# Stellar Power(Luminosity) Flux

So I was applying some mathematical techniques I learned to physics, and one thing that captured my interest, is the power or luminosity flux of a star. So modeling the situation, taking the scalar field of Luminosity($L$), and the vector surface(stellar surface), $S$, I began by setting up a Surface (Flux, in this case) integral of

$$\iint_{S}L\,dS$$

Where,

$$S=\langle r\sin\theta \cos\phi,r\sin\theta \sin\phi,r\cos\theta \rangle \;,\; \theta \, \epsilon \, [0,\pi] \wedge \phi \, \epsilon \, [0,2\pi]$$ $$L = 4\pi r^{2}\sigma T^{4}$$

$r$ in this model is the invariant stellar radius. Following logic, since $L$ has no explicit dependency on $\theta$ or $\phi$, I rearranged as:

$$4\pi r^{2} \sigma T^{4} \iint_{S} dS$$

Since the integral of dS over S is merely surface area, and our surface is a sphere of radius $r$, this expression reduces to:

$$16\pi^{2} r^{4} \sigma T^{4}$$

The dimensional analysis for this breaks down to $W \cdot m^{2}$. Isn't flux supposed to be quantity per area?

• Why would you integrate luminosity over area? Luminosity is energy per time. Flux is energy per area per time. You could integrate flux over area to get luminosity. But I don't understand what you're trying to do here. – kleingordon Dec 7 '13 at 9:26
• If this is the case, then how should the problem of finding the flux of luminosity be stated initially? – Doryan Miller Dec 8 '13 at 7:49
• Using conventional definitions, there is no "flux of luminosity." Again, consider the units. Luminosity is energy per time. Flux is energy per area per time. The two are related because you can integrate flux over area to obtain luminosity. For a blackbody emitter, the flux is $\sigma T^4$. And if the emitter is spherically symmetric, then the luminosity is equal to the spherical surface area times the flux: $L = 4 \pi r^2 \sigma T^4$. – kleingordon Dec 8 '13 at 10:54
• Ah! That makes sense. So the essence of what I was doing was finding the flux of the flux, which is meaningless in this case. Aye? – Doryan Miller Dec 9 '13 at 5:19
• More or less. You were taking something with units of energy per time, and then integrating over a surface to get something with units of energy per time multiplied by area, which does not have much physical relevance. I think what you really wanted to do was to start with something with units of flux (energy per time per area), like $\sigma T^4$, and then integrate over a surface to get something with units of energy per time, or luminosity. The relevant relation that emerges is $L = 4 \pi r^2 \sigma T^4$, which you quoted. I think that is the final result you are proving here. – kleingordon Dec 9 '13 at 5:58

$L = \int \int {\bf F} \cdot d{\bf s}$
is where you should start, where $F$ is the flux in units of Watts/m$^2$.
Blackbody flux is given by $\sigma T^4$ and hence an isotropic flux integrated over a sphere
$$L = \int^{2\pi}_{0} \int^{\pi}_{0} \sigma T^4 r^2 \sin \theta\, d\theta\,d\phi = 4\pi r^2 \sigma T^4$$
$L$, the term for Luminousity, IS the power flux across a stellar surface. I've been trying to find what I already had. Doing so then resulted in multiplying the flux in essence by distance units $m$, but in the fourth order of magnitude.