Why do we sometimes have to divide uncertainty of the measurement by $\sqrt{3}$?

For example we have the uncertainty of the measurement with a ruler with the smallest scale of 0.01 cm. This 0.01 cm is not our uncertainty of measurement, but we first divide it by $\sqrt{3}$ and then the result is our final uncertainty value.

Why is that this way?


I have found out the answer.

When you do a measurement (one measurement) you have many uncertainty sources. But if you want to have the combined uncertainty, you don't add like $1 + 1$, because that would give you uncertainty of $2$, but that doesn't have to be the case. The true value of the measurement may lie in between the uncertainty regions but it may lie on the left or on the right of the range.

As we don't know where exactly is the measurement we have to treat that situation like a rectangular distribution. We have $a=x-\Delta x$ and $b=x+\Delta x$, where $\Delta x$ is the uncertainty.

Graph representation

$$E(X) = \int_a^b xp(x)\mathrm{d}x = \int_a^b \frac{x}{b - a}\mathrm{d}x = \left.\frac{1}{2}\frac{x^2}{b - a}\right|_a^b = \frac{1}{2}\frac{b^2 - a^2}{b - a} = \frac{b + a}{2}$$

$$\begin{align} \sigma^2 &= \int_a^b \bigl(x - E(x)\bigr)^2 p(x)\mathrm{d}x \\ &= \int_a^b \biggl(x - \frac{b + a}{2}\biggr)^2\frac{1}{b - a}\mathrm{d}x \\ &= \left.\frac{1}{3(b - a)}\biggl(x - \frac{b + a}{2}\biggr)^3\right|_a^b \\ &= \frac{1}{3(b - a)}\Biggl[\biggl(\frac{b - a}{2}\biggr)^3 - \biggl(\frac{a - b}{2}\biggr)^3\Biggr] \\ &= \frac{1}{3}\biggl(\frac{b - a}{2}\biggr)^2 \end{align}$$

So we see that the standard deviation of this distribution is exactly our uncertainty divided by $\sqrt3$. Without that our uncertainty would be bigger than it needs to be. And so the final uncertainty of the single measurement is then: $u(x)=\Delta x/\sqrt3$

  • 5
    $\begingroup$ Except if the ruler has tick marks every $b-a$ and you are saying all you know is that the true value is uniformly in this range, the uncertainty is $(b-a)/\sqrt{12}$. This is called discretization error and the $\sqrt{12}$ is a standard result. Saying it's $\sqrt{3}$ off from $(b-a)/2$ is, well, bizarre, since $(b-a)/2$ is meaningless. $\endgroup$ – user10851 Dec 8 '15 at 3:41

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