4
$\begingroup$

I have two methods which allow for a detection of a signal that is at most of magnitude x. If method 2 can detect signal that are half a magnitude larger, i.e. fainter, let's say x.5 mag, is it correct to say that method 2 can detect signals that are about 60% fainter than what method 1 can detect?

Is the 60% correct? I got this number by converting the magnitudes $m_i$ to flux $f_i$, i.e. $$100^{(m_1-m_2)/2.5}=\frac{f_2}{f_1}\approx 0.40$$ $$\Rightarrow f_2=0.40 f_1\sim 40\% f_1$$

Thus, $f_2$ is about $40\% f_1$ which means about $60\%$ fainter. I have never worked with magnitudes before, so is this a valid statement?

$\endgroup$
  • $\begingroup$ Completely correct, except you should divide by 2.5, not by 5. But your result is correct, so I guess it's just a typo. $\endgroup$ – pela Jul 9 '19 at 9:50
  • 1
    $\begingroup$ Of course, you are right! I actually calculated it the wrong way, but I have updated the numbers so now it should be 60% fainter. Thanks a lot for your reply! $\endgroup$ – Philipp Jul 9 '19 at 11:00
  • 1
    $\begingroup$ @pela are you sure? $100^{1/5} = 10^{1/2.5}$ $\endgroup$ – Mike G Jul 9 '19 at 13:40
  • 1
    $\begingroup$ @MikeG Oh I'm sorry, you're absolutely right. I read "10" instead of "100", since that's usually how you write it. +1 for your answer below! $\endgroup$ – pela Jul 9 '19 at 20:43
5
$\begingroup$

Textbooks typically express the flux vs. magnitude relation something like this:

$$m_2 - m_1 = -2.5 \log_{10} \frac{f_2}{f_1}$$

which we can transform into this:

$$\frac{f_2}{f_1} = 10^{(m_1 - m_2) / 2.5}$$

so the original version of your formula was correct.

If m2 - m1 = 0.5, then f2 / f1 = 0.63. Some readers find comparisons like "x% fainter" confusing, so I would say method 2 is more sensitive than method 1 by a factor of 1.58.

| improve this answer | |
$\endgroup$

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.