Earth science curriculum tailored to fit wavering fundamentalists

That's completely wrong. The amount of CO2 in the atmosphere has nothing at all to do with the production rate of C-14 from Nitrogen-14. The large amount of carbon can only buffer the decay rate of C-14. I suppose that this is a good argument that C-14 is generally not in equilibrium.Edited by NoNukes, : No reason given.

Perhaps I misunderstood you, but you helped. To wit: Whether or not such a calculation is acceptable depends on whether you care about or choose to ignore the inaccuracies such a calculation produces. If you care about 20% errors then you have to parse out that 'nearly'. It is not clear to me what they meant. I'm asking what Libby meant and what the non-experts meant when they had their dispute with Libby. Well here you are just wrong.In fact the amount of C-14 in the atmosphere does affect decay rate. The decay rate of C-14 is proportional to the product of the amount of C-14 and the decay constant of C-14. The decay constant lambda is of course constant, and as you've argued, the amount of variation in the amount of C-14 is swamped by the amount already in the atmosphere.Just what do you mean when you refer to the decay rate?Edited by NoNukes, : No reason given.

Except that the assumption of constant production rate is incorrect. And, as you have indicated the assumption does not account for the depression of C-14 by the release of old carbon. You've only begun to tell the story here. The variations in the earth's magnetic field are not periodic.What is important is that over the range of time, the amount of C-14 in the atmosphere, when averaged over quite some time, has not fluctuated by huge amounts over 50,000 years. But even over 10,000 year periods, the amount has varied enough to need correcting. My question is simply what definition of equilibrium did Libby argue for and was he correct or incorrect. Why would I want to skew the data? The truth is that even if you use real data, the production rate of C-14 in the atmosphere often does not match the current rate of depletion. In fact, given the long half life of C-14 compared to say, the solar cycle, the product rate and the removal rate may be different more often than not.Henry Morris (and Hovind) argued that the earth could not be old because equilibrium would have been reached long ago. Morris's argument is wrong not because equilibrium has been reached, but because there are good reasons why equilibrium does not exist. Hovind of course is a convicted liar and felon.Edited by NoNukes, : No reason given.

Nonetheless, given that this was a controversy about equilibrium in Libby's time that I want to understand, I am still asking about equilibrium. I understand the calibration of dates. Those things have been expounded on here at length in this very thread. I am not challenging the concept of C-14 dating.Telling me that equilibrium is irrelevant for dating is not helpful here. It is also the case the expounding on the basic physics of nuclear decay and glossing over variations by talking about 'average rates', might be reasonable in another context. But that is not helpful to me either. I understand the physics behind the situation at a level approaching your own. Apparently there are a few things about the subject where I know more than you do. My reason for asking is that I was pretty sure you were making that mistake.I recommend that you check the definitions. Decay rate means the number of disintegrations over a period of time. The half life, time constant and decay constant are actually constants although some of those things (namely the decay constant) have reciprocal relationships to the half life. But the decay rate depends on how large the sample is.The decay rate is constant per gram of material. But for an amount of material, such as the C-14 in the atmosphere, the rate of removal, which is essentially the decay rate is given by -N * lambda, where lambda is the decay constant and N is the amount. It's also pretty clear that N*lambda is the value that must be in equilibrium with the production rate, which ought to be enough to remove any ambiguity from what I posted.Perhaps a source of confusion is that lambda is sometimes called 'the decay rate constant' because it is the constant factor used to calculate the decay rate.http://formulas.tutorvista.com/...rate-of-decay-formula.html

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"The rate of decay is the number of nuclei that decay each second. It is measured in becquerel(Bq) where 1Bq = 1 decay/s."

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Radioactive Decay Rates - Chemistry LibreTexts

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Decay Rates
Due to the smaller size of the nucleus compared to the atom and the enormity of electromagnetic forces, it is impossible to predict radioactive decay. The atomic nucleus which is in the center of the atom is buffered by surrounding electrons and external conditions. Because of this, the study of decay is independent of the element's environment. In other words, the decay rate is independent of an element's physical state such as surrounding temperature and pressure. For a given element, the decay or disintegration rate is proportional to the number of atoms and the activity measured in terms of atoms per unit time. If "A" represents the disintegration rate and "N" is number of radioactive atoms, then the direct relationship between them can be shown as below:

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A∝N
or mathematically speaking

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A=λN
where

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A is the Total activity and is the number of decays per unit time of a radioactive sample.
N is the total number of particles in the sample.
λ is the constant of proportionality or decay constant.

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Hope this helps...Edited by NoNukes, : No reason given.Edited by NoNukes, : Provide references

Well given that your contribution to addressing my question is to pretend that there was no question at all, I don't see that you've been very helpful. Apparently if I want to know what Libby and his detractors are arguing about, I'll have to look it up.And I didn't claim to know more than everyone else. Just more than you, a person who feels the need to correct me when I am 'technically correct' and then complains about the fact that he is in a semantic argument.

More precisely, Libby defines equilibrium based on the average rate of production of C-14 and provides some evidence/argument that the the C-14 level was consistent with equilibrium (within 10%) at the point in the time Libby made the argument. The last couple of posts by kbertsche pretty much nail that down at least to my satisfaction.By contrast, people like Morris were making the claim that the current levels of C-14 were well away from equilibrium. Whether or not Morris was using good information or bad information about the production rate is part of the problem. The other part of the problem is that the sources Morris was relying were already at least speculating on the fact that C-14 production varied with the earth's magnetic field which meant that simple determinations of equilibrium, unlike the one's Libby apparently used, were not valid.

Edited by NoNukes, : No reason given.

Most people would see that as an argument against the Flood.

I am tutoring several young folks in chemistry this semester, so perhaps I am overly sensitive, but I think your definition is just a tiny bit narrower than the one used in chemistry. From the wikipedia artilce (emphasis added by me).Precipitation - Wikipedia(chemistry)

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Precipitation is the creation of a solid. When the reaction occurs in a liquid solution, the solid formed is called the 'precipitate'.

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Sometimes the formation of a precipitate indicates the occurrence of a chemical reaction. If silver nitrate solution is poured into a solution of sodium chloride, a chemical reaction occurs forming a white precipitate of silver chloride. When potassium iodide solution reacts with lead nitrate solution, a yellow precipitate of lead iodide is formed.

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Precipitation may occur if the concentration of a compound exceeds its solubility (such as when mixing solvents or changing their temperature). Precipitation may occur rapidly from a supersaturated solution.

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As I understand it, even a non-ionic substance, when coming out of a solution due to change in solubility or temperature may be precipitation. However in the case of a non-ionic substance no chemical reaction has taken place.

There is no question that Faith's statement was imprecise. However Admin's correction lacked some precision as well (cleared up for the most part in his subsequent posts). I think petrophysics1 could rightly take issue with the following: What does in our dirt refer to? It appears to me not to refer to just the grand canyon, but to sedimentation in general. But even if we limit it to the GC, surely we can find gypsum, limestone and metal carbonates and sulphates in places in the Canyon. For example the Redwall and Muav includes limestone layers which are in part chemical sediments.

I can find images of the AZ lab results with dates, but those images do not include the notations about contamination. The warnings given by the University of Arizona lab about contamination are described in some references but without any images. There may be a note about it in the link that PaulK provided.

You might want to consider visiting a local university library if there is one in your area. The public and private universities nearest my area are all open to the public, and I believe most of them allow you to check out materials and use their computers if you pay a nominal annual fee. (for example UNC-Chapel Hill charges about 25 dollars per year)