I want to calculate the Luminosity of a star e.g. Vega but can be any star and came across the formula for Luminosity at the University of Oregon (http://pages.uoregon.edu/soper/Light/luminosity.html).

I'm not an expert on Maths, could anyone help me with the figures to put into the formula.

$$L = (4\pi d^2)b$$

$d$ is the distance but the page doesn't say Parsecs or Light Years.

$b$ is the apparent brightness.

I'm thinking

$$L = ((4\times3.141) \times (25.04^2) ) \times 0.026 = 204.81 \mathrm{(light\ years)}$$

$$L = ((4\times3.141) \times (7.68^2)) \times 0.026 = 19.26 \mathrm{(parsecs)}$$

Which gives me 204.81 which is different to Luminosity which is quoted (40.12) on the Wikipedia page. Is this the right formula for what I'm after. Please no scientific equations like what you see on the page linked above. Any calculations would need to be the type you would put into a non-scientific calculator.

  • $\begingroup$ using your rational, the Luminosity of the Sun would be $-50.89\times10^9$ and that isn't right at all... $\endgroup$ – LaserYeti Jan 28 '17 at 20:31
  • $\begingroup$ What is $b$ measured in? Your units need to be self consistent. $\endgroup$ – ProfRob Jan 28 '17 at 20:32
  • 3
    $\begingroup$ The luminosity isn't "40.12"; it has units. Units are important. $\endgroup$ – ProfRob Jan 28 '17 at 20:33
  • $\begingroup$ As to what b is measured in, I don't know, the link I saw didn't say what it was so I'm making assumptions. Im assuming apparent brightness and apparent magnitude are the same, yes? $\endgroup$ – MiscellaneousUser Jan 28 '17 at 21:42
  • $\begingroup$ That formula is to find luminosity from brightness in units of flux per unit area. A star's magnitude is related to the flux per unit area, but magnitude is a logarithmic scale and the brightness required for that formula is not. $\endgroup$ – antlersoft Jan 29 '17 at 3:42

The $b$ in your equation, brightness, is what we call a "Flux". That's the amount of energy per unit area at a certain distance. Flux can be calculated using the Stefan-Boltzmann law.
$$ J=\sigma T^4 $$ where $\sigma=5.670373\times10^{-8}\space W\space m^{-2}\space K^{-4}$is the Stefan-Boltzmann constant, This is measured in Watts per square meter. That means that we will need our distance to be in meters, and we will get a luminosity in Watts. We can go a little further and divide our answer by $3.828\times10^{26} \space W$ to get our final value in solar luminosities (L$_\odot$).

That will make our final equation: $$ L=\frac{4\pi R^2 \sigma T^4}{L_\odot} $$ Because we're using the Stefan-Boltzmann equation, instead of the distance to the star, we have to use its radius. Vega's radius is $2.362 \space R_\odot$, which is $1.64\times10^9$ meters. Its surface temperature is 9,600 K. Plugging in those numbers yields a luminosity of: $$ L=42.7L_\odot $$ This value is slightly different from the value quoted by Wikipedia, due to rounding, but is well within the acceptable range.


The root of your problem is not the maths, it is choosing the correct units. The relevant equation is $$L = 4\pi d^2 b,$$ Where $b$ is the power per unit area received at Earth from Vega, and $4\pi d^2$ is the surface area of an enormous sphere constructed around Vega, with radius equal to the distance to the earth and which would catch all the light emitted.

To get a luminosity in Watts, we need to express $b$ in Watts per square metre and the area in square metres. It helps to know that $d =19.26 \times 3.08\times 10^{16}$ m. The value of $b$ you have used is the apparent visual magnitude of Vega and I'm afraid that is not the power per unit area received at the Earth. A bit of reverse engineering tells me that actually $b = 3.5\times 10^{-9}$ W/m$^2$.

To obtain a final answer in terms of a multiple of solar luminosities, you divide your result by the solar luminosity in Watts, which is $L_{\odot}= 3.83\times 10^{26}$ W.

However, I think you want to be able to calculate the stellar luminosity from the distance and apparent magnitude. In which case, I refer you to James K's answer, noting that the bolometric correction depends on what type of star you are looking at.


This calculation attempts to estimate the luminosity from the distance and apparent brightness, instead of modelling the star as a black body.

I notice that you have "b=0.026", this is not the brightness, but instead the apparent magnitude of Vega. If you are trying to estimate the luminosity from distance and apparent magnitude, you can proceed as follows.

The absolute magnitude $M$ of a star can be found from the apparent magnitude $m$ and distance $d$ (in parsecs). $$M = m - 5 \left(\log_{10}d-1\right)$$

The luminosity is related to the absolute bolometric magnitude, which considers all the energy at all wavelengths (including infrared and ultraviolet). Bolometric magnitude differs most from visible magnitude when the star is very cool or very hot.

Vega has a absolute visible magnitude of 0.57, and a bolometric magnitude only slightly brighter (ie smaller) than that.

The luminosity, L is found from the bolometric magnitude as: $$L=10^{\frac{71.197-M_\mathrm{bol}}{2.5}}$$

Putting the value 0.55 into this formula give a luminosity of $1.8\times10^{28}$Watts, or about $46L_\odot$, which differs slightly from the published value, due I think to not properly adjusting for bolometric magnitude.

Combining these gives

L = (10^26.5 * d^2)/2.51^m / 3.28e26

Formulae from wikipedia: https://en.wikipedia.org/wiki/Absolute_magnitude

  • $\begingroup$ Does not use $b$. $\endgroup$ – ProfRob Jan 29 '17 at 9:50
  • $\begingroup$ That is intentional. The value of "b" that the OP is using is 0.026, which makes me think that they are actually using the apparent magnitude. From the context it seems that they want to calculate luminiosity (relative to the sun) from apparent magnitude. I;ve clarifed the answer $\endgroup$ – James K Jan 29 '17 at 12:15
  • $\begingroup$ So for the last formula that James said, d => 7.68 (parsec) and m -> 0.582 ? I get 33. $\endgroup$ – MiscellaneousUser Jan 29 '17 at 16:16
  • $\begingroup$ m is apparent magnitude, not absolute. $\endgroup$ – James K Jan 29 '17 at 16:30
  • $\begingroup$ Using 0.026, I get the figure 55.50, a much larger number in the opposite way. $\endgroup$ – MiscellaneousUser Jan 29 '17 at 19:57

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