Error propagation and symmetric errors

[Malamala]

Hello! I am a bit confused about how to interpret symmetric error when doing error propagation. For example, if I have $E = \frac{mv^2}{2}$, and I do error propagation I get $\frac{dE}{E} = 2\frac{dv}{v}$. Which implies that if I have v being normally distributed, and hence having a symmetric error (by this I mean that the upper and lower limit of the uncertainty interval in the value is the same), the same is true for E, as that formula is predicting just one $dE$ and people would quote (at least in most of the papers I read) it as $E \pm dE$. However if I have say, $E = 1000$ eV, $dE = 100$ eV, $m = 100$ amu and I calculate the associated speed, I get, for $E = 1000$ eV, $v = 43926$ m/s, for $E = 1100$ eV, $v = 46070$ m/s and for $E = 900$ eV, $v = 41672$ m/s and as you can see, the associated uncertainties on v are not symmetric. So how should I think of the symmetric uncertainty given when doing error propagation, as when plugging in the numbers this is not the case. Doesn't it mean that if I have a symmetric (gaussian) error on $v$ the error on $E$ will not be symmetric? And is this not the case most of the time, except when there is a linear relationship between the variables? Thank you!

 

[Dale]

First, you need to decide if you are going to analyze and report relative uncertainties or uncertainties. You are confusing things by mixing them. $\sigma_v$ is the uncertainty in $v$ and $\frac{\sigma_v}{v}$ is the relative uncertainty in $v$. If one is symmetric then the other is usually not, so you need to be clear about what you are saying is symmetric.

By the way, I would not use $dv$ for an uncertainty, I would use $\sigma_v$. Your notation risks lots of confusion with derivatives.