Unwarranted claim of higher degree of accuracy in zircon geochronology












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$begingroup$


The uncertainty in the half life of uranium-238 is stated at 0.05% [1]. The same paper gives the date 251.941 myr +- 31kyr.
251.941 X 0.05% = 125 kyr.



How are the authors justified in claiming an accuracy of +- 31 kyr when the uncertainty in the half life alone is +- 125 kyr? On top of that, they list two other kinds of analytical uncertainties, which would only increase the overall uncertainty.



[1] ​ Burgess, S. ​et al​, “High-precision timeline for Earth’s most severe extinction,” Proceedings of the National Academy of Sciences USA​, Volume 111, 2014.
https://www.pnas.org/content/pnas/111/9/3316.full.pdf






measurements mass-extinction geochronology uranium







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    $begingroup$


    The uncertainty in the half life of uranium-238 is stated at 0.05% [1]. The same paper gives the date 251.941 myr +- 31kyr.
    251.941 X 0.05% = 125 kyr.



    How are the authors justified in claiming an accuracy of +- 31 kyr when the uncertainty in the half life alone is +- 125 kyr? On top of that, they list two other kinds of analytical uncertainties, which would only increase the overall uncertainty.



    [1] ​ Burgess, S. ​et al​, “High-precision timeline for Earth’s most severe extinction,” Proceedings of the National Academy of Sciences USA​, Volume 111, 2014.
    https://www.pnas.org/content/pnas/111/9/3316.full.pdf






    measurements mass-extinction geochronology uranium







    share|improve this question







    New contributor




    sidharth chhabra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      $begingroup$


      The uncertainty in the half life of uranium-238 is stated at 0.05% [1]. The same paper gives the date 251.941 myr +- 31kyr.
      251.941 X 0.05% = 125 kyr.



      How are the authors justified in claiming an accuracy of +- 31 kyr when the uncertainty in the half life alone is +- 125 kyr? On top of that, they list two other kinds of analytical uncertainties, which would only increase the overall uncertainty.



      [1] ​ Burgess, S. ​et al​, “High-precision timeline for Earth’s most severe extinction,” Proceedings of the National Academy of Sciences USA​, Volume 111, 2014.
      https://www.pnas.org/content/pnas/111/9/3316.full.pdf






      measurements mass-extinction geochronology uranium







      share|improve this question







      New contributor




      sidharth chhabra is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      $endgroup$




      The uncertainty in the half life of uranium-238 is stated at 0.05% [1]. The same paper gives the date 251.941 myr +- 31kyr.
      251.941 X 0.05% = 125 kyr.



      How are the authors justified in claiming an accuracy of +- 31 kyr when the uncertainty in the half life alone is +- 125 kyr? On top of that, they list two other kinds of analytical uncertainties, which would only increase the overall uncertainty.



      [1] ​ Burgess, S. ​et al​, “High-precision timeline for Earth’s most severe extinction,” Proceedings of the National Academy of Sciences USA​, Volume 111, 2014.
      https://www.pnas.org/content/pnas/111/9/3316.full.pdf






      measurements mass-extinction geochronology uranium




      measurements mass-extinction geochronology uranium






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      sidharth chhabrasidharth chhabra

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          1 Answer
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          $begingroup$

          I can't be entirely sure but I'll make an informed guess:



          That value doesn't come for a single measurement. Therefore, if the error in the age of a single sample is $pm125$ kyr, you just need to average 16 samples to get it down to $pm31$ kyr.



          The uncertainty in the addition (or substraction) of two or more quantities is equal to the square root of the addition of the squares of the uncertainties of each quantity (assuming they arise from random errors). For example, if we have quantity A with uncertainty $sigma_a$ and quantity B with uncertainty $sigma_b$, the error in the quantity $C=A+B$ would be:



          $sigma_c=sqrt{sigma_a^2+sigma_b^2}$



          And if we call M to the average between A and B. The uncertainty in the average is



          $sigma_m=frac{sqrt{sigma_a^2+sigma_b^2}}{2}$



          So if we average 16 samples with $sigma=125$ kyr, the uncertainty in the average would be



          $sigma_m=frac{sqrt{16 sigma^2}}{16}=frac{sqrt{16 times 125^2}}{16}=31$ kyr



          Uncertainty propagation can be seen in the abstract of the article you refer to:




          The extinction occurred between 251.941 ± 0.037 and 251.880 ± 0.031
          Mya, an interval of 60 ± 48 ka.




          Where the ±48 ka comes from?



          $sqrt{37^2+31^2}=48$






          share|improve this answer









          $endgroup$













          • $begingroup$
            As far as I know, this only applies if the uncertainties are independent. If the uncertainty is in the length of your measuring stick (the half-life of U-238), the uncertainties are correlated and you can't reduce them by making more measurements.
            $endgroup$
            – Mark
            11 mins ago











          Your Answer





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          1 Answer
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          active

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          active

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          2












          $begingroup$

          I can't be entirely sure but I'll make an informed guess:



          That value doesn't come for a single measurement. Therefore, if the error in the age of a single sample is $pm125$ kyr, you just need to average 16 samples to get it down to $pm31$ kyr.



          The uncertainty in the addition (or substraction) of two or more quantities is equal to the square root of the addition of the squares of the uncertainties of each quantity (assuming they arise from random errors). For example, if we have quantity A with uncertainty $sigma_a$ and quantity B with uncertainty $sigma_b$, the error in the quantity $C=A+B$ would be:



          $sigma_c=sqrt{sigma_a^2+sigma_b^2}$



          And if we call M to the average between A and B. The uncertainty in the average is



          $sigma_m=frac{sqrt{sigma_a^2+sigma_b^2}}{2}$



          So if we average 16 samples with $sigma=125$ kyr, the uncertainty in the average would be



          $sigma_m=frac{sqrt{16 sigma^2}}{16}=frac{sqrt{16 times 125^2}}{16}=31$ kyr



          Uncertainty propagation can be seen in the abstract of the article you refer to:




          The extinction occurred between 251.941 ± 0.037 and 251.880 ± 0.031
          Mya, an interval of 60 ± 48 ka.




          Where the ±48 ka comes from?



          $sqrt{37^2+31^2}=48$






          share|improve this answer









          $endgroup$













          • $begingroup$
            As far as I know, this only applies if the uncertainties are independent. If the uncertainty is in the length of your measuring stick (the half-life of U-238), the uncertainties are correlated and you can't reduce them by making more measurements.
            $endgroup$
            – Mark
            11 mins ago
















          2












          $begingroup$

          I can't be entirely sure but I'll make an informed guess:



          That value doesn't come for a single measurement. Therefore, if the error in the age of a single sample is $pm125$ kyr, you just need to average 16 samples to get it down to $pm31$ kyr.



          The uncertainty in the addition (or substraction) of two or more quantities is equal to the square root of the addition of the squares of the uncertainties of each quantity (assuming they arise from random errors). For example, if we have quantity A with uncertainty $sigma_a$ and quantity B with uncertainty $sigma_b$, the error in the quantity $C=A+B$ would be:



          $sigma_c=sqrt{sigma_a^2+sigma_b^2}$



          And if we call M to the average between A and B. The uncertainty in the average is



          $sigma_m=frac{sqrt{sigma_a^2+sigma_b^2}}{2}$



          So if we average 16 samples with $sigma=125$ kyr, the uncertainty in the average would be



          $sigma_m=frac{sqrt{16 sigma^2}}{16}=frac{sqrt{16 times 125^2}}{16}=31$ kyr



          Uncertainty propagation can be seen in the abstract of the article you refer to:




          The extinction occurred between 251.941 ± 0.037 and 251.880 ± 0.031
          Mya, an interval of 60 ± 48 ka.




          Where the ±48 ka comes from?



          $sqrt{37^2+31^2}=48$






          share|improve this answer









          $endgroup$













          • $begingroup$
            As far as I know, this only applies if the uncertainties are independent. If the uncertainty is in the length of your measuring stick (the half-life of U-238), the uncertainties are correlated and you can't reduce them by making more measurements.
            $endgroup$
            – Mark
            11 mins ago














          2












          2








          2





          $begingroup$

          I can't be entirely sure but I'll make an informed guess:



          That value doesn't come for a single measurement. Therefore, if the error in the age of a single sample is $pm125$ kyr, you just need to average 16 samples to get it down to $pm31$ kyr.



          The uncertainty in the addition (or substraction) of two or more quantities is equal to the square root of the addition of the squares of the uncertainties of each quantity (assuming they arise from random errors). For example, if we have quantity A with uncertainty $sigma_a$ and quantity B with uncertainty $sigma_b$, the error in the quantity $C=A+B$ would be:



          $sigma_c=sqrt{sigma_a^2+sigma_b^2}$



          And if we call M to the average between A and B. The uncertainty in the average is



          $sigma_m=frac{sqrt{sigma_a^2+sigma_b^2}}{2}$



          So if we average 16 samples with $sigma=125$ kyr, the uncertainty in the average would be



          $sigma_m=frac{sqrt{16 sigma^2}}{16}=frac{sqrt{16 times 125^2}}{16}=31$ kyr



          Uncertainty propagation can be seen in the abstract of the article you refer to:




          The extinction occurred between 251.941 ± 0.037 and 251.880 ± 0.031
          Mya, an interval of 60 ± 48 ka.




          Where the ±48 ka comes from?



          $sqrt{37^2+31^2}=48$






          share|improve this answer









          $endgroup$



          I can't be entirely sure but I'll make an informed guess:



          That value doesn't come for a single measurement. Therefore, if the error in the age of a single sample is $pm125$ kyr, you just need to average 16 samples to get it down to $pm31$ kyr.



          The uncertainty in the addition (or substraction) of two or more quantities is equal to the square root of the addition of the squares of the uncertainties of each quantity (assuming they arise from random errors). For example, if we have quantity A with uncertainty $sigma_a$ and quantity B with uncertainty $sigma_b$, the error in the quantity $C=A+B$ would be:



          $sigma_c=sqrt{sigma_a^2+sigma_b^2}$



          And if we call M to the average between A and B. The uncertainty in the average is



          $sigma_m=frac{sqrt{sigma_a^2+sigma_b^2}}{2}$



          So if we average 16 samples with $sigma=125$ kyr, the uncertainty in the average would be



          $sigma_m=frac{sqrt{16 sigma^2}}{16}=frac{sqrt{16 times 125^2}}{16}=31$ kyr



          Uncertainty propagation can be seen in the abstract of the article you refer to:




          The extinction occurred between 251.941 ± 0.037 and 251.880 ± 0.031
          Mya, an interval of 60 ± 48 ka.




          Where the ±48 ka comes from?



          $sqrt{37^2+31^2}=48$







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 59 mins ago









          Camilo RadaCamilo Rada

          12.3k33882




          12.3k33882












          • $begingroup$
            As far as I know, this only applies if the uncertainties are independent. If the uncertainty is in the length of your measuring stick (the half-life of U-238), the uncertainties are correlated and you can't reduce them by making more measurements.
            $endgroup$
            – Mark
            11 mins ago


















          • $begingroup$
            As far as I know, this only applies if the uncertainties are independent. If the uncertainty is in the length of your measuring stick (the half-life of U-238), the uncertainties are correlated and you can't reduce them by making more measurements.
            $endgroup$
            – Mark
            11 mins ago
















          $begingroup$
          As far as I know, this only applies if the uncertainties are independent. If the uncertainty is in the length of your measuring stick (the half-life of U-238), the uncertainties are correlated and you can't reduce them by making more measurements.
          $endgroup$
          – Mark
          11 mins ago




          $begingroup$
          As far as I know, this only applies if the uncertainties are independent. If the uncertainty is in the length of your measuring stick (the half-life of U-238), the uncertainties are correlated and you can't reduce them by making more measurements.
          $endgroup$
          – Mark
          11 mins ago










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