I have the following formula for diffusivity: D=$\frac{\omega_n \Delta r^2}{2(\ln(\gamma))^2}$. I have found $\Delta r$ in lab, and it has an uncertainty of $0.086$. My question is how do I calculate the uncertainty on the diffusivity? Usually I would either go for adding the fractional errors in quadrature, or with the derivative method. My choice would thus be between $\Delta D=D\sqrt{(\frac{\delta r}{r})^2+(\frac{\delta r}{r})^2}$ or $\Delta D=\frac{\partial D}{\partial r}\delta r=2\frac{\omega_n}{2(\ln(\gamma))^2} \delta r$. These give pretty different results and I do not know which one I should pick. I feel extremely confused.

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    $\begingroup$ Have you tried to look up "error propagation" of a random variable? The different formulas are e.g. explained here physics.stackexchange.com/questions/587041/… $\endgroup$
    – Semoi
    Feb 11, 2021 at 18:06
  • $\begingroup$ Mhm I see. So I guess since I am dealing with random errors I should use the first formula, right? $\endgroup$
    – Agnese
    Feb 11, 2021 at 18:54
  • $\begingroup$ You have only one parameter with an uncertainty, how would you have several partial errors to cumulate in quadrature ? I dont get your first formula. $\endgroup$
    – manu190466
    Feb 11, 2021 at 22:16
  • $\begingroup$ That is exactly why I would have used the derivative formula. I have been suggested adding in quadrature by a friend because he said each of the $\Delta r$ would be counted as a variable. But since they are dependent on one another it wouldn't necessarily make sense to use that, right? $\endgroup$
    – Agnese
    Feb 12, 2021 at 9:42
  • $\begingroup$ You're right, "each of the Δr" doesn't make sense because there is only one, elevated to the square, so a derivation is required. $\endgroup$
    – manu190466
    Feb 16, 2021 at 11:26


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