Flux that reaches a certain radius in a star

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SUMMARY

The discussion centers on the calculation of radiative flux in a star's atmosphere, governed by the equation dF/dr = -opacity * F * density. The user seeks clarification on integrating this first-order ordinary differential equation (ODE) to derive the flux at a given radius, specifically F(r) = F0 * e^(opacity * (1/r - 1/r0)). The user correctly identifies that the density decreases with radius as 1/r^2 and demonstrates the integration process leading to the exponential term in the solution. The confusion regarding the appearance of r^-2 is acknowledged, indicating a need for further understanding of the integration steps.

PREREQUISITES
  • Understanding of ordinary differential equations (ODEs)
  • Familiarity with radiative transfer concepts
  • Knowledge of isothermal atmospheres in astrophysics
  • Basic calculus, particularly integration techniques
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  • Study the derivation of radiative transfer equations in stellar atmospheres
  • Learn about the role of opacity in astrophysical contexts
  • Explore advanced integration techniques for solving ODEs
  • Investigate the implications of isothermal models in stellar physics
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Astronomy students, astrophysicists, and researchers interested in stellar atmospheres and radiative transfer processes will benefit from this discussion.

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I know the radiative flux traveling through the atmosphere of a star obeys the equation:

dF/dr=-opacity*F*density

If we have an isothermal atmosphere the density decreases with radius such that density is proportional to 1/r^2.

If the flux entering such an atmosphere from the core of the star at radius r0 is F0 show that the flux that reaches radius r is

F(r)=F0 *e^opacity*(r^-2 -r0^-2) where the opacity is a constant.

I do not understand I'm guessing I need to integrate the 1st order ODE but how can I do this when flux is on both sides of the equation and where does the exponential come from in the answer?

Help!
Thanks :) Note: for some reason the symbols tab is not working on my laptop
 
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if you divide your first eqn by F and multiply by dr you get 1/F dF = - opacity * density dr

which gives 1/F dF α -opacity * 1/r^2 dr where α means proportional to, which since opacity is a constant you can write (i think)

1/F dF = -opacity * 1/r^2 dr

integral both sides to give ln(F/Fo)=-opacity [1/r0 - 1/r]

do exp of both sides

F = F0 exp (opacity (1/r - 1/r0))

not sure why it gives r^-2 , maybe I've done it wrong, I am guessing that's where the exponential comes from anyway!
 

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