Relation between spectral radiance and density of cavity

[Jhordan]

Hi, my teacher showed us how we can derive de relation between spectral radiance and density of cavity (of a black hole), but I have a doubt.
This is the equation of the energy that are coming from definited directions by the intervals of angles θ and Φ with frequency in a determined interval:

$$
\left[\rho\left(\nu\right)d\nu\right]\cdot\left[\frac{\sin\left(\theta\right)d\theta d\phi}{4\pi}\right]
$$

But if we integrade θ from 0 to π/2 and Φ from 0 to 2π so we has:
$$
\frac{\rho\left(\nu\right)d\nu}{2}
$$

I think that this are because if the body are in thermal equilibriu, same quantity energy that are coming to in are coming out from, so how we onlye need want the nergy that are being irradiated by the black body, we have half of this total of density of energy. I am correct?

Thank you

Link for complete derivation [in portuguese]: https://www.if.ufrgs.br/~betz/iq_XX_A/radTerm/aRadTermAd_1.htm

 

[DrClaude]

But if we integrade θ from 0 to π/2 and Φ from 0 to 2π so we has:
$$
\frac{\rho\left(\nu\right)d\nu}{2}
$$

Why are you integrating $\theta$ from 0 to $\pi/2$? If you want to consider the entire radiation from a piece of the blackbody cavity, you need to integrate up to $\pi$.

The reason why the limit is $\pi/2$ in the derivation is that a hole in the cavity will only get radiation from one side, the inside of the cavity.

 

[Jhordan]

Well, this is exact the fact that i am trying confirm, maybe i made some confusion betweem the entire black body and this cavity. So, i am correct when I say that i integrate θ only up to π/2 π, because I only want get radiation from inside of the cavity?

The other half are radiation coming from external of the cavity?

Thanks again.

 

[DrClaude]

Well, this is exact the fact that i am trying confirm, maybe i made some confusion betweem the entire black body and this cavity. So, i am correct when I say that i integrate θ only up to π/2 π, because I only want get radiation from inside of the cavity?

The point of the calculation is to figure out the emission of a blackbody by considering a blackbody cavity with a hole in it, and calculating the radiation that leaks out, which will be the same as the radiation from the blackbody itself.

The other half are radiation coming from external of the cavity?

It depends on the situation you are considering. If you just want the emission from the blackbody, then there is nothing on the other side of the hole.

 

[Jhordan]

So, we have the equation of the energy that are coming from definited directions by the intervals of angles θ and Φ with frequency in a determined interval, that are:
$$\left[\rho\left(\nu\right)d\nu\right]\cdot\left[\frac{\sin\left(\theta\right)d\theta d\phi}{4\pi}\right]$$
And at this moment, don't matter if the energy are coming from black body or infinite, or whatever. If we integrate over a sphere, we have the radiation coming from all directions to our element of volume.

But more later, in the derivation, we want only the energy that are coming from inside of the cavity, the emission of blackbody. Thus we integrate over the half of the sphere, that is, the outer directions to our infinite plane - the black body - of the hole.

 

[DrClaude]

And at this moment, don't matter if the energy are coming from black body or infinite, or whatever. If we integrate over a sphere, we have the radiation coming from all directions to our element of volume.

Well, I don't like the way you put it. What bothers me is that the $\rho(\nu)$ in the equation is the energy density inside a blackbody cavity. It is true that $\rho(\nu)$ doesn't have to be specified at that point, but it is assumed to be uniform.

I am used to the presentation of the problem being the reverse of the way you present it here. Looking again at the link you provided (my Portuguese is limited), it appears to be also presented in the way I know: you start by considering a little element of the cavity, and see how much of the radiation contained therein reaches the hole. Then, you integrate over all elements inside the cavity that reach the hole at a given time. This is the reason why the integration is over only a hemisphere.

 

[Jhordan]

So, the cylinder are the little element, and then this equation multiplied by the volume of this element give me the energy contained in this element of the cavity. The next step are see how much of this radiation reaches the hole, then we integrate over all elements inside the cavity that reach the hole, and here that we want only a hemisphere.

 

[Jhordan]

I think that I understand now. Thanks, if you want close the topic, it's okay for me.

Thank you again.