Uncertainty Calculation: Applying Product Rule instead of Power Rule

Question

I use $\delta$ to represent absolute uncertainty. The power rule for the calculation of relative uncertainty in $t^2$ is $$\frac{\delta (t^2)}{(t^2)}=2\left(\frac{\delta t}{t}\right).$$ But if I treat the power as a product and apply the product rule, I have $$\frac{\delta (t \times t)}{(t \times t)} = \sqrt{\left(\frac{\delta t}{t}\right)^2 + \left(\frac{\delta t}{t}\right)^2} = \sqrt{2\left(\frac{\delta t}{t}\right)^2} = \sqrt{2}\left(\frac{\delta t}{t}\right).$$ Am I making a mistake? If not, how is this inconsistency reconciled?

Answer 1

The product rule assumes that the things being multiplied vary independently of one another, which is clearly not the case when multiplying something by itself. As such, the power rule is the correct one here.

Answer 2

As was pointed out by Sandejo's answer, uncertainties only add in quadrature if the two quantities being measured are uncorrelated. The formulas for propagation of uncertainty have additional terms that must be included if the variables' uncertainties are correlated.

Specifically, suppose we have two quantities $A$ and $B$ with uncertainties $\sigma_A$ and $\sigma_B$ and covariance $\sigma_{AB}$. Then the uncertainty in $f = AB$ is [given by] $$ \frac{\sigma_f}{f} = \sqrt{\frac{\sigma_A^2}{A^2} + \frac{\sigma_B^2}{B^2} + 2 \frac{\sigma_{AB}}{AB}}. $$ Uncorrelated uncertainties have $\sigma_{AB} = 0$, and so you end up with the "addition in quadrature" formula you're familiar with.

On the other hand, suppose that $A$ and $B$ are perfectly correlated (as they would be if they were secretly the same variable, which you're calling $t$.) In this case, it can be shown that $\sigma_{AB} = \sigma_A \sigma_B$, and so $$ \frac{\sigma_f}{f} = \sqrt{\frac{\sigma_A^2}{A^2} + \frac{\sigma_B^2}{B^2} + 2 \frac{\sigma_{A}\sigma_{B}}{AB}} = \frac{\sigma_A}{A} + \frac{\sigma_B}{B}. $$ In particular, if $A = B = t$, then you have $$ \frac{\sigma_f}{f} = 2 \frac{\sigma_t}{t}, $$ as one would expect from the power rule.