My question is the following: what is the uncertainty propagated to area of a square?
The usual answer is as follows. Let $(l\pm\sigma_l)$ denote an experimental measure of the sides of the square, with $l$ being the size of the sides of the square and $\sigma_l$ its associated uncertainty. The area of the square is given by the function $A_{\textrm{square}}(l)=l^2$, with the corresponding uncertainty given by the formula for the propagation of uncertainty in indirect measures:
$\sigma_{A_{\textrm{square}}} = \sqrt{\left(\frac{dA}{dl}\right)^2 \sigma_l^2} = 2 l \sigma_l.$
Now let us think on the uncertaint propagated to the area of a rectangle with sides $(l_1\pm\sigma_{l_1})$ and $(l_2\pm\sigma_{l_2})$. The area is given by the function $A_{\textrm{rectangle}}(l_1,l_2)=l_1 l_2$ and the associated uncertainty propagated to the area of the rectangle is:
$\sigma_{A_{\textrm{rectangle}}} = \sqrt{\left(\frac{\partial A}{\partial l_1}\right)^2 \sigma_{l_1}^2 + \left(\frac{\partial A}{\partial l_2}\right)^2 \sigma_{l_2}^2} = \sqrt{l_2^2 \sigma_{l_1}^2 + l_1^2 \sigma_{l_2}^2}.$
These are the usual answers. However, I can think on a square as a limiting case of a rectangle where $l_1 = l_2 \equiv l$. Correspondingly, it follows that in this limiting case, $A_{\textrm{rectangle}}(l,l) = l^2 = A_{\textrm{square}}(l)$. Then, I would expect that the uncertainty of the area of a rectangle would reduce to that of the area of a square, when I take $l_1 = l_2 \equiv l$, but that is not what happens:
$\sigma_{A_{\textrm{rectangle}}}\Bigr\rvert_{l_1=l_2\equiv l} = \sqrt{l^2 \sigma_{l}^2 + l^2 \sigma_{l}^2} = \sqrt{2} l \sigma_l,$
and this is different than the first formula obtained above for the uncertainty propagated to the area of a square! What is the source of this paradox???
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EDIT: the user Henry has solved the problem! I will only provide below the details on the solution.
The point is that the formula for the propagation of uncertainty used above assumes that the two sides of the rectangle are uncorrelated. This is no longer the case when we take both sides to be equal, in which case they are obviously correlated (that was my mistake!). In this case, the formula for the propagation of uncertainty to the area of the rectangle with both sides equal is:
$\sigma_{A_{\textrm{rectangle}}} = \sqrt{\left(\frac{\partial A}{\partial l_1}\right)^2 \sigma_{l_1}^2 + \left(\frac{\partial A}{\partial l_2}\right)^2 \sigma_{l_2}^2 + 2 \frac{\partial A}{\partial l_1}\frac{\partial A}{\partial l_2} \sigma_{l_1 l_2}} = \sqrt{l_2^2 \sigma_{l_1}^2 + l_1^2 \sigma_{l_2}^2 + 2 l_1 l_2 \sigma_{l_1 l_2}},$
where $\sigma_{l_1 l_2}$ is the sample covariance between $l_1$ and $l_2$,
$\sigma_{l_1 l_2} = \sum_{i=1}^N \frac{(l_{1_i}-\langle l_1\rangle) (l_{2_i}-\langle l_2\rangle)}{N-1}$.
When $l_1 = l_2 \equiv l$, the above formula reduces to the sample variance, $\sigma_{l_1 l_2} \to \sigma_l^2$, therefore,
$\sigma_{A_{\textrm{rectangle}}}\Bigr\rvert_{l_1=l_2\equiv l} = \sqrt{l^2 \sigma_{l}^2 + l^2 \sigma_{l}^2 + 2 l l \sigma_l^2} = 2 l \sigma_l = \sigma_{A_{\textrm{square}}},$
as expected!
Best Answer
It seems that in the case of the rectangle you have assumed that errors on one side are independent of errors on the other: this leads you to add the variances before you take the square root.
In the case of the square you have assumed that errors on one side are equal to errors on the other: it is this dependency (implicitly the associated covariance) that leads to the higher propagation of error into the area.