1
$\begingroup$

Let's say we know the "real" flux of a star in a certain bandwidth because we consider it a black body of known temperature. That's just the integer of the power density, so we find the real flux $F\left[=\right]\frac{\mathrm{W}}{\mathrm{m}^2}$.

How do we find the portion of the flux that reaches us? Let's say $R_{\text{star}}$ is the radius of the star and $D_{\text{star}}$ is the distance of the star from Earth; is the apparent flux then$$ A\left(f\right)~=~ \left(\frac{R_\text{star}}{D_\text{star}}\right)^2 \,F \,,$$or am I doing something wrong here?

Improve this question
$\endgroup$
4
  • $\begingroup$ What are the units of the final answer you want? $\endgroup$
    – probably_someone
    Jul 24, 2017 at 22:33
  • $\begingroup$ Watt per square meter is fine, I just want to know if I am correct to multiply by the square of the radius and divide by the square of the distance $\endgroup$
    – user2083840
    Jul 24, 2017 at 22:39
  • $\begingroup$ You appear to want to multiply the flux given in W/(m^2 sr) by a certain solid angle (as the inverse-square law only makes sense in this context). What's your detector area? $\endgroup$
    – probably_someone
    Jul 24, 2017 at 22:49
  • $\begingroup$ 80 cm, but what I am really trying to do here is compare this flux with the magnitude of the star in a certain bandwidth. Does this make sense? $\endgroup$
    – user2083840
    Jul 24, 2017 at 22:54

2 Answers 2

Reset to default
1
$\begingroup$

I haven't seen the term 'apparent flux' before. Flux is always 'apparent' in the sense that it depends on the distance from you to the source. Your equation for flux received $A(f) = \frac{FR^2}{D^2}$ is only true if $F$ is the flux at the surface of the star.

Improve this answer
$\endgroup$
0
$\begingroup$

Start with the Stefan-Boltzmann law, which gives you luminosity as a function of temperature $T$ and radius $R$:

$$L=4\pi R^2 \sigma T^4$$

where the units of luminosity are W/m^2. This luminosity is integrated over all solid angle and over the entire EM spectrum, so, to find the flux passing through a certain detector of area $a$, we must multiply $L$ by the fraction of the total solid angle taken up by the detector*:

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

which is where the inverse-square law comes in. This is the bolometric flux (i.e. it's still integrated over the entire EM spectrum), so to obtain the flux in a particular band, you must multiply by the blackbody spectrum intensity $I(\lambda,T)$ for the detector's particular wavelength bandwidth $d\lambda$:

$$F_{\lambda,d\lambda} = I(\lambda,T)d\lambda\frac{a}{D^2}L$$

Apparent bolometric magnitude is defined as:

$$m=-2.5\log_{10}\left(\frac{F}{F_0}\right)$$

where $F_0$ is a particular standard bolometric magnitude. Likewise, for magnitude in a particular band, we have:

$$m_{\lambda,d\lambda}=-2.5\log_{10}\left(\frac{F_{\lambda,d\lambda}}{F_{0_{\lambda,d\lambda}}}\right)$$

where $F_{0_{\lambda,d\lambda}}$ is the analogous magnitude standard.

*This assumes that the flux is isotropic, i.e. it doesn't depend on direction.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Thanks for helping but I don't understand what you're doing with the Stefan-Boltzmann law. When obtaining the flux in a particular band, you're multiplying L by the blackbody spectrum intensity, and dimensionally that's Watt squared.... $\endgroup$
    – user2083840
    Jul 25, 2017 at 10:44
  • $\begingroup$ It should be: $$F_{\lambda,d\lambda} = I(\lambda,T)d\lambda\frac{\pi a}{D^2}$$ Either way, it doesn't work. Consider a star with 11.4 visible magnitude, you can easily calculate the flux in W/m^2 because a star with zero visible magnitude has a flux of 3.64 * 10^(-23) W/m^2 . So the flux from the 11.4 mag star should be something like 10^(-27) W/m^2, while with mine and your formula we're off by a long shot. $\endgroup$
    – user2083840
    Jul 25, 2017 at 11:09
  • $\begingroup$ I was referring to Tres-2 by the way, so you can make your own calc! $\endgroup$
    – user2083840
    Jul 25, 2017 at 11:17
  • $\begingroup$ @user2083840 The confusing thing about flux is that it can either be flux over the whole sky or flux per unit solid angle. The units for these two quantities are the same, so we must always specify which we mean. Explaining the inverse square law requires us to introduce the flux per unit solid angle to compare the area of the detector with the area of the whole sky. $\endgroup$
    – probably_someone
    Aug 2, 2017 at 7:21
  • $\begingroup$ @user2083840 The formula does work, though. An apparent magnitude $m=11.4$ implies that $\log_{10}(F/F_0)=-4.56$. This means that $F=10^{-4.56}F_0=10^{-27}$ W/m^2. $\endgroup$
    – probably_someone
    Aug 2, 2017 at 7:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.