0
$\begingroup$

I have a block of glass. The glass is tilted at some angle $\alpha$ with respect to a fixed axis. I rotate the glass and $\alpha$ changes. The way I am rotating the glass is by placing it onto a stick that serves as a handle (like rotating a frying pan). Unfortunately, the handle is a bit broken. With the handle I measure the angle $\alpha$. I have a way to set the glass perfectly at $\alpha=0$. When I do this, I measure the angle and it is not $0$, it is some other value $\alpha_0$.

I begin to tilt the glass, and take some measurements $\alpha_i$. They are all supposedly shifted from 0 by $\alpha_0$. This would be a systematic error. However when I set the glass at $\alpha=0$ again, I measure and I get some other value $\alpha_{0-}$.

If I have some uncertainty $u$ for my measurements (coming from the precision that I'm allowed to measure) then if I had a systematic error I would write my data as:

$$(\alpha_i - \alpha_0)\pm u.$$

But given that I have two different values for my $0$, should I present my data as

$$\alpha_i \pm \sqrt{u^2+\lvert\alpha_0-\alpha_{0-}\rvert^2}?$$

$\endgroup$
2
$\begingroup$

Let me avoid the term "systematic error", because the question of what that actually means is a complicated one.

The measurement error of the device, by which I mean the difference between the true value and the reading on the device, can be considered a random variable. It has some standard deviation, which we typically take to be the uncertainty, and it has some mean, which we typically take to be zero (because if it's not, you can adjust the device to compensate for that). We determine the mean and standard deviation by making a bunch of measurements of a known reference.

What you're proposing here is basically using two samples to determine the mean and standard deviation of the device's measurement error. But that doesn't make much sense, does it? The solution is to make many more measurements. Set the glass perfectly at $\alpha = 0$, make a measurement, twist it around a bit, set it back to $0$, make another measurement, and repeat until you have a sufficiently large set of samples. Assuming the distribution is not too weird, you can then use the mean $\bar{\alpha}_0$ and standard deviation $\sigma_0$ of this sample set to write your results as $$(\alpha - \bar{\alpha}_0) \pm \sqrt{u^2 + \sigma_0^2}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.