# Calculations of apparent magnitude

I was attempting to do some calculations of apparent magnitude to help solidify my understanding of the topic, but have been running into some confusion.

According to Wikipedia, the apparent magnitude can be given as:

$m_x = -2.5\log_{10}(F_x/F^0_x)$

where $F_x$ is the observed flux and $F^0_x$ is a reference flux (in other words, this equation provides the difference of apparent magnitude between two observed values). Also, this is assuming that the same wavelength band is used in both flux measurements.

Flux, in turn, can be calculated as:

$F = \frac{L}{A}$

where $L$ is the star's luminosity and $A$ is the flux density. Since stars act as point sources, this can be simplified to:

$F = \frac{L}{4\pi r^2}$

where $r$ is the distance to the star.

Since, historically, Vega has been used as the reference zero-point (having an apparent magnitude around 0.03), I tried doing a simple calculation to find out the apparent magnitude of Fomalhaut using the values for luminosity and distance given in Wikipedia for both of them.

First, the flux of Vega:

$F_{Vega} = \frac{37\,L_\odot}{4\pi (25.3\,ly)^2}$

$F_{Vega} = 4.5999\times 10^{-3}\,L_\odot/ly^2$

Next, the flux of Fomalhaut:

$F_{Fomalhaut} = \frac{17.66\,L_\odot}{4\pi (25\,ly)^2}$

$F_{Fomalhaut} = 2.2485\times 10^{-3}\,L_\odot/ly^2$

Now, to calculate the apparent magnitude:

$m_{Fomalhaut} = -2.5\log_{10}(\frac{2.2485\times 10^{-3}\,L_\odot/ly^2}{4.5999\times 10^{-3}\,L_\odot/ly^2})$

$m_{Fomalhaut} = 0.7777$

Huh?? Fomalhaut's apparent magnitude is supposed to be 1.16. Even correcting for Vega's offset of 0.03, we still come up with 0.8077. Why are the calculations failing? I don't think I've made a mistake in the mathematics. Am I using the wrong values?

-

Thanks for asking this question. It is something we all assume to be obviously trivial and often skip. Your question made me think and I wasn't sure whether the values for luminosities listed in Wikipedia were in the optical range, or the bolometric luminosity i.e. the luminosity over all wavelengths.

A little bit of googling led me to this page, where this question seems to have been discussed well and also resolved.

-

I think the real problem will be in the errors in measurements. Recall that the error in a function of $n$ variables, $f(x_1, x_2, ..., x_n)$ with associated errors for each variable $\sigma_1$, $\sigma_2$, ... , $\sigma_n$ is given by $$\sigma_{f}=\sqrt{\sum_i^n \left(\sigma_i\frac{df(x_1, x_2, ..., x_n)}{dx_i}\right)^2}$$

In your case, the function is $F(L, r)=\frac{L}{4\pi r^2}$, so the error propagation formula is

$$\sigma_{F}=\sqrt{\left(\frac{\sigma_{L}}{4 \pi r^2}\right)^2 + \left(\frac{-3\sigma_{r}}{4 \pi r^3}\right)^2}$$

For Vega, $\sigma_{L}=3L_{\odot}$, $\sigma_r=0.1LY$. That gives $$\sigma_{F_{Vega}}=3.77\times10^{-4}$$

For Fomalhaut, it was a bit trickier to track down since wikipedia doesn't give the error in luminosity, but in the article it's given: $\sigma_{L}=0.82L_{\odot}$, $\sigma_r=0.1LY$. That gives $$\sigma_{F_{Fomalhaut}}=1.05\times10^{-4}$$

Using the standard error propagation formula again, the error associated with the apparent magnitude is

$$\sigma_{m}=\sqrt{\left(\frac{-2.5\sigma_{F_{Fomalhaut}}}{F_{Fomalhaut}}\right)^2 + \left(\frac{2.5\sigma_{F_{Vega}}}{F_{Vega}}\right)^2}$$

If you plug in all the values used above, I get $$\sigma_{m}=0.237$$ which means that your estimate is off by only $1.5\sigma$. That's pretty good, considering that these luminosities are probably bolometric rather than the visual band alone, yet your apparent magnitude is only in the V-band which makes it only a fraction of all the light from the star.

-
Good point on the bolometric luminosities - the B-V index of Vega is indeed quite low (listed as 0.00 on Wikipedia, actually). –  voithos Jun 13 '11 at 17:21

The calculations look correct. I didn't check your input values, but I think that the error arises because you're overestimating the accuracy of the luminosity and distance values for Vega and Fomalhaut. Both of those things are notoriously hard to measure in astronomy and sometimes require the specification of caveats. Apparent magnitude in some waveband is relatively easy to measure since it's just a relative brightness measurement. The luminosities that you've quoted may or may not be bolometric (the sum of all wavebands) and there's also the issue of whether or not the luminosities have been corrected for extinction (aka reddening) along the line of sight.

-