hits). Also, it turns out a number of those hits, e.g. [1] are almost word-for-word identical to the radiogeology article - presumably this is not how things are supposed to work here? Lastly, I can't see that there is any content currently in radiogeology that should not also be in this article, and if this article is seen to be deficient then it would be better to fix it than to fork it (this goes for the "controversy" fork above too - I think it should be merged in and edited as part of this article because if any evidence of a controversy can be found then it should obviously be presented here). Actinide 06:00, 23 February 2006 (UTC)

Well, I can see that consensus is against keeping the articles seperate. I'll just merge them. --Primetime 07:19, 23 February 2006 (UTC)

All done. --Primetime 07:29, 23 February 2006 (UTC)

Thank you Wikipedia for such a wonderful article. You always know what is going on and share everything you know

Addition

There should be a section on this page entitled "Problems with radiometric dating". Scorpionman 19:18, 24 April 2006 (UTC)

If you can find a reliable source in a serious, peer reviewed journal that says there are problems, and it isn't a quote mine, you are welcome to discuss putting it in. Otherwise, please stop trolling all the science pages that make you feel uncomfortable and try to actually make some productive edits in article space. Thanks. JoshuaZ 21:26, 24 April 2006 (UTC)

there arent any problems with radiometric dating! i do it every day!

ezkerraldean

Coarse dating

Please explain granularity of one half-life staement in the section. Vsmith 03:19, 26 June 2006 (UTC)

Thanks for asking. It meant granularity used as in physics and statistics, similar to resolution in graphics. Anyhow, it was incorrect to apply it to the method described. The expression was a remnant of a previous attempt to describe an intuitive method. I have eliminated the section. Jclerman 05:47, 26 June 2006 (UTC)

Course dating simply means unrefined dating as when no calibration is applied. Granularity refers to the degree of coarseness as in significant digits of a measurement. In essense refering to a whole half-life versus any decimal fraction thereof. ...IMHO (Talk) 05:22, 26 June 2006 (UTC)

The date equation

The carbon-14 stuff has been replaced by a more general section valid for all parent-daughter decays. Carbon 14: (a) is only one of many isotopes used, and (b) is treated extensively in several other articles. Jclerman 20:34, 27 June 2006 (UTC)

Would this equation have problems if the daughter is unstable? Dan Watts 20:57, 27 June 2006 (UTC)

See [2]. Jclerman 22:06, 27 June 2006 (UTC)

replace constant with variable number of half lives

Please, Pce3, don't introduce variable number of half lives where they don't belong to. Jclerman 11:39, 29 June 2006 (UTC)

Equation as published produces incorrect results

I know this formula came from the USGA website but it was published only as a means of showing the relationship of the variables and not for the purpose of calculations even though the variable definitions are included as the disclaimer above the so called age equation states it is the:

"...mathematical expression that relates radioactive decay to geologic time..."

For age computation you need to replace the constant for a single half-life with a variable to represent the number of half-lives.

Thus

$t={\frac {1}{\lambda }}{\times }{\ln \left(1+{\frac {D}{P}}\right)}$

becomes

$t={\frac {n}{\lambda }}{\times }{\ln \left(1+{\frac {D}{P}}\right)}$

where $n$ is the number of half-lives.

Do a few calculations to see.

...IMHO (Talk) 19:58, 29 June 2006 (UTC)

The burden of proof is on you. Show your calculations. --Jclerman 20:16, 29 June 2006 (UTC)

Since Wiki math is not functional yet I can't offer equations here that you can plug values into. How about giving me some values for P and D etc. and I will give you comparative examples in return based upon the values you give. Please limit the number of sets to about five. That should be plenty enough proof. ...IMHO (Talk) 21:49, 29 June 2006 (UTC)

I didn't mean equations. I am expecting you show your calculations, or your work as they called that in high-school. Plug in numbers in the formulas and show your work. Just one or if you want two examples should be enough. No computer needed, not even a calculator. The formula is so simple that can be tested inside our heads. Perhaps the problem is that you chose inappropriate numbers for your tests? Show them, or choose two new sets.--Jclerman 22:46, 29 June 2006 (UTC)

I'm happy to show you my calculations using my values but in order to eliminate the idea that my values somehow defy the laws of physics and mathematics as an issue right up front I would prefer to use your valuues to start. I'll then do a separate set of calculations with my values if you are not convinced or if I have some issue with your values. ...IMHO (Talk) 02:17, 30 June 2006 (UTC)

OK. Use D = 16, P = 4, 1/λ= 8033 (for C-14 using 5568 yr/HL) plug 'em in and crank out the results. The relationship is a mathematical equation and directly usable as is. What is the solution? And yes, C-14 measurements aren't done that way, but the equation works. Vsmith 03:55, 30 June 2006 (UTC)

Okay I have done the calculations using these values and have obtained the results. However before showing them to you please allow me to express the following concerns. Since 14C is the parent isotope and 14N is the daughter isotope and their numbers of atoms are represented by P and by D respectively their values would be a bit higher in the real world. Also allow me to point out that their values would be identical since their transformation is based upon Beta decay. 14C is usually shown as having a half-life of 5700 plus of minus 30 years. [Note: In the table below the decay constant λ value of 0.000124 has been obtained as the inverse of the value for half-life you have given above and the alternate value of 5568 yr/HL not used.]

Even with these values, however, the results of the calculations are the same. I will now proceed to "package" the calculations for display here which may take a little time since the HTML and Wiki markup code has to be manually reconfigured since I do not yet have a macro to do the conversion of the code. Should not be more than an hour. Thanks for your patience and I'll be back with you then. ...IMHO (Talk) 04:28, 30 June 2006 (UTC)

Okay now just to be sure we are on the same page please verify that I have correctly matched your values with the appropriate designations in the table below:

Number of atoms parent isotope Carbon-14	Number of atoms daughter isotope Nitrogen-14	Percent parent isotope found in archaeological sample compared with modern sample "p = 1/(2^n)"	Number of half-lives of parent isotope "half-life = ln(1/p)/ln(2)"	Half-life value of parent isotope	Decay constant of parent isotope
P	D	p	n	half-life	λ
4	16	1	0	8033	0.000124

A few hints:

Your physics is incorrect. D and P are not equal. It is the decrease in P that equals the increase in N.

So then you (or someone else in one of the previous discussions (I'll go back and try and see who it was that rejected this idea) no longer reject the idea of the transition in numbers of atoms from 14C to 14N as the result of Beta decay which I was showing in table provided on the half-life discussion page? If not then their is no need for recalculation. ...IMHO (Talk) 08:21, 30 June 2006 (UTC)

You do not need n to apply the USGS date equation.
You do not need p to apply the USGS date equation.
You might want to use p to verify the results from the USGS date equation with an alternate formula. Then, first check your arithmetic because for the given D and P there is not 1 percent of parent isotope in the sample as you indicate, incorrectly, in your table.

The table column format is the only problem here. The format is set for decimal rather than percentage display. The 1 actually represents 100% in decimal format. ...IMHO (Talk) 08:21, 30 June 2006 (UTC)

The about 8000 yrs value is OK, but notice that it is called mean lifetime.

Now, plug the numbers in the equations and get the answers.

--Jclerman 07:27, 30 June 2006 (UTC)

So long as you make it clear for the laymen that the division of D by P is not the same as adding D and P (which total is always equal to the same value) then I think this will solve the problem. ...IMHO (Talk) 08:21, 30 June 2006 (UTC)

Another misunderstanding for laymen which this equation may encourage is that age of a sample as derived by the ratio of 14C is not a ratio of P to D but rather the ratio of archaeological 14C to current 14C. So long as this misunderstanding is cleared up then these writings may be comprehendible for the layman as well. ...IMHO (Talk) 08:50, 30 June 2006 (UTC)

First:

Your table, which was eventually called Transition of Carbon-14 to Nitrogen-14, reflected a major contradiction between its numbers and the narrative, precisely concerning the beta decay.

E.g.: According to the table,

(a) between Cycle 1 and Cycle 2:

5E+307 atoms of C14 decay into 1 single atom of N14. Where are the N14 missing atoms which amount to (5E+307)-1 ?

...

Okay I see the problem now. No this is not what the table shows. The table shows that for each atom of 14C that decays via Beta emission a corresponding atom of 14N is created to take is place. If you add a third column which I realise from what you are saying that I should have done to show the total nomber of atoms then you get the identical value throughout the third column. My point here was just to show that atoms are indivisible and all that is happening is that the atom is only undergoing a change in name and characteristics. I think I made this clarification somewhere but I'll look for it again. ...IMHO (Talk) 11:35, 30 June 2006 (UTC)

(b) between Cycle 1024 and Cycle 1025:

1 single atom of C14 decays into 1E+307 atoms of N14. Which process creates 1E+307 atoms of N14 out of a single atom of C14?

Needles to repeat it again and again, as it has been said already, the number of C14-decayed + N14-created atoms at each cycle should remain equal and constant throughout the whole table if it is to reflect the correct physics. (From: Jclerman 16:18, 25 June 2006 (UTC))

Second:

The P and D values suggested to you, i.e. 4 and 16, neither imply nor evaluate to 100% parent isotope.

These values were suggested by Vsmith not me. I have since used them as an index ranging from 0 to 20 to allow for a high number of atoms to be represented but if you only have 20 total atoms then use of 4 for the parent and 16 for the daughter as just one of twenty sets of values then it works out okay. ...IMHO (Talk) 11:35, 30 June 2006 (UTC)

Third:

The layman is expected to know arithmetic, hence that D/P is not equal to D+P.

That assumes that the layman was not raised on a fishing vessel and was not required to attend any school or that the layman did not attend his grandmother's funeral on the day when this was explained in class. You have to pretend like the laymen knows absolutely nothing and then try to find a way to cover all of the bases for everyone. That is what makes the wiki such an important medium of publication because there is virtually an unlimited number of contributors from all different kinds of expertise and understanding and misunderstanding backgrounds who's interaction results naturally in a publication that covers all needs. This is why you have to tolerate other people editing your stuff. Its not easy I know but it is the way of the wiki. ...IMHO (Talk) 11:35, 30 June 2006 (UTC)

Fourth:

Now, can you plug 4, 16, and 8000, in the formulas and obtain the resulting age? --Jclerman 10:53, 30 June 2006 (UTC)

I'm constructing several different tables based on the various interpretations and differences which we have discussed but before going further let me be sure we have the correct understanding now. You accept a total of 20 atoms starting with 20 14C and 0 14N and going to 0 14C and 20 14N for the values of P and D respectively, correct? ...IMHO (Talk) 11:35, 30 June 2006 (UTC)

Let me suggest once more that if you do understand the formulas and the meaning of P, D, and lambda, you should be able to plug in 4, 16, and 8000, and obtain an age. No tables are requested or needed. Just one age obtained from each one of the formulas that you wish to evaluate for their correctness. Can you now plug 4, 16, and 8000, in those formulas and show the results you obtain? --Jclerman 12:05, 30 June 2006 (UTC)

I think you are missing the point here. You have not clarified any of this in the article. You need to first clarify this in the article so that anyone who reads the article can comprehend the full meaning, relationship and purpose of P and D. Unitl you understand this it will serve no constructive purpose for me to follow through. ...IMHO (Talk) 12:22, 30 June 2006 (UTC)

No, you are missing the point. You claimed an error in the equation, yet you seem incapable of simply plugging in numbers to check. No again, anyone reading the article may not comprehend the full meaning. A reasonable background is assumed. My 7 year old granddaughter no doubt would not comprehend. And you apparently don't comprehend, but the equation and article are accurate and reasonable. Wikipedia is not written for 2nd graders - a reasonable background is assumed for most articles, especially those of a technical nature.

I assume that you lack the math ability to evaluate the equation with the simple numbers provided - as you have not done so and attempt to confuse the issue with unneeded tables and computer code. This obfustication of yours is getting tiresome. Oh, and 5568 is the value for the Libby half life of C-14 in case you were interested. Vsmith 13:47, 30 June 2006 (UTC)

Disection of age equation

We must complete a few more preliminaries before calculations can proceed.

First let me ask what role in your mind the following portion of the age equation ${\ln \left(1+{\frac {D}{P}}\right)}$ plays?

What is the role played by the portion ${1+{\frac {D}{P}}}$ and what is the purpose of taking it's natural logarithm?.

What in your mind is the role ${\frac {D}{P}}$ plays since age may also be calculated by the following?

$age={half\ life}\times {number\ of\ half\ lives\ }$

where

${number\ of\ half\ lives\ }={\frac {\ln \left({\frac {1}{percent}}\right)}{\ln \left({2}\right)}}$

and inversely

$percent={\frac {1}{\left({2^{number\ of\ half\ lives\ }}\right)}}$

where percent is defined by

$percent={\frac {{14C}_{archaeological}}{{14C}_{recent}}}$

Once you have explained each of the parts of the age equation requested above and commented on the merits of selecting or using one age equation over another I have several more questions regarding the relationship P has with D before calculations can be presented. ...IMHO (Talk) 16:21, 30 June 2006 (UTC)

Yor absurd waffle, We must complete a few more preliminaries before calculations can proceed.

Nonsense. Just plug in the numbers and crunch, what is so difficult about that? To verify the answer, you need to realize that a D/P ratio of 4 (16/4) can be restated as a fraction remaining f of 0.2 for checking with the other equation (n=log 0.2/log 0.5). The equation is valid and referenced to a reliable source, you original charge Equation as published produces incorrect results is without merit. Stop the waffling and obfustication and use the numbers provided per your request. Vsmith 22:43, 30 June 2006 (UTC)

Your statement that a ratio of D/P or (16/4) or 4 can be restated as p = 0.2 and used to check the other equation is incorrect. When we are talking about the ratio of D/P we are talking about apples whereas when we are talking about the ratio of

{\frac {P_{archaeological}}{P_{recent}}}

are talking about oranges.

Jclerman thought the values you provided for P and D were provided by me. There remain other misunderstandings between Jclerman and myself that need to be cleared up before we can proceed. Please stop injecting your comments so that we can proceed to resolve any misunderstandings we have between us without interruptions, interference or side bars. ...IMHO (Talk) 02:08, 1 July 2006 (UTC)

The ratio D/P of 16/4 means that we started with 16 + 4 = 20 of P and now we have only 4 remaining. And 4/20 = 1/5 = 0.2 = f the fraction remaining for the equation (n=log 0.2/log 0.5). Both of these relationships apply to radiometric dating and are based on and use the same data (they are not apples and oranges).

Jclerman knew where the numbers in question came from, read the discussion above again. Thanks for removing the butt out comment. This is a public discussion, if you want a private discussion then move it to your talk page please.

Now, have you done the simple calculations? Vsmith 20:44, 1 July 2006 (UTC)

Here is the problem: In a previous discussion ~~(which I am still searching for)~~ Nice try. But... Jclerman said that 14C continues to decay Ad infinitum. The not-so conclusive radiocarbon dating I know this is a misconception and so do many others so where might someone as obviously knowledgeable as Jclerman get the idea that 14C continues to decay Ad infinitum? To find out requires a little bit of detective work and in this case a little bit of interaction in hope of revealing the source of his misunderstanding. The only purpose of the table I created was to demonstrate that 14C can in fact decay to a zero value. I also explained this in response to the example provided by Melchior: (10 atoms). Then by chance Jclerman replaced the age equation I had submitted

age=h\times {\frac {\ln \left({\frac {1}{\frac {{14C}_{archaeological}}{{14C}_{recent}}}}\right)}{\ln \left({2}\right)}}

with

t={\frac {1}{\lambda }}{\times }{\ln \left(1+{\frac {D}{P}}\right)}

.

Ah Ha! The source of the misunderstanding at last! Now if only there was a way to expose it. All of this was very subconscious mind you. I did not plan to replace the constant 1 with the variable n but rather it just happened like a subconscious shot in the dark and up to the point of your interruption Jclerman and I were getting closer to resolving his misunderstanding. Unfortunately your interruption brought this opportunity to an end so that the only choice I have now is to return to using a direct approach which might not give Jclerman the opportunity he needs to realize for himself the basis for his misunderstanding.

I have prepared such a direct explanation but I was hoping that Jclerman might pickup and continue the discussion at the point prior to your injection but I have run out of time. If Jclerman still believes that 14C continues to decay Ad infinitum then I won't argue with him further. I'll just simply post what I feel is the reason for his misunderstanding here although I would have preferred that Jclerman be given and would have taken the opportunity to come to this understanding for himself. ...IMHO (Talk) 21:47, 1 July 2006 (UTC)

Conclusion

Please note at the bottom of the table that while ${\left(1+{\frac {D}{P}}\right)}$ can not be mathematically determined as the result of a divide by zero error when the value of P is zero (but is presumed to equal infinity) that P is in fact equal to zero when D is equal to 20. (BTW I no longer advocate replacement of the constant 1 with the variable n making calculations for it a mute point.) ...IMHO (Talk) 16:56, 2 July 2006 (UTC)

stop nonsense, please

Attn Pce3@ij.net aka IMHO:

You have been discussing the big bang and the biblical experiments on radioactive natural reactors and have been rebuked by everybody in many help desks and discussion pages. You have been dscussing also the scientists who hide intermediate results and those in the elitist towers of ivory, with about the same negative comments. Did you ever think that "others" might be correct in their assessments of your admittedly very creative approaches? BTW, we used to teach the hourglass model for radiocarbon dating already in the 1970s. See also comments below. --Jclerman 16:50, 5 July 2006 (UTC)

Attn unaware readers:

Follow the trail through the wikipedia help desks, policy boards, and the dicussion/talk pages of many related articles. They can be located by surfing the contribs list of User:Pce3@ij.net (who also signs as IMHO) during several weeks. Extensive discussions about dating formulas have ignored concepts such as asymptotes and exponentials, and the logical fact that nature's behavior can not be dictated by approximate and/or faulty formulas with, at best, restricted domain of validity. Physical phenomena can be approximated by mathematical expressions; however, mathematical expressions can not tell to the physical world how it is to behave. A layman's presentation of the hourglass model can be found in a web site of the USGS. Just search for "hourglass radiocarbon". See also comments above. --Jclerman 16:50, 5 July 2006 (UTC)

You have tried to make connections which do not exist and you have failed to recognize the deliberate and extensive effort I have made to give you (and anyone else with your misconceptions) the benefit of understanding the basis for your miscomprehensions. Not only have you failed to appreciate this extraordinary effort but to deny that any miscomprehension exists. I suggest that you STOP STALKING, GROW UP and GO BACK TO SCHOOL! ...IMHO (Talk) 18:09, 5 July 2006 (UTC)

Pce... please stop the nonsense. And please be be civil, your shouting of childish taunts do not belong in a civil discussion. The rather absurd calculations that you are trying to insert into the article fall under Wikipedia:Original research and for you to continue to attempt to push your OR POV by edit warring amounts to little more than vandalism. If you can find published concerns regarding the equation you are attempting to discredit, post the sources (note: as the equations and theory of radiometric dating have been well established in the scientific literature, only peer reviewed sources will carry much if any weight). --Vsmith 23:15, 5 July 2006 (UTC)

Straw poll

This straw poll is being conducted to determine which of the following points of view are favored by users so as to reach a neutral point of view.

In a sealed container where an unstable radioactive isotope is not replenished the isotope is never completely gone. It decays exponentially ad infinitum.
All atoms of an unstable radiocative isotope in a sealed container without benifit of replenishment which undergo the process of Beta decay will eventually be gone.
When the exponential decay calculation gives a result of less than one atom, the value calculated can be used to give a statistical probability that the last atom will have finally decayed (since random events do not follow a strict timeline.)
This last point was added by Wdanwatts, but even it's not really correct. The point at which the average number of atoms remaining drops below 1 is not the same as the point at which the probability of zero atoms remaining surpasses 50%. The two are, unfortunately, not easily tied together (as far as I know). This problem emerges because the standard half-life model weights probabilities with the number of atoms. A sample calculation of this can be found in an earlier iteration of this discussion at [3]; note that while this case shows a >50% chance of no atoms, the expected number of atoms is not particularly close to a "good-looking guess" of 1 atom remaining or 0.5 atoms remaining. — Lomn | Talk 02:55, 7 July 2006 (UTC)

This last part is entirely irrelevant because all phenomena in science must be observable, and without sounding too pedantic and querrulous, there's Heisenberg's Principle to consider when you get into arguments about whose equation better represents the probability of one atom which you can never measure in any circumstance in nature, disappearing from the universe.

Similarly, any equation of radiometric age determination in which one atom is used as the parent nucleide has no information to give on the subject anywa because, as alluded to above, this one atome could have been hanging on as the last example of its species within the sample for an indeterminate length of time (any number of half-lives), rendering this whole discussion moot. You don't even need to check the math, it makes logical sense. Rolinator 00:17, 10 July 2006 (UTC)

Please sign your name using four tildes (~~~~) under the position you support, preferably adding a brief comment. Since these are opposing points of view please do not sign your name to more than one.

Those who favor the "never gone" point of view (POV number one)

Those who favor the "eventually gone" point of view (POV number two)

I favor the expression of the second point of view as being neutral for the following reason:

Since atoms are indivisible then the number of parent isotopes atoms remaining when the variable P in the age equation $t={\frac {1}{\lambda }}{\times }{\ln \left(1+{\frac {D}{P}}\right)}$ is less than one will be zero. ...IMHO (Talk) 22:47, 5 July 2006 (UTC)

Those who are undecided

Those who favor the idea that under practical circumstances unstable atoms do not decay although they could decay and are indivisible units despite the mathematics used to determine their number as reliant upon decimal versus integer variables.

Those who recognize the irrelevancy of this straw poll

Irrelevant due to being based on a complete misunderstanding and attempted mis-representation of the issue by User:Pce3@ij.net. - Vsmith 00:36, 6 July 2006 (UTC)
- This is the only issue as far as I am concerned. However, please be mindful of the fact that if you think this issue is irrelevant then this is your opportunity to state exactly why you think that way instead of trying to dodge the issue by just making the claim that you think this issue is irrelevant. ...IMHO (Talk) 00:59, 6 July 2006 (UTC)
Read what I said. Pce3@ij has been trying to push is own unsourced calculations into the article. These calculations and this poll show his lack of comprehension of the statistical nature of radioactive decay and the meaning of exponential equations. Vsmith 01:20, 6 July 2006 (UTC)
An unfortunate misunderstanding by Pce3 that there must be a physical correspondence between the predictions of a mathematical model and the real world at all times and under all circumstances. TenOfAllTrades(talk) 01:33, 6 July 2006 (UTC)
- Only the need to clarify Nice try. But... User:Jclerman's statement that 14C continues to decay Ad infinitum while other references such as The not-so conclusive radiocarbon dating and A "New" Date for Mount Mazama's Climactic Eruption By Ron Mastrogiuseppe and Steve Mark seem to differ. ...IMHO (Talk) 01:41, 6 July 2006 (UTC)
My opinion is that you have misunderstood what has been explained to you on many pages for the last couple of weeks, and continuing to try to discuss it here is pointless. You're an otherwise productive editor who has done some very useful things—please, work on those things. TenOfAllTrades(talk) 01:50, 6 July 2006 (UTC)

This is a valid question. Please show good faith by commiting to one side of it or the other instead of continuing to evade. ...IMHO (Talk) 01:56, 6 July 2006 (UTC)
I've tried explaining things – Wikipedia and scientific – to you in the past. I've done so with good humour and patience. My assistance has been ignored when not treated with contempt, and that patience has been exhausted. You have refused to hear any answer from anyone that isn't what you want to hear. Invoking the assumption of good faith while you've been edit warring with Jclerman (and calling him a vandal) over his identification of you as the creator of this pointless straw poll takes a remarkable amount of chutzpah. Trying to order me to participate in your waste of time is unwelcome and absurd. TenOfAllTrades(talk) 02:07, 6 July 2006 (UTC)
The place for identification is at the bottom of the straw poll where it is not embeded in the introductory text of the straw poll which is an obvious attempt on the part of Jclerman to disrupt this straw poll. Again you need to commit to one side of the question or the other instead of making personal attacks. ...IMHO (Talk) 02:21, 6 July 2006 (UTC)
This OR has been discussed extensively, receiving negative comments in many forums. Regretfully, User:Pce3@ij.net does not accept that "others" might be correct in their assessments of his, admitedly, very creative approaches, e.g., the one that he published as an unreviewed Wikibook. The topics that originated the current "poll" can best be followed on a trail threaded along the wikipedia help desks, policy boards, and the dicussion/talk pages of many related articles. They can be located by surfing the contribs list of User:Pce3@ij.net (who also signs as IMHO) during the recent several weeks. Such trail shows extensive and obfuscating discussions about dating formulas, being rebutted by a host of users: (a) because he ignores concepts such as asymptotes, exponentials, and division by 0, and (b) because he does not understand the fact that nature's behavior can not be dictated by formulas. True, physical phenomena can be approximated by mathematical expressions; however, the reverse is not true: mathematical expressions can not and do not dictate to the physical world how it is to behave. --Jclerman 02:05, 6 July 2006 (UTC)
- This has been and is my point exactly. However, it is you who have missed this point but are now slowly and reluctantly coming around. It should not be long now before you realise that a value of a function need only be less than one and not reach an asymptote or zero when integers and whole numbers and whole atoms are involved before question number two above is qualified. Once you realize this your problem and the one you have caused everyone else will finally be resolved. ...IMHO (Talk) 02:14, 6 July 2006 (UTC)
Pointless timesink that ought to be abandoned. It seems transparently clear to me that deliberately inflammatory interpretations of that "nice try" thing are being taken.
Yes, we all agree that atoms decay as units. Yes, we all agree that all units of X could decay. However, we can also say that, under all practical circumstances, complete decay doesn't occur, and we should be able to accept that generalization. Similarly, nobody complains when a falling-body calculation ignores the attractive force that a bowling ball exerts on the Earth, even though it's physically real. — Lomn | Talk 03:56, 6 July 2006 (UTC)
My purpose here is to determine the basis of support for a particular point of view such as expressed by Jclerman in his "nice try" comment and to be sure the article has benefit of the rationale behind such thinking. Otherwise the reader may be left in doubt as to the validity or impartiality of the article. In that regard you need to give the reason why you feel that complete decay does not occur other than the fact that the above age function neither permits P to reach a zero asymptote or for P to be regarded as an integer rather than as a decimal value whenever it has an actual value that is less than one. Is there some lab data or other reason you can site that might serve as the basis of your opinion that isotopes continue to decay ad infinitum? ...IMHO (Talk) 09:32, 6 July 2006 (UTC)

This is exactly my point. I have specifically stated that atomic decay does not proceed to infinity (only that it's a process that can be adequately generalized as such). However, rather than looking to resolve a dispute, you're looking to continue and exacerbate one. Further, as noted elsewhere, your notions of how and where radioactive decay ends ("value of a function need only be less than one") is completely wrong (see User:Lomn/Sandbox for my experimental data on radioactive decay).

The math in question, though, is a sidepoint. It's perfectly acceptable to be wrong or mistaken, or even to be right and still leave an article unchanged, particularly when the content under dispute makes up such a small part of the article(s) in question. What's unacceptable is your combative nature of demanding others show good faith while ignoring it yourself. — Lomn | Talk 13:14, 6 July 2006 (UTC)
The article is entitled Radiometric dating yet shows only one method of computation. The equation that is used fails to clarify whether radioactive isotope decay continues ad infinitum or terminates when the number of isotope atoms is less than one. I think this is a relevant point and one which needs to be addressed and answered in the article for the benefit of readers. Please limit you comments to that concern. Thanks. ...IMHO (Talk) 14:58, 6 July 2006 (UTC)
Alright, then, to that concern: your understanding of the process (or at least your means of presenting it) is flawed and should not be included. A mathematical model reaching a given value does not dictate the actual behavior of the physical process modeled. Note my link above: of 6 trial runs, only once does decay terminate when you say it should. As has been explained to you before, the actual physical process is probabilistic, and as such, no firm statement can be made that radioactive decay of a given sample will or will not cease at a given point—that is, the potential for decay may be present ad infinitum, though such a case is unlikely. Decay may instead proceed exactly per the model, exactly halving the sample at every half-life, but this case is also unlikely.

Anyway, it looks like consensus is pretty clear, so this will be my last reply on this poll. — Lomn | Talk 15:49, 6 July 2006 (UTC)
All that you have said regarding an explanation for the reader should be included in the article but it is not and you need to include the measure of unlikelihood or probability of the potential for decay being ad infinitum in the article. As with DNA evidence presented to juries most persons now understand high values of unlikelihoodedness such as 10 billion to one. You need to cite such unlikelihoodedness in the article. ...IMHO (Talk) 00:41, 7 July 2006 (UTC)

Unlikelihoodedness...I like that word. It is so long, involved, obfuscatory and redundant. Its like your whole argument, in microcosm.Rolinator 00:45, 10 July 2006 (UTC)

Immediately following the age equation section is the limitation of techniques section, which states: For instance, carbon-14 has a half-life of less than 6000 years. After an organism has been dead for 60,000 years, so little carbon-14 is left in it that accurate dating becomes impossible. Does that not make it clear that unstable isotopes reduce in number over time to the point where you lose statistical significance? I don't understand your obsession with these equations. —Bradley 22:52, 6 July 2006 (UTC)
The reason I have no problem with the following equation is that the value of the amount of isotope in the recent sample does not change in correspondence with the amount of isotope in the archaeological sample (initial and terminal amount of isotope in recent sample remain the same) and if a zero value of isotope is encountered in the recent sample all that is suggested is that there may be a problem with the recent sample rather than with the mathematics, i.e, $age=h\times {\frac {\ln \left({\frac {1}{\frac {{14C}_{archaeological}}{{14C}_{recent}}}}\right)}{\ln \left({2}\right)}}$

Although an initial value of zero for P in the following equation will also indicate an error in the amount of isotope in the archaeological sample the following equation does not permit a terminal value of zero for P even if a terminal value of zero for P is physically possible. Thus leading to a deception of the fact that it is possible for a terminal value of zero for P, regardless of how improbable, to be reached in the real world. $t={\frac {1}{\lambda }}{\times }{\ln \left(1+{\frac {D}{P}}\right)}$

...IMHO (Talk) 01:22, 7 July 2006 (UTC)
You truly don't see it do you? Is (2/3)^-1 not equal to 3/2? Think about that a bit. Both equations above suffer the same irrelevant problem. Vsmith 01:54, 7 July 2006 (UTC)

Your equation is the same. You do know that the natural log of zero is undefined, right? That the limit as you approach zero is negative infinity, but that infinity, positive or negative, is not a number? Do you also realize that every single mathematical model used to describe things in the universe has limits to its applications? The mathematical model for the spring force assumes the spring does not undergo plastic deformation. If I were to not assume good faith, I would think you are a young-Earth creationist looking to "disprove" radiocarbon dating—if I were to not assume good faith, of course. The article states that after long enough insufficent parent isotope remains. The article makes it abundently clear that other isotopes then have to be used. Also, this is all based on the law of large numbers—if you have too small of a sample, the randomness overwhelms the situation and you cannot use the statistical model. —Bradley 22:13, 7 July 2006 (UTC)

That's great for everyone except laymen and students unless you explain this for them in the article text. But then I realize that you are not editing for all readers but primarily only for yourselves... ...IMHO (Talk) 18:03, 8 July 2006 (UTC)

I assume that only an inebriate would use a zero value of Carbon-14 isotope i.e, $0=14C_{recent}\$ measured in a recent sample to attempt to find the percentage of Carbon-14 isotope measured in an archaeological sample using the following equation. $percent={\frac {{14C}_{archaeological}}{{14C}_{recent}}}$ . If they did they would get a divide by zero error and the reason for this error would be attributed to a faulty sample or faulty measure rather than to an inappropriate mathematical expression. This is also true when modeling the equation.
I likewise assume that only an inebriate would use a zero value of Carbon-14 isotope i.e., $0={archaeological_{Carbon-14}}\$ measured in an archaeological sample to attempt to find the proportion of daughter to parent isotopes using the following relation: ${\frac {archaeological_{D}}{archaeological_{P}}}$ or ${\frac {archaeological_{Nitrogen-14}}{archaeological_{Carbon-14}}}$ or ${\frac {archaeological_{D}}{0}}$ or ${\frac {archaeological_{Nitrogen-14}}{0}}$ Should they do this they would get a divide by zero error and the reason for this error would likewise be attributed to a faulty sample or faulty measure rather than to an inappropriate mathematical expression.

However, when one models the process of radioactive decay (which students will often do as a learning aid) beginning with a value for an archaeological sample that is greater than zero, i.e., $0<{archaeological_{Carbon-14}}\$ and then decrements the value of the number of parent isotope atoms in the model until they are zero then a divide by zero error will occur which instead of being attributed to a faulty sample suggests that the number of parent isotope atoms can never have a value of zero when in reality a value of zero number of parent isotope atoms is possible although possibly not 100% probable. The results of modeling the equation can therefore be misleading in this regard unless these facts are clearly and fully explained. ...IMHO (Talk) 09:14, 7 July 2006 (UTC)

Of course it is possible to have a zero number of atoms of the parent nucleide. Since we are talking C-14, lets assume you analyze a sample of zircon, which is several billion years old. You probably had some C-14 in there at some stage. Maybe one part per billion Carbon, and some trillionth of C-14. Several billion years pass. You can no longer measure any carbon, or C-14. Does this make the whole process of radiometric dating irrelevant, or does it mean you simply cannot physically measure the lone remaining C-14 atom?

The other problem you have with your brain is the incapacity to recognise that arguing about this is irrelevant because you do no understand probability maths. Probability assumptions work in the real world on large number populations or samples; depending on the variability of the dataset, "large" is a woolly figure but we can assume that 1.6X10^23 is a large number of atoms. Most atoms are present in astoundingly large numbers, which renders the inherent probabilistic decay of each individual atom meaningless to an equation.

In other words, not like you're actually paying attention, when you have enough atoms, the probability chances of decay and the randomness of the actuall occurrences even out and approximate the decay equation, which humans use with all their frailties, to derive ages which tell us that Moses was a liar. But I digress.

When you reduce the population of atoms to a small enough sample, the individual probailities of the atoms overwhelm the mathematical utility of averaging the probability of a whole population and you are left with a situation where, due to the extreme unlikelihood of a decay happening at any exact time but the overwhelming likelihood of the atom decaying eventually, the assumptions inherent in the radioisotope decay equation are invalid for describing the behaviour of that populaton.

Ergo, yes, the equation is inappropriate for use on small populations of parent nucleides. We all admit that. This is what other people call "approximation" and "fudge factor". Because, aside from some very bright theoretical mathematicians, no one really cares. And in the end, if the equation does not describe the behaviour properly and should not be used why are you so worried? No one here would say "Oh, i have 4 C-14 atoms and X number of N daughter atoms, how old is it? Oh, wait, a C-14 atom decayed a second ago and made my sample 75% older! You dipshit. Rolinator 00:45, 10 July 2006 (UTC)

What Vsmith said. Melchoir 04:26, 6 July 2006 (UTC)

Summary statement

The following facts are known and have been established: Atoms of Carbon-14 decay as units and all units of Carbon-14 can decay completely.Lomn It is calculated that a mole of Carbon-14 atoms would decay to less than one atom within about a half million years. The age equation represents a statistical prediction of the rate of decay rather than an infinite curve.Woodstone Suppose that we could take the entire atmosphere and put it in a perfectly-sealed container for a million years. The atmosphere has a mass of about 5*10^21 g of which about 0.053% is CO2 Density and Mass. Thus the atmosphere contains about 6*10^16 mol C, and so roughly 4*10^28 atoms of 14C. Consider any one of those atoms: after 5730 years, the probability that it has not decayed is 0.5; so after a million years, the probability that it has not decayed is about 3*10^(-53). The probabilities for the different atoms decaying are independent. Thus the probability that even one atom of 14C is left in the container after a million years is about 10^(-24), i.e. effectively zero.Daphne A. There are two equations that may be used to calculate age.imho The first equation is based upon the percent of isotope in the archaeological sample as compared with the amount of isotope in a recent sample as being equal to 100%. Whether utilizing or modeling this equation a zero amount of isotope in the recent sample is not accepted and attributed to sample error. Although the second equation is based upon the proportion of daughter to parent isotope in the archaeological sample and an ‘’initial’’ zero amount of parent isotope would likewise be rejected in the case of ‘’’’’modeling’’’’’ the equation a value of parent isotope greater than zero amount would be accepted and then decremented. However, since a divide by zero error occurs whenever the value of the parent isotope amount reaches zero in the equation making it appear that the number of parent isotopes can never reach a zero amount it must be clarified when this equation is used in the article that the number of parent isotope atoms will equal zero whenever the parent isotope amount in the equation is less than one.imho ...IMHO (Talk) 18:24, 9 July 2006 (UTC)
Since the development of personal computers and sophisticated software packages such as Excel spreadsheet the ability of the student or laymen to model equations has become commonplace and routine. Therefore it has become necessary to clarify the results of such modeling when applied to the age equations… ...IMHO (Talk) 10:59, 8 July 2006 (UTC)
This hardly seem to "best summarize and integrate the results of this straw poll." The result of the straw poll appears to be that this is a non-issue and needs no further consideration. —Bradley 14:51, 8 July 2006 (UTC)

If you look very closely at the three little continuation periods at the end of last sentence in the paragraph above and read the edit summary statement that accompanied this post you will see that it was offered simply as an introductory suggestion for anyone who might want to come down off their high horse and actually make a contribution to the Wikipedia rather than being a complete summary unto itself. ...IMHO(Talk) 18:05, 8 July 2006 (UTC)

Concur with Bradley—the only editor who seems to see that this is an issue is Pce3. TenOfAllTrades(talk) 17:35, 8 July 2006 (UTC)

Mine and that of three other teachers and forty-five of our students who are discouraged that this article and others like it are being dominated by old timers who have no clue what the use of computers to model equations is all about. ...IMHO (Talk) 18:10, 8 July 2006 (UTC)

Well, we are all suitably castigated that you and your ill-informed lackeys used a computer to do this equation and you got a !DIV/0 in your excel spreadsheet, thus proving the rest of us wrong. You know you can't argue with computers guys! So everyone stop, lay down your pens and papers and abaci and go home. His computer proved you wrong when it spat out a !DIV/0.

May I suggest closing and archiving this section?--Jclerman 20:31, 7 July 2006 (UTC)

This subsection is for the purpose of providing a summary statement for the benefit of readers who may be confused by this point and not the place for you to reiterate your inability to comprehend the problem from a lay or student point of view. Hiding information from students will not solve the problem but only make them question why the problem has not been addressed in the past only to learn that it was hidden away as if that would make it forever go away. Please give others who are confronted by this question the right and the opportunity to review previous coverage. Thanks. ...IMHO (Talk) 00:56, 8 July 2006 (UTC)

...IMHO (Talk) 00:24, 6 July 2006 (UTC

I would now like to thank everyone for their participation in this straw poll and ask that each of you post a statement beneath the "Summary statement" subsection above that will in their opinon best summarize and integrate the results of this straw poll into the article text proper for the benefit of readers by July 10, 2006 at 21:00 EST or July 11, 2006 at 00:00 UTC. Thanks again to everyone for your participation. The experience has been great! ...IMHO (Talk) 15:30, 7 July 2006 (UTC)

This year the class was granted a choice between donating the money they earned along with the sponsorships they received to the Wikimedia Foundation or to pay for a field trip to a nuclear facility where they could talk with nuclear scientist and engineers. Rather than waiting until the last day of the straw poll the class unanimously voted to spend 100% of its funds on the field trip. Thank you for helping our students reach their decision and good luck next year. ...IMHO (Talk) 18:33, 9 July 2006 (UTC)