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I have a value for the relative flux $F_2/F_1$. With some uncertainty value $a$.

If I use the equation to get relative magnitude: $m_1 - m_2 = 2.5 \log_{10}(F_2/F_1)$

How do you now calculate the uncertainty value $a$ for this equation? Do you just do the same equation but applied to the uncertainty value or is there some other alteration you have to make to uncertainty values?

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You need to do what is called “propagation of uncertainties”. You can search to get more information on that, but briefly if you have some function $f(x)$ that depends on variable $x$, then the uncertainty $\sigma_x$ on the quantity $x$ is related to the uncertainty on $f$ by

$$ \sigma_f = \sqrt{\left(\frac{\partial f}{\partial x}\right)^2 \sigma_x^2} $$

Here $f$ is your relative magnitude $m_1 - m_2$, and $x$ is your flux ratio. So taking that partial derivative will give you the factor that relates the two uncertainties.

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    $\begingroup$ This literally works to first order only. I don't think this is sufficient as an answer unless you explain how it breaks down once the uncertainties are large, as they are regularly in astronomy. For the OP's case if it's spectrophotometry and errors are of order say 0.1 mag then okay. But if there are larger errors or if there are multiple steps, then you should use a better "propagation of errors" than this very simple approximation. $\endgroup$
    – uhoh
    Apr 6, 2021 at 4:09
  • $\begingroup$ @WDUK no, not yet perfect, or perhaps we can say "perfect to first order only"? $\endgroup$
    – uhoh
    Apr 6, 2021 at 4:11
  • $\begingroup$ @uhoh What is the expression for the second order term here? It would be interesting to compare with some real-world numbers and see what errors are too large. $\endgroup$
    – Eric Jensen
    Apr 6, 2021 at 12:34
  • $\begingroup$ Square both sides, remember that the square root of the squares is just a fancy absolute value sign, keep track of that for later. Then try to see it as the Taylor series that it is. There should be thoroughly documented better methods. I'm no expert but I know a very short Taylor series when I see one! $\endgroup$
    – uhoh
    Apr 6, 2021 at 12:47
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    $\begingroup$ So we’re each saying, no you do the math. :-) Truncation of series approximations is an effective tool for a reason - those other terms get small! Empirically, I can’t think of an example where I’ve ever seen anything but the above expression used in actual practice. That doesn’t mean it’s 100% correct, but perhaps those higher-order terms truly don’t matter most of the time. $\endgroup$
    – Eric Jensen
    Apr 6, 2021 at 13:08

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