Physics Asked by philosophytophysics on December 2, 2020
I need to calculate the components of the radiative flux tensor and the pressure tensor, but I am having a little trouble figuring out what the integral should be. The flux is defined as
$$F_v = 4{pi}H_v$$ where $$H_v = 1/4{pi} int I_v {hat k} d{Omega}$$
and
$${hat k_i} = (sin{theta}cos{phi}, sin{theta}sin{phi}, cos{theta})$$
I know I need to integrate over the solid angle, and so the limits of integration are 0 to 2pi but I need to determine the components for an observer distance d away. Do I need a second integral that involves d in some way, or do I just integrate over the solid angle?
The same question holds for the pressure tensor, which is defined similarly, except
$$K_v = 1/4{pi}int I_v {hat k_i}{hat k_j}d{Omega}$$
I’m pretty sure I need to something like
$$1/4{pi} int_{0}^{2{pi}} d{phi} int {hat k_i}d{theta}$$
but I don’t know what the limits of integration are for an observer distance d from the center of the sphere.
First I start with the component formulation of the (monochromatic) radiative flux tensor $mathbf{F_nu}$, i.e. $left( mathbf{F_nu} right)^{i} equiv F_nu^{;i}$,
$$ F_nu^{;i} := oint_Omega I_nu(mathbf{k}) , k^{i} , dOmega quad. $$
If we express the direction vector $mathbf{k}$ in cartesian coordiantes, i.e. $mathbf{k} = (k^i)_{i in {x,y,z}} = (k^x, k^y, k^z)$ we can re-write1 the radiation flux from above, we get
$$ left( F_nu^{;x},, F_nu^{;y}, , F_nu^{;z} right) = left( oint_Omega I_nu(mathbf{k}) , k^{x} , dOmega, , oint_Omega I_nu(mathbf{k}) , k^{y} , dOmega, , oint_Omega I_nu(mathbf{k}) , k^{z} , dOmega right) quad,$$ with
$$ mathbf{k} = left( sin (theta) cos(phi),, sin (theta) sin(phi),, cos(theta) right) quad. $$
A useful resource for orientation about the definition of the solid angle $Omega$ and how to apply certain ranges of integration is the nrao.edu website which also reminds that the solid angle increment $dOmega$ in spherical coordiantes is given by $dOmega = sintheta , dtheta , dphi$. Furthermore, as you asked as second part of your question, how to tackle ways to take a certain distance $d$ from the source of emission to the observer into account, the solid angle $Omega$ needs to be changed accordingly based on how large the detector will be, see also the nrao.edu website and slide no. 5 here.
For the (monochromatic) pressure tensor $mathbf{P_nu}$ you need to be careful not to confuse the former with the $K$ integral or angularmoment of order two $mathbf{K_nu}$. However, both of which are directly related through,
$$ P_nu^{; ij} equiv frac{4pi}{c_0} K_nu^{; ij} = frac{1}{c_0} oint I_nu(mathbf{k}) , k^i , k^j , dOmega quad. $$
Addendum: (more background info, in case you need it)
For any star we know that the effective temperature $T_{text{eff}}$ of the radiation source depends on the total, actual radiation flux, integrated over all frequences if we can assume the star's spectrum to be reasonably close (or identical) to a black body,
$$ F_{tot.} = int_{0}^{infty}dnu , int_Omega I_nu(mathbf{k}) , k^{i} , dOmega equiv sigma , T_{text{eff}} overset{!}{=} frac{L}{4 pi r^2} quad, $$
which finally links the flux density to the luminosity of the source.
Although the source intensity $I_nu$ is independent of the distance $d$ between the source and the observer, the flux density is not independent of source distance, and scales due to the solid angle $Omega$ with
$$ int_Omega ldots dOmega propto frac{1}{d^2}, $$
see also the nrao.edu site.
I hope this synopsis helps as a starting point.
Footnotes & References:
1 Hubeny I. & Mihalas, D.: Theory of Stellar Atmospheres. Princeton University Press, 2015, p. 72ff.
Correct answer by Diazenylium on December 2, 2020
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