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Let say we have some stars and their spectral classification in the MK index. (not all HD stars)

  • HD xxxx - B3
  • HD yyyy - O7
  • HD zzzz - F0
  • ....

If the stars share the same spectral type, then they share the same superficial temperature $T$. Also, the $LRT$ relation holds

$$\boxed{L=4\pi \sigma R^2 T^4} $$

for $L$ absolute luminosity, $\sigma$ the Stefan-Boltzmann constant and $R$ the stellar radius. By definition of $L$

$$L=\ell 4\pi r^2$$

where $\ell$ is the apparent luminosity and $r$ is the distance of the star from the Earth. Let's take a look at the distance

$$r^2=\frac{L}{\ell 4 \pi}=\frac{4\pi \sigma R^2T^4}{\ell 4 \pi}=\frac{\sigma R^2 T^4}{\ell}$$


Do I need further information to say anything about the distance and the apparent magnitude of the stars, in the form of their stellar radius $R$ and the apparent luminosity $\ell$.

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Haven't you already answered your own question? Are you just asking for confirmation that your answer is correct? – Ben Crowell Jun 23 '13 at 16:52
I'm asking for confirmation - further comments about it – Jorge Jun 23 '13 at 16:53
up vote 1 down vote accepted

In general, your relation among $r$, $R$, $T$, and $\ell$ holds. Knowing any three can determine the fourth.

That said, I've never seen "apparent luminosity" defined like that, where it doesn't have the same units as luminosity. Really I've never seen "apparent luminosity" defined. In fact, the flux matches that definition: $$ F = \frac{L}{4\pi r^2}. $$ Then your relation could be written $$ r^2 = \frac{\sigma R^2T^4}{F}. $$ $F$ is measured from photometry and $T$ is measured from spectroscopy. $R$ can be determined from the spectrum with stellar modeling, but those models are calibrated on stars with known values of $r$ obtained from parallax.

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Apparent luminosity is a different name for flux. Yes, it's somewhat a misnomer, like many other things in astronomy :-) – Pulsar Jun 24 '13 at 9:16

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