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(Following the definitions here: Teff and log g Determinations)

What is the "stellar angular diameter", as measured by astronomers specializing in stellar astrophysics?

Using the Stefan-Boltzmann law, the effective temperature of a star, $\text{T}_\text{eff}$, is defined as:

$\sigma\text{T}_\text{eff}^4=\int^{\infty}_0 F_{\nu}d\nu=\text{F}_{*}=\frac{\text{L}}{4\pi R^2}$

where luminosity is denoted $\text{L}$ and total radiant power per unit area at the stellar surface is $\text{F}_{*}$. Stellar luminosity is defined as $\text{L}_{*}=4\pi R^2_{*}\sigma\text{T}_{*}^4$.

The integrated radiative flux at the stellar surface is $\text{F}=\int^{\infty}_0F_{\nu}d\nu$.

Denoting the total observed flux at earth to be $f_{\oplus}$, the total flux of the star is determined by

$\text{F}_{*}=\frac{\theta^2}{4}f_{\oplus}$.

The stellar angular diameter is $\theta$. What is this?

Apparently, $\theta$ is measured using speckle photometry, interferometry, and lunar occultations, and indirectly measured from eclipsing binary systems of known distances.

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    $\begingroup$ I'm guessing it is the diameter of the star as measured from a telescope from Earth? $\endgroup$ Aug 28, 2015 at 23:35
  • $\begingroup$ What in the Wikipedia article on angular diameter doesn't answer your question? $\endgroup$
    – Warrick
    Sep 3, 2015 at 7:59
  • $\begingroup$ @Warrick Yes, why is the quantity squared and divided by 4? $\endgroup$ Sep 4, 2015 at 1:46

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The angular diameter (in radians) is the physical diameter of the star's photosphere divided by the distance to the star. i.e. $$\theta = \frac{2R}{d}\ ,$$ where $R$ is the stellar radius and $d$ is the distance to the star.

The flux at the Earth is $$f_\oplus = \frac{4\pi R^2 F_*}{4\pi d^2} = \frac{R^2}{d^2}F_* = \frac{\theta^2}{4}F_*\ . $$

i.e. I think you have your relationship back to front.

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