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I am reading the book "Experimentation : an introduction to measurement theory and experiment design" by D. C. Baird and I have come across a paragraph where it says that the absolute uncertainty in an indirect measurement cannot have more significant figures than that of the original direct measurements. Since there are different ways to achieve an estimation of the final uncertainty (and the author has previously explained a calculus-based approach to do so) I wonder why this additional rule regarding significant figures is needed.

I leave the aforementioned paragraph down below. For further clarification, I feel that there is nothing wrong with the expression $R = 9.06 \, \pm \, 0.59 \, \Omega$ because the original uncertainties in $V$ and $I$ have already been taken into consideration when computing the $0.59 \, \Omega$ uncertainty.

Because computations tend to produce answers consisting of long strings of numbers, we must be careful to quote the final answer sensibly. If, for example, we are given the voltage across a resistor as $15.4 ± 0.1 $ V and the current as $1.7 ± 0.1 $ A, we can calculate a value for the resistance. The ratio $V/I$ comes out on my calculator as $9.0588235 \, \Omega$. Is this the answer? Clearly not. A brief calculation shows that the absolute uncertainty in the resistance is close to $0.59 \, \Omega$. So, if the first two places of decimals in the value for the resistance are uncertain, the rest are clearly meaningless. A statement like $R = 9.0588235 ± 0.59 \, \Omega$ is, therefore, nonsense. We should quote our results in such a way that the answer and its uncertainty are consistent, perhaps something like $R = 9.06 ± 0.59 \, \Omega$. But is even this statement really valid? Remember that the originally quoted uncertainties for $V$ and $I$ had the value $±0.1$, containing one significant figure. If we do not know these uncertainties any more precisely than that, we have no right to claim two significant figures for the uncertainty in $R$. Our final, valid, and self-consistent statement is, therefore, $R = 9.1 ± 0.6 \, \Omega$. Only if we had a good reason to believe that our original uncertainty was accurate to two significant figures, could we lay claim to two significant figures in the final uncer­ tainty and a correspondingly more precisely quoted value for $R$.

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Short story: I assume you have an idea of propagation of errors, then the relative error on the resistance's value is the sum of both relative errors, hence:

$$\Delta R = ((0.1 / 15.4) + (0.1 / 1.7)) \cdot 9.0588235 \Omega \approx 0.59 \Omega$$

So the final answer $R = 9.06\Omega \pm 0.59 \Omega \:$ is perfectly reasonable. You dont need (more over, you shouldn't) to do some obscure magic tricks to manage to get the final answer rounded off to make the decimals fit with the initial raw data. In fact, the decimal digits aren't anything more than a representation of a number, that doesn't mean you should round them off blindly.

Long story: Uncertainties can be a difficult subject if you don't have enough mathematical knowledge. Often, the matter is treated with a very poor focus based on simple but sometimes non-sense rules. I'd recommend you to read this article to get a better understanding. But if this is some sort of homework, ask to your superiors if what that book says is the reciepe they want you to follow and just do that.

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    $\begingroup$ thanks for the answer and the article, it looks interesting. This is not homework, and I although I am not a mathematician, I have some maths background, including some probability and statistics and its application to estimate uncertainties. I am reading this book only for interest, so I hope to have a deeper insight after reading the article you provided $\endgroup$
    – Javi
    Feb 13, 2023 at 15:14
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The first thing to note is that, in general one estimates the error, ie there will be an error in the error.

The next thing is to consider, what benefit there is in writing the value as $9.06 ± 0.59$ instead of $9.1 ± 0.6$?
The way it works is that you believe that if you took lots and lots and lots and lots (infinite number?) of readings you would get what is called a population mean (actual/true value).
The (mean) value that you found from a smaller number of readings is called the sample mean.
When you are considering errors what you are trying to do is to estimate how close you are to the population mean.
Does it really make much difference to your estimate of the population mean if you use $\pm 0.59$ or $\pm 0.6$?
Most often the answer is "No" and so you are advised to quote the value in its simplest form $9.1 ± 0.6$.

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  • $\begingroup$ so to be sure, your opinion is that the number of digits in -the estimation of- the uncertainty should be chosen according to simplicity, and fewer digits make a simpler value. And it has nothing to do with a comparison between the original uncertainties and the final one, as the author proposes. Did I get your point? $\endgroup$
    – Javi
    Feb 13, 2023 at 15:10
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    $\begingroup$ It is not simplicity, it is practicality at this level. $\endgroup$
    – Farcher
    Feb 13, 2023 at 19:04
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We should best address this by full calculations and by propagating relative rather than absolute uncertainties. Let's do the actual numbers.

$$ R = \frac{V}{I} = \frac{15.4}{1.7} = 9.05882...$$

The current has two significant digits (sig figs). By the rules to propagate sig figs, the resistance is therefore limited to have only two sig figs.

$$ R = 9.1 \Omega$$

In multiplication or division, we propagate relative uncertainties, not absolute uncertainties.

$$ \delta_r V = \frac{0.1}{15.4} = 0.0064935...$$

$$ \delta_r I = \frac{0.1}{1.7} = 0.05882...$$

We see that the uncertainty in current will contribute the greatest to the overall uncertainty. By linear propagation, we use the rule of quadrature (summation of variances) to propagate division uncertainties.

$$ \Delta_r^2 R = \delta_r^2 V + \delta_r^2 I = 0.003502...$$

$$ \Delta_r R = 0.05918...$$

The absolute uncertainty in $R$ becomes

$$ \Delta R = R\ \Delta_r R = (9.05882)(0.05918) = 0.5361$$

The uncertainties each had one sig fig. The relative uncertainties therefore each have one sig fig. The final total uncertainty therefore can only have one sig fig.

$$ R = 9.1 \pm 0.5\ \Omega $$

As you see, the true total uncertainty is not the estimated value $\pm 0.59$ (to two sig figs) that we obtain by simply multiplying the resistance by the sum of the two relative uncertainties. The true absolute total uncertainty in resistance using proper linear propagation methods is $\pm 0.54$ (also to two sig figs). The two values, estimated and proper, disagree in the second digit. As you also see, the proper treatment of the rules to propagate sig figs through the relative uncertainties would have each of the relative uncertainties with only one sig fig each, meaning the final calculation after the multiplication can only have one sig fig. Finally, the proper approach to linear propagation shows that the true uncertainty is smaller than the estimated uncertainty. This is a general rule. The true result by summing relative variances (rule of quadrature) is always smaller than the estimate result by summing relative deviations.

We could also write the form below

$$ R = 9.06 \pm 0.5(4)\ \Omega $$

to emphasize that we do not trust the second digit in the uncertainty.

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  • $\begingroup$ Thanks for providing a different insight. First, after some thought I believe there is no "true" uncertainty but different ways to get an estimation of it, e.g. in this case you propose the sum of variances rule to characterize the uncertainty using variance, but I think this rule involves, in general, a linearisation of the indirect measured value as a function of the direct measured values. Obviously some estimations are better than other. Secondly, I see that in your derivation you apply "the rules to prpagate sig figs" which is precisely what my question is about. I am looking for a ... $\endgroup$
    – Javi
    Feb 14, 2023 at 19:37
  • $\begingroup$ ... more logically consistent approach to handle uncertainty propagation, by questioning this kind of rules. I don't see any logical need to keep the same number of digits in $R$ as in $I$ (they don't even have the same dimension so it is strange to compare the number of digits.) Finally, you also force the relative uncertainty to keep only 1 fig. but again, this is just convention, I believe there is no logical need to this, right? $\endgroup$
    – Javi
    Feb 14, 2023 at 19:42
  • $\begingroup$ In continuous distributions where inputs are uncorrelated and random uncertainties follow a normal distribution, we sum variances and the result is a true metric. Otherwise, we apply weightings or matrix correlation terms that still use variances not standard deviations. Discrete distributions also use additional terms (student t-factor). Conventions to propagate sig figs are derived from a consistent mathematical approach. The "logical needs" are rigor and consistency. You are free to ignore the conventions, but do not be surprised then to be ignored as an inconsistent (naive) outlier. $\endgroup$ Feb 15, 2023 at 14:55
  • $\begingroup$ well, since I am studying this just by interest, I don't mind my error propagation being ignored. But it would really helpful if could point me to some source about that mathematical approach to sig figs propagation convention $\endgroup$
    – Javi
    Feb 16, 2023 at 22:57
  • $\begingroup$ Propagation methods for sig figs are given in introductory textbooks on general chemistry and physics. Rules are precise for adding/subtracting as well as for multiplying/dividing. $\endgroup$ Feb 18, 2023 at 0:52

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