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I have measured $k_1$ and $k_2$ in two measurements and then I calculated $\Delta k_1$ and $\Delta k_2$. Now I want to calculate $k$ and $\Delta k$.

$k$ is just the mean of $k_1$ and $k_2$. I thought that I would need to square-sum the errors together, like so:

$$ \Delta k = \sqrt{(\Delta k_1)^2 + (\Delta k_2)^2} $$

But if I measure $k_n$ $n$ times, $\Delta k$ would become greater and greater, not smaller. So I need to divide the whole root by some power of $n$, but I am not sure whether $1/n$ or $1/\sqrt n$. Which is it?

Qmechanic
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Martin Ueding
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2 Answers2

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The formula you've specified

$$ \Delta k = \sqrt{(\Delta k_1)^2 + (\Delta k_2)^2} $$

is the formula to obtain error of quantity $k$, as being dependent on $k_1$ and $k_2$ according to the following expression

$$ k = k_1 + k_2.$$

Generally, to calculate experimental error of a dependent quantity (and the expression stated in your question), you start with the expression for dependent quantity

$$k = f(k_1, k_2, ...)$$

and use statistical expression

$$\Delta k = \sqrt{\sum_i \left(\frac{\partial f}{\partial k_i} \Delta k_i \right)^2}.$$

If

$$k = \frac{k_1 + k_2}{2}$$

then

$$ \Delta k = \frac{\sqrt{(\Delta k_1)^2 + (\Delta k_2)^2}}{2} $$

So the generalized answer might be: you have to divide by $n$ and not $\sqrt{n}$.

However, bear in mind that the statistical expression is applicable only when the measured quantities are "independent" of each other. If $k_1$ and $k_2$ are the same quantity measured in two measurements, this is not exactly true, and the exact statistical expression is much more complicated.

MichaelMaggs
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Pygmalion
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2

$k$ is just the mean of $k_1$ and $k_2$

No, the best value of k is calculated using a weighted mean, weighting by the reciprocals of the squares of the respective individual uncertainty values. An accurate measurement must contribute more to the best value than an inaccurate measurement.

I thought that I would need to square-sum the errors together, like so:

$$ \Delta k = \sqrt{(\Delta k_1)^2 + (\Delta k_2)^2} $$

No, instead $$ (\Delta k)^{-1} = \sqrt{(\Delta k_1)^{-2} + (\Delta k_2)^{-2}} $$

Intuitively, a very uncertain value must make little contribution. The uncertainty in k must always be less than or equal to the smallest of the individual uncertainties. Also, multiple, equally accurate measurements must decrease uncertainty.

see the example at the bottom of page 4 of this reference for the general case of n measurements:

http://www.physics.umd.edu/courses/Phys261/F06/ErrorPropagation.pdf

Community
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DavePhD
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