# Calculating pressure and flux tensors in radiative astrophysics

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.

• I am not completely sure what you are looking for since your question seems to be twofold, 1) looking for the component expressions of the (monochromatic) radiation flux tensor $\left( F_\nu \right)^{i} = F_\nu^{i}$ and the component expression of the radiation pressure tensor $\left(K_\nu\right)^{ij} = K_\nu^{ij}$ as well as 2) a way to determine those quantities for an observer at a distance $d$ away from the radiation source, right? I think for the second part you probably need to know whether the source of interest is point-like or extended to make reasonable simplifications. Sep 8, 2020 at 20:07
• Thanks for answering. Yea, I think I need both parts, since presumably the answer the first informs the answer to the second. The source is a uniformly bright sphere, which can be treated as point-like, I think. I don't have any other information if we are to treat it as an extended system. Sep 8, 2020 at 21:41

First I start with the component formulation of the (monochromatic) radiative flux tensor $$\mathbf{F_\nu}$$, i.e. $$\left( \mathbf{F_\nu} \right)^{i} \equiv F_\nu^{\;i}$$,

$$F_\nu^{\;i} := \oint_\Omega I_\nu(\mathbf{k}) \, k^{i} \, d\Omega \quad.$$

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

$$\left( F_\nu^{\;x},\, F_\nu^{\;y}, \, F_\nu^{\;z} \right) = \left( \oint_\Omega I_\nu(\mathbf{k}) \, k^{x} \, d\Omega, \, \oint_\Omega I_\nu(\mathbf{k}) \, k^{y} \, d\Omega, \, \oint_\Omega I_\nu(\mathbf{k}) \, k^{z} \, d\Omega \right) \quad,$$ 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 $$d\Omega$$ in spherical coordiantes is given by $$d\Omega = sin\theta \, d\theta \, d\phi$$. 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,

$$P_\nu^{\; ij} \equiv \frac{4\pi}{c_0} K_\nu^{\; ij} = \frac{1}{c_0} \oint I_\nu(\mathbf{k}) \, k^i \, k^j \, d\Omega \quad.$$

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,

$$F_{tot.} = \int_{0}^{\infty}d\nu \, \int_\Omega I_\nu(\mathbf{k}) \, k^{i} \, d\Omega \equiv \sigma \, T_{\text{eff}} \overset{!}{=} \frac{L}{4 \pi r^2} \quad,$$

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

$$\int_\Omega \ldots d\Omega \propto \frac{1}{d^2},$$

• Unfortunately, I had a hard time to go into more specific details about how to compute the integrals and how to set the $d\Omega$ properly, since I do not know much about your problem at hand from what you have provided in your question. But I am happy to amend and revamp my answer based on your feedback. Sep 9, 2020 at 21:12